mm-4839 === Subject: Re: Algebraic Equations I want to get into the field of Software Engineering because I'm more interested in video games, computers, websites, etc, more than anything else. Here's the link to the practice test, which covers everything that will be on my test. http://www.centennialcollege.ca/adx/aspx/adxGetMedia.aspx?DocID=2580 I think you need Microsoft Office, or another similar program to view it. It's all easy stuff up to section 3, Algebraic Equations. I can potentially pass this test, if section 3 was not present, but unfortunately, it's present. === Subject: Re: Algebraic Equations http://mathforum.org/kb/message.jspa?messageID=6622668 > I want to get into the field of Software Engineering > because I'm more interested in video games, computers, > websites, etc, more than anything else. Here's the link to the practice test, which covers > everything that will be on my test. http://www.centennialcollege.ca/adx/aspx/adxGetMedia.aspx?DocID=2580 I think you need Microsoft Office, or another similar program > to view it. It's all easy stuff up to section 3, Algebraic Equations. I can potentially pass this test, if section 3 was not > present, but unfortunately, it's present. Well, you're still here after my rant. I'll tell you what, instead of answering the questions, I'll describe what you need to do for each of them. 48. If T = (1/3)rm^2, then r = r is being divided by 3 (multiplied by 1/3) and r is being multiplied by m^2, so you can isolate r on one side of the equation by multiplying both sides by 3 and then dividing both sides by m^2 (or, in this case, you can perform these equation operations in the opposite order). 49. Solve for b: 3b = 5(2-b) - 4(1-3b) Expand (use distributive property) the two expressions on the right hand side. Then move all terms involving b to one side of the equation and move the other terms to the other side of the equation. Combine the terms involving b, and you should get an equation of the form (some number)b = 'terms not involving b'. Now divide both sides of this last equation by the coefficient of b. 50. 8x - 3 = 4x - 15, x = Same as in #49, except you don't have to expand any expressions first and, instead of 'b', the variable you're solving for is 'x'. 51. If bx = c - x, x = Same as #50, except you have an unspecified value 'b' instead of a specified value such as 8. The same situation occurs in computer tasks, where you have to understand and work with something without always being given the input parameters explicitly. I'll do this one for you. First, move all the terms involving x to one side and put the remaining terms on the other side. bx + x = c Now add the coefficients of x. Alternatively, this can be accomplished by factoring out x. (b + 1)x = c You can think of it like this: b cows plus 1 cow is b+1 cows. Here I'm using cow for 'x'. Now divide both sides by the coefficient of x. (b + 1)x/(b + 1) = c/(b + 1) x = c/(b + 1) 52. Which is equal to x^2 - 2x - 3? This is trinomial factoring with a leading coefficient of 1, so you want two numbers with a sum equal to -2 and a product equal to -3. Note, if it's of any interest, that you can immediately eliminate (c) and (d) because the given expression is equal to -3 when x = 0, while (c) and (d) are equal to (-3)(-1) = 3 and (3)(1) = 3, respectively, when x = 0. 53. (3 - y)(3 + y) = Google: FOIL + algebra 54. a^2/16 - 1/9 = Get a common denominator and then combine into a single fraction. Factor the numerator, which will be a difference of squares form. Note that by considering a = 0, you can by sight eliminate both (b) and (d). [See my comments in #52 above.] The next three are absolute value evaluations that you should be able to do. After this, it looks like Section IV begins, although it's not labeled as such. Dave L. Renfro === Subject: Re: Algebraic Equations How the heck do you divide both sides by M^2? It don't make sense to me. I'm about ready to give up on this algebra. === Subject: Re: Algebraic Equations http://mathforum.org/kb/message.jspa?messageID=6624066 > I'm about ready to give up on this algebra. Maybe these web pages will help: http://math.about.com/od/algebraworksheets/a/basicsimp.htm http://math.about.com/od/algebraworksheets/Algebra_Worksheets.htm http://www.chem.tamu.edu/class/fyp/mathrev/mr-algeb.html http://www.math.uakron.edu/~dpstory/mpt_home.html http://www.mathleague.com/help/algebra/algebra.htm http://math.uww.edu/~mcfarlat/141/solve1f.htm http://www.hood.edu/documents/pdf/placement_basicAlgebraReview.pdf http://www.math.sc.edu/~jlane/math122/ws.2.pdf http://www.openmathtext.org/lecture_notes/elementary_algebra_book.pdf Dave L. Renfro === Subject: Re: Algebraic Equations http://mathforum.org/kb/message.jspa?messageID=6623953 > 48. If T = (1/3)rm^2, then r = r is being divided by 3 (multiplied by 1/3) and > r is being multiplied by m^2, so you can isolate > r on one side of the equation by multiplying both > sides by 3 and then dividing both sides by m^2 > (or, in this case, you can perform these equation > operations in the opposite order). http://mathforum.org/kb/message.jspa?messageID=6624066 > How the heck do you divide both sides by M^2? > It don't make sense to me. I'm about ready to give up on this algebra. T = (1/3)rm^2 T divided by m^2 = (1/3)rm^2 divided by m^2 T/m^2 = [(1/3)rm^2]/m^2 T/m^2 = (1/3)*(r/1)*(m^2/1)*(1/m^2) T/m^2 = (1/3)*(r/1)*(m^2/m^2) T/m^2 = (1/3)*(r/1)*1 T/m^2 = (1/3)*(r/1) T/m^2 = (1/3)r Dave L. Renfro === Subject: Re: Algebraic Equations ...Still makes no sense. === Subject: Re: Algebraic Equations > ...Still makes no sense. Which step makes no sense (to you)? === Subject: Re: Algebraic Equations http://mathforum.org/kb/message.jspa?messageID=6624066 > How the heck do you divide both sides by M^2? > It don't make sense to me. I'm about ready to give up on this algebra. http://mathforum.org/kb/message.jspa?messageID=6624131 > T = (1/3)rm^2 T divided by m^2 = (1/3)rm^2 divided by m^2 [snip rest] http://mathforum.org/kb/message.jspa?messageID=6624156 > ....Still makes no sense. See these web pages, then: Dave L. Renfro === Subject: Re: Algebraic Equations > How the heck do you divide both sides by M^2? It don't make sense to me. I'm about ready to give up on this algebra. T = (1/3)*r*m^2 T/(m^2) = (1/3)*r*(m^2)/(m^2 ) Dividing both sides by m^2 T/(m^2) = (1/3)*r The two m^2's on the righthand 3*T/(m^2) = 3*(1/3)*r Multiply both sides by 3 === Subject: Re: Algebraic Equations Mhm, I'm still here. And thank you kindly for those tips. I'm gonna see if I can finally learn this wacky thing you call algebra. === Subject: Re: Algebraic Equations Distribution: world > I want to get into the field of Software Engineering > because I'm more interested in video games, computers, > websites, etc, more than anything else. Here's the link to the practice test, which covers > everything that will be on my test. > It's all easy stuff up to section 3, Algebraic Equations. I can potentially pass this test, if section 3 was not > present, but unfortunately, it's present. Well, you're still here after my rant. I'll tell you >what, instead of answering the questions, I'll describe >what you need to do for each of them. >51. If bx = c - x, x = Same as #50, except you have an unspecified value >'b' instead of a specified value such as 8. The >same situation occurs in computer tasks, where >you have to understand and work with something >without always being given the input parameters >explicitly. I'll do this one for you. First, move all the terms involving x to one side >and put the remaining terms on the other side. bx + x = c Now add the coefficients of x. Alternatively, >this can be accomplished by factoring out x. (b + 1)x = c You can think of it like this: b cows plus 1 cow >is b+1 cows. Here I'm using cow for 'x'. Now divide both sides by the coefficient of x. I'd just like to point out to our aspiring programmer that what Mr. Renfro suggests can only be done when (b+1) is not zero. >(b + 1)x/(b + 1) = c/(b + 1) x = c/(b + 1) What values can't b have for this to be valid? -- Michael F. Stemper #include 2 + 2 = 5, for sufficiently large values of 2 === Subject: Change of coordinate of a wave PDE I have the following boundary value problem on the wave PDE: (Problem #4 on Page 87 of Strauss') U_tt = c^2 U_xx-r*U_x where 0on Page 87 of Strauss') U_tt = c^2 U_xx-r*U_x where 0Directly using U(x,t)=X(x) T(t) will give me a 2nd order ODE in X(x), >which is c^2 X''-rX'+lambda X = 0 and I cannot fit the boundary condition nicely to the eigen functions of >this ODE. I think I would follow the method of change of coordinate, but > don't know how to proceed. Any pointer on how to apply the change of >coordinate on this problem would be appreciated. >Kevin I think you just need to keep pushing. Write it as X'' - (r / c^2)X' + (p / c^2) X = 0 with boundary conditions X(0) = X(l) = 0. where I have used p instead of lambda. For simplicity, lets let a = - r / c^2 and b = p / c^2 X'' + aX' + bX = 0 with characteristic equation m^2 + am + b = 0 Complete the square: m^2 + am + a^2 / 2 + (b - a^2 / 2) = 0 (m + a / 2)^2 + (b - a^2/2) = 0 Now you have three cases, where (b - a^2/2) is positive, negative, or zero. So call (b - a^2/2) k^2, 0, or -k^2 accordingly. I think you will find the negative and zero cases yield only the trivial solution (check it). But the positive case will yield solutions like: X(x) = A exp(-ax/2) cos(kx) + Bexp(-ax/2) sin(kx) and when you put your boundary conditions in you should get some eigenvalues and nontrivial eigenfunctions. Of course you will have to back substitute your variables when you are done. Can you take it from there? --Lynn http://math.asu.edu/~kurtz === Subject: induction Does the recurrence relations T(n) = 2 -- if n = 2 T(n) = 2 * T(n/2) + n -- if n = 2^k, for k > 1 will have the close form of -- T(n) = n lg n Why T(n) become n lg n ( only if n is an exact power of 2 )? can mathematical induction show the above theorem? === Subject: Re: induction posting-account=O9zR9AkAAACmp918j6u5m5plppeILcze Filter 1.2.0.72; GTB5; .NET CLR 1.0.3705; .NET CLR 1.1.4322; Media Center PC 4.0; .NET CLR 2.0.50727; .NET CLR 3.0.04506.648; .NET CLR 3.5.21022; \ .NET CLR 3.0.4506.2152; .NET CLR 3.5.30729),gzip(gfe),gzip(gfe) > Does the recurrence relations T(n) = 2 -- if n = 2 > T(n) = 2 * T(n/2) + n -- if n = 2^k, for k > 1 will have the close form of -- > T(n) = æn lg n Why T(n) become n lg n ( only if n is an exact power of 2 )? can mathematical æinduction show the above theorem? Sure. When k = 1, 2^k lg 2^k = 2^1 * lg 2^1 = 2 * lg 2 = 2. Check. Let k >= 1, and assume that T(2^k) = 2^k lg 2^k = k * 2^k. To show that T(2^(k+1)) = (k + 1) * 2^(k+1). Then T(2^(k+1)) = 2 * T(2^k) + 2^(k+1) = 2 * (k * 2^k) + 2^(k+1) = k * 2^(k+1) + 2^(k+1) = (k + 1) * 2^(k+1). Check. Proving that T(n) = n lg n when n = 2^k. Dave === Subject: Re: induction posting-account=suWj4AkAAADE1IvGmj55Nmq3f98qb17e .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; InfoPath.2; .NET CLR 3.5.21022; Tablet PC 2.0; .NET CLR 1.1.4322),gzip(gfe),gzip(gfe) > Does the recurrence relations T(n) = 2 -- if n = 2 > T(n) = 2 * T(n/2) + n -- if n = 2^k, for k > 1 will have the close form of -- > T(n) = æn lg n Why T(n) become n lg n ( only if n is an exact power of 2 )? can mathematical æinduction show the above theorem? ****************************************************************************\ * According to your definition, T(4) = 2*ln(4/2) + 4 = 2*ln(2) + 4 = ln 4 + 4, which is not the same as 4 ln 4, so T(n) isn't n ln n even for 2^2...unless, of course, I misunderstood something. Tonio === Subject: Re: induction > According to your definition, T(4) = 2*ln(4/2) + 4 = 2*ln(2) + 4 = ln > 4 + 4, which is not the > same as 4 ln 4, so T(n) isn't n ln n even for 2^2...unless, of > course, I misunderstood something. lg = log base 2 [not ln = log base e], but he should have said so since it is not a standard usage (in general mathematics) === Subject: Re: induction posting-account=Jz4DtgkAAAAZkdWvJAd__jMF7l1N5_1V CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) Does the recurrence relations T(n) = 2 -- if n = 2 > T(n) = 2 * T(n/2) + n -- if n = 2^k, for k > 1 will have the close form of -- > T(n) = æn lg n Why T(n) become n lg n ( only if n is an exact power of 2 )? can mathematical æinduction show the above theorem? ***************************************************************************[\ CapitalEth]** According to your definition, T(4) = 2*ln(4/2) + 4 = 2*ln(2) + 4 = ln > 4 + 4, which is not the > same as 4 ln 4, so T(n) isn't æn ln n æeven for 2^2...unless, of > course, I misunderstood something. Tonio Not quite: (a) lg() is the base 2 logarithim, ~0.693... times the natural logarithm (b) you lost a factor of 2 in your statement You should have T(4) = 2*T(2) + 4 = 2*2 + 4 = 8 or T(4) = 2 * (4/2) * lg (4/2) + 4 = 2*lg(4) + 4 = 8 and 8 = 4*2 = 4*lg(4) In general and inductively T(2^k) = 2*T(2^(k-1)) + 2^k = 2*2^(k-1)*(k-1) + 2^k = 2^k*k = 2^k*lg(2^k) === Subject: upper / lower bounds What is it mean by - asymptotic upper and lower bounds for each of following recurrence relations? T(n) = 2 * T(n/2) + n^3 T(n) = T(9n/10) + n T(n) = 16 * T(n/4) + n^2 T(n) = 7 * T(n/3) + n^2 is the asymptotic upper and lower bounds are functions that's always bigger / smaller than above recurrence relations? To solve the upper / lower bound, I do the following: 1) subsitute T(n) for (n-1) and multiply by (n-2)^2 to get (A) 2) multiply T(n) by n^2 to get (B) 3) perform (B) - (A) In other word, perform n^2 * T(n) - (n-1)^2 * T(n-1) Why I need to (B) - (A)? What is - n^2 * T(n) - (n-1)^2 * T(n-1) used for? How can I found the upper / lower bound from the formula? What is (B) and (A) used for? How to deduce 2 bounding functions from (B) and (A)? === Subject: newton method and complex equations system hello, I have no experience with mathematica, but I want (have to) solve the following problem: The problem is a complex system of equations with 31 equations and 31 unknowns (the unknown are delta_i i=1,...,31). The equation is as follows. (beta and alpha are parameters and set to 0.1 and 0.05). X_i= beta Y_i + alpha SUM(j=1,...,31; j notequal i) (delta_i/gamma_ij)*e^((S_ij / delta_i)-1)^2 + Z_i i=1,\.83\.8331 For X_i, Y_i and Z_i I have empirical values. Also for the matrices gamma_ij and S_ij, I have the empirical values. I only need to figure out the deltas by using Newton method. Can give me anybody an advice how to start my problem. Remember I have no experience with Mahtematica. === Subject: Re: newton method and complex equations system > hello, I have no experience with mathematica, but I want (have to) solve the following problem: The problem is a complex system of equations with 31 equations and 31 unknowns (the unknown are delta_i i=1,...,31). The equation is as follows. (beta and alpha are parameters and set to 0.1 and 0.05). X_i= beta Y_i + alpha SUM(j=1,...,31; j notequal i) (delta_i/gamma_ij)*e^((S_ij / delta_i)-1)^2 + Z_i i=1,..31 For X_i, Y_i and Z_i I have empirical values. > Also for the matrices gamma_ij and S_ij, I have the empirical values. I only need to figure out the deltas by using Newton method. > Can give me anybody an advice how to start my problem. Remember I have no experience with Mahtematica. > Is the final Z_i term inside the summation? === Subject: Re: newton method and complex equations system no, Z_i is not inside the summation. === Subject: Re: A really pretty rose grows in the fertilizer of JSH > Let's look at these for d = 2, f1 = f2 = 1. Then we have: x(v) = (-1 + 2 v - 3 v^2) / (-1 + 2 v + v^2) > y(v) = (2 (-1 + v) v) / (-1 + 2 v + v^2) Let's take the positive solutions of the Pell equation, x^2 - 2 y^2 = 1, > and see what values we have to set v to in order to get those solutions. ... If Pi/Qi is the i'th convergent, the v's are P1/P2 > Q2/Q3 > P3/P4 > Q4/Q5 > ... I took a look at this again, but instead of JSH's parameterization, I used the simplest one that has been proposed: x(v) = (v^2 + d) / (v^2 - d) y(v) = (2v) / (v^2 - d) That, of course, gives a different sequence of v values for d = 2: 3,2 {{v->2}} 17,12 {{v->3/2}} 99,70 {{v->10/7}} 577,408 {{v->17/12}} 3363,2378 {{v->58/41}} 19601,13860 {{v->99/70}} 114243,80782 {{v->338/239}} 665857,470832 {{v->577/408}} 3880899,2744210 {{v->1970/1393}} 22619537,15994428 {{v->3363/2378}} But again, the v's are tied to the convergents for sqrt(2) in a simple pattern: 2 Q1 / P1 P2 / Q2 2 Q3 / P3 P4 / Q4 2 Q5 / P5 ... That pattern isn't as interesting, in my opinion, as the one that arises out of JSH's parameterization, but, on the other hand, I didn't see any obvious pattern past d = 2 with the JSH form. With the simpler form, there is an obvious pattern. Here's d - 5: Convergents: 2, 9/4, 38/17, 161/72, 682/305, 2889/1292, 12238/5473, 51841/23184, 219602/98209, 930249/416020, 3940598/1762289, 16692641/7465176, 70711162/31622993, 299537289/133957148, 1268860318/567451585, 5374978561/2403763488 Values of v that give Pell solutions: 9,4 {{v->5/2}} 161,72 {{v->9/4}} 2889,1292 {{v->85/38}} 51841,23184 {{v->161/72}} 930249,416020 {{v->1525/682}} 16692641,7465176 {{v->2889/1292}} 299537289,133957148 {{v->27365/12238}} 5374978561,2403763488 {{v->51841/23184}} 96450076809,43133785636 {{v->491045/219602}} 1730726404001,774004377960 {{v->930249/416020}} Relationship of v to convergents: 5 Q1 / P1 P2 / Q2 5 Q3 / P3 P4 / Q4 5 Q5 / P5 ... Checking a few other values of d, it appears that in general, for positive integers d where d is not a perfect square, then these are the values of v that give the successive solutions to x^2 - d y^2 = 1: d Q1 / P1 P2 / Q2 d Q3 / P3 P4 / Q4 ... -- --Tim Smith === Subject: incomplete magic square I have a simple ( and incomplete ) magic ( maybe not so magic ) square where \ the only element we are given is the one at the middle 35 and 4 more integers forming a cross around 35. Clockwise from the top the numbers are 32, 46, 38 and 24. we can see that 2*35=(32+38) and 2*35=(24+46). Is there a general way to solve this if the only number given is the one in \ the middle? === Subject: an apology to some t.tao or somebody hello, firstly sorry about the immigration thing..if i could only get it to \ make sense....and also i am in the middle of writing a letter...as the internet is a construct of satan himselves.... on the up-side i have found perpetual motion, although you are forbidden to \ discuss and ruin any chances whatsoever of an awardings type stuff. and finally,as this is a mathematical page...i reckon that the notion of impossibility of certain actions in disciplines involving proof has been misunderstood. i say this because a) all avenues cannot possibly have been investigated and b)legend has it that Bretty has really gone and done it \ with perpetual motion!! the real thing! does anybody know the source of the notion of impossibility? was it Godel? or was it å? === Subject: Re: Nomenclature for various groups Distribution: world posting-account=_l4K0QkAAAC09JhOoK_ZfoJKXOmr_jZf Gecko/2009020911 Ubuntu/8.04 (hardy) Firefox/3.0.6,gzip(gfe),gzip(gfe) On Feb 24, 12:10æpm, mstem...@walkabout.empros.com (Michael Stemper) > I'm aware that Z is symbol for the cyclic group of integers (with > addition). Actually, it depends. I once had a couple of referees object to my using the blackboard bold or the bold Z to denote the additive infinite cyclic group of integers, on the grounds that the symbol is reserved for the ->ring<- structure... > I also know that nZ represents the sub-group of Z that > is obtained by taking every n-th element of Z (starting with the > identity). It is better to think of it as the subgroup obtained by considering the set of elements of the form n*x for every x in Z. The reason is that, technically, your description does NOT give the corresponding subgroup, it just gives a subsemigroup consisting of the nonnegative multiples of n, and in addition it presupposes some ordering on Z, which while it exists is not part of the cyclic group structure. (It ->is<- part of its structure as an ->ordered<- group, but you are not specifying it as an ordered group). Does this nomenclature generalize? Remember that for an additive group, we define n*g for g in G and integer n, inductively as follows 0*g = 0 for all g in G (where the zero on the left is the integer 0, and the zero on the right is the additive identity of G); (n+1)*g = n*g + g for all nonnegative n in Z; and (-n)*g = - (n*g) for all nonnegative n in Z. In a multiplicative group, we use multiplicative notation and define g^0 = 1 for all g in G, g^(n+1) = g^n*g for all g in G, nonnegative n in Z; and g^(-n) = (g^n)^(-1) for all gin G, nonnegative n in Z. Then, in for any integer n, we define in the additive case (which is assumed to be abelian): nG = {n*g : g in G} and in the multiplicative case (which may or may not be abelian) we define G^{n} = {g^n : g in G} and G^n = , the subgroup generated by all the nth powers of elements of G. Note that in the abelian case, G^{n}=G^n, as G^{n} is already a subgroup, but in the noncommutative case the nth powers may fail to form a subgroup. If I'm referring to Z 12 (modular addition, mod 12), is it correct to > use 3Z 12 to refer to the sub-group of Z 12 consisting of {0,3,6,9}? Yes.. see above. That said, again: many object to the notation Z n for the cyclic group of order n or for the integers under modular addition modulo n; particularly, those who come from number theory, where Z p has a different meaning. Many recomment/suggest/demand using Z/nZ for the additive group modulo n (though again you may run into objections from those who consider the boldface Z to be reserved for the ring structure). Alas, we cannot make everyone happy. > If not, what is the standard naming to use -- or isn't there one? Note that your questions are not really about nomenclature or naming, but rather about notation... -- Arturo Magidin, sans .sig === Subject: Re: Nomenclature for various groups Distribution: world > I'm aware that Z is symbol for the cyclic group of integers (with > addition). I also know that nZ represents the sub-group of Z that > is obtained by taking every n-th element of Z (starting with the > identity). Does this nomenclature generalize? If I'm referring to Z_12 (modular addition, mod 12), is it correct to > use 3Z_12 to refer to the sub-group of Z_12 consisting of {0,3,6,9}? If not, what is the standard naming to use -- or isn't there one? For an additive group G and non-negative integer n: nG = { ng : g is in G }. -- Paul Sperry Columbia, SC (USA) === Subject: Re: Nomenclature for various groups Distribution: world > If I'm referring to Z_12 (modular addition, mod 12), is it correct to > use 3Z_12 to refer to the sub-group of Z_12 consisting of {0,3,6,9}? If not, what is the standard naming to use -- or isn't there one? For an additive group G and non-negative integer n: nG = { ng : g is in G }. -- Michael F. Stemper #include 2 + 2 = 5, for sufficiently large values of 2 === Subject: ||xy^T||_2 = ||x||_2*||y||_2 (2-norm problem) posting-account=Qq_GEwoAAAD0Kc2k70q6xwuZdqx8sCzT 4334.5003; Windows NT 5.1; SV1; .NET CLR 2.0.50727; .NET CLR 3.0.04506.648; .NET CLR 3.5.21022),gzip(gfe),gzip(gfe) spider-mtc-tf06.proxy.aol.com[400C70A6] (Prism/1.2.1), HTTP/1.1 cache-mtc-ab12.proxy.aol.com[400C744C] (Traffic-Server/6.1.5 [uScM]) Can you show this for x an (m x 1) vector and y an (n x 1) vector? birdhaldol (1998-2009) 11 yrs old. === Subject: || x y^T || = || x ||_2 * || y ||_2 (2-norms) posting-account=Qq_GEwoAAAD0Kc2k70q6xwuZdqx8sCzT 4334.5003; Windows NT 5.1; SV1; .NET CLR 2.0.50727; .NET CLR 3.0.04506.648; .NET CLR 3.5.21022),gzip(gfe),gzip(gfe) spider-mtc-tf06.proxy.aol.com[400C70A6] (Prism/1.2.1), HTTP/1.1 cache-mtc-ab12.proxy.aol.com[400C744C] (Traffic-Server/6.1.5 [uScM]) Can you show || x y^T || = || x ||_2 * || y ||_2 ? sincerely birdhaldol === Subject: || x y^T ||_2 = || x ||_2 * || y ||_2 [2-norm proof] posting-account=Qq_GEwoAAAD0Kc2k70q6xwuZdqx8sCzT 4334.5003; Windows NT 5.1; SV1; .NET CLR 2.0.50727; .NET CLR 3.0.04506.648; .NET CLR 3.5.21022),gzip(gfe),gzip(gfe) spider-mtc-tf06.proxy.aol.com[400C70A6] (Prism/1.2.1), HTTP/1.1 cache-mtc-ab12.proxy.aol.com[400C744C] (Traffic-Server/6.1.5 [uScM]) x is an (m x 1) vector and y is an (n x 1) vector. The 2-norm relationship to show is: || x y^T ||_2 = || x ||_2 * || y ||_2 i give up. i'd be grateful to know how. birdhaldol (1998-2009) === Subject: Re: || x y^T ||_2 = || x ||_2 * || y ||_2 [2-norm proof] On 2009-02-25 16:32:46 -0400, birdhaldol@yahoo.com said: > x is an (m x 1) vector and y is an (n x 1) vector. The 2-norm relationship to show is: || x y^T ||_2 = || x ||_2 * || y ||_2 i give up. i'd be grateful to know how. birdhaldol (1998-2009) Some hints only ... What is the rank of matrices of form x y^T? What do you know about || M ||_2 in terms of singular values? Did you really mean to apply a matrix norm to a nonsquare matrix? Plausible but not usual. Assuming the || M ||_2 is the usual induced matrix norm for || v ||_2 of vectors. How did you come to be asking these questions? They are more typical of early undergraduate courses. === Subject: Re: || x y^T ||_2 = || x ||_2 * || y ||_2 [2-norm proof] <2009022517070716807-gsande@worldnetattnet> posting-account=Qq_GEwoAAAD0Kc2k70q6xwuZdqx8sCzT 4334.5003; Windows NT 5.1; SV1; .NET CLR 2.0.50727; .NET CLR 3.0.04506.648; .NET CLR 3.5.21022),gzip(gfe),gzip(gfe) spider-mtc-tf06.proxy.aol.com[400C70A6] (Prism/1.2.1), HTTP/1.1 cache-mtc-aa10.proxy.aol.com[400C740E] (Traffic-Server/6.1.5 [uScM]) > On 2009-02-25 16:32:46 -0400, birdhal...@yahoo.com said: x is an (m x 1) vector and y is an (n x 1) vector. The 2-norm relationship to show is: || x y^T || 2 = || x || 2 * || y || 2 i give up. ?i'd be grateful to know how. birdhaldol(1998-2009) Some hints only ... What is the rank of matrices of form x y^T? What do you know about || M || 2 in terms of singular values? Did you really mean to apply a matrix norm to a nonsquare matrix? > Plausible but not usual. Assuming the || M || 2 is the usual induced > matrix norm for || v || 2 of vectors. How did you come to be asking these questions? They are more typical > of early undergraduate courses. linear algebra. yes the 2-norm of a matrix i only know for a square diagonal one is the maximum diag element. we've been taught svd so maybe that will come into it === Subject: Re: A special case of weighted mean Originator: bergv@math.uiuc.edu (Maarten Bergvelt) During some current research I came across something I can't say why > it should be true (or false). Namely: Let (a(n)) and (b(n)) be 2 bounded integer sequences satisfying a(n) >=1 and b(n)>=1. Let A(n)=sum_{k=1}^{n}a(k) and B(n)=sum_{k=1}^{n}b(k). What I know: lim n-->infty A(n)/n=L>0 exists and B(B(n))=sum_{k=1}^{n}a > (k)b(k)+O(1) My claim is: If (a) and (b) satisfy above conditions we have : lim n-->infty B(B > (n))/B(n)=L I'm not confident since experiments yield a slow and erratic > convergence. BTW, I checked a lot of sequences and I always observe a > tendancy which supports the claim. I used for (a) periodic sequences, > arithmetical functions, differences of Beatty sequences... Perhaps > some hypothesis must be added but > It's not true in general. For example: a(n) = 5 if n is odd, 1 if n is even b(n) = 1 if n is odd, 3 if n is even B(n) = 2 n - 1 if n is odd, 2 n if n is even B(B(n)) = 4n - 3 if n is odd, 4n if n is even sum_{k=1}^n a(k) b(k) = 4n + 1 if n is odd, 4 n if n is even = B(B(n)) + O(1) A(n) = 3n + 2 if n is odd, 3n if n is even So L = 3 but B(B(n))/B(n) -> 2 as n -> infinity. -- Robert Israel israel@math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: question on model categories Originator: bergv@math.uiuc.edu (Maarten Bergvelt) let C be a model category. If f:X-->Y is a map in C between objects X > and Y which are both fibrant and cofibrant. Then f is a weak > equivalence iff f has a homotopy inverse. This is the Whitehead > theorem. I wonder if the implication æf has a homotopy inverse => f is a weak > equivalence is true for any map f:X-->Y, i.e. for non necessarily > fibrant and cofibrant X and Y. You are right: you don't need X or Y to be fibrant or cofibrant and your argument is quite correct, but I think it needs to be clarified a little bit. > It would be true, I think, if there is no mistake in the following > argument. > As usual, two maps are called left-homotopic if there exists a (good) > cylinder object and a homotopy such that the folding map factorizes. > This is not a equivalence relation. If two maps f, f' are left- > homotopic then h(f)=h(f') where h:C-->HoC denotes the usual functor > into the homotopy category. By the homotopy category, do you mean the quotient of C by the (equivalence relation generated by) the left homotopy relation? Then this is indeed a tautology. But then you would have to prove your next claim: > The morphism h(f) is a isomorphism in HoC > iff f is a weak equivalence. So, maybe your HoC is Quillen's HoC, the Quillen homotopy category, the localization of C with respect to weak equivalences. Then this second claim is a tautology, but your first implication f left homotopic to f' ==> h(f) = h(f') does not follow so immediately. > Now let f:X-->Y, g:Y-->X be two maps such > that gf is left-homotopic to the identity and fg is left-homotopic to > the identity. Then h(g)h(f)=id and h(f)h(g)=id and therefore f is an > isomorphism in HoC. Then it follows that f is a weak equivalence. Tony I would rather argue like this. By definition, a cylinder object for X means that you have an object Cyl(X) and three morfhisms i, j : X ---> Cyl(X) and p: Cyl(X) ---> X where p is a weak equivalence (and a fibatrion), and such that pi = 1 and pj = 1 , where 1 means the identity of X . So, by the 2 out 3 property of model categories, i and j are weak equivalences too. Equivalently, because if h: C ---> HoC is Quillen's localization functor (and HoC is Quillen's homotopy category), then h(p) h(i) = 1 and, h(p) being an isomorphism, implies that so is h(i) and then, because a morphism is a weak equivalence if and only if its image by h becomes an isomorphism, i is a weak equivalence. (This property holds on what Quillen called _closed_ model categories, but nowadays all model categories are closed.) Next, two maps f, g : X ---> Y are left homotopic if there exists a morphism H : Cyl(X) ---> Y such that Hi = f and Hj = g . Let now f: X ---> Y be a homotopy equivalence and g : Y ---> X a homotopy inverse of f . You will have then two homotopies G and H such that Gi_X = gf , Gj_X = 1_X and Hi_Y = fg , Hj_Y = 1_Y . So in HoC , you will obtain first h(G) h(j_X) = 1 and h(H) h(j_Y) = 1 . Thus, h(G) and h(H) are both isomorphisms. Therefore G and H are weak equivalences. So h(G) h(i_X) = h(g) h(f) and h(H) h(i_Y) = h(f) h(g) imply that both compositions h(g) h(f) and h(f) h(g) are automorphisms. So, both h(f) and h(g) are isomorphisms and f and g weak equivalences. Agusti Roig === Subject: Re: question on model categories Thread-Topic: question on model categories Thread-Index: AcmWop42cHK93l7dQcSceyVAck3fTQ== Accept-Language: en-GB > let C be a model category. If f:X-->Y is a map in C between objects X > and Y which are both fibrant and cofibrant. Then f is a weak > equivalence iff f has a homotopy inverse. This is the Whitehead > theorem. I wonder if the implication æf has a homotopy inverse => f is a weak > equivalence is true for any map f:X-->Y, i.e. for non necessarily > fibrant and cofibrant X and Y. You are right: you don't need X or Y to be fibrant or cofibrant and > your argument is quite correct, but I think it needs to be clarified a > little bit. It would be true, I think, if there is no mistake in the following > argument. > As usual, two maps are called left-homotopic if there exists a (good) > cylinder object and a homotopy such that the folding map factorizes. > This is not a equivalence relation. If two maps f, f' are left- > homotopic then h(f)=h(f') where h:C-->HoC denotes the usual functor > into the homotopy category. By the homotopy category, do you mean the quotient of æC æby the > (equivalence relation generated by) the left homotopy relation? Then > this is indeed a tautology. But then you would have to prove your next > claim: The morphism h(f) is a isomorphism in HoC > iff f is a weak equivalence. So, maybe your æHoC æis Quillen's HoC, the Quillen homotopy > category, the localization of æC æwith respect to weak equivalences. > Then this second claim is a tautology, but your first implication æ æ f left homotopic to f' æ ==> æ h(f) = h(f') does not follow so immediately. Now let f:X-->Y, g:Y-->X be two maps such > that gf is left-homotopic to the identity and fg is left-homotopic to > the identity. Then h(g)h(f)=id and h(f)h(g)=id and therefore f is an > isomorphism in HoC. Then it follows that f is a weak equivalence. Tony I would rather argue like this. By definition, a cylinder object for > X æmeans that you have an object æCyl(X) æand three morfhisms æ æ æ i, j : X ---> Cyl(X) æ æ and æ æp: Cyl(X) ---> X where æp æis a weak equivalence (and a fibatrion), and such that æ æ æ æpi = 1 æ æand æ æ æpj = 1 , where æ1 æmeans the identity of æX . So, by the 2 out 3 property of > model categories, æi æand æj æare weak equivalences too. Equivalently, > because if æh: C ---> HoC æis Quillen's localization functor (and > HoC æis Quillen's homotopy category), then æh(p) h(i) = 1 æand, æh(p) > being an isomorphism, implies that so is æh(i) æand then, because a > morphism is a weak equivalence if and only if its image by æh æbecomes > an isomorphism, æi æis a weak equivalence. (This property holds on > what Quillen called _closed_ model categories, but nowadays all model > categories are closed.) Next, two maps æf, g : X ---> Y æare left homotopic if there exists a > morphism æH : Cyl(X) ---> æY æsuch that æHi = f æand æHj = g . Let now æf: X ---> Y æbe a homotopy equivalence and æg : Y ---> X æa > homotopy inverse of æf . You will have then two homotopies æG æand æH > such that æ æ æ æGi_X = gf , Gj_X = 1_X æ æand æ æ Hi_Y = fg , Hj_Y = 1_Y . So in æHoC , you will obtain first æ æ æ æ h(G) h(j_X) = 1 æ æ æ and æ æ æh(H) h(j_Y) = 1 . Thus, æh(G) æand æh(H) æare both isomorphisms. Therefore æG æand æH > are weak equivalences. So æ æ æ h(G) h(i_X) = h(g) h(f) æ æand æ æ æ æh(H) h(i_Y) = h(f) h(g) imply that both compositions æh(g) h(f) æand æh(f) h(g) æare > automorphisms. So, both æh(f) æand æh(g) æare isomorphisms and æf > and æg æweak equivalences. Agusti Roig Is it true that homotopy equivalence is an equivalence relation on the objects of C? This is, two objects X and Y are homotopy equivalent iff there exists a homotopy equivalence f:X-->Y. Weak equivalence between objects (defined in the same way) is not an equivalence relation in general, I think. Tony === Subject: Re: question on model categories Thread-Topic: question on model categories Thread-Index: AcmW8eJTJnlcIv4cSk2ATP0Efoibdg== Accept-Language: en-GB Is it true that homotopy equivalence is an equivalence relation on > the objects of C? This is, two objects X and Y are homotopy equivalent > iff there exists a homotopy equivalence f:X-->Y. Weak equivalence > between objects (defined in the same way) is not an equivalence > relation in general, I think. Tony As for weak equivalences: if by X and Y being weakly equivalent you mean that there exists a weak equivalence f: X ---> Y , then you probably have the doubt that this relation wouldn't have simmetry, right? (Reflexivity follows from the fact that the class of weak equivalences in a model category includes isomorphisms, hence identities, because the class of we is necessarily saturated; transitivity is obvious.) Well, it's true that in general you can't produce a weak inverse of f in the same category C where f lives, so in general, with this definition, you probably won't have simmetry. But if you are willingly ready to change this previous definition of weakly equivalent objects by being isomorphic in the localized (Quillen homotopy) categoy HoC then the problem disappears. As for homotopy equivalences, you don't have problems with reflexivity (because pi = 1 and pj = 1 implies that identities are homotopy equivalences, the homotopy being p ), nor with simmetry (because the definition of homotopy equivalent objects demands the existence of a homotopy inverse). So we are left with transitivity. Here the problem is the compatibility between homotopy relation and composition of morphisms. This means that we would like to have: (r) f sim g ==> fu sim gu and (l) f sim g ==> vf sim vg . If we had conditions (r) and (l) verified for every u and v and every pair of homotopic maps f and g , then transitivity would follow immediately. But, in general, in a Quillen model category, we only have (r) for right homotopies (defined with path objects) and (l) for left homotopies (defined with cylinder objects). Nevertheless there is an easy way to have both at the same time: you only need homotopy relation defined by a _functorial_ path or cylinder object. (By easy, I mean that is not strange to have such functorial objects; e.g., the usual cylinder objects for topological spaces and chain complexes, or the known path object for dgc algebras, operads...) Alternatively, you can take the subcategory of cofibrant-fibrant objects, where left and right homotopy coincide, are equivalence relations and compatible with composition. Agusti Roig PS. You'll find most of the results in Model categories and its localizations, by P.S. Hirschhorn. === Subject: Re: areas of intersecting circles Originator: bergv@math.uiuc.edu (Maarten Bergvelt) > On 21 Feb, 20:00, Mariano Su.87rez-Alvarez But I think the OP is interested in having many circles. > Your approach would be incredibly complicated... No more complicated than any other proceedure. For a pair of circles > we can calculate theta as follows. We have a triangle sides r1, r2, X where X is the distance between the > centers. We know a^2 = b^2 + c^2 -2bcCos A This gives A as Cos^-1((r1^2 + X^2- r2^2)/2Xr1) Similarly for B This is an extremely simple computation. I can't see how you can get > it any simpler. Area = 0.5(r2^2(A-Sin(A))+r1^2(B-Sin(B)) A program to calculate it can be written in a very few lines. Did you not read the post you were replying to? > The OP is interested in the situation inwhich there are > *many* circles! It is quite clear that for two your > procedure works nicely. Now, can you give us the > formulas for the area of the intersection of 25 circles? -- m- Hide quoted text - - Show quoted text - Sorry. I don't think there is a deep theory of this. It is a question > of taking each problem as it comes. The algorithmic proceedure I would > use is the following. 1) Find circles which are completely inside each other. > Defined as being when d(O1 - O2) < |r1-r2|. You take the smaller > circle if in first group, the larger circle if in second. If a first > is within a second the problem is impossible. 2) Find d(nm)in all the first group. d(mn) = d(On - Om)if rn+rm it is impossible (they never intersect) 3) Take smallest circle of first group. Find all points of > intersection. Find common arc. You will now have reduced the problem > to just 3 circles (first group). The smallest circle, and the circles > that cross at the two ends of the common arc. This will give you a > figure to integrate. 4) You then need to check the points of intersection of your second > set of circles to see whether the area defined by the 3 arcs is > totally outside, totally inside of partly outside partly inside the > second set of circles. Proving impossibility is quite efficient computationally. The hardest > case is when your 3 arcs are partly inside and partly outside the > second set of circles. But this approach is in no way related to the approach you used for the case of two circles, and in fact is based on exactly the same idea than Robert's, but made more complicated. I can't see what motivated your I think there is an even simpler way of looking at this in response to Robert :-) -- m === Subject: Re: graph Originator: bergv@math.uiuc.edu (Maarten Bergvelt) > Let G be a(p,q) graph determine a sharp bound f(p) such that if q>f > (p) then G is disconnected? If number of vertices = 2 + number of edges then G is disconnected. If number of vertices = 1 + number of edges then G may be connected. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Looking for a paper by V.A.Lebesgue Originator: bergv@math.uiuc.edu (Maarten Bergvelt) sinuslog@gmail.com a .8ecrit : > For months I've been trying to find an old paper on Number Theory. I > am retired and have no access to specialized sources. I would be very > scan it if necessary and email it to me: - V. A. Lebesgue, Nouv. Ann. Math., 8, 1849, 350 I found the reference in L.E.Dickson's History of the Theory of > Numbers, vol.1, p.332:, ref. 35 > http://ia310813.us.archive.org/0/items/historyoftheoryo01dick/historyoftheor\ y o01dick.pdf > (Caution: 35 Meg) On page 332, Dickson gives 2 alternate references, both with V. A. > Lebesgue as the author: > - Introduction a la theorie des nombres, 1862, 51-53 > - Exercices d'analyse numerique, 1859, 31-32, 118-9 There are 2 more references in german and one in dutch on page 335, > and although I can't read these languages, I could probably find my > way if I have to: > - E. Dintz, Zeitschrift fur das Realschulwesen, Wien, 27. 1902, 654-9, > 722 > ^_- Merten's Vorlesungen uber Zahlentheorie > - De Jough, Nieuw Archief voor Wiskunde, (2), 5, 1901, 262-7 > prints : Exercices d'analyse num.8erique by Lebesgue is there too : Hoping it helped, Raymond === Subject: Re: Looking for a paper by V.A.Lebesgue Originator: bergv@math.uiuc.edu (Maarten Bergvelt) > For months I've been trying to find an old paper on Number > Theory. I am retired and have no access to specialized > sources. I would be very grateful to the person who would > and email it to me: - V. A. Lebesgue, Nouv. Ann. Math., 8, 1849, 350 Most every old mathematics journal is on the internet now, either at archiving sites for math journals or V. A. Lebesgue, Extrait des exercises d'analyse num.8erique, Nouvelles Annales de Mathematiques (1) 8 (1849), 347-353. Dave L. Renfro === Subject: Call for Contribution Thread-Topic: Call for Contribution Thread-Index: AcmXWc2SOTTvii8NTMy4jrzdY1lmjQ== Accept-Language: en-GB Annals. Computer Science Series is an international electronic journal publishing original scientific research papers presented in the framework of the 5th International Conference Actualities and Perspectives in Hardware and Software, event organized under the high patronage of the Romanian Academy. Annals. Computer Science Series intends to stimulate the research activity and to establish interactions between Romanian and foreign researchers, teachers, B. Sc. and M. Sc. students and specialists, through its on-line version: http://anale-informatica.tibiscus.ro/?page=00_primapagina&lang=en The topics of interest are: [Eth] Algorithms & Complexity [Eth] Computational Mathematics [Eth] Graphics & Multimedia [Eth] Fundamentals of Programming [Eth] Operating Systems [Eth] Software Engineering & Industrial Applications [Eth] Informational Society [Eth] Applied Computer Science in Medicine [Eth] Applied Computer Science in Economics This list is not intended to be exhaustive. We invite submissions, before March 27th , 2009, in the form of: [Eth] full research papers [Eth] short papers (work in progress) [Eth] systems\.89 descriptions and software demonstrations For further information and enquiries you might have, please contact the Editorial Board at conference.fcia@tibiscus.ro Assist. Prof. Alexandra Fortis, Ph. D. Candidate Assistant Editor-in-Chief, Annals. Computer Science Series \.8bTibiscus\.8a University of Timisoara, Romania Faculty of Computers and Applied Computer Science === Subject: A special case of weighted mean Accept-Language: en-GB [...] generalised too much. Conditions were actually more restrictive. Namely: (a(n)) and (b(n)) are bounded positive integer sequences satisfying for n>=1: (i) b(1)=1 and b(k+1)=a(n) for B(n-1)< k <= B(n) (ii) B(B(n))=1-a(n)+ sum_{k=1}^{n}a(k)b(k) (iii) Limit n--> infty A(n)/n=L>0 exists Under these assumptions my guess is: B(B(n))/B(n)-->L as n->infty Note (i) and (ii) are simply related and my hope was to find a generalised version of (ii) implying directly B(B(n))/B(n)-->L as n->infty. Benoit Traxler === Subject: Re: Are the integers as normed vector space complete? > Yes. Every discrete metric space is complete. Careful. Every uniformly discrete metric space is complete, but not every metric space that is discrete (as a topological space) is complete. That is: a metric space (X,d) is (topologically) discrete if for each x in X there is epsilon > 0 such that d(y,x) < epsilon implies y=x. It is uniformly discrete if there is a single epsilon > 0 such that for all x and y, d(y,x) < epsilon implies y = x. If the metric space is a group with a metric invariant under the group operations, the same epsilon that works for one x works for all of them, so in this case discreteness does imply completeness. -- Robert Israel israel@math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Are the integers as normed vector space complete? posting-account=_l4K0QkAAAC09JhOoK_ZfoJKXOmr_jZf Gecko/2009020911 Ubuntu/8.04 (hardy) Firefox/3.0.6,gzip(gfe),gzip(gfe) On Feb 25, 3:40æpm, Robert Israel > Yes. Every discrete metric space is complete. Careful. æEvery uniformly discrete metric space is complete, but not > every metric space that is discrete (as a topological space) is complete. -- Arturo Magidin, sans .sig === Subject: Re: How to prove this <27107722.1235573020209.JavaMail.jakarta@nitrogen.mathforum.org>, > Certainly the thing being summed, which in my version is > ---> arctan(1/sqrt(k)) > and in your version is > ---> arcsin(1/sqrt(k)) > could likely be shown to be irrational. > I would even bet it could be shown to be an > irrational multiple of Pi. Well, not always - for k = 1, they're both rational multiples of pi. arctan(1 / sqrt 3) and arcsin(1 / sqrt 4) are also rational multiples of pi. > But neither of these (if shown) would imply that > the *sum* of the arctans or arcsins could not be > a multiple of pi. What he said. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Alain's T-approximation does not work (was Re: Continuous iteration of another quadratic map) posting-account=fwSgtAkAAACFnX70ssKwbvm9_oCZVHrx Gecko/2009011913 Firefox/3.0.6,gzip(gfe),gzip(gfe) > On Feb 23, 2:50æam, alainvergh...@gmail.com > Am 22.02.2009 23:20 schrieb mike3: What I'd be more curious about though is how you got that > series you posted in the message you gave in terms of the > height. As I'd be curious about a similar case for f(0), i.e. > g(-phi). (phi = golden ratio) Then I could see what that period > 2 orbit looks like when done continuously. > First you build the matrix-operator for the function. The function is æ æ g(x) = a*x + 1*x^2 æ æ æ æ æ æ æ æ where æ æ a = (1+sqrt(5)) = 2*phi For iteration you need g(x) and its powers g(x)^2, g(x)^3,... To compute these powers it suffices to operate on the coefficients only > (square,cube,... the formal powerseries (which is here only a polynomial)) For iteration you combine these coefficients in the same way as you would > do a matrix-multiplication. So write the coefficients of the powers > of g(x) into a matrix: æ g(x)^0 æ = 1 æ+ æ 0*x æ+ æ 0*x^2 + æ0*x^3 + 0*x^4 + ... > æ g(x)^1 æ = 0 æ+ æ a*x æ+ æ 1*x^2 + æ0*x^3 + 0*x^4 + ... > æ g(x)^2 æ = 0 æ+ æ 0*x æ+ a^2*x^2 + 2a*x^3 + 1*x^4 + ... > æ ... > up to a reasonable dimension (to get well approximated results > for fractional iterates) The matrix G has then only the coefficients of the formal powerseries > (and I write it in transposed form) æG = æ 1 æ. æ . æ æ. æ . > æ æ æ æ. æa æ . æ æ. æ . > æ æ æ æ. æ1 æa^2 æ . æ . > æ æ æ æ. æ. 2a æ æa^3 æ. > æ æ æ æ.... ---------------------------------------------- > Second, the use of the matrix G Then a vector V(x) = [1,x,x^2,x^3,...] up to the selected dimension gives > using æ V(x) * G = [ 1, g(x), g(x)^2, g(x)^3 ,... ] In Pari/GP write (for a given dimension n) gp> æV(x) = vector(n,c,x^(c-1)) gp> æY = V(1) * G > æ then Y = [1, g(1) , g(1)^2, g(1)^3,...] correctly up to a certain column > æ The remaining columns are wrong due to truncation to finite size. Since the form of the result-vector is again like V(), a vector of consecutive > powers, the iteration can be performed by the formal relation æ V(x) æ æ* G = V(g(x)) > æ V(g(x)) * G = V(g(g(x)) > æ ... > and so on. However: this is exactly valid only in the limit, where we > assume infinite dimension. For integer iterates k this can be written as æV(gÁk(x)) = V(x) * G^k æ æ æ(theoretically) gp> æ æ æ æY = V(x) * G^k æ æ æ(practically) > æ where Y ~ V(gÁk(x)) æ æ æ æ æonly approximately which is true for the case of infinite dimension, and only approximate > with finite dimension. Just try this in Pari/GP. ----------------------------------------------------------- > 3) fractional iterates Next step is fractional iterate. It is an obvious idea to use > fractional powers of G for this. Fractional powers of G can either be determined by binomial-expansion > with G and I as arguments or by diagonalization. > The general formula for diagonalization of a matrix is æ æG æ = W * D * W^-1 æ æ æ æ æ æ æ// D is diagonal > and > æ æG^h = W * D^h * W^-1 So to find W and D in Pari/GP there is the function mateigen, which > gives a version of W first. gp> æ W = mateigen(G) Then compute WInv as inverse of W gp> æ WInv = W^-1 Then compute D as gp> æ D = WInv * G * W Then you can define a pari-function for arbitrary powers of D: gp> æ ædpow(x) = matrix(n,n,r,c,if(r==c,D[r,c]^x)) and then write gp> æ æGPow = W * dpow(0.5) * WInv to store G^0.5 in a constant matrix. Then you use gp> æ æ æY = V(x) * GPow æ æand Y is approximately ~ V(gÁh(x)) or just define the scalar function, using the relevant coefficients of GPow gp> æ gpow(x) = sum(k=0,n-1, x^k * GPow[1+k,2]) ----------------------------------------- > 4) Remarks: However, the approximation may be poor. I use dimension n=64 for > standard discussion and n=128 if the approximation is difficult. > The mateigen-procedure in Pari/GP may perform poorly because it > doesn't exploit the easiness of the solution for triangular matrices, > for which we can even get exact solutions. I've a version for > triangular matrices of the special form here (matrix-operators on V(x)) > implemented, which gives the eigensystem-decomposition of n=64x64 > or 128x128 triangular-matrices in a second or two and even symbolically > if the matrix-dimension is small. > I could mail it to you, but I think I've already posted it in the > tetration-forum in the matrix-method-thread. Also I use Euler-summation for convergence-acceleration. But this > needs some more explanations - try this first to get used to it. Gottfried Bonjour, I do believe you're using heavy tools. > What about approximating x^2-1 by a conjugate of > a known iterate :Ex (x^2-1) = f-1(f(x))^2 - 2) ,T(x)=x^2-2 > or æ (x^2-1) =~f^-1 o T o f (x) > and (x^2-1)^[r] =~f^-1 o T^r o f (x) The trouble is it only seems to work when x is large. What about small > x? There's even more trouble. The approximation does not work at all! :( Try r = 2, x = 4. You get ~ 222.9955156499789347277396687. Which is close to 223 -- and that's (4^2 - 1)^2 - 2 not (4^2 - 1)^2 - 1. It's even worse for close to 49727 -- and *that* is just ((4^2 - 1)^2 - 2)^2 - 2 not ((4^2 - 1)^2 - 1)^2 - 1. It is not approaching f^n(x) where f(x) = x^2 - 1, it's approaching f^(n-1)(x^2 - 1) where f(x) = x^2 - 2! === Subject: Re: Alain's T-approximation does not work (was Re: Continuous iteration of another quadratic map) posting-account=06BQLAoAAADoC7Y4z9FWcUwGvMa7xMG9 7.4),gzip(gfe),gzip(gfe) On Feb 23, 2:50æam, alainvergh...@gmail.com > Am 22.02.2009 23:20 schrieb mike3: What I'd be more curious about though is how you got that > series you posted in the message you gave in terms of the > height. As I'd be curious about a similar case for f(0), i.e. > g(-phi). (phi = golden ratio) Then I could see what that period > 2 orbit looks like when done continuously. > First you build the matrix-operator for the function. The function is æ æ g(x) = a*x + 1*x^2 æ æ æ æ æ æ æ æ where æ æ a = (1+sqrt(5)) = 2*phi For iteration you need g(x) and its powers g(x)^2, g(x)^3,... To compute these powers it suffices to operate on the coefficients only > (square,cube,... the formal powerseries (which is here only a polynomial)) For iteration you combine these coefficients in the same way as you would > do a matrix-multiplication. So write the coefficients of the powers > of g(x) into a matrix: æ g(x)^0 æ = 1 æ+ æ 0*x æ+ æ 0*x^2 + æ0*x^3 + 0*x^4 + ... > æ g(x)^1 æ = 0 æ+ æ a*x æ+ æ 1*x^2 + æ0*x^3 + 0*x^4 + ... > æ g(x)^2 æ = 0 æ+ æ 0*x æ+ a^2*x^2 + 2a*x^3 + 1*x^4 + ... > æ ... > up to a reasonable dimension (to get well approximated results > for fractional iterates) The matrix G has then only the coefficients of the formal powerseries > (and I write it in transposed form) æG = æ 1 æ. æ . æ æ. æ . > æ æ æ æ. æa æ . æ æ. æ . > æ æ æ æ. æ1 æa^2 æ . æ . > æ æ æ æ. æ. 2a æ æa^3 æ. > æ æ æ æ.... ---------------------------------------------- > Second, the use of the matrix G Then a vector V(x) = [1,x,x^2,x^3,...] up to the selected dimension gives > using æ V(x) * G = [ 1, g(x), g(x)^2, g(x)^3 ,... ] In Pari/GP write (for a given dimension n) gp> æV(x) = vector(n,c,x^(c-1)) gp> æY = V(1) * G > æ then Y = [1, g(1) , g(1)^2, g(1)^3,...] correctly up to a certain column > æ The remaining columns are wrong due to truncation to finite size. Since the form of the result-vector is again like V(), a vector of consecutive > powers, the iteration can be performed by the formal relation æ V(x) æ æ* G = V(g(x)) > æ V(g(x)) * G = V(g(g(x)) > æ ... > and so on. However: this is exactly valid only in the limit, where we > assume infinite dimension. For integer iterates k this can be written as æV(gÁk(x)) = V(x) * G^k æ æ æ(theoretically) gp> æ æ æ æY = V(x) * G^k æ æ æ(practically) > æ where Y ~ V(gÁk(x)) æ æ æ æ æonly approximately which is true for the case of infinite dimension, and only approximate > with finite dimension. Just try this in Pari/GP. ----------------------------------------------------------- > 3) fractional iterates Next step is fractional iterate. It is an obvious idea to use > fractional powers of G for this. Fractional powers of G can either be determined by binomial-expansion > with G and I as arguments or by diagonalization. > The general formula for diagonalization of a matrix is æ æG æ = W * D * W^-1 æ æ æ æ æ æ æ// D is diagonal > and > æ æG^h = W * D^h * W^-1 So to find W and D in Pari/GP there is the function mateigen, which > gives a version of W first. gp> æ W = mateigen(G) Then compute WInv as inverse of W gp> æ WInv = W^-1 Then compute D as gp> æ D = WInv * G * W Then you can define a pari-function for arbitrary powers of D: gp> æ ædpow(x) = matrix(n,n,r,c,if(r==c,D[r,c]^x)) and then write gp> æ æGPow = W * dpow(0.5) * WInv to store G^0.5 in a constant matrix. Then you use gp> æ æ æY = V(x) * GPow æ æand Y is approximately ~ V(gÁh(x)) or just define the scalar function, using the relevant coefficients of GPow gp> æ gpow(x) = sum(k=0,n-1, x^k * GPow[1+k,2]) ----------------------------------------- > 4) Remarks: However, the approximation may be poor. I use dimension n=64 for > standard discussion and n=128 if the approximation is difficult. > The mateigen-procedure in Pari/GP may perform poorly because it > doesn't exploit the easiness of the solution for triangular matrices, > for which we can even get exact solutions. I've a version for > triangular matrices of the special form here (matrix-operators on V(x)) > implemented, which gives the eigensystem-decomposition of n=64x64 > or 128x128 triangular-matrices in a second or two and even symbolically > if the matrix-dimension is small. > I could mail it to you, but I think I've already posted it in the > tetration-forum in the matrix-method-thread. Also I use Euler-summation for convergence-acceleration. But this > needs some more explanations - try this first to get used to it. Gottfried Bonjour, I do believe you're using heavy tools. > What about approximating x^2-1 by a conjugate of > a known iterate :Ex (x^2-1) = f-1(f(x))^2 - 2) ,T(x)=x^2-2 > or æ (x^2-1) =~f^-1 o T o f (x) > and (x^2-1)^[r] =~f^-1 o T^r o f (x) The trouble is it only seems to work when x is large. What about small > x? There's even more trouble. The approximation does not work at all! :( Try r = 2, x = 4. You get ~ 222.9955156499789347277396687. Which > is close to 223 -- and that's (4^2 - 1)^2 - 2 not (4^2 - 1)^2 - 1. > It's even worse for > close to 49727 -- and *that* is just ((4^2 - 1)^2 - 2)^2 - 2 not > ((4^2 - 1)^2 - 1)^2 - 1. It is not approaching f^n(x) where f(x) = x^2 > - 1, > it's approaching f^(n-1)(x^2 - 1) where f(x) = x^2 - 2!- Masquer le texte \ des messages pr.8ec.8edents - - Afficher le texte des messages pr.8ec.8edents - Bonjour Mike, I am not sure I've made my idea very clear : t1(x)=x^2-1 which has to be iterated, t2(x)=x^2-2 continuous iterate supposed known, We may search a conjugaison: t1 = h^[-1] o t2 o h or x^2-1 = h^[-1] ( h(x)^2-2) = x^2-2 I try h(x) = x+1/(2*x) as a first approximation , going far away from zero ... Alain === Subject: Re: Alain's T-approximation does not work (was Re: Continuous iteration of another quadratic map) posting-account=fwSgtAkAAACFnX70ssKwbvm9_oCZVHrx Gecko/2009011913 Firefox/3.0.6,gzip(gfe),gzip(gfe) On Feb 26, 3:25æam, alainvergh...@gmail.com > On Feb 23, 2:50æam, alainvergh...@gmail.com > Am 22.02.2009 23:20 schrieb mike3: What I'd be more curious about though is how you got that > series you posted in the message you gave in terms of the > height. As I'd be curious about a similar case for f(0), i.e. > g(-phi). (phi = golden ratio) Then I could see what that period > 2 orbit looks like when done continuously. > First you build the matrix-operator for the function. The function is æ æ g(x) = a*x + 1*x^2 æ æ æ æ æ æ æ æ where æ æ a = (1+sqrt(5)) = 2*phi For iteration you need g(x) and its powers g(x)^2, g(x)^3,... To compute these powers it suffices to operate on the coefficients only > (square,cube,... the formal powerseries (which is here only a polynomial)) For iteration you combine these coefficients in the same way as you would > do a matrix-multiplication. So write the coefficients of the powers > of g(x) into a matrix: æ g(x)^0 æ = 1 æ+ æ 0*x æ+ æ 0*x^2 + æ0*x^3 + 0*x^4 + ... > æ g(x)^1 æ = 0 æ+ æ a*x æ+ æ 1*x^2 + æ0*x^3 + 0*x^4 + ... > æ g(x)^2 æ = 0 æ+ æ 0*x æ+ a^2*x^2 + 2a*x^3 + 1*x^4 + ... > æ ... > up to a reasonable dimension (to get well approximated results > for fractional iterates) The matrix G has then only the coefficients of the formal powerseries > (and I write it in transposed form) æG = æ 1 æ. æ . æ æ. æ . > æ æ æ æ. æa æ . æ æ. æ . > æ æ æ æ. æ1 æa^2 æ . æ . > æ æ æ æ. æ. 2a æ æa^3 æ. > æ æ æ æ.... ---------------------------------------------- > Second, the use of the matrix G Then a vector V(x) = [1,x,x^2,x^3,...] up to the selected dimension gives > using æ V(x) * G = [ 1, g(x), g(x)^2, g(x)^3 ,... ] In Pari/GP write (for a given dimension n) gp> æV(x) = vector(n,c,x^(c-1)) gp> æY = V(1) * G > æ then Y = [1, g(1) , g(1)^2, g(1)^3,...] correctly up to a certain column > æ The remaining columns are wrong due to truncation to finite size. Since the form of the result-vector is again like V(), a vector of consecutive > powers, the iteration can be performed by the formal relation æ V(x) æ æ* G = V(g(x)) > æ V(g(x)) * G = V(g(g(x)) > æ ... > and so on. However: this is exactly valid only in the limit, where we > assume infinite dimension. For integer iterates k this can be written as æV(gÁk(x)) = V(x) * G^k æ æ æ(theoretically) gp> æ æ æ æY = V(x) * G^k æ æ æ(practically) > æ where Y ~ V(gÁk(x)) æ æ æ æ æonly approximately which is true for the case of infinite dimension, and only approximate > with finite dimension. Just try this in Pari/GP. ----------------------------------------------------------- > 3) fractional iterates Next step is fractional iterate. It is an obvious idea to use > fractional powers of G for this. Fractional powers of G can either be determined by binomial-expansion > with G and I as arguments or by diagonalization. > The general formula for diagonalization of a matrix is æ æG æ = W * D * W^-1 æ æ æ æ æ æ æ// D is diagonal > and > æ æG^h = W * D^h * W^-1 So to find W and D in Pari/GP there is the function mateigen, which > gives a version of W first. gp> æ W = mateigen(G) Then compute WInv as inverse of W gp> æ WInv = W^-1 Then compute D as gp> æ D = WInv * G * W Then you can define a pari-function for arbitrary powers of D: gp> æ ædpow(x) = matrix(n,n,r,c,if(r==c,D[r,c]^x)) and then write gp> æ æGPow = W * dpow(0.5) * WInv to store G^0.5 in a constant matrix. Then you use gp> æ æ æY = V(x) * GPow æ æand Y is approximately ~ V(gÁh(x)) or just define the scalar function, using the relevant coefficients of GPow gp> æ gpow(x) = sum(k=0,n-1, x^k * GPow[1+k,2]) ----------------------------------------- > 4) Remarks: However, the approximation may be poor. I use dimension n=64 for > standard discussion and n=128 if the approximation is difficult. > The mateigen-procedure in Pari/GP may perform poorly because it > doesn't exploit the easiness of the solution for triangular matrices, > for which we can even get exact solutions. I've a version for > triangular matrices of the special form here (matrix-operators on V(x)) > implemented, which gives the eigensystem-decomposition of n=64x64 > or 128x128 triangular-matrices in a second or two and even symbolically > if the matrix-dimension is small. > I could mail it to you, but I think I've already posted it in the > tetration-forum in the matrix-method-thread. Also I use Euler-summation for convergence-acceleration. But this > needs some more explanations - try this first to get used to it. Gottfried Bonjour, I do believe you're using heavy tools. > What about approximating x^2-1 by a conjugate of > a known iterate :Ex (x^2-1) = f-1(f(x))^2 - 2) ,T(x)=x^2-2 > or æ (x^2-1) =~f^-1 o T o f (x) > and (x^2-1)^[r] =~f^-1 o T^r o f (x) The trouble is it only seems to work when x is large. What about small > x? There's even more trouble. The approximation does not work at all! :( Try r = 2, x = 4. You get ~ 222.9955156499789347277396687. Which > is close to 223 -- and that's (4^2 - 1)^2 - 2 not (4^2 - 1)^2 - 1. > It's even worse for > close to 49727 -- and *that* is just ((4^2 - 1)^2 - 2)^2 - 2 not > ((4^2 - 1)^2 - 1)^2 - 1. It is not approaching f^n(x) where f(x) = x^2 > - 1, > it's approaching f^(n-1)(x^2 - 1) where f(x) = x^2 - 2!- Masquer le texte des messages pr.8ec.8edents - - Afficher le texte des messages pr.8ec.8edents - Bonjour Mike, I am not sure I've made my idea very clear : t1(x)=x^2-1 æwhich has to be iterated, > t2(x)=x^2-2 æcontinuous iterate supposed known, > We may search a conjugaison: > æ æt1 = h^[-1] o t2 o h > or x^2-1 = h^[-1] ( h(x)^2-2) = x^2-2 > I try æh(x) = x+1/(2*x) as a first approximation , > going far away from zero ... Alain Hmm. This seems to work better, although the superfunction identity (see my post) does not seem to transform it correctly to get iteration on smaller x-values. === Subject: sum of fibonaccis: = -1 ? The sum of the fibonacci-numbers is infinite; the partial sums diverge. However if I use the geometric series-formula I get sum = - 1 It fits also the pattern, which emerges, if I sum with some integer m>1 sum k=1,inf fib(k)/m^k = m/(m^2 - m -1) Is this appropriate? Gottfried Helms === Subject: Re: sum of fibonaccis: = -1 ? Am 25.02.2009 23:40 schrieb Gottfried Helms: > Is this appropriate? Gottfried Helms Hi - that I can defend that assignment of -1 to the sum by refering to the analytic continuation of the powerseries 1 + x + x^2 + ... and the closed-form-formula for the geometric series. (is the same way appropriate as...) However, Bill's invention makes me lough... :-) Very nice! Well, I took the result from the two geometric series, which occur, if I chose the Lucas-representation for the fibonacci numbers, and the sum of two analytic continued terms should be acceptable as closed-form formula too. In another context I had the *infinite* sum (though converging) of terms which all result from analytic continued geometric series - but that seems inappropriate: a concurring method gave different result which was better founded. Can we pinpoint the reason for an error/the error in the case of infinitely many values taken by analytic-continuation, or possibly have a means to quantify it/describe it by a function? Gottfried === Subject: Re: sum of fibonaccis: = -1 ? posting-account=AFsgCgkAAAA3VOfxqn2cTB2LbLN3nbER Gecko/20070319,gzip(gfe),gzip(gfe) One can also get it this way... Sumfib = 1 + 3 + 8 + 21 + 55 + ... : 1 + 2 + 5 + 13 + 34 + ... thus... : Sumfib +1 =(1+1)+ 3 + 8 + 21 + 55 + ... : (1 + 2)+ 5 + 13 + 34 + ... : : = (2 + 3) + 8 + 21 + 55 + ... : + (3 + 5) + 13 + 34 + ... : : = 0 + (5 + 8) + 21 + 55 + ... : 0 + (8 + 13) + 34 + ... : : = 0 + 0 +(13 + 21) + 55 + ... : 0 + 0 + (21 + 34) + ... : : = 0 + 0 + 0 + (34 + 55) + ... : 0 + 0 + 0 + (55 + ... : : = 0 + 0 = 0 : thus Sumfib = -1 . This is OC a totally Illegal Telescoping Cancellation (Infinite), (ITSI), but this itsi bitsi sort of calculation can be made rigorous by most of the summability methods Robert Israel alluded to (& Ben Goddard used); details in Hardy's Divergent Series. -- Wittering William * What an intellectual! * - Magdalen College Oxford, King's College Cambridge, * Imperial College London, Trinity College Dublin, * the Sorbonne, Harvard, MIT - he's heard of them all! === Subject: Re: sum of fibonaccis: = -1 ? The sum of the fibonacci-numbers is infinite; the partial sums > diverge. However if I use the geometric series-formula I get > sum = - 1 It fits also the pattern, which emerges, if I sum with some > integer m>1 sum k=1,inf fib(k)/m^k = m/(m^2 - m -1) Is this appropriate? It's appropriate as long as you make it clear that this is not literally a sum, but rather the result of some specified summability method. -- Robert Israel israel@math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: sum of fibonaccis: = -1 ? @mid.dfncis.de: The sum of the fibonacci-numbers is infinite; the partial sums > diverge. However if I use the geometric series-formula I get > sum = - 1 It fits also the pattern, which emerges, if I sum with some > integer m>1 sum k=1,inf fib(k)/m^k = m/(m^2 - m -1) Is this appropriate? It's as appropriate as sum_{k=0}^{inf} 2^k = 1/(1-2) = -1. B. -- Cheerfully resisting change since 1959. === Subject: Re: sum of fibonaccis: = -1 ? posting-account=Z3AipgkAAABkoMfyNwddSxsYhXHi5CDt MathPlayer 2.10d; .NET CLR 1.1.4322; PeoplePal 3.0),gzip(gfe),gzip(gfe) > @mid.dfncis.de: The sum of the fibonacci-numbers is infinite; the partial sums > diverge. However if I use the geometric series-formula I get > æsum = æ- 1 It fits also the pattern, which emerges, if I sum with some > integer m>1 æ sum k=1,inf fib(k)/m^k æ= m/(m^2 - m -1) Is this appropriate? It's as appropriate as sum {k=0}^{inf} æ2^k = 1/(1-2) = -1. B. But that sum is completely appropriate! It is a true infinite sum converging to the limit you give. At least it does if you are in the filed of 2-adic numbers. I can't think of a similar justification for the sum of the Fibonacci numbers, since we are raising units to powers and so there is no obvious way to create a justification. Achava === Subject: Re: sum of fibonaccis: = -1 ? Am 26.02.2009 06:19 schrieb Achava Nakhash, the Loving Snake: > (...) > I can't think of a similar justification for the sum of the Fibonacci > numbers, since we are raising units to powers and so there is no > obvious way to create a justification. Hi Achava - I don't understand this. raising units to powers - where do we this here? Would you mind to explain further? Btw, in my other post I said, that I got the sum by considering the lucas form fib(n) = (phi^n - (1-phi)^n)/sqrt(5) I just took the geometric series fibsum1 = sum {n>0} phi^n = 1/(1 - phi) // by anal.cont. of geometric series = - phi fibsum2 = sum {n>0} (1-phi)^n = 1/(1 - (1-phi)) // by anal.cont. of geometric series = 1/phi = phi-1 and fibsum = (fibsum1 - fibsum2)/sqrt(5) = -(2 phi - 1)/sqrt(5) = -1 Gottfried === Subject: Re: JSH: Factorization and the global economy Bull. Orders of Magnitude. You have no idea at all. >Ah. And you know this how??? >If the algorithms that you claim are 'far more secure' are known to >the >public, then they become available to all. e.g. Baton. See the NSA's >website. Look up their type 1 crypto algorithms. >If, however, they are secret, and you are privy to them, then your >revealing their existence and their security levels is a criminal >security violation. I know clearances and responcabilities very well, thank you. Just do a lot \ of reading and searching and you can run into several of the systems, public \ with some interesting info, watered down, not specific, to what is really there. And how would anyone know that ? NSA website; Cool they have newly declassified crypto from 1938. a lot of other good reading for sci.crypt for background they do have a declassified TS on cold war, interesting they would post such on the internet, known generic stuff, but why show a real declassified? ln 1968 a Soviet Golf-class nuclear submarine on patrol in the Pacific mysteriously went to the bottom with all hands. The Soviets could not locate the wreck, but the U.S. Navy could, and the U.S. began to study the feasibility of capturing it. (all rest white-ed out) They have some code patches for lenux too. If you factored some large #s that is quite impressive, and I wont bother you. I know of another person Brittish who was doing it, was obsessed with it and went off the deep end for good about 5 years ago, but he had factored \ some really big ones, not the blind search stuff either. === Subject: Re: JSH: Factorization and the global economy > Another moron. Do you know who you are talking to?? æHave you actually > done any crypto work for the NSA or DOD?? > Other methods are NOT 'far more secure'. > And the only method that is impossible to break is > a true OTP. æAnd key establishment problems make it > totally impractical to implement. æOne needs a secondary > communication channel for key establishment. Yes, but. æ What about quantum key distribution? It has not yet practical except over short distances. >(a few 10's of miles). Maintaining coherence over >longer distances has been a problem. I expect that these >difficulties will be solved in time. But it is not yet >ready for 'prime time'. > James, you said that you knew next to nothing about cryptography, so I will amplify some of the points Bob makes here. >However, this isn't the only problem. The key must be >truly random, not just pseudo-random, No, you cannot just use rand(). You need TRUE random data which is difficult to generate. A great deal of effort has been expended on this problem - true random data is definitely a problem. Google Yarrow or Fortuna for a couple of software approaches to the problem. For a hardware approach have a look at http://www.av8n.com/turbid/paper/turbid.htm >and it must be as long as the text it is encrypting. That means you need lots and lots of true random data. You do not just have a problem of data quality, you also have a problem of data quantity. >Furthermore, it can only be used ONCE. This is the big one. Once means once. A two time pad is as secure as transmitting the plaintext - not secure at all. Read up on the Venona break of Soviet cyphers: http://en.wikipedia.org/wiki/Venona_project Used only once can impose a lot of restrictions. For example, if your computer operating system can write a running program to a swapfile on disk then you might have to go as far as terminating your hard disk with extreme prejudice (and a sledgehemmer) and dissolving the result in acid. Once means once and once only. These issues are nothing to do with QKD, but are inherent in the use of a One Time Pad. The mathematical proof of its unbreakability depends critically on having TRUE random data used ONCE. If either condition is not met then the mathematical proof is not applicable and the resulting system is breakable. rossum >Thus, if you are encrypting a 10MB >message, you need a random 10MB that must be generated, >communicated, used ONCE, then thrown away. This makes >key management one royal pain in the *ss. Apparently, you, Pubkeybreaker, are known to be The > Real Thing to others in sci.crypt. It seems like too > good an opportunity to pass up, to find out whether > quantum key distribution is The Real Thing too. It is real. It works. But not yet on a large enough scale to >be useful. It can't yet be used if one wants to (say) >encrypt a message with OTP going from LA to (say) Miami. > Apparently, you also enjoy a good flame. I don't mind receiving flames. I have an asbestos suit :-) >Flame away. > === Subject: Re: JSH: Factorization and the global economy James, you said that you knew next to nothing about cryptography, so I > will amplify some of the points Bob makes here. However, this isn't the only problem. The key must be >truly random, not just pseudo-random, No, you cannot just use rand(). You need TRUE random data which is > difficult to generate. A great deal of effort has been expended on > this problem - true random data is definitely a problem. Google > Yarrow or Fortuna for a couple of software approaches to the > problem. For a hardware approach have a look at > http://www.av8n.com/turbid/paper/turbid.htm and it must be as long as the text it is encrypting. That means you need lots and lots of true random data. You do not > just have a problem of data quality, you also have a problem of data > quantity. But you're generating all these lovely quantum measurements anyway. Can't the measurements that are part of the quantum bit-passing also be the source of the one time pad? >Furthermore, it can only be used ONCE. This is the big one. Once means once. A two time pad is as secure as > transmitting the plaintext - not secure at all. Read up on the Venona > break of Soviet cyphers: http://en.wikipedia.org/wiki/Venona_project Used only once can impose a lot of restrictions. For example, if your > computer operating system can write a running program to a swapfile on > disk then you might have to go as far as terminating your hard disk > with extreme prejudice (and a sledgehemmer) and dissolving the result > in acid. Once means once and once only. These issues are nothing to do with QKD, but are inherent in the use > of a One Time Pad. The mathematical proof of its unbreakability > depends critically on having TRUE random data used ONCE. If either > condition is not met then the mathematical proof is not applicable and > the resulting system is breakable. I have a vague appreciation of the importance of using a One Time Pad only once (meaning really, really, really once -- no fooling). It's about on the level of More information good. I can't begin to imagine what I would do with, for example, two messages that I suspected were XORed with the same OTP. I don't understand why I might need a sledge hammer to be certain some files were not used again. Assuming all this stuff is software-mediated (the only way I can imagine me using the stuff), surely, the software is not going to re-use an old OTP accidentally, is it? It's possible someone could break into my computer and get the old OTPs laying around unerased and unsmashed, but in that case, they should just go ahead and read my mail. Jim Burns === Subject: Re: JSH: Factorization and the global economy >I have a vague appreciation of the importance of using a >One Time Pad only once (meaning really, really, really once >-- no fooling). It's about on the level of More information >good. I can't begin to imagine what I would do with, for example, >two messages that I suspected were XORed with the same OTP. Tim has answered this, you XOR the two cyphertexts together which removes all trace of the key, just leaving the XOR of the plaintexts. That is easily decryptable as an English plaintext is going to contain substrings like the , and etc. Using them as trial keys in all possible positions will start to bring out chunks of the other plaintext in the appropriate positions. Other methods are possible. >I don't understand why I might need a sledge hammer to be >certain some files were not used again. Assuming all this >stuff is software-mediated (the only way I can imagine me >using the stuff), surely, the software is not going to >re-use an old OTP accidentally, is it? If the adversary gets hold of your OTP after you have used it ... I have seen stories of the CIA doing exactly that with some of their old mainframe disc packs. When they failed they were terminated rather than let them out of the building for repair. As with all security, you need to judge the level of security you use to be commensurate with the resources available to your potential attackers. rossum === Subject: Re: The Real Purpose Of Math <19796970.1235316588589.JavaMail.jakarta@nitrogen.mathforum.org> <8763j2ffhf.fsf@phiwumbda.org> posting-account=KroL8woAAAACDGpRprxyFi_gYw4Un8Xt 2.0.50727; yplus 5.1.05b),gzip(gfe),gzip(gfe) > The idea is clearly religious, so why not take it to a æforum for religious discussion? What is totally religious about a system of Governing Equations? Yes, what *is* religious about the following inference? æ Governing Equation > existence of a Governor > existence of God > æ (topologically identified with the Governor) That's just topology, that is. æAnyone who denies it is simply > prejudiced against simple mathematics. You've convinced me, I tell you what. > -- > But what if I'm right [and have solved the factoring problem]? æAnd > what if I can't contain the problem and, oh, in six months from now, > you are desperately trying to find food as you run from humans who > have turned to cannibalism to survive? æ-- James S. Harris (12/21/09) A preliminary conceptualization of the Governing Equation (actually an inter-locking system of partial differential equations) is that it would contain the equations associated with the Standard Model and the best and most elegant description of gravitation as special cases. This set of equations possibly will utilize a generalization of complex variables to higher dimensional commutative algebras. http://www.intelrap.com/lt1.html Complex Variables span only the plane but in order to fully understand a higher dimensional brane we need a more powerful math tool. But if we do not believe that a Creator does exist who created many of the things for which we wish then we will never be motivated to search for the rule... the rule of the highest law of the Universe which may be metaphorically described as a United Verse that beckons all those with a genuine thirst to seek an understanding and to seek it first. If a man has a genuine creative wonder then he understands that he stands under a power much greater than himself. Then he may be lifted upon the shoulders of giants to learn the wonders of math and science and the knowledge that when lies are deleted only the truth is left. Newton stood on the shoulders of giants who now seek more new wondering clients to reveal more mysteries of a magnificent creation in which man plays a principal role in the observation. www.intelrap.com === Subject: Re: Proof of Fermat's Last Theorem posting-account=brOM-AoAAAChaAJEiH5z610-YOTfECd9 Gecko/20070530 Fedora/1.5.0.12-1.fc5 Firefox/1.5.0.12,gzip(gfe),gzip(gfe) > According to Fermat's Last Theorem, a cube can not be divided into > two smaller cubes, a fourth power can not be divided into two smaller > fourth powers, and so on. In other words, if a,b,c, and n are integers and a^n+b^n=c^n and > abcn <> 0, then the set of possible solutions for n is {-2, -1, 1, 2}. What is the proof for this? It's pretty elegant, but there isn't enough space here to write it. Michael === Subject: Re: number theory for children The problem is that number theory makes assertions requiring little background to understand, but proving things, which after all is what number theorists do, requires a whole armamentarium of stuff ranging widely over mathematics. OTOH, sure, show children some of the fascinating conclusions. Just be prepared to run for cover if the child asks How do they know that? === Subject: Re: Congrunces and perfect squares This number, when factorised, gives (10^(k+1) - 1)^2 which is a > perfect square. However, is it possible to show that the original number is a perfect > square using congruences from number theory and not by factorising it? Perhaps you mean 10^(2k+2) - 2(10^(k+1)) + 1 > instead of 10^(2k+1) - 2(10^(k+1)) + 1 Oops - I missed that. But yes, as you indicate, the first term was clearly meant to be (10^(2k+1))^2 or equivalently, 10^(2k+2) For the OP: Using the simple substitution x = 10^(k+1) your expression becomes x^2 + 2x + 1 which clearly factors as (x + 1)^2 The substitution makes the factorization more obvious. quasi === Subject: Re: Congrunces and perfect squares > This number, when factorised, gives (10^(k+1) - 1)^2 which is a > perfect square. However, is it possible to show that the original number is a perfect > square using congruences from number theory and not by factorising it? Perhaps you mean 10^(2k+2) - 2(10^(k+1)) + 1 > instead of 10^(2k+1) - 2(10^(k+1)) + 1 Oops - I missed that. But yes, as you indicate, the first term was clearly meant to be (10^(2k+1))^2 or equivalently, 10^(2k+2) For the OP: Using the simple substitution x = 10^(k+1) your expression becomes x^2 + 2x + 1 The above should be x^2 - 2x + 1 >which clearly factors as (x + 1)^2 And hence the above should be (x - 1)^2 >The substitution makes the factorization more obvious. quasi === Subject: Re: Congrunces and perfect squares This number, when factorised, gives (10^(k+1) - 1)^2 which is a > perfect square. However, is it possible to show that the original number is a perfect > square using congruences from number theory and not by factorising it? Perhaps you mean 10^(2k+2) - 2(10^(k+1)) + 1 > instead of 10^(2k+1) - 2(10^(k+1)) + 1 Oops - I missed that. But yes, as you indicate, the first term was clearly meant to be (10^(2k+1))^2 Ugh -- how careless I am. The above should be 10^(k+1) >or equivalently, 10^(2k+2) For the OP: Using the simple substitution x = 10^(k+1) your expression becomes x^2 + 2x + 1 The above should be x^2 - 2x + 1 which clearly factors as (x + 1)^2 And hence the above should be (x - 1)^2 The substitution makes the factorization more obvious. To the OP: I made a mess of things in my last 2 replies, but with the current corrections, it should now be ok. quasi === Subject: Re: Congrunces and perfect squares > This number, when factorised, gives (10^(k+1) - 1)^2 which is a > perfect square. However, is it possible to show that the original number is a perfect > square using congruences from number theory and not by factorising it? Perhaps you mean 10^(2k+2) - 2(10^(k+1)) + 1 > instead of 10^(2k+1) - 2(10^(k+1)) + 1 Oops - I missed that. But yes, as you indicate, the first term was clearly meant to be (10^(2k+1))^2 Ugh -- how careless I am. The above should be 10^(k+1) Ugh again -- I meant (10^(k+1))^2 >or equivalently, 10^(2k+2) For the OP: Using the simple substitution x = 10^(k+1) your expression becomes x^2 + 2x + 1 The above should be x^2 - 2x + 1 which clearly factors as (x + 1)^2 And hence the above should be (x - 1)^2 The substitution makes the factorization more obvious. To the OP: I made a mess of things in my last 2 replies, but with the current >corrections, it should now be ok. My last 3 replies -- all with careless errors. But now it's OK. quasi === Subject: Re: (m-1)x^2-2x+1-m=0 has two real solutions... posting-account=DSjMzAoAAAAISwlGnTY2c6bnmUVmdFNf AppleWebKit/528.16 (KHTML, like Gecko) Version/4.0 Safari/528.16,gzip(gfe),gzip(gfe) > Hy, I'm having troubles with the following problem: Prove that the following equation has two real solutions, for any m, > if m is not 1: > (m-1)x^2-2x+1-m=0 so, to find out how many solutions I apply the quadratic formula: ( 2 +/- sqr(4 - 4 * (m-1) * (1-m) ) / 2 (m-1) And take out the part that matters in this case: 16 - 4 * (m-1) * (1-m) And, since here I'm at a lost. I tried solving this last part, so: > 4 - 4 * (m-1)*(1-m) = 4 - 4 *(m-m^2-1+m)=4-4*(-m^2+2m-1)=4+4m^2-8m > +4=4m^2-8m+8 > This solves in my calculator to 1+1i or 1-1i. Which I have no use for > (to my knowledge). Now the book's answer is: > 4+4(m-1)^2 > 0 > witch I understand proves the point, althouhgt I don't know how I'm > supposed to get from (m-1)*(1-m) to (m-1)^2 (even if I'm feeling it > should be something rather obvious....) The discriminant b^2 - 4ac equals 4 - 4 (m-1)(1-m) which equals 4 + 4 (m-1)(m-1) (what is -(1-m) ? ). So you have a positive number plus a square; the sum is positive, therefore two solutions. But there is the division by 2a = 2 (m-1) which doesn't work out when m = 1. In that case the complete equation is -2x = 0 with solution x = 0 only. === Subject: Re: In need of a closed form for this constant! >How about using an root finding algorithm which >converges faster? >For example, if one start with a close enough initial >value, >Newton-Raphson method ( >http://en.wikipedia.org/wiki/Newton%27s_method > ) >usually converges quadratically (ie. double your >precision in >every iteration). Funny you should ask because I just discovered > a faster convergence by modifying my original > algorithm. The original which produces a slow convergence of (n)-- (3 - (log((sqrt(pi+1))*pi^2))) + pi = n > (3 - (log((sqrt(n+1))*n^2))) + n =n_1 > (3 - (log((sqrt(n_1+1))*n_1^2))) + n_1 = n_2 > (3 - (log((sqrt(n_2+1))*n_2^2))) + n_2 = n_3 > (3 - (log((sqrt(n_3+1))*n_3^2))) + n_3 = n_4 > etc. Modifying for a much faster convergence of (n) (3 - (log((sqrt(pi+1))*pi^2))) *((1/pi)+1) + pi = n > (3 - (log((sqrt(n+1))*n^2))) *((1/n)+1) + n = n_1 > (3 - (log((sqrt(n_1+1))*n_1^2))) *((1/n_1)+1) + n_1 = n_2 > (3 - (log((sqrt(n_2+1))*n_2^2))) *((1/n_2)+1) + n_2 = n_3 > (3 - (log((sqrt(n_3+1))*n_3^2))) *((1/n_3)+1) + n_3 = n_4 > etc. Probably more than twice as fast as my original version. Dan Your iteration converges linearly. I think you need around > 350 iterations to get a 1000 digit precision. In contrast, Newton-Raphson on f(x) = x^5 + x^4 - e^6 need > only about 9 iteration to achieve same accuracy ( with the > additional benefit you don't need to deal with the slower > multi-precision log() function ). BTW, does anyone know what is the relative complexity of > computing multiple precison log vs multiply and divide???? Denote by AGM the aritmethic-geometric mean. All elementary transcendental function can be computed with bit complexity O_B(M(n).log(n)) where M(n) is the bit complexity of multiplication. Borwein and Borwein, Pi_and_the_AGM. Denote by K(k) the complete elliptic integral of the first kind. K(k') (1) pi.----- = log(-) K(k) {q) K(k) 1 ----- = - [theta_3(q)]^2. pi 2 [theta_2(q)]^2 k = ---------------- [theta_3(q)]^2 theta_3(q) = sum_{-oo < n < oo} q^{n^2} theta_2(q) = sum_{-oo < n < oo} q^{(n + 1/2)^2} Fix q, 0 < a < q < b < 1. Step 1: Calculate K(k)/pi Step 2: Calculate k. Step 3: Calculate K(k') from the AGM. Step 4: Calculate log(1/q). -- Michael Press === Subject: Re: In need of a closed form for this constant! <28197189.1235495408224.JavaMail.jakarta@nitrogen.mathforum.org> posting-account=SvltewoAAAAi7TTYrD3mAaLUHzDiF2d1 Gecko/2009011913 Firefox/3.0.6,gzip(gfe),gzip(gfe) BTW, does anyone know what is the relative complexity of > computing multiple precison log vs multiply and divide???? Denote by AGM the aritmethic-geometric mean. All elementary > transcendental function can be computed with bit complexity æ æ O B(M(n).log(n)) where M(n) is the bit complexity of multiplication. æ > Borwein and Borwein, Pi and the AGM. Michael Press Cool, one more item on my bed-side to-read list (growing very fast in past 3 weeks since I start to re-read sci.math ;-p) === Subject: countable collection of dense open subsets of reals with intersection having 0 measure posting-account=2kYL5QoAAAAmH1e_g3WSyZ5Z9zXmRQX3 .NET CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) How can I construct a sequence of dense open subsets of IR such that the intersection is Lesbesque null set? === Subject: Re: countable collection of dense open subsets of reals with intersection having 0 measure > How can I construct a sequence of dense open subsets of IR such that > the intersection is Lesbesque null set? Am I reading this right? Isn't any dense, open subset of the reals (and hence a countable intersection) cocountable? === Subject: Re: countable collection of dense open subsets of reals with intersection having 0 measure posting-account=AdyLXQoAAABgRay99CKv1O8Y_7jjivwq InfoPath.1; .NET CLR 2.0.50727; .NET CLR 3.0.04506.30; .NET CLR 3.0.04506.648),gzip(gfe),gzip(gfe) > Am I reading this right? Isn't any dense, open subset of > the reals (and hence a countable intersection) cocountable? If C is a closed nowhere dense subset of the reals, then the complement of C is a dense open set. Now note that C can be uncountable, for example when C is a Cantor set. Dave L. Renfro === Subject: Re: countable collection of dense open subsets of reals with intersection having 0 measure > How can I construct a sequence of dense open subsets of IR such that > the intersection is Lesbesque null set? Cover the rationals by small open sets. === Subject: Re: countable collection of dense open subsets of reals with intersection having 0 measure posting-account=2kYL5QoAAAAmH1e_g3WSyZ5Z9zXmRQX3 .NET CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) How can I construct a sequence of dense open subsets of IR such that > the intersection is Lesbesque null set? Cover the rationals by small open sets. How can I construct a sequence of dense open subsets of IR such that > the intersection is Lesbesque null set? Cover the rationals by small open sets. Ok, here is my attempt (your hint was very good). Enumerate the rationals Q = {r k} for k=1,2,... Let G n = / ( r k - 1/(n*2*2^k) , r k + 1/(n*2*2^k) ) over n=1,2,... now by subadditivity, m(G n) <= SUM over k of( 1/n*2^k ) = 1/n (1/2 + 1/4 + 1/8 + ...) = 1/n Define G := /G n over n = 1,2,... Now, clearly G 1 superset of G 2 superset of G 3 ... Thus m(G) = lim{n->inf} m(G n) <= lim{n->inf} 1/n = 0 So m(G) = 0. === Subject: Coogi Men's Jeans - discount posting-account=uz1cQgoAAACe_NoiQ9omKJf2BvOAUW4x Mozilla/4.0 (compatible; MSIE 6.0; Windows NT 5.1; SV1) ; Embedded \ Web Browser from: http://bsalsa.com/),gzip(gfe),gzip(gfe) Where can go to find Coogi Men's Jeans? now We have a good collection of Coogi Men's Jeans in different styles and different colors. Please see them : http://www.luxury-fashion.org/static/Apparels/Coogi-Mens-Jeans-61.html http://www.luxury-fashion.org/static/Apparels/Coogi-Mens-Jeans-60.html Our Coogi Men's Jeans are fine quality And find more new fashion cloth,shoes and apparels please view : http://www.luxury-fashion.org Welcome check our other pages or feel free contact us. You can find what do you want here! === Subject: Re: Alternating Vertical and Horizontal Moves In Grid posting-account=ey-PjQoAAAD_Gi3gK0ZWKofsmviYOWhV Gecko/20081217 Firefox/2.0.0.20,gzip(gfe),gzip(gfe) Hi! This is an interesting problem, that I have never seen before. I will make the assumption that at any point a vertical move may be U or D, and a horizontal move may be L or R. In both the start-at-a-corner and general case, a(1) = 1, a(2) = 4, and a(3) = 6, for starters. Now I will focus only on the start-at-a-corner case, and it may as well be the upper left. Obviously n^2 is an upper bound for a(n), and if we try to establish some reasonable lower bounds, by considering a boustrophedonic traversal of the grid, there is a natural division into cases based on the value of n mod 4. In particular, I conjecture that the optimal algorithm for sufficiently large n = 1 mod 4 is to traverse the grid according to the following rules: 1) Start with down 1) Choose up over down whenever at least 4 more moves can be made as a result of doing so. 2) Alternate between choosing right n - 1 times in a row, then left n - 1 times in a row, then right, etc. To see what I mean by this, the 9 x 9 grid would be traversed like: DRURDRURDRURDRDRDLDLULULDLULDLDLDRDRURURDRDRURDRDLDLULDLULDLULDLU. With a little work this establishes: n mod 4 = 1 -> [ n^2 - 2n + 3 <= a(n) ] For the other cases some slight modifications of the algorithm above produce pretty tight lower bounds as well. In the above case, I would expect equality probably holds, but proving it is another story, it seems like it would be tricky. dc === Subject: Re: Alternating Vertical and Horizontal Moves In Grid >What is the sequence {a(k)} that is defined as follows? Start with an n-by-n grid. Move from square to adjacent square by >moving vertically then horizontally then vertically then horizontally, >following each horizontal move with a vertical move, and following >each vertical move with a horizontal move. >No square may be visited more than once. a(n) = the maximum number of squares visitable in this way on an n-by- >n grid. By brute force search, here are the values of a(n) for n = 1, ..., 8: 1, 4, 7, 14, 19, 32, 39, 58 >A variation: You must start in a corner square. Then the first 8 values are: 1, 4, 6, 14, 18, 32, 38, 58 Note that, so far, if n is even, starting in a corner is just as good a starting place as any other location, while if n is odd and n > 1, the value of a(n) when starting in a corner appears to be just 1 less than that for an optimum starting place. >(It does not matter if the first move is vertical or horizontal, as >far as the values of each a(n) are concerned, of course.) I am guessing that this sequence is really easy to calculate, maybe a >(n) is as simple as a quadratic polynomial in terms of n. Or is the >best way to calculate each a(n) is to do a brute-force search of >possible paths through the grid? The function a(n) is definitely not a polynomial in n. Clearly a(n) is positive and bounded above by n^2, hence, if a(n) was a polynomial in n, it would have to be at most of degree 2. But the calculated values of a(n) as posted above contradict that. So what is the nature of a(n)? Consider the sequence b(n) = n^2 - a(n) Clearly b(n) must nonnegative. I suspect it would not be too hard to prove that b(n) must be positive, for n > 2. We can ask whether the sequence b(n) is bounded above? I suspect not, but if, in fact, b(n) is bounded above, then I would guess that the sequence b(n) repeats. Other than brute force search, you might try for recursive relationships, or at least recursive bounds. quasi === Subject: Re: Alternating Vertical and Horizontal Moves In Grid What is the sequence {a(k)} that is defined as follows? Start with an n-by-n grid. Move from square to adjacent square by >moving vertically then horizontally then vertically then horizontally, >following each horizontal move with a vertical move, and following >each vertical move with a horizontal move. >No square may be visited more than once. a(n) = the maximum number of squares visitable in this way on an n-by- >n grid. By brute force search, here are the values of a(n) for n = 1, ..., 8: 1, 4, 7, 14, 19, 32, 39, 58 A variation: You must start in a corner square. Then the first 8 values are: 1, 4, 6, 14, 18, 32, 38, 58 Note that, so far, if n is even, starting in a corner is just as good >a starting place as any other location, while if n is odd and n > 1, >the value of a(n) when starting in a corner appears to be just 1 less >than that for an optimum starting place. (It does not matter if the first move is vertical or horizontal, as >far as the values of each a(n) are concerned, of course.) I am guessing that this sequence is really easy to calculate, maybe a >(n) is as simple as a quadratic polynomial in terms of n. Or is the >best way to calculate each a(n) is to do a brute-force search of >possible paths through the grid? The function a(n) is definitely not a polynomial in n. Clearly a(n) is >positive and bounded above by n^2, hence, if a(n) was a polynomial in >n, it would have to be at most of degree 2. But the calculated values >of a(n) as posted above contradict that. So what is the nature of a(n)? Consider the sequence b(n) = n^2 - a(n) Clearly b(n) must nonnegative. I suspect it would not be too hard to >prove that b(n) must be positive, for n > 2. We can ask whether the sequence b(n) is bounded above? I suspect not, >but if, in fact, b(n) is bounded above, then I would guess that the >sequence b(n) repeats. Other than brute force search, you might try for recursive >relationships, or at least recursive bounds. Ok, there is a simple pattern that I missed. When n is even, it appears that a(n) = n^2 - n + 2 quasi === Subject: Re: Alternating Vertical and Horizontal Moves In Grid >What is the sequence {a(k)} that is defined as follows? Start with an n-by-n grid. Move from square to adjacent square by >moving vertically then horizontally then vertically then horizontally, >following each horizontal move with a vertical move, and following >each vertical move with a horizontal move. >No square may be visited more than once. a(n) = the maximum number of squares visitable in this way on an n-by- >n grid. By brute force search, here are the values of a(n) for n = 1, ..., 8: 1, 4, 7, 14, 19, 32, 39, 58 A variation: You must start in a corner square. Then the first 8 values are: 1, 4, 6, 14, 18, 32, 38, 58 Note that, so far, if n is even, starting in a corner is just as good >a starting place as any other location, while if n is odd and n > 1, >the value of a(n) when starting in a corner appears to be just 1 less >than that for an optimum starting place. (It does not matter if the first move is vertical or horizontal, as >far as the values of each a(n) are concerned, of course.) I am guessing that this sequence is really easy to calculate, maybe a >(n) is as simple as a quadratic polynomial in terms of n. Or is the >best way to calculate each a(n) is to do a brute-force search of >possible paths through the grid? The function a(n) is definitely not a polynomial in n. Clearly a(n) is >positive and bounded above by n^2, hence, if a(n) was a polynomial in >n, it would have to be at most of degree 2. But the calculated values >of a(n) as posted above contradict that. So what is the nature of a(n)? Consider the sequence b(n) = n^2 - a(n) Clearly b(n) must nonnegative. I suspect it would not be too hard to >prove that b(n) must be positive, for n > 2. We can ask whether the sequence b(n) is bounded above? I suspect not, >but if, in fact, b(n) is bounded above, then I would guess that the >sequence b(n) repeats. Other than brute force search, you might try for recursive >relationships, or at least recursive bounds. Ok, there is a simple pattern that I missed. When n is even, it appears that a(n) = n^2 - n + 2 Moreover, it's fairly easy to show that, for even n, a(n) >= n^2 - n + 2 Assuming n is even, a valid path through n^2 - n + 2 vertices can be explicitly constructed as follows ... Start at (0,0). Go right, up, left, up. Repeat until reaching the top. Since n is even, the last 2 moves must have been right, up, ending at (1,n-1). Now go right, down, ending at (2,n-2). We have so far visited 2n + 1 vertices. Now recursively do the same in the n-2 by n-2 square with upper left corner vertex (2,n-2) and lower right corner vertex (n-1,1). The path described above adheres to the restrictions, and it's easily proved, by induction, that it has exactly n^2 - n + 2 vertices. quasi === Subject: Re: Alternating Vertical and Horizontal Moves In Grid What is the sequence {a(k)} that is defined as follows? Start with an n-by-n grid. Move from square to adjacent square by >moving vertically then horizontally then vertically then horizontally, >following each horizontal move with a vertical move, and following >each vertical move with a horizontal move. >No square may be visited more than once. a(n) = the maximum number of squares visitable in this way on an n-by- >n grid. By brute force search, here are the values of a(n) for n = 1, ..., 8: 1, 4, 7, 14, 19, 32, 39, 58 A variation: You must start in a corner square. Then the first 8 values are: 1, 4, 6, 14, 18, 32, 38, 58 Note that, so far, if n is even, starting in a corner is just as good >a starting place as any other location, while if n is odd and n > 1, >the value of a(n) when starting in a corner appears to be just 1 less >than that for an optimum starting place. (It does not matter if the first move is vertical or horizontal, as >far as the values of each a(n) are concerned, of course.) I am guessing that this sequence is really easy to calculate, maybe a >(n) is as simple as a quadratic polynomial in terms of n. Or is the >best way to calculate each a(n) is to do a brute-force search of >possible paths through the grid? The function a(n) is definitely not a polynomial in n. Clearly a(n) is >positive and bounded above by n^2, hence, if a(n) was a polynomial in >n, it would have to be at most of degree 2. But the calculated values >of a(n) as posted above contradict that. So what is the nature of a(n)? Consider the sequence b(n) = n^2 - a(n) Clearly b(n) must nonnegative. I suspect it would not be too hard to >prove that b(n) must be positive, for n > 2. We can ask whether the sequence b(n) is bounded above? I suspect not, >but if, in fact, b(n) is bounded above, then I would guess that the >sequence b(n) repeats. Other than brute force search, you might try for recursive >relationships, or at least recursive bounds. Ok, there is a simple pattern that I missed. When n is even, it appears that a(n) = n^2 - n + 2 Moreover, it's fairly easy to show that, for even n, a(n) >= n^2 - n + 2 Assuming n is even, a valid path through n^2 - n + 2 vertices can be >explicitly constructed as follows ... Start at (0,0). Go right, up, left, up. Repeat until reaching the top. Since n is even, the last 2 moves must have been right, up, ending at >(1,n-1). Now go right, down, ending at (2,n-2). We have so far visited 2n + 1 vertices. Now recursively do the same in the n-2 by n-2 square with upper left >corner vertex (2,n-2) and lower right corner vertex (n-1,1).| To clarify: At each successive lower level in the recursion, switch up and down. Also, when you finish the 2 x 2 level, stop. >The path described above adheres to the restrictions, and it's easily >proved, by induction, that it has exactly n^2 - n + 2 vertices. quasi === Subject: Re: Alternating Vertical and Horizontal Moves In Grid <4utbq4dhhldvts8o5hl3ld180tan2tbdmq@4ax.com> posting-account=ey-PjQoAAAD_Gi3gK0ZWKofsmviYOWhV Gecko/20081217 Firefox/2.0.0.20,gzip(gfe),gzip(gfe) What is the sequence {a(k)} that is defined as follows? Start with an n-by-n grid. Move from square to adjacent square by >moving vertically then horizontally then vertically then horizontally, >following each horizontal move with a vertical move, and following >each vertical move with a horizontal move. >No square may be visited more than once. a(n) = the maximum number of squares visitable in this way on an n-by- >n grid. By brute force search, here are the values of a(n) for n = 1, ..., 8: 1, 4, 7, 14, 19, 32, 39, 58 A variation: You must start in a corner square. Then the first 8 values are: 1, 4, 6, 14, 18, 32, 38, 58 Note that, so far, if n is even, starting in a corner is just as good >a starting place as any other location, while if n is odd and n > 1, >the value of a(n) when starting in a corner appears to be just 1 less >than that for an optimum starting place. (It does not matter if the first move is vertical or horizontal, as >far as the values of each a(n) are concerned, of course.) I am guessing that this sequence is really easy to calculate, maybe a >(n) is as simple as a quadratic polynomial in terms of n. Or is the >best way to calculate each a(n) is to do a brute-force search of >possible paths through the grid? The function a(n) is definitely not a polynomial in n. Clearly a(n) is >positive and bounded above by n^2, hence, if a(n) was a polynomial in >n, it would have to be at most of degree 2. But the calculated values >of a(n) as posted above contradict that. So what is the nature of a(n)? Consider the sequence b(n) = n^2 - a(n) Clearly b(n) must nonnegative. I suspect it would not be too hard to >prove that b(n) must be positive, for n > 2. We can ask whether the sequence b(n) is bounded above? I suspect not, >but if, in fact, b(n) is bounded above, then I would guess that the >sequence b(n) repeats. Other than brute force search, you might try for recursive >relationships, or at least recursive bounds. Ok, there is a simple pattern that I missed. When n is even, it appears that a(n) = n^2 - n + 2 Moreover, it's fairly easy to show that, for even n, a(n) >= n^2 - n + 2 Assuming n is even, a valid path through n^2 - n + 2 vertices can be >explicitly constructed as follows ... Start at (0,0). Go right, up, left, up. Repeat until reaching the top. Since n is even, the last 2 moves must have been right, up, ending at >(1,n-1). Now go right, down, ending at (2,n-2). We have so far visited 2n + 1 vertices. Now recursively do the same in the n-2 by n-2 square with upper left >corner vertex (2,n-2) and lower right corner vertex (n-1,1).| To clarify: At each successive lower level in the recursion, switch up and > down. Also, when you finish the 2 x 2 level, stop. The path described above adheres to the restrictions, and it's easily >proved, by induction, that it has exactly n^2 - n + 2 vertices. quasi this seems legitimate to me. quasi, by the sounds of it, you have some kind of a computer program capable of a brute force test? can you test my conjecture above for n = 9, 13? i'm thinking we are going tol end up here with a quadratic answer, but different quadratics depending on the value of n mod 4. === Subject: Re: Alternating Vertical and Horizontal Moves In Grid What is the sequence {a(k)} that is defined as follows? Start with an n-by-n grid. Move from square to adjacent square by >moving vertically then horizontally then vertically then horizontally, >following each horizontal move with a vertical move, and following >each vertical move with a horizontal move. >No square may be visited more than once. a(n) = the maximum number of squares visitable in this way on an n-by- >n grid. By brute force search, here are the values of a(n) for n = 1, ..., 8: 1, 4, 7, 14, 19, 32, 39, 58 A variation: You must start in a corner square. Then the first 8 values are: 1, 4, 6, 14, 18, 32, 38, 58 Note that, so far, if n is even, starting in a corner is just as good >a starting place as any other location, while if n is odd and n > 1, >the value of a(n) when starting in a corner appears to be just 1 less >than that for an optimum starting place. (It does not matter if the first move is vertical or horizontal, as >far as the values of each a(n) are concerned, of course.) I am guessing that this sequence is really easy to calculate, maybe a >(n) is as simple as a quadratic polynomial in terms of n. Or is the >best way to calculate each a(n) is to do a brute-force search of >possible paths through the grid? The function a(n) is definitely not a polynomial in n. Clearly a(n) is >positive and bounded above by n^2, hence, if a(n) was a polynomial in >n, it would have to be at most of degree 2. But the calculated values >of a(n) as posted above contradict that. So what is the nature of a(n)? Consider the sequence b(n) = n^2 - a(n) Clearly b(n) must nonnegative. I suspect it would not be too hard to >prove that b(n) must be positive, for n > 2. We can ask whether the sequence b(n) is bounded above? I suspect not, >but if, in fact, b(n) is bounded above, then I would guess that the >sequence b(n) repeats. Other than brute force search, you might try for recursive >relationships, or at least recursive bounds. Ok, there is a simple pattern that I missed. When n is even, it appears that a(n) = n^2 - n + 2 Moreover, it's fairly easy to show that, for even n, a(n) >= n^2 - n + 2 Assuming n is even, a valid path through n^2 - n + 2 vertices can be >explicitly constructed as follows ... Start at (0,0). Go right, up, left, up. Repeat until reaching the top. Since n is even, the last 2 moves must have been right, up, ending at >(1,n-1). Now go right, down, ending at (2,n-2). We have so far visited 2n + 1 vertices. Now recursively do the same in the n-2 by n-2 square with upper left >corner vertex (2,n-2) and lower right corner vertex (n-1,1).| To clarify: At each successive lower level in the recursion, switch up and > down. Also, when you finish the 2 x 2 level, stop. The path described above adheres to the restrictions, and it's easily >proved, by induction, that it has exactly n^2 - n + 2 vertices. quasi this seems legitimate to me. quasi, by the sounds of it, you have some >kind of a computer program capable of a brute force test? can you test >my conjecture above for n = 9, 13? i'm thinking we are going tol end >up here with a quadratic answer, but different quadratics depending on >the value of n mod 4. Yes, that seems very plausible -- 3 quadratics, one for even n, one for n = 1 mod 4, and one for n = 3 mod 4. I do have a brute force test, but it wasn't designed for a heavy load. For n > 8, it struggles. I'll let it run for a while -- hopefully, it will at least compute a(9). quasi === Subject: Re: Alternating Vertical and Horizontal Moves In Grid <4utbq4dhhldvts8o5hl3ld180tan2tbdmq@4ax.com> posting-account=ey-PjQoAAAD_Gi3gK0ZWKofsmviYOWhV Gecko/20081217 Firefox/2.0.0.20,gzip(gfe),gzip(gfe) no, i changed my mind about the mod 4 thing, it looks a lot like your formula is right whenever n is even, and my formula is right (by a similar argument) when n is odd. of course, we have only established them as lower bounds with certainty. === Subject: Re: Alternating Vertical and Horizontal Moves In Grid >no, i changed my mind about the mod 4 thing, it looks a lot like your >formula is right whenever n is even, and my formula is right (by a >similar argument) when n is odd. of course, we have only established >them as lower bounds with certainty. Yes, if we require corner starting point, and assuming n > 1, I think you may be right. Thus, for a corner starting point, and n > 1, the conjecture is: a(n) = n^2 - 2n + 3, if n is odd n^2 - n + 2, if n is even quasi PS -- My Maple program is still working on a(9), assuming a required corner starting point. === Subject: Re: Alternating Vertical and Horizontal Moves In Grid no, i changed my mind about the mod 4 thing, it looks a lot like your >formula is right whenever n is even, and my formula is right (by a >similar argument) when n is odd. of course, we have only established >them as lower bounds with certainty. Yes, if we require corner starting point, and assuming n > 1, I think >you may be right. Thus, for a corner starting point, and n > 1, the conjecture is: a(n) = > n^2 - 2n + 3, if n is odd > n^2 - n + 2, if n is even quasi PS -- My Maple program is still working on a(9), assuming a required >corner starting point. Ok, Maple completed a(9) and is now working on a(11). Good news -- a(9) = 66, as expected. Note, for convenience, we are now always making the assumption of a corner starting point. For even n, it apparently doesn't matter. For odd n > 1, allowing a variable starting point appears to make a(n) increase by 1. quasi === Subject: Re: Alternating Vertical and Horizontal Moves In Grid <4utbq4dhhldvts8o5hl3ld180tan2tbdmq@4ax.com> <4p7cq45og61mehq711tc9qm7ir9avhp4gl@4ax.com> posting-account=ey-PjQoAAAD_Gi3gK0ZWKofsmviYOWhV Gecko/20081217 Firefox/2.0.0.20,gzip(gfe),gzip(gfe) no, i changed my mind about the mod 4 thing, it looks a lot like your >formula is right whenever n is even, and my formula is right (by a >similar argument) when n is odd. of course, we have only established >them as lower bounds with certainty. Yes, if we require corner starting point, and assuming n > 1, I think >you may be right. Thus, for a corner starting point, and n > 1, the conjecture is: a(n) = > n^2 - 2n + 3, if n is odd > n^2 - n + 2, if n is even quasi PS -- My Maple program is still working on a(9), assuming a required >corner starting point. Ok, Maple completed a(9) and is now working on a(11). Good news -- a(9) = 66, as expected. Note, for convenience, we are now always making the assumption of a > corner starting point. For even n, it apparently doesn't matter. For > odd n > 1, allowing a variable starting point appears to make a(n) > increase by 1. quasi this makes sense to me, i am starting to work on a proof of the bound for the even case, as i think it will be the easier one. i have some ideas, but nothing concrete enough to comment on yet. === Subject: Re: Newton's Method: Cute trick. posting-account=DSjMzAoAAAAISwlGnTY2c6bnmUVmdFNf AppleWebKit/528.16 (KHTML, like Gecko) Version/4.0 Safari/528.16,gzip(gfe),gzip(gfe) I have never seen this trick explicitly mentioned anywhere. Of course if f (p) < 0, f (q) > 0, f continuous on [p, q] then there is a root of f in (p, q), and if you use _any_ iterative algorithm, then clamping the next iterated result to [p, q] will get you closer to the root. Some things I noted: (1) If you find x0 such that f (x0) is small enough, but x0 is not in [p, q] then clamping x0 to [p, q] might give a new value where f (x) is _not_ small enough anymore, so care must be taken. (2) Clamping can actually only happen on the first Newton-Raphson step if your convergence criterion is actually met, because you proved that each iteration gets closer to the root of F (x). If you are not 100% sure that the convergence criterion is met, clamping might not help. (3) This doesn't solve the problem of finding x0 close enough to the root. Of course it is no big deal if a and b are known. === Subject: Re: Help!How to solve the equation === Subject: Re: Are the integers as normed vector space complete? > Yes. Every discrete metric space is complete. Careful. Every uniformly discrete metric space is complete, but not every metric space that is discrete (as a topological space) is complete. That is: a metric space (X,d) is (topologically) discrete if for each x in X there is epsilon > 0 such that d(y,x) < epsilon implies y=x. It is uniformly discrete if there is a single epsilon > 0 such that for all x and y, d(y,x) < epsilon implies y = x. If the metric space is a group with a metric invariant under the group operations, the same epsilon that works for one x works for all of them, so in this case discreteness does imply completeness. -- Robert Israel israel@math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Are the integers as normed vector space complete? posting-account=_l4K0QkAAAC09JhOoK_ZfoJKXOmr_jZf Gecko/2009020911 Ubuntu/8.04 (hardy) Firefox/3.0.6,gzip(gfe),gzip(gfe) On Feb 25, 3:40æpm, Robert Israel > Yes. Every discrete metric space is complete. Careful. æEvery uniformly discrete metric space is complete, but not > every metric space that is discrete (as a topological space) is complete. -- Arturo Magidin, sans .sig === Subject: Re: How to prove this <27107722.1235573020209.JavaMail.jakarta@nitrogen.mathforum.org>, > Certainly the thing being summed, which in my version is > ---> arctan(1/sqrt(k)) > and in your version is > ---> arcsin(1/sqrt(k)) > could likely be shown to be irrational. > I would even bet it could be shown to be an > irrational multiple of Pi. Well, not always - for k = 1, they're both rational multiples of pi. arctan(1 / sqrt 3) and arcsin(1 / sqrt 4) are also rational multiples of pi. > But neither of these (if shown) would imply that > the *sum* of the arctans or arcsins could not be > a multiple of pi. What he said. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Alain's T-approximation does not work (was Re: Continuous iteration of another quadratic map) posting-account=fwSgtAkAAACFnX70ssKwbvm9_oCZVHrx Gecko/2009011913 Firefox/3.0.6,gzip(gfe),gzip(gfe) > On Feb 23, 2:50æam, alainvergh...@gmail.com > Am 22.02.2009 23:20 schrieb mike3: What I'd be more curious about though is how you got that > series you posted in the message you gave in terms of the > height. As I'd be curious about a similar case for f(0), i.e. > g(-phi). (phi = golden ratio) Then I could see what that period > 2 orbit looks like when done continuously. > First you build the matrix-operator for the function. The function is æ æ g(x) = a*x + 1*x^2 æ æ æ æ æ æ æ æ where æ æ a = (1+sqrt(5)) = 2*phi For iteration you need g(x) and its powers g(x)^2, g(x)^3,... To compute these powers it suffices to operate on the coefficients only > (square,cube,... the formal powerseries (which is here only a polynomial)) For iteration you combine these coefficients in the same way as you would > do a matrix-multiplication. So write the coefficients of the powers > of g(x) into a matrix: æ g(x)^0 æ = 1 æ+ æ 0*x æ+ æ 0*x^2 + æ0*x^3 + 0*x^4 + ... > æ g(x)^1 æ = 0 æ+ æ a*x æ+ æ 1*x^2 + æ0*x^3 + 0*x^4 + ... > æ g(x)^2 æ = 0 æ+ æ 0*x æ+ a^2*x^2 + 2a*x^3 + 1*x^4 + ... > æ ... > up to a reasonable dimension (to get well approximated results > for fractional iterates) The matrix G has then only the coefficients of the formal powerseries > (and I write it in transposed form) æG = æ 1 æ. æ . æ æ. æ . > æ æ æ æ. æa æ . æ æ. æ . > æ æ æ æ. æ1 æa^2 æ . æ . > æ æ æ æ. æ. 2a æ æa^3 æ. > æ æ æ æ.... ---------------------------------------------- > Second, the use of the matrix G Then a vector V(x) = [1,x,x^2,x^3,...] up to the selected dimension gives > using æ V(x) * G = [ 1, g(x), g(x)^2, g(x)^3 ,... ] In Pari/GP write (for a given dimension n) gp> æV(x) = vector(n,c,x^(c-1)) gp> æY = V(1) * G > æ then Y = [1, g(1) , g(1)^2, g(1)^3,...] correctly up to a certain column > æ The remaining columns are wrong due to truncation to finite size. Since the form of the result-vector is again like V(), a vector of consecutive > powers, the iteration can be performed by the formal relation æ V(x) æ æ* G = V(g(x)) > æ V(g(x)) * G = V(g(g(x)) > æ ... > and so on. However: this is exactly valid only in the limit, where we > assume infinite dimension. For integer iterates k this can be written as æV(gÁk(x)) = V(x) * G^k æ æ æ(theoretically) gp> æ æ æ æY = V(x) * G^k æ æ æ(practically) > æ where Y ~ V(gÁk(x)) æ æ æ æ æonly approximately which is true for the case of infinite dimension, and only approximate > with finite dimension. Just try this in Pari/GP. ----------------------------------------------------------- > 3) fractional iterates Next step is fractional iterate. It is an obvious idea to use > fractional powers of G for this. Fractional powers of G can either be determined by binomial-expansion > with G and I as arguments or by diagonalization. > The general formula for diagonalization of a matrix is æ æG æ = W * D * W^-1 æ æ æ æ æ æ æ// D is diagonal > and > æ æG^h = W * D^h * W^-1 So to find W and D in Pari/GP there is the function mateigen, which > gives a version of W first. gp> æ W = mateigen(G) Then compute WInv as inverse of W gp> æ WInv = W^-1 Then compute D as gp> æ D = WInv * G * W Then you can define a pari-function for arbitrary powers of D: gp> æ ædpow(x) = matrix(n,n,r,c,if(r==c,D[r,c]^x)) and then write gp> æ æGPow = W * dpow(0.5) * WInv to store G^0.5 in a constant matrix. Then you use gp> æ æ æY = V(x) * GPow æ æand Y is approximately ~ V(gÁh(x)) or just define the scalar function, using the relevant coefficients of GPow gp> æ gpow(x) = sum(k=0,n-1, x^k * GPow[1+k,2]) ----------------------------------------- > 4) Remarks: However, the approximation may be poor. I use dimension n=64 for > standard discussion and n=128 if the approximation is difficult. > The mateigen-procedure in Pari/GP may perform poorly because it > doesn't exploit the easiness of the solution for triangular matrices, > for which we can even get exact solutions. I've a version for > triangular matrices of the special form here (matrix-operators on V(x)) > implemented, which gives the eigensystem-decomposition of n=64x64 > or 128x128 triangular-matrices in a second or two and even symbolically > if the matrix-dimension is small. > I could mail it to you, but I think I've already posted it in the > tetration-forum in the matrix-method-thread. Also I use Euler-summation for convergence-acceleration. But this > needs some more explanations - try this first to get used to it. Gottfried Bonjour, I do believe you're using heavy tools. > What about approximating x^2-1 by a conjugate of > a known iterate :Ex (x^2-1) = f-1(f(x))^2 - 2) ,T(x)=x^2-2 > or æ (x^2-1) =~f^-1 o T o f (x) > and (x^2-1)^[r] =~f^-1 o T^r o f (x) The trouble is it only seems to work when x is large. What about small > x? There's even more trouble. The approximation does not work at all! :( Try r = 2, x = 4. You get ~ 222.9955156499789347277396687. Which is close to 223 -- and that's (4^2 - 1)^2 - 2 not (4^2 - 1)^2 - 1. It's even worse for close to 49727 -- and *that* is just ((4^2 - 1)^2 - 2)^2 - 2 not ((4^2 - 1)^2 - 1)^2 - 1. It is not approaching f^n(x) where f(x) = x^2 - 1, it's approaching f^(n-1)(x^2 - 1) where f(x) = x^2 - 2! === Subject: Re: Alain's T-approximation does not work (was Re: Continuous iteration of another quadratic map) posting-account=06BQLAoAAADoC7Y4z9FWcUwGvMa7xMG9 7.4),gzip(gfe),gzip(gfe) On Feb 23, 2:50æam, alainvergh...@gmail.com > Am 22.02.2009 23:20 schrieb mike3: What I'd be more curious about though is how you got that > series you posted in the message you gave in terms of the > height. As I'd be curious about a similar case for f(0), i.e. > g(-phi). (phi = golden ratio) Then I could see what that period > 2 orbit looks like when done continuously. > First you build the matrix-operator for the function. The function is æ æ g(x) = a*x + 1*x^2 æ æ æ æ æ æ æ æ where æ æ a = (1+sqrt(5)) = 2*phi For iteration you need g(x) and its powers g(x)^2, g(x)^3,... To compute these powers it suffices to operate on the coefficients only > (square,cube,... the formal powerseries (which is here only a polynomial)) For iteration you combine these coefficients in the same way as you would > do a matrix-multiplication. So write the coefficients of the powers > of g(x) into a matrix: æ g(x)^0 æ = 1 æ+ æ 0*x æ+ æ 0*x^2 + æ0*x^3 + 0*x^4 + ... > æ g(x)^1 æ = 0 æ+ æ a*x æ+ æ 1*x^2 + æ0*x^3 + 0*x^4 + ... > æ g(x)^2 æ = 0 æ+ æ 0*x æ+ a^2*x^2 + 2a*x^3 + 1*x^4 + ... > æ ... > up to a reasonable dimension (to get well approximated results > for fractional iterates) The matrix G has then only the coefficients of the formal powerseries > (and I write it in transposed form) æG = æ 1 æ. æ . æ æ. æ . > æ æ æ æ. æa æ . æ æ. æ . > æ æ æ æ. æ1 æa^2 æ . æ . > æ æ æ æ. æ. 2a æ æa^3 æ. > æ æ æ æ.... ---------------------------------------------- > Second, the use of the matrix G Then a vector V(x) = [1,x,x^2,x^3,...] up to the selected dimension gives > using æ V(x) * G = [ 1, g(x), g(x)^2, g(x)^3 ,... ] In Pari/GP write (for a given dimension n) gp> æV(x) = vector(n,c,x^(c-1)) gp> æY = V(1) * G > æ then Y = [1, g(1) , g(1)^2, g(1)^3,...] correctly up to a certain column > æ The remaining columns are wrong due to truncation to finite size. Since the form of the result-vector is again like V(), a vector of consecutive > powers, the iteration can be performed by the formal relation æ V(x) æ æ* G = V(g(x)) > æ V(g(x)) * G = V(g(g(x)) > æ ... > and so on. However: this is exactly valid only in the limit, where we > assume infinite dimension. For integer iterates k this can be written as æV(gÁk(x)) = V(x) * G^k æ æ æ(theoretically) gp> æ æ æ æY = V(x) * G^k æ æ æ(practically) > æ where Y ~ V(gÁk(x)) æ æ æ æ æonly approximately which is true for the case of infinite dimension, and only approximate > with finite dimension. Just try this in Pari/GP. ----------------------------------------------------------- > 3) fractional iterates Next step is fractional iterate. It is an obvious idea to use > fractional powers of G for this. Fractional powers of G can either be determined by binomial-expansion > with G and I as arguments or by diagonalization. > The general formula for diagonalization of a matrix is æ æG æ = W * D * W^-1 æ æ æ æ æ æ æ// D is diagonal > and > æ æG^h = W * D^h * W^-1 So to find W and D in Pari/GP there is the function mateigen, which > gives a version of W first. gp> æ W = mateigen(G) Then compute WInv as inverse of W gp> æ WInv = W^-1 Then compute D as gp> æ D = WInv * G * W Then you can define a pari-function for arbitrary powers of D: gp> æ ædpow(x) = matrix(n,n,r,c,if(r==c,D[r,c]^x)) and then write gp> æ æGPow = W * dpow(0.5) * WInv to store G^0.5 in a constant matrix. Then you use gp> æ æ æY = V(x) * GPow æ æand Y is approximately ~ V(gÁh(x)) or just define the scalar function, using the relevant coefficients of GPow gp> æ gpow(x) = sum(k=0,n-1, x^k * GPow[1+k,2]) ----------------------------------------- > 4) Remarks: However, the approximation may be poor. I use dimension n=64 for > standard discussion and n=128 if the approximation is difficult. > The mateigen-procedure in Pari/GP may perform poorly because it > doesn't exploit the easiness of the solution for triangular matrices, > for which we can even get exact solutions. I've a version for > triangular matrices of the special form here (matrix-operators on V(x)) > implemented, which gives the eigensystem-decomposition of n=64x64 > or 128x128 triangular-matrices in a second or two and even symbolically > if the matrix-dimension is small. > I could mail it to you, but I think I've already posted it in the > tetration-forum in the matrix-method-thread. Also I use Euler-summation for convergence-acceleration. But this > needs some more explanations - try this first to get used to it. Gottfried Bonjour, I do believe you're using heavy tools. > What about approximating x^2-1 by a conjugate of > a known iterate :Ex (x^2-1) = f-1(f(x))^2 - 2) ,T(x)=x^2-2 > or æ (x^2-1) =~f^-1 o T o f (x) > and (x^2-1)^[r] =~f^-1 o T^r o f (x) The trouble is it only seems to work when x is large. What about small > x? There's even more trouble. The approximation does not work at all! :( Try r = 2, x = 4. You get ~ 222.9955156499789347277396687. Which > is close to 223 -- and that's (4^2 - 1)^2 - 2 not (4^2 - 1)^2 - 1. > It's even worse for > close to 49727 -- and *that* is just ((4^2 - 1)^2 - 2)^2 - 2 not > ((4^2 - 1)^2 - 1)^2 - 1. It is not approaching f^n(x) where f(x) = x^2 > - 1, > it's approaching f^(n-1)(x^2 - 1) where f(x) = x^2 - 2!- Masquer le texte \ des messages pr.8ec.8edents - - Afficher le texte des messages pr.8ec.8edents - Bonjour Mike, I am not sure I've made my idea very clear : t1(x)=x^2-1 which has to be iterated, t2(x)=x^2-2 continuous iterate supposed known, We may search a conjugaison: t1 = h^[-1] o t2 o h or x^2-1 = h^[-1] ( h(x)^2-2) = x^2-2 I try h(x) = x+1/(2*x) as a first approximation , going far away from zero ... Alain === Subject: Re: Alain's T-approximation does not work (was Re: Continuous iteration of another quadratic map) posting-account=fwSgtAkAAACFnX70ssKwbvm9_oCZVHrx Gecko/2009011913 Firefox/3.0.6,gzip(gfe),gzip(gfe) On Feb 26, 3:25æam, alainvergh...@gmail.com > On Feb 23, 2:50æam, alainvergh...@gmail.com > Am 22.02.2009 23:20 schrieb mike3: What I'd be more curious about though is how you got that > series you posted in the message you gave in terms of the > height. As I'd be curious about a similar case for f(0), i.e. > g(-phi). (phi = golden ratio) Then I could see what that period > 2 orbit looks like when done continuously. > First you build the matrix-operator for the function. The function is æ æ g(x) = a*x + 1*x^2 æ æ æ æ æ æ æ æ where æ æ a = (1+sqrt(5)) = 2*phi For iteration you need g(x) and its powers g(x)^2, g(x)^3,... To compute these powers it suffices to operate on the coefficients only > (square,cube,... the formal powerseries (which is here only a polynomial)) For iteration you combine these coefficients in the same way as you would > do a matrix-multiplication. So write the coefficients of the powers > of g(x) into a matrix: æ g(x)^0 æ = 1 æ+ æ 0*x æ+ æ 0*x^2 + æ0*x^3 + 0*x^4 + ... > æ g(x)^1 æ = 0 æ+ æ a*x æ+ æ 1*x^2 + æ0*x^3 + 0*x^4 + ... > æ g(x)^2 æ = 0 æ+ æ 0*x æ+ a^2*x^2 + 2a*x^3 + 1*x^4 + ... > æ ... > up to a reasonable dimension (to get well approximated results > for fractional iterates) The matrix G has then only the coefficients of the formal powerseries > (and I write it in transposed form) æG = æ 1 æ. æ . æ æ. æ . > æ æ æ æ. æa æ . æ æ. æ . > æ æ æ æ. æ1 æa^2 æ . æ . > æ æ æ æ. æ. 2a æ æa^3 æ. > æ æ æ æ.... ---------------------------------------------- > Second, the use of the matrix G Then a vector V(x) = [1,x,x^2,x^3,...] up to the selected dimension gives > using æ V(x) * G = [ 1, g(x), g(x)^2, g(x)^3 ,... ] In Pari/GP write (for a given dimension n) gp> æV(x) = vector(n,c,x^(c-1)) gp> æY = V(1) * G > æ then Y = [1, g(1) , g(1)^2, g(1)^3,...] correctly up to a certain column > æ The remaining columns are wrong due to truncation to finite size. Since the form of the result-vector is again like V(), a vector of consecutive > powers, the iteration can be performed by the formal relation æ V(x) æ æ* G = V(g(x)) > æ V(g(x)) * G = V(g(g(x)) > æ ... > and so on. However: this is exactly valid only in the limit, where we > assume infinite dimension. For integer iterates k this can be written as æV(gÁk(x)) = V(x) * G^k æ æ æ(theoretically) gp> æ æ æ æY = V(x) * G^k æ æ æ(practically) > æ where Y ~ V(gÁk(x)) æ æ æ æ æonly approximately which is true for the case of infinite dimension, and only approximate > with finite dimension. Just try this in Pari/GP. ----------------------------------------------------------- > 3) fractional iterates Next step is fractional iterate. It is an obvious idea to use > fractional powers of G for this. Fractional powers of G can either be determined by binomial-expansion > with G and I as arguments or by diagonalization. > The general formula for diagonalization of a matrix is æ æG æ = W * D * W^-1 æ æ æ æ æ æ æ// D is diagonal > and > æ æG^h = W * D^h * W^-1 So to find W and D in Pari/GP there is the function mateigen, which > gives a version of W first. gp> æ W = mateigen(G) Then compute WInv as inverse of W gp> æ WInv = W^-1 Then compute D as gp> æ D = WInv * G * W Then you can define a pari-function for arbitrary powers of D: gp> æ ædpow(x) = matrix(n,n,r,c,if(r==c,D[r,c]^x)) and then write gp> æ æGPow = W * dpow(0.5) * WInv to store G^0.5 in a constant matrix. Then you use gp> æ æ æY = V(x) * GPow æ æand Y is approximately ~ V(gÁh(x)) or just define the scalar function, using the relevant coefficients of GPow gp> æ gpow(x) = sum(k=0,n-1, x^k * GPow[1+k,2]) ----------------------------------------- > 4) Remarks: However, the approximation may be poor. I use dimension n=64 for > standard discussion and n=128 if the approximation is difficult. > The mateigen-procedure in Pari/GP may perform poorly because it > doesn't exploit the easiness of the solution for triangular matrices, > for which we can even get exact solutions. I've a version for > triangular matrices of the special form here (matrix-operators on V(x)) > implemented, which gives the eigensystem-decomposition of n=64x64 > or 128x128 triangular-matrices in a second or two and even symbolically > if the matrix-dimension is small. > I could mail it to you, but I think I've already posted it in the > tetration-forum in the matrix-method-thread. Also I use Euler-summation for convergence-acceleration. But this > needs some more explanations - try this first to get used to it. Gottfried Bonjour, I do believe you're using heavy tools. > What about approximating x^2-1 by a conjugate of > a known iterate :Ex (x^2-1) = f-1(f(x))^2 - 2) ,T(x)=x^2-2 > or æ (x^2-1) =~f^-1 o T o f (x) > and (x^2-1)^[r] =~f^-1 o T^r o f (x) The trouble is it only seems to work when x is large. What about small > x? There's even more trouble. The approximation does not work at all! :( Try r = 2, x = 4. You get ~ 222.9955156499789347277396687. Which > is close to 223 -- and that's (4^2 - 1)^2 - 2 not (4^2 - 1)^2 - 1. > It's even worse for > close to 49727 -- and *that* is just ((4^2 - 1)^2 - 2)^2 - 2 not > ((4^2 - 1)^2 - 1)^2 - 1. It is not approaching f^n(x) where f(x) = x^2 > - 1, > it's approaching f^(n-1)(x^2 - 1) where f(x) = x^2 - 2!- Masquer le texte des messages pr.8ec.8edents - - Afficher le texte des messages pr.8ec.8edents - Bonjour Mike, I am not sure I've made my idea very clear : t1(x)=x^2-1 æwhich has to be iterated, > t2(x)=x^2-2 æcontinuous iterate supposed known, > We may search a conjugaison: > æ æt1 = h^[-1] o t2 o h > or x^2-1 = h^[-1] ( h(x)^2-2) = x^2-2 > I try æh(x) = x+1/(2*x) as a first approximation , > going far away from zero ... Alain Hmm. This seems to work better, although the superfunction identity (see my post) does not seem to transform it correctly to get iteration on smaller x-values. === Subject: sum of fibonaccis: = -1 ? The sum of the fibonacci-numbers is infinite; the partial sums diverge. However if I use the geometric series-formula I get sum = - 1 It fits also the pattern, which emerges, if I sum with some integer m>1 sum k=1,inf fib(k)/m^k = m/(m^2 - m -1) Is this appropriate? Gottfried Helms === Subject: Re: sum of fibonaccis: = -1 ? Am 25.02.2009 23:40 schrieb Gottfried Helms: > Is this appropriate? Gottfried Helms Hi - that I can defend that assignment of -1 to the sum by refering to the analytic continuation of the powerseries 1 + x + x^2 + ... and the closed-form-formula for the geometric series. (is the same way appropriate as...) However, Bill's invention makes me lough... :-) Very nice! Well, I took the result from the two geometric series, which occur, if I chose the Lucas-representation for the fibonacci numbers, and the sum of two analytic continued terms should be acceptable as closed-form formula too. In another context I had the *infinite* sum (though converging) of terms which all result from analytic continued geometric series - but that seems inappropriate: a concurring method gave different result which was better founded. Can we pinpoint the reason for an error/the error in the case of infinitely many values taken by analytic-continuation, or possibly have a means to quantify it/describe it by a function? Gottfried === Subject: Re: sum of fibonaccis: = -1 ? posting-account=AFsgCgkAAAA3VOfxqn2cTB2LbLN3nbER Gecko/20070319,gzip(gfe),gzip(gfe) One can also get it this way... Sumfib = 1 + 3 + 8 + 21 + 55 + ... : 1 + 2 + 5 + 13 + 34 + ... thus... : Sumfib +1 =(1+1)+ 3 + 8 + 21 + 55 + ... : (1 + 2)+ 5 + 13 + 34 + ... : : = (2 + 3) + 8 + 21 + 55 + ... : + (3 + 5) + 13 + 34 + ... : : = 0 + (5 + 8) + 21 + 55 + ... : 0 + (8 + 13) + 34 + ... : : = 0 + 0 +(13 + 21) + 55 + ... : 0 + 0 + (21 + 34) + ... : : = 0 + 0 + 0 + (34 + 55) + ... : 0 + 0 + 0 + (55 + ... : : = 0 + 0 = 0 : thus Sumfib = -1 . This is OC a totally Illegal Telescoping Cancellation (Infinite), (ITSI), but this itsi bitsi sort of calculation can be made rigorous by most of the summability methods Robert Israel alluded to (& Ben Goddard used); details in Hardy's Divergent Series. -- Wittering William * What an intellectual! * - Magdalen College Oxford, King's College Cambridge, * Imperial College London, Trinity College Dublin, * the Sorbonne, Harvard, MIT - he's heard of them all! === Subject: Re: sum of fibonaccis: = -1 ? The sum of the fibonacci-numbers is infinite; the partial sums > diverge. However if I use the geometric series-formula I get > sum = - 1 It fits also the pattern, which emerges, if I sum with some > integer m>1 sum k=1,inf fib(k)/m^k = m/(m^2 - m -1) Is this appropriate? It's appropriate as long as you make it clear that this is not literally a sum, but rather the result of some specified summability method. -- Robert Israel israel@math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: sum of fibonaccis: = -1 ? @mid.dfncis.de: The sum of the fibonacci-numbers is infinite; the partial sums > diverge. However if I use the geometric series-formula I get > sum = - 1 It fits also the pattern, which emerges, if I sum with some > integer m>1 sum k=1,inf fib(k)/m^k = m/(m^2 - m -1) Is this appropriate? It's as appropriate as sum_{k=0}^{inf} 2^k = 1/(1-2) = -1. B. -- Cheerfully resisting change since 1959. === Subject: Re: sum of fibonaccis: = -1 ? posting-account=Z3AipgkAAABkoMfyNwddSxsYhXHi5CDt MathPlayer 2.10d; .NET CLR 1.1.4322; PeoplePal 3.0),gzip(gfe),gzip(gfe) > @mid.dfncis.de: The sum of the fibonacci-numbers is infinite; the partial sums > diverge. However if I use the geometric series-formula I get > æsum = æ- 1 It fits also the pattern, which emerges, if I sum with some > integer m>1 æ sum k=1,inf fib(k)/m^k æ= m/(m^2 - m -1) Is this appropriate? It's as appropriate as sum {k=0}^{inf} æ2^k = 1/(1-2) = -1. B. But that sum is completely appropriate! It is a true infinite sum converging to the limit you give. At least it does if you are in the filed of 2-adic numbers. I can't think of a similar justification for the sum of the Fibonacci numbers, since we are raising units to powers and so there is no obvious way to create a justification. Achava === Subject: Re: sum of fibonaccis: = -1 ? Am 26.02.2009 06:19 schrieb Achava Nakhash, the Loving Snake: > (...) > I can't think of a similar justification for the sum of the Fibonacci > numbers, since we are raising units to powers and so there is no > obvious way to create a justification. Hi Achava - I don't understand this. raising units to powers - where do we this here? Would you mind to explain further? Btw, in my other post I said, that I got the sum by considering the lucas form fib(n) = (phi^n - (1-phi)^n)/sqrt(5) I just took the geometric series fibsum1 = sum {n>0} phi^n = 1/(1 - phi) // by anal.cont. of geometric series = - phi fibsum2 = sum {n>0} (1-phi)^n = 1/(1 - (1-phi)) // by anal.cont. of geometric series = 1/phi = phi-1 and fibsum = (fibsum1 - fibsum2)/sqrt(5) = -(2 phi - 1)/sqrt(5) = -1 Gottfried === Subject: Re: JSH: Factorization and the global economy Bull. Orders of Magnitude. You have no idea at all. >Ah. And you know this how??? >If the algorithms that you claim are 'far more secure' are known to >the >public, then they become available to all. e.g. Baton. See the NSA's >website. Look up their type 1 crypto algorithms. >If, however, they are secret, and you are privy to them, then your >revealing their existence and their security levels is a criminal >security violation. I know clearances and responcabilities very well, thank you. Just do a lot \ of reading and searching and you can run into several of the systems, public \ with some interesting info, watered down, not specific, to what is really there. And how would anyone know that ? NSA website; Cool they have newly declassified crypto from 1938. a lot of other good reading for sci.crypt for background they do have a declassified TS on cold war, interesting they would post such on the internet, known generic stuff, but why show a real declassified? ln 1968 a Soviet Golf-class nuclear submarine on patrol in the Pacific mysteriously went to the bottom with all hands. The Soviets could not locate the wreck, but the U.S. Navy could, and the U.S. began to study the feasibility of capturing it. (all rest white-ed out) They have some code patches for lenux too. If you factored some large #s that is quite impressive, and I wont bother you. I know of another person Brittish who was doing it, was obsessed with it and went off the deep end for good about 5 years ago, but he had factored \ some really big ones, not the blind search stuff either. === Subject: Re: JSH: Factorization and the global economy > Another moron. Do you know who you are talking to?? æHave you actually > done any crypto work for the NSA or DOD?? > Other methods are NOT 'far more secure'. > And the only method that is impossible to break is > a true OTP. æAnd key establishment problems make it > totally impractical to implement. æOne needs a secondary > communication channel for key establishment. Yes, but. æ What about quantum key distribution? It has not yet practical except over short distances. >(a few 10's of miles). Maintaining coherence over >longer distances has been a problem. I expect that these >difficulties will be solved in time. But it is not yet >ready for 'prime time'. > James, you said that you knew next to nothing about cryptography, so I will amplify some of the points Bob makes here. >However, this isn't the only problem. The key must be >truly random, not just pseudo-random, No, you cannot just use rand(). You need TRUE random data which is difficult to generate. A great deal of effort has been expended on this problem - true random data is definitely a problem. Google Yarrow or Fortuna for a couple of software approaches to the problem. For a hardware approach have a look at http://www.av8n.com/turbid/paper/turbid.htm >and it must be as long as the text it is encrypting. That means you need lots and lots of true random data. You do not just have a problem of data quality, you also have a problem of data quantity. >Furthermore, it can only be used ONCE. This is the big one. Once means once. A two time pad is as secure as transmitting the plaintext - not secure at all. Read up on the Venona break of Soviet cyphers: http://en.wikipedia.org/wiki/Venona_project Used only once can impose a lot of restrictions. For example, if your computer operating system can write a running program to a swapfile on disk then you might have to go as far as terminating your hard disk with extreme prejudice (and a sledgehemmer) and dissolving the result in acid. Once means once and once only. These issues are nothing to do with QKD, but are inherent in the use of a One Time Pad. The mathematical proof of its unbreakability depends critically on having TRUE random data used ONCE. If either condition is not met then the mathematical proof is not applicable and the resulting system is breakable. rossum >Thus, if you are encrypting a 10MB >message, you need a random 10MB that must be generated, >communicated, used ONCE, then thrown away. This makes >key management one royal pain in the *ss. Apparently, you, Pubkeybreaker, are known to be The > Real Thing to others in sci.crypt. It seems like too > good an opportunity to pass up, to find out whether > quantum key distribution is The Real Thing too. It is real. It works. But not yet on a large enough scale to >be useful. It can't yet be used if one wants to (say) >encrypt a message with OTP going from LA to (say) Miami. > Apparently, you also enjoy a good flame. I don't mind receiving flames. I have an asbestos suit :-) >Flame away. > === Subject: Re: JSH: Factorization and the global economy James, you said that you knew next to nothing about cryptography, so I > will amplify some of the points Bob makes here. However, this isn't the only problem. The key must be >truly random, not just pseudo-random, No, you cannot just use rand(). You need TRUE random data which is > difficult to generate. A great deal of effort has been expended on > this problem - true random data is definitely a problem. Google > Yarrow or Fortuna for a couple of software approaches to the > problem. For a hardware approach have a look at > http://www.av8n.com/turbid/paper/turbid.htm and it must be as long as the text it is encrypting. That means you need lots and lots of true random data. You do not > just have a problem of data quality, you also have a problem of data > quantity. But you're generating all these lovely quantum measurements anyway. Can't the measurements that are part of the quantum bit-passing also be the source of the one time pad? >Furthermore, it can only be used ONCE. This is the big one. Once means once. A two time pad is as secure as > transmitting the plaintext - not secure at all. Read up on the Venona > break of Soviet cyphers: http://en.wikipedia.org/wiki/Venona_project Used only once can impose a lot of restrictions. For example, if your > computer operating system can write a running program to a swapfile on > disk then you might have to go as far as terminating your hard disk > with extreme prejudice (and a sledgehemmer) and dissolving the result > in acid. Once means once and once only. These issues are nothing to do with QKD, but are inherent in the use > of a One Time Pad. The mathematical proof of its unbreakability > depends critically on having TRUE random data used ONCE. If either > condition is not met then the mathematical proof is not applicable and > the resulting system is breakable. I have a vague appreciation of the importance of using a One Time Pad only once (meaning really, really, really once -- no fooling). It's about on the level of More information good. I can't begin to imagine what I would do with, for example, two messages that I suspected were XORed with the same OTP. I don't understand why I might need a sledge hammer to be certain some files were not used again. Assuming all this stuff is software-mediated (the only way I can imagine me using the stuff), surely, the software is not going to re-use an old OTP accidentally, is it? It's possible someone could break into my computer and get the old OTPs laying around unerased and unsmashed, but in that case, they should just go ahead and read my mail. Jim Burns === Subject: Re: JSH: Factorization and the global economy >I have a vague appreciation of the importance of using a >One Time Pad only once (meaning really, really, really once >-- no fooling). It's about on the level of More information >good. I can't begin to imagine what I would do with, for example, >two messages that I suspected were XORed with the same OTP. Tim has answered this, you XOR the two cyphertexts together which removes all trace of the key, just leaving the XOR of the plaintexts. That is easily decryptable as an English plaintext is going to contain substrings like the , and etc. Using them as trial keys in all possible positions will start to bring out chunks of the other plaintext in the appropriate positions. Other methods are possible. >I don't understand why I might need a sledge hammer to be >certain some files were not used again. Assuming all this >stuff is software-mediated (the only way I can imagine me >using the stuff), surely, the software is not going to >re-use an old OTP accidentally, is it? If the adversary gets hold of your OTP after you have used it ... I have seen stories of the CIA doing exactly that with some of their old mainframe disc packs. When they failed they were terminated rather than let them out of the building for repair. As with all security, you need to judge the level of security you use to be commensurate with the resources available to your potential attackers. rossum === Subject: Re: The Real Purpose Of Math <19796970.1235316588589.JavaMail.jakarta@nitrogen.mathforum.org> <8763j2ffhf.fsf@phiwumbda.org> posting-account=KroL8woAAAACDGpRprxyFi_gYw4Un8Xt 2.0.50727; yplus 5.1.05b),gzip(gfe),gzip(gfe) > The idea is clearly religious, so why not take it to a æforum for religious discussion? What is totally religious about a system of Governing Equations? Yes, what *is* religious about the following inference? æ Governing Equation > existence of a Governor > existence of God > æ (topologically identified with the Governor) That's just topology, that is. æAnyone who denies it is simply > prejudiced against simple mathematics. You've convinced me, I tell you what. > -- > But what if I'm right [and have solved the factoring problem]? æAnd > what if I can't contain the problem and, oh, in six months from now, > you are desperately trying to find food as you run from humans who > have turned to cannibalism to survive? æ-- James S. Harris (12/21/09) A preliminary conceptualization of the Governing Equation (actually an inter-locking system of partial differential equations) is that it would contain the equations associated with the Standard Model and the best and most elegant description of gravitation as special cases. This set of equations possibly will utilize a generalization of complex variables to higher dimensional commutative algebras. http://www.intelrap.com/lt1.html Complex Variables span only the plane but in order to fully understand a higher dimensional brane we need a more powerful math tool. But if we do not believe that a Creator does exist who created many of the things for which we wish then we will never be motivated to search for the rule... the rule of the highest law of the Universe which may be metaphorically described as a United Verse that beckons all those with a genuine thirst to seek an understanding and to seek it first. If a man has a genuine creative wonder then he understands that he stands under a power much greater than himself. Then he may be lifted upon the shoulders of giants to learn the wonders of math and science and the knowledge that when lies are deleted only the truth is left. Newton stood on the shoulders of giants who now seek more new wondering clients to reveal more mysteries of a magnificent creation in which man plays a principal role in the observation. www.intelrap.com === Subject: Re: Proof of Fermat's Last Theorem posting-account=brOM-AoAAAChaAJEiH5z610-YOTfECd9 Gecko/20070530 Fedora/1.5.0.12-1.fc5 Firefox/1.5.0.12,gzip(gfe),gzip(gfe) > According to Fermat's Last Theorem, a cube can not be divided into > two smaller cubes, a fourth power can not be divided into two smaller > fourth powers, and so on. In other words, if a,b,c, and n are integers and a^n+b^n=c^n and > abcn <> 0, then the set of possible solutions for n is {-2, -1, 1, 2}. What is the proof for this? It's pretty elegant, but there isn't enough space here to write it. Michael === Subject: Re: number theory for children The problem is that number theory makes assertions requiring little background to understand, but proving things, which after all is what number theorists do, requires a whole armamentarium of stuff ranging widely over mathematics. OTOH, sure, show children some of the fascinating conclusions. Just be prepared to run for cover if the child asks How do they know that? === Subject: Re: Congrunces and perfect squares This number, when factorised, gives (10^(k+1) - 1)^2 which is a > perfect square. However, is it possible to show that the original number is a perfect > square using congruences from number theory and not by factorising it? Perhaps you mean 10^(2k+2) - 2(10^(k+1)) + 1 > instead of 10^(2k+1) - 2(10^(k+1)) + 1 Oops - I missed that. But yes, as you indicate, the first term was clearly meant to be (10^(2k+1))^2 or equivalently, 10^(2k+2) For the OP: Using the simple substitution x = 10^(k+1) your expression becomes x^2 + 2x + 1 which clearly factors as (x + 1)^2 The substitution makes the factorization more obvious. quasi === Subject: Re: Congrunces and perfect squares > This number, when factorised, gives (10^(k+1) - 1)^2 which is a > perfect square. However, is it possible to show that the original number is a perfect > square using congruences from number theory and not by factorising it? Perhaps you mean 10^(2k+2) - 2(10^(k+1)) + 1 > instead of 10^(2k+1) - 2(10^(k+1)) + 1 Oops - I missed that. But yes, as you indicate, the first term was clearly meant to be (10^(2k+1))^2 or equivalently, 10^(2k+2) For the OP: Using the simple substitution x = 10^(k+1) your expression becomes x^2 + 2x + 1 The above should be x^2 - 2x + 1 >which clearly factors as (x + 1)^2 And hence the above should be (x - 1)^2 >The substitution makes the factorization more obvious. quasi === Subject: Re: Congrunces and perfect squares This number, when factorised, gives (10^(k+1) - 1)^2 which is a > perfect square. However, is it possible to show that the original number is a perfect > square using congruences from number theory and not by factorising it? Perhaps you mean 10^(2k+2) - 2(10^(k+1)) + 1 > instead of 10^(2k+1) - 2(10^(k+1)) + 1 Oops - I missed that. But yes, as you indicate, the first term was clearly meant to be (10^(2k+1))^2 Ugh -- how careless I am. The above should be 10^(k+1) >or equivalently, 10^(2k+2) For the OP: Using the simple substitution x = 10^(k+1) your expression becomes x^2 + 2x + 1 The above should be x^2 - 2x + 1 which clearly factors as (x + 1)^2 And hence the above should be (x - 1)^2 The substitution makes the factorization more obvious. To the OP: I made a mess of things in my last 2 replies, but with the current corrections, it should now be ok. quasi === Subject: Re: Congrunces and perfect squares > This number, when factorised, gives (10^(k+1) - 1)^2 which is a > perfect square. However, is it possible to show that the original number is a perfect > square using congruences from number theory and not by factorising it? Perhaps you mean 10^(2k+2) - 2(10^(k+1)) + 1 > instead of 10^(2k+1) - 2(10^(k+1)) + 1 Oops - I missed that. But yes, as you indicate, the first term was clearly meant to be (10^(2k+1))^2 Ugh -- how careless I am. The above should be 10^(k+1) Ugh again -- I meant (10^(k+1))^2 >or equivalently, 10^(2k+2) For the OP: Using the simple substitution x = 10^(k+1) your expression becomes x^2 + 2x + 1 The above should be x^2 - 2x + 1 which clearly factors as (x + 1)^2 And hence the above should be (x - 1)^2 The substitution makes the factorization more obvious. To the OP: I made a mess of things in my last 2 replies, but with the current >corrections, it should now be ok. My last 3 replies -- all with careless errors. But now it's OK. quasi === Subject: Re: (m-1)x^2-2x+1-m=0 has two real solutions... posting-account=DSjMzAoAAAAISwlGnTY2c6bnmUVmdFNf AppleWebKit/528.16 (KHTML, like Gecko) Version/4.0 Safari/528.16,gzip(gfe),gzip(gfe) > Hy, I'm having troubles with the following problem: Prove that the following equation has two real solutions, for any m, > if m is not 1: > (m-1)x^2-2x+1-m=0 so, to find out how many solutions I apply the quadratic formula: ( 2 +/- sqr(4 - 4 * (m-1) * (1-m) ) / 2 (m-1) And take out the part that matters in this case: 16 - 4 * (m-1) * (1-m) And, since here I'm at a lost. I tried solving this last part, so: > 4 - 4 * (m-1)*(1-m) = 4 - 4 *(m-m^2-1+m)=4-4*(-m^2+2m-1)=4+4m^2-8m > +4=4m^2-8m+8 > This solves in my calculator to 1+1i or 1-1i. Which I have no use for > (to my knowledge). Now the book's answer is: > 4+4(m-1)^2 > 0 > witch I understand proves the point, althouhgt I don't know how I'm > supposed to get from (m-1)*(1-m) to (m-1)^2 (even if I'm feeling it > should be something rather obvious....) The discriminant b^2 - 4ac equals 4 - 4 (m-1)(1-m) which equals 4 + 4 (m-1)(m-1) (what is -(1-m) ? ). So you have a positive number plus a square; the sum is positive, therefore two solutions. But there is the division by 2a = 2 (m-1) which doesn't work out when m = 1. In that case the complete equation is -2x = 0 with solution x = 0 only. === Subject: Re: In need of a closed form for this constant! >How about using an root finding algorithm which >converges faster? >For example, if one start with a close enough initial >value, >Newton-Raphson method ( >http://en.wikipedia.org/wiki/Newton%27s_method > ) >usually converges quadratically (ie. double your >precision in >every iteration). Funny you should ask because I just discovered > a faster convergence by modifying my original > algorithm. The original which produces a slow convergence of (n)-- (3 - (log((sqrt(pi+1))*pi^2))) + pi = n > (3 - (log((sqrt(n+1))*n^2))) + n =n_1 > (3 - (log((sqrt(n_1+1))*n_1^2))) + n_1 = n_2 > (3 - (log((sqrt(n_2+1))*n_2^2))) + n_2 = n_3 > (3 - (log((sqrt(n_3+1))*n_3^2))) + n_3 = n_4 > etc. Modifying for a much faster convergence of (n) (3 - (log((sqrt(pi+1))*pi^2))) *((1/pi)+1) + pi = n > (3 - (log((sqrt(n+1))*n^2))) *((1/n)+1) + n = n_1 > (3 - (log((sqrt(n_1+1))*n_1^2))) *((1/n_1)+1) + n_1 = n_2 > (3 - (log((sqrt(n_2+1))*n_2^2))) *((1/n_2)+1) + n_2 = n_3 > (3 - (log((sqrt(n_3+1))*n_3^2))) *((1/n_3)+1) + n_3 = n_4 > etc. Probably more than twice as fast as my original version. Dan Your iteration converges linearly. I think you need around > 350 iterations to get a 1000 digit precision. In contrast, Newton-Raphson on f(x) = x^5 + x^4 - e^6 need > only about 9 iteration to achieve same accuracy ( with the > additional benefit you don't need to deal with the slower > multi-precision log() function ). BTW, does anyone know what is the relative complexity of > computing multiple precison log vs multiply and divide???? Denote by AGM the aritmethic-geometric mean. All elementary transcendental function can be computed with bit complexity O_B(M(n).log(n)) where M(n) is the bit complexity of multiplication. Borwein and Borwein, Pi_and_the_AGM. Denote by K(k) the complete elliptic integral of the first kind. K(k') (1) pi.----- = log(-) K(k) {q) K(k) 1 ----- = - [theta_3(q)]^2. pi 2 [theta_2(q)]^2 k = ---------------- [theta_3(q)]^2 theta_3(q) = sum_{-oo < n < oo} q^{n^2} theta_2(q) = sum_{-oo < n < oo} q^{(n + 1/2)^2} Fix q, 0 < a < q < b < 1. Step 1: Calculate K(k)/pi Step 2: Calculate k. Step 3: Calculate K(k') from the AGM. Step 4: Calculate log(1/q). -- Michael Press === Subject: Re: In need of a closed form for this constant! <28197189.1235495408224.JavaMail.jakarta@nitrogen.mathforum.org> posting-account=SvltewoAAAAi7TTYrD3mAaLUHzDiF2d1 Gecko/2009011913 Firefox/3.0.6,gzip(gfe),gzip(gfe) BTW, does anyone know what is the relative complexity of > computing multiple precison log vs multiply and divide???? Denote by AGM the aritmethic-geometric mean. All elementary > transcendental function can be computed with bit complexity æ æ O B(M(n).log(n)) where M(n) is the bit complexity of multiplication. æ > Borwein and Borwein, Pi and the AGM. Michael Press Cool, one more item on my bed-side to-read list (growing very fast in past 3 weeks since I start to re-read sci.math ;-p) === Subject: countable collection of dense open subsets of reals with intersection having 0 measure posting-account=2kYL5QoAAAAmH1e_g3WSyZ5Z9zXmRQX3 .NET CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) How can I construct a sequence of dense open subsets of IR such that the intersection is Lesbesque null set? === Subject: Re: countable collection of dense open subsets of reals with intersection having 0 measure > How can I construct a sequence of dense open subsets of IR such that > the intersection is Lesbesque null set? Am I reading this right? Isn't any dense, open subset of the reals (and hence a countable intersection) cocountable? === Subject: Re: countable collection of dense open subsets of reals with intersection having 0 measure posting-account=AdyLXQoAAABgRay99CKv1O8Y_7jjivwq InfoPath.1; .NET CLR 2.0.50727; .NET CLR 3.0.04506.30; .NET CLR 3.0.04506.648),gzip(gfe),gzip(gfe) > Am I reading this right? Isn't any dense, open subset of > the reals (and hence a countable intersection) cocountable? If C is a closed nowhere dense subset of the reals, then the complement of C is a dense open set. Now note that C can be uncountable, for example when C is a Cantor set. Dave L. Renfro === Subject: Re: countable collection of dense open subsets of reals with intersection having 0 measure > How can I construct a sequence of dense open subsets of IR such that > the intersection is Lesbesque null set? Cover the rationals by small open sets. === Subject: Re: countable collection of dense open subsets of reals with intersection having 0 measure posting-account=2kYL5QoAAAAmH1e_g3WSyZ5Z9zXmRQX3 .NET CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) How can I construct a sequence of dense open subsets of IR such that > the intersection is Lesbesque null set? Cover the rationals by small open sets. How can I construct a sequence of dense open subsets of IR such that > the intersection is Lesbesque null set? Cover the rationals by small open sets. Ok, here is my attempt (your hint was very good). Enumerate the rationals Q = {r k} for k=1,2,... Let G n = / ( r k - 1/(n*2*2^k) , r k + 1/(n*2*2^k) ) over n=1,2,... now by subadditivity, m(G n) <= SUM over k of( 1/n*2^k ) = 1/n (1/2 + 1/4 + 1/8 + ...) = 1/n Define G := /G n over n = 1,2,... Now, clearly G 1 superset of G 2 superset of G 3 ... Thus m(G) = lim{n->inf} m(G n) <= lim{n->inf} 1/n = 0 So m(G) = 0. === Subject: Re: Alternating Vertical and Horizontal Moves In Grid posting-account=ey-PjQoAAAD_Gi3gK0ZWKofsmviYOWhV Gecko/20081217 Firefox/2.0.0.20,gzip(gfe),gzip(gfe) Hi! This is an interesting problem, that I have never seen before. I will make the assumption that at any point a vertical move may be U or D, and a horizontal move may be L or R. In both the start-at-a-corner and general case, a(1) = 1, a(2) = 4, and a(3) = 6, for starters. Now I will focus only on the start-at-a-corner case, and it may as well be the upper left. Obviously n^2 is an upper bound for a(n), and if we try to establish some reasonable lower bounds, by considering a boustrophedonic traversal of the grid, there is a natural division into cases based on the value of n mod 4. In particular, I conjecture that the optimal algorithm for sufficiently large n = 1 mod 4 is to traverse the grid according to the following rules: 1) Start with down 1) Choose up over down whenever at least 4 more moves can be made as a result of doing so. 2) Alternate between choosing right n - 1 times in a row, then left n - 1 times in a row, then right, etc. To see what I mean by this, the 9 x 9 grid would be traversed like: DRURDRURDRURDRDRDLDLULULDLULDLDLDRDRURURDRDRURDRDLDLULDLULDLULDLU. With a little work this establishes: n mod 4 = 1 -> [ n^2 - 2n + 3 <= a(n) ] For the other cases some slight modifications of the algorithm above produce pretty tight lower bounds as well. In the above case, I would expect equality probably holds, but proving it is another story, it seems like it would be tricky. dc === Subject: Re: Alternating Vertical and Horizontal Moves In Grid >What is the sequence {a(k)} that is defined as follows? Start with an n-by-n grid. Move from square to adjacent square by >moving vertically then horizontally then vertically then horizontally, >following each horizontal move with a vertical move, and following >each vertical move with a horizontal move. >No square may be visited more than once. a(n) = the maximum number of squares visitable in this way on an n-by- >n grid. By brute force search, here are the values of a(n) for n = 1, ..., 8: 1, 4, 7, 14, 19, 32, 39, 58 >A variation: You must start in a corner square. Then the first 8 values are: 1, 4, 6, 14, 18, 32, 38, 58 Note that, so far, if n is even, starting in a corner is just as good a starting place as any other location, while if n is odd and n > 1, the value of a(n) when starting in a corner appears to be just 1 less than that for an optimum starting place. >(It does not matter if the first move is vertical or horizontal, as >far as the values of each a(n) are concerned, of course.) I am guessing that this sequence is really easy to calculate, maybe a >(n) is as simple as a quadratic polynomial in terms of n. Or is the >best way to calculate each a(n) is to do a brute-force search of >possible paths through the grid? The function a(n) is definitely not a polynomial in n. Clearly a(n) is positive and bounded above by n^2, hence, if a(n) was a polynomial in n, it would have to be at most of degree 2. But the calculated values of a(n) as posted above contradict that. So what is the nature of a(n)? Consider the sequence b(n) = n^2 - a(n) Clearly b(n) must nonnegative. I suspect it would not be too hard to prove that b(n) must be positive, for n > 2. We can ask whether the sequence b(n) is bounded above? I suspect not, but if, in fact, b(n) is bounded above, then I would guess that the sequence b(n) repeats. Other than brute force search, you might try for recursive relationships, or at least recursive bounds. quasi === Subject: Re: Alternating Vertical and Horizontal Moves In Grid What is the sequence {a(k)} that is defined as follows? Start with an n-by-n grid. Move from square to adjacent square by >moving vertically then horizontally then vertically then horizontally, >following each horizontal move with a vertical move, and following >each vertical move with a horizontal move. >No square may be visited more than once. a(n) = the maximum number of squares visitable in this way on an n-by- >n grid. By brute force search, here are the values of a(n) for n = 1, ..., 8: 1, 4, 7, 14, 19, 32, 39, 58 A variation: You must start in a corner square. Then the first 8 values are: 1, 4, 6, 14, 18, 32, 38, 58 Note that, so far, if n is even, starting in a corner is just as good >a starting place as any other location, while if n is odd and n > 1, >the value of a(n) when starting in a corner appears to be just 1 less >than that for an optimum starting place. (It does not matter if the first move is vertical or horizontal, as >far as the values of each a(n) are concerned, of course.) I am guessing that this sequence is really easy to calculate, maybe a >(n) is as simple as a quadratic polynomial in terms of n. Or is the >best way to calculate each a(n) is to do a brute-force search of >possible paths through the grid? The function a(n) is definitely not a polynomial in n. Clearly a(n) is >positive and bounded above by n^2, hence, if a(n) was a polynomial in >n, it would have to be at most of degree 2. But the calculated values >of a(n) as posted above contradict that. So what is the nature of a(n)? Consider the sequence b(n) = n^2 - a(n) Clearly b(n) must nonnegative. I suspect it would not be too hard to >prove that b(n) must be positive, for n > 2. We can ask whether the sequence b(n) is bounded above? I suspect not, >but if, in fact, b(n) is bounded above, then I would guess that the >sequence b(n) repeats. Other than brute force search, you might try for recursive >relationships, or at least recursive bounds. Ok, there is a simple pattern that I missed. When n is even, it appears that a(n) = n^2 - n + 2 quasi === Subject: Re: Alternating Vertical and Horizontal Moves In Grid >What is the sequence {a(k)} that is defined as follows? Start with an n-by-n grid. Move from square to adjacent square by >moving vertically then horizontally then vertically then horizontally, >following each horizontal move with a vertical move, and following >each vertical move with a horizontal move. >No square may be visited more than once. a(n) = the maximum number of squares visitable in this way on an n-by- >n grid. By brute force search, here are the values of a(n) for n = 1, ..., 8: 1, 4, 7, 14, 19, 32, 39, 58 A variation: You must start in a corner square. Then the first 8 values are: 1, 4, 6, 14, 18, 32, 38, 58 Note that, so far, if n is even, starting in a corner is just as good >a starting place as any other location, while if n is odd and n > 1, >the value of a(n) when starting in a corner appears to be just 1 less >than that for an optimum starting place. (It does not matter if the first move is vertical or horizontal, as >far as the values of each a(n) are concerned, of course.) I am guessing that this sequence is really easy to calculate, maybe a >(n) is as simple as a quadratic polynomial in terms of n. Or is the >best way to calculate each a(n) is to do a brute-force search of >possible paths through the grid? The function a(n) is definitely not a polynomial in n. Clearly a(n) is >positive and bounded above by n^2, hence, if a(n) was a polynomial in >n, it would have to be at most of degree 2. But the calculated values >of a(n) as posted above contradict that. So what is the nature of a(n)? Consider the sequence b(n) = n^2 - a(n) Clearly b(n) must nonnegative. I suspect it would not be too hard to >prove that b(n) must be positive, for n > 2. We can ask whether the sequence b(n) is bounded above? I suspect not, >but if, in fact, b(n) is bounded above, then I would guess that the >sequence b(n) repeats. Other than brute force search, you might try for recursive >relationships, or at least recursive bounds. Ok, there is a simple pattern that I missed. When n is even, it appears that a(n) = n^2 - n + 2 Moreover, it's fairly easy to show that, for even n, a(n) >= n^2 - n + 2 Assuming n is even, a valid path through n^2 - n + 2 vertices can be explicitly constructed as follows ... Start at (0,0). Go right, up, left, up. Repeat until reaching the top. Since n is even, the last 2 moves must have been right, up, ending at (1,n-1). Now go right, down, ending at (2,n-2). We have so far visited 2n + 1 vertices. Now recursively do the same in the n-2 by n-2 square with upper left corner vertex (2,n-2) and lower right corner vertex (n-1,1). The path described above adheres to the restrictions, and it's easily proved, by induction, that it has exactly n^2 - n + 2 vertices. quasi === Subject: Re: Alternating Vertical and Horizontal Moves In Grid What is the sequence {a(k)} that is defined as follows? Start with an n-by-n grid. Move from square to adjacent square by >moving vertically then horizontally then vertically then horizontally, >following each horizontal move with a vertical move, and following >each vertical move with a horizontal move. >No square may be visited more than once. a(n) = the maximum number of squares visitable in this way on an n-by- >n grid. By brute force search, here are the values of a(n) for n = 1, ..., 8: 1, 4, 7, 14, 19, 32, 39, 58 A variation: You must start in a corner square. Then the first 8 values are: 1, 4, 6, 14, 18, 32, 38, 58 Note that, so far, if n is even, starting in a corner is just as good >a starting place as any other location, while if n is odd and n > 1, >the value of a(n) when starting in a corner appears to be just 1 less >than that for an optimum starting place. (It does not matter if the first move is vertical or horizontal, as >far as the values of each a(n) are concerned, of course.) I am guessing that this sequence is really easy to calculate, maybe a >(n) is as simple as a quadratic polynomial in terms of n. Or is the >best way to calculate each a(n) is to do a brute-force search of >possible paths through the grid? The function a(n) is definitely not a polynomial in n. Clearly a(n) is >positive and bounded above by n^2, hence, if a(n) was a polynomial in >n, it would have to be at most of degree 2. But the calculated values >of a(n) as posted above contradict that. So what is the nature of a(n)? Consider the sequence b(n) = n^2 - a(n) Clearly b(n) must nonnegative. I suspect it would not be too hard to >prove that b(n) must be positive, for n > 2. We can ask whether the sequence b(n) is bounded above? I suspect not, >but if, in fact, b(n) is bounded above, then I would guess that the >sequence b(n) repeats. Other than brute force search, you might try for recursive >relationships, or at least recursive bounds. Ok, there is a simple pattern that I missed. When n is even, it appears that a(n) = n^2 - n + 2 Moreover, it's fairly easy to show that, for even n, a(n) >= n^2 - n + 2 Assuming n is even, a valid path through n^2 - n + 2 vertices can be >explicitly constructed as follows ... Start at (0,0). Go right, up, left, up. Repeat until reaching the top. Since n is even, the last 2 moves must have been right, up, ending at >(1,n-1). Now go right, down, ending at (2,n-2). We have so far visited 2n + 1 vertices. Now recursively do the same in the n-2 by n-2 square with upper left >corner vertex (2,n-2) and lower right corner vertex (n-1,1).| To clarify: At each successive lower level in the recursion, switch up and down. Also, when you finish the 2 x 2 level, stop. >The path described above adheres to the restrictions, and it's easily >proved, by induction, that it has exactly n^2 - n + 2 vertices. quasi === Subject: Re: Alternating Vertical and Horizontal Moves In Grid <4utbq4dhhldvts8o5hl3ld180tan2tbdmq@4ax.com> posting-account=ey-PjQoAAAD_Gi3gK0ZWKofsmviYOWhV Gecko/20081217 Firefox/2.0.0.20,gzip(gfe),gzip(gfe) What is the sequence {a(k)} that is defined as follows? Start with an n-by-n grid. Move from square to adjacent square by >moving vertically then horizontally then vertically then horizontally, >following each horizontal move with a vertical move, and following >each vertical move with a horizontal move. >No square may be visited more than once. a(n) = the maximum number of squares visitable in this way on an n-by- >n grid. By brute force search, here are the values of a(n) for n = 1, ..., 8: 1, 4, 7, 14, 19, 32, 39, 58 A variation: You must start in a corner square. Then the first 8 values are: 1, 4, 6, 14, 18, 32, 38, 58 Note that, so far, if n is even, starting in a corner is just as good >a starting place as any other location, while if n is odd and n > 1, >the value of a(n) when starting in a corner appears to be just 1 less >than that for an optimum starting place. (It does not matter if the first move is vertical or horizontal, as >far as the values of each a(n) are concerned, of course.) I am guessing that this sequence is really easy to calculate, maybe a >(n) is as simple as a quadratic polynomial in terms of n. Or is the >best way to calculate each a(n) is to do a brute-force search of >possible paths through the grid? The function a(n) is definitely not a polynomial in n. Clearly a(n) is >positive and bounded above by n^2, hence, if a(n) was a polynomial in >n, it would have to be at most of degree 2. But the calculated values >of a(n) as posted above contradict that. So what is the nature of a(n)? Consider the sequence b(n) = n^2 - a(n) Clearly b(n) must nonnegative. I suspect it would not be too hard to >prove that b(n) must be positive, for n > 2. We can ask whether the sequence b(n) is bounded above? I suspect not, >but if, in fact, b(n) is bounded above, then I would guess that the >sequence b(n) repeats. Other than brute force search, you might try for recursive >relationships, or at least recursive bounds. Ok, there is a simple pattern that I missed. When n is even, it appears that a(n) = n^2 - n + 2 Moreover, it's fairly easy to show that, for even n, a(n) >= n^2 - n + 2 Assuming n is even, a valid path through n^2 - n + 2 vertices can be >explicitly constructed as follows ... Start at (0,0). Go right, up, left, up. Repeat until reaching the top. Since n is even, the last 2 moves must have been right, up, ending at >(1,n-1). Now go right, down, ending at (2,n-2). We have so far visited 2n + 1 vertices. Now recursively do the same in the n-2 by n-2 square with upper left >corner vertex (2,n-2) and lower right corner vertex (n-1,1).| To clarify: At each successive lower level in the recursion, switch up and > down. Also, when you finish the 2 x 2 level, stop. The path described above adheres to the restrictions, and it's easily >proved, by induction, that it has exactly n^2 - n + 2 vertices. quasi this seems legitimate to me. quasi, by the sounds of it, you have some kind of a computer program capable of a brute force test? can you test my conjecture above for n = 9, 13? i'm thinking we are going tol end up here with a quadratic answer, but different quadratics depending on the value of n mod 4. === Subject: Re: Alternating Vertical and Horizontal Moves In Grid What is the sequence {a(k)} that is defined as follows? Start with an n-by-n grid. Move from square to adjacent square by >moving vertically then horizontally then vertically then horizontally, >following each horizontal move with a vertical move, and following >each vertical move with a horizontal move. >No square may be visited more than once. a(n) = the maximum number of squares visitable in this way on an n-by- >n grid. By brute force search, here are the values of a(n) for n = 1, ..., 8: 1, 4, 7, 14, 19, 32, 39, 58 A variation: You must start in a corner square. Then the first 8 values are: 1, 4, 6, 14, 18, 32, 38, 58 Note that, so far, if n is even, starting in a corner is just as good >a starting place as any other location, while if n is odd and n > 1, >the value of a(n) when starting in a corner appears to be just 1 less >than that for an optimum starting place. (It does not matter if the first move is vertical or horizontal, as >far as the values of each a(n) are concerned, of course.) I am guessing that this sequence is really easy to calculate, maybe a >(n) is as simple as a quadratic polynomial in terms of n. Or is the >best way to calculate each a(n) is to do a brute-force search of >possible paths through the grid? The function a(n) is definitely not a polynomial in n. Clearly a(n) is >positive and bounded above by n^2, hence, if a(n) was a polynomial in >n, it would have to be at most of degree 2. But the calculated values >of a(n) as posted above contradict that. So what is the nature of a(n)? Consider the sequence b(n) = n^2 - a(n) Clearly b(n) must nonnegative. I suspect it would not be too hard to >prove that b(n) must be positive, for n > 2. We can ask whether the sequence b(n) is bounded above? I suspect not, >but if, in fact, b(n) is bounded above, then I would guess that the >sequence b(n) repeats. Other than brute force search, you might try for recursive >relationships, or at least recursive bounds. Ok, there is a simple pattern that I missed. When n is even, it appears that a(n) = n^2 - n + 2 Moreover, it's fairly easy to show that, for even n, a(n) >= n^2 - n + 2 Assuming n is even, a valid path through n^2 - n + 2 vertices can be >explicitly constructed as follows ... Start at (0,0). Go right, up, left, up. Repeat until reaching the top. Since n is even, the last 2 moves must have been right, up, ending at >(1,n-1). Now go right, down, ending at (2,n-2). We have so far visited 2n + 1 vertices. Now recursively do the same in the n-2 by n-2 square with upper left >corner vertex (2,n-2) and lower right corner vertex (n-1,1).| To clarify: At each successive lower level in the recursion, switch up and > down. Also, when you finish the 2 x 2 level, stop. The path described above adheres to the restrictions, and it's easily >proved, by induction, that it has exactly n^2 - n + 2 vertices. quasi this seems legitimate to me. quasi, by the sounds of it, you have some >kind of a computer program capable of a brute force test? can you test >my conjecture above for n = 9, 13? i'm thinking we are going tol end >up here with a quadratic answer, but different quadratics depending on >the value of n mod 4. Yes, that seems very plausible -- 3 quadratics, one for even n, one for n = 1 mod 4, and one for n = 3 mod 4. I do have a brute force test, but it wasn't designed for a heavy load. For n > 8, it struggles. I'll let it run for a while -- hopefully, it will at least compute a(9). quasi === Subject: Re: Alternating Vertical and Horizontal Moves In Grid <4utbq4dhhldvts8o5hl3ld180tan2tbdmq@4ax.com> posting-account=ey-PjQoAAAD_Gi3gK0ZWKofsmviYOWhV Gecko/20081217 Firefox/2.0.0.20,gzip(gfe),gzip(gfe) no, i changed my mind about the mod 4 thing, it looks a lot like your formula is right whenever n is even, and my formula is right (by a similar argument) when n is odd. of course, we have only established them as lower bounds with certainty. === Subject: Re: Alternating Vertical and Horizontal Moves In Grid >no, i changed my mind about the mod 4 thing, it looks a lot like your >formula is right whenever n is even, and my formula is right (by a >similar argument) when n is odd. of course, we have only established >them as lower bounds with certainty. Yes, if we require corner starting point, and assuming n > 1, I think you may be right. Thus, for a corner starting point, and n > 1, the conjecture is: a(n) = n^2 - 2n + 3, if n is odd n^2 - n + 2, if n is even quasi PS -- My Maple program is still working on a(9), assuming a required corner starting point. === Subject: Re: Alternating Vertical and Horizontal Moves In Grid no, i changed my mind about the mod 4 thing, it looks a lot like your >formula is right whenever n is even, and my formula is right (by a >similar argument) when n is odd. of course, we have only established >them as lower bounds with certainty. Yes, if we require corner starting point, and assuming n > 1, I think >you may be right. Thus, for a corner starting point, and n > 1, the conjecture is: a(n) = > n^2 - 2n + 3, if n is odd > n^2 - n + 2, if n is even quasi PS -- My Maple program is still working on a(9), assuming a required >corner starting point. Ok, Maple completed a(9) and is now working on a(11). Good news -- a(9) = 66, as expected. Note, for convenience, we are now always making the assumption of a corner starting point. For even n, it apparently doesn't matter. For odd n > 1, allowing a variable starting point appears to make a(n) increase by 1. quasi === Subject: Re: Alternating Vertical and Horizontal Moves In Grid <4utbq4dhhldvts8o5hl3ld180tan2tbdmq@4ax.com> <4p7cq45og61mehq711tc9qm7ir9avhp4gl@4ax.com> posting-account=ey-PjQoAAAD_Gi3gK0ZWKofsmviYOWhV Gecko/20081217 Firefox/2.0.0.20,gzip(gfe),gzip(gfe) no, i changed my mind about the mod 4 thing, it looks a lot like your >formula is right whenever n is even, and my formula is right (by a >similar argument) when n is odd. of course, we have only established >them as lower bounds with certainty. Yes, if we require corner starting point, and assuming n > 1, I think >you may be right. Thus, for a corner starting point, and n > 1, the conjecture is: a(n) = > n^2 - 2n + 3, if n is odd > n^2 - n + 2, if n is even quasi PS -- My Maple program is still working on a(9), assuming a required >corner starting point. Ok, Maple completed a(9) and is now working on a(11). Good news -- a(9) = 66, as expected. Note, for convenience, we are now always making the assumption of a > corner starting point. For even n, it apparently doesn't matter. For > odd n > 1, allowing a variable starting point appears to make a(n) > increase by 1. quasi this makes sense to me, i am starting to work on a proof of the bound for the even case, as i think it will be the easier one. i have some ideas, but nothing concrete enough to comment on yet. === Subject: Re: Newton's Method: Cute trick. posting-account=DSjMzAoAAAAISwlGnTY2c6bnmUVmdFNf AppleWebKit/528.16 (KHTML, like Gecko) Version/4.0 Safari/528.16,gzip(gfe),gzip(gfe) I have never seen this trick explicitly mentioned anywhere. Of course if f (p) < 0, f (q) > 0, f continuous on [p, q] then there is a root of f in (p, q), and if you use _any_ iterative algorithm, then clamping the next iterated result to [p, q] will get you closer to the root. Some things I noted: (1) If you find x0 such that f (x0) is small enough, but x0 is not in [p, q] then clamping x0 to [p, q] might give a new value where f (x) is _not_ small enough anymore, so care must be taken. (2) Clamping can actually only happen on the first Newton-Raphson step if your convergence criterion is actually met, because you proved that each iteration gets closer to the root of F (x). If you are not 100% sure that the convergence criterion is met, clamping might not help. (3) This doesn't solve the problem of finding x0 close enough to the root. Of course it is no big deal if a and b are known. === Subject: Re: Help!How to solve the equation === Subject: Re: imaginary exponents -- I *want* to believe... Something's bugged me about imaginary exponents ever since learning in >high school that e^ix = cos(x) + isin(x). Now that I am passing on my >alleged wisdom to high school students, I'd like to make sure it is at >least mostly right. After learning that x^n for natural numbers n means x times itself n >times, it seemed like the natural way to extend this to negative >numbers and fractions would be to do it in a way that gets you back >to the natural numbers, where you already know the answer. So if x^(a >+b) = x^a*x^b for natural numbers, then x^(-1) would be defined such >that x^(-1)*x^3 = x^2, which leads you to x^(-1) = 1/x. And x^(1/2) >must be such that x^(1/2)*x^(1/2) = x^(1/2+1/2) = x, so x^(1/2) = +/- >sqrt(x). But that strategy doesn't work for imaginary exponents, since the way >to get back to the integers would be through multiplication, so you >could try to start with (x^i)^i = x^(i*i) = x^(-1) -- but that doesn't >tell you what x^i is. Meanwhile, the book shows how you can start with y = cosx + isinx and >do some implicit differentiation and end up with y = e^ix, but I >couldn't make myself *believe* it. I can follow the math that ends >with y = e^ix, but it seemed just like a neat trick, and what I >couldn't convince myself is that someone else might find an equally >neat trick to prove that, say, e^ix = tan(x) + i*2^x or something >equally exotic. It seems much less convincing than the argument that >x^(-1) = 1/x, which seems to me to forcefully drive home the point >that you could never come up with a trick to show that x^(-1) equals >anything other than 1/x. If the book proves that exp(ix) = cos(x) + i sin(x) without > first giving a _definition_ of exp(ix) then the book is doing > very naughty things - before you can prove anything about > whatever you need to define whatever. There would be nothing _wrong_ in any absolute mathematical > sense in defining e^(ix) = 27 for all x. But if we did that then > the function e^(ix) would not have any of the properties that > we _want_ it to have. Is there a cure for my skepticism? :) Find that a huge number of things work out very nicely if > we define e^(ix) = cos(x) + i sin(x). One that seems very compelling to me: We know > that for real x, (*) e^x = 1 + x + x^2/2! + x^3/x! + ... . Now suppose we decide we _want_ (*) to hold > for complex values of x as well - that would be nice, > wouldn't it? If you substitute ix for x in (*) (with > x real) you get exactly cos(x) + i sin(x). There is an IMO Complex Made Simple-r way to see this. The gist of the argument is. First prove that any function F with the property that F(x+y) = F(x).F(y) is an exponential function e^(c.x) Then prove that cos(x) + i.sin(x) has this property and therefore it must be a function of the form e^(c.x) . At last find out that c = i The argument below will be without any (set theoretic) embellishments, but I believe it's possible to fill in remaining details for oneself. I'd like to hear, though, if there are serious omissions. Suppose that all we know there is a neat (everywhere differentiable) function F such that for all x and y : F(x+y) = F(x).F(y) Theorem: F(x-y) = F(x)/F(y) Proof: F(x) = F(x-y+y) = F(x-y).F(y) Theorem: F(0) = 1 Proof: F(x-x) = F(x)/F(x) Theorem: F'(x) = c.F(x) where c is a constant Proof: F(x+h) - F(x) F(h) - F(0) F'(x) = ------------- = F(x) ----------- = F(x).F'(0) ; h-> 0 h h Theorem: F(x) = e^(c.x) Proof: F'(x) / F(x) = c -> integral dF/F = integral c.dx -> ln(F) = d + c.x -> F(x) = C.e^(c.x) But F(0) = 1 -> C = 1 . Now consider: G(x) = cos(x) + i.sin(x) . Theorem: G(x+y) = G(x).G(y) ; thus G is a function like F above. Proof: [cos(x) + i.sin(x)].[cos(y) + i.sin(y)] = [cos(x).cos(y) - sin(x).sin(y)] + i.[sin(x).cos(y) + cos(x).sin(y)] But: cos(x).cos(y) - sin(x).sin(y) = cos(x+y) : according to well-known sin(x).cos(y) + cos(x).sin(y) = sin(x+y) trigonometric formulas Hence: [cos(x) + i.sin(x)].[cos(y) + i.sin(y)] = cos(x+y) + i.sin(x+y) Therefore: cos(x) + i.sin(x) = e^(c.x) It remains to show that c = i . Differentiate with respect to x : - sin(x) + i.cos(x) = c.e^(c.x) -> for x = 0 : i = c . Quod Erat Demonstrandum. Han de Bruijn === Subject: Re: imaginary exponents -- I *want* to believe... > >Something's bugged me about imaginary exponents ever since learning in >high school that e^ix = cos(x) + isin(x). Now that I am passing on my >alleged wisdom to high school students, I'd like to make sure it is at >least mostly right. After learning that x^n for natural numbers n means x times itself n >times, it seemed like the natural way to extend this to negative >numbers and fractions would be to do it in a way that gets you back >to the natural numbers, where you already know the answer. So if x^(a >+b) = x^a*x^b for natural numbers, then x^(-1) would be defined such >that x^(-1)*x^3 = x^2, which leads you to x^(-1) = 1/x. And x^(1/2) >must be such that x^(1/2)*x^(1/2) = x^(1/2+1/2) = x, so x^(1/2) = +/- >sqrt(x). But that strategy doesn't work for imaginary exponents, since the way >to get back to the integers would be through multiplication, so you >could try to start with (x^i)^i = x^(i*i) = x^(-1) -- but that doesn't >tell you what x^i is. Meanwhile, the book shows how you can start with y = cosx + isinx and >do some implicit differentiation and end up with y = e^ix, but I >couldn't make myself *believe* it. I can follow the math that ends >with y = e^ix, but it seemed just like a neat trick, and what I >couldn't convince myself is that someone else might find an equally >neat trick to prove that, say, e^ix = tan(x) + i*2^x or something >equally exotic. It seems much less convincing than the argument that >x^(-1) = 1/x, which seems to me to forcefully drive home the point >that you could never come up with a trick to show that x^(-1) equals >anything other than 1/x. If the book proves that exp(ix) = cos(x) + i sin(x) without >first giving a _definition_ of exp(ix) then the book is doing >very naughty things - before you can prove anything about >whatever you need to define whatever. There would be nothing _wrong_ in any absolute mathematical >sense in defining e^(ix) = 27 for all x. But if we did that then >the function e^(ix) would not have any of the properties that >we _want_ it to have. Is there a cure for my skepticism? :) Find that a huge number of things work out very nicely if >we define e^(ix) = cos(x) + i sin(x). One that seems very compelling to me: We know >that for real x, (*) e^x = 1 + x + x^2/2! + x^3/x! + ... . Now suppose we decide we _want_ (*) to hold >for complex values of x as well - that would be nice, >wouldn't it? If you substitute ix for x in (*) (with >x real) you get exactly cos(x) + i sin(x). There is an IMO Complex Made Simple-r way to see this. The gist of the argument is. First prove that any function F with the >property that F(x+y) = F(x).F(y) is an exponential function e^(c.x) This is assuming we've already _defined_ e^z for complex z, > of course. What _definition_ do you have in mind? Uhm, well .. yours, of course: > (The definition by the power series seems like the simplest > to me.) Then prove that cos(x) + i.sin(x) has this property and therefore it >must be a function of the form e^(c.x) . At last find out that c = i The argument below will be without any (set theoretic) embellishments, >but I believe it's possible to fill in remaining details for oneself. >I'd like to hear, though, if there are serious omissions. Suppose that all we know there is a neat (everywhere differentiable) >function F such that for all x and y : F(x+y) = F(x).F(y) Theorem: F(x-y) = F(x)/F(y) Proof: F(x) = F(x-y+y) = F(x-y).F(y) Of course we're assuming F never vanishes here. Uhm .. can we prove then F(0) = 1 _without_ F(x-y) = F(x)/F(y) ? Let's see. If F(x) is zero somewhere, then it is zero everywhere, because: F(x+y) = F(x).F(y) = 0.F(y) = 0 . Therefore assume (yes!) that F(x) is _zero nowhere_. Then we see that: F(x) = F(x-x+x) = F(x-x).F(x) = F(0).F(x) -> F(0) = 1 . Right ? >Theorem: F(0) = 1 Proof: F(x-x) = F(x)/F(x) Theorem: F'(x) = c.F(x) where c is a constant Proof: F(x+h) - F(x) F(h) - F(0) > F'(x) = ------------- = F(x) ----------- = F(x).F'(0) ; h-> 0 > h h Theorem: F(x) = e^(c.x) Proof: F'(x) / F(x) = c -> integral dF/F = integral c.dx -> ln(F) = d + c.x -> F(x) = C.e^(c.x) Yiles. Since we're talking about complex numbers > bringing in the logarithm opens many cans of worms. > A better proof is this: Show that F(x)e^(-cx) is > constant by calculating the derivative. F'(x).e^(-cx) + F(x).(-c).e^(-cx) = F(x).(c - c).e^(-cx) = 0 . That feels much better, indeed. > But F(0) = 1 -> C = 1 . Now consider: G(x) = cos(x) + i.sin(x) . Theorem: G(x+y) = G(x).G(y) ; thus G is a function like F above. _If_ we're going to be bringing differential equations into > the picture it seems simpler to forget about F(x+y) = etc > and G(x+y) = etc. If G(x) = cos(x) + i sin(x) then > G' = iG and we're done by the previous theorem. Uhm .. you're quite right. >Proof: [cos(x) + i.sin(x)].[cos(y) + i.sin(y)] = > [cos(x).cos(y) - sin(x).sin(y)] > + i.[sin(x).cos(y) + cos(x).sin(y)] But: cos(x).cos(y) - sin(x).sin(y) = cos(x+y) : according to well-known > sin(x).cos(y) + cos(x).sin(y) = sin(x+y) trigonometric formulas Hence: [cos(x) + i.sin(x)].[cos(y) + i.sin(y)] = cos(x+y) + i.sin(x+y) Therefore: cos(x) + i.sin(x) = e^(c.x) It remains to show that c = i . Differentiate with respect to x : - sin(x) + i.cos(x) = c.e^(c.x) -> for x = 0 : i = c . Quod Erat Demonstrandum. You can _prefer_ this argument if you want (in fact > a few weeks ago I _defined_ exp, cos and sin in > advanced calculus as the solution to certain differential > equations). But saying it's simpler than just that one > substitution into that power series seems a little much. shortcomings. I've just asked for it, though: > I'd like to hear, though, if there are serious omissions. Han de Bruijn === Subject: Re: imaginary exponents -- I *want* to believe... sha1:XASw+xIDFmI26+QXiMxMsRV/o4s= >The gist of the argument is. First prove that any function F with the >property that F(x+y) = F(x).F(y) is an exponential function e^(c.x) This is assuming we've already _defined_ e^z for complex z, > of course. What _definition_ do you have in mind? Perhaps he means that any function with that property is, when restricted to R, an exponential function e^cx:R -> R? -- Jesse F. Hughes The future is a fascinating thing, and so is history. And you people are a fascinating part of history, for those in the future. -- James S. Harris is fascinating, too === Subject: Re: imaginary exponents -- I *want* to believe... The gist of the argument is. First prove that any function F with the >property that F(x+y) = F(x).F(y) is an exponential function e^(c.x) This is assuming we've already _defined_ e^z for complex z, >of course. What _definition_ do you have in mind? Perhaps he means that any function with that property is, when > restricted to R, an exponential function e^cx:R -> R? Oh no, Ullrich's comment is quite right. And R -> C is needed at least. Han de Bruijn === Subject: Re: imaginary exponents -- I *want* to believe... posting-account=brOM-AoAAAChaAJEiH5z610-YOTfECd9 Gecko/20070530 Fedora/1.5.0.12-1.fc5 Firefox/1.5.0.12,gzip(gfe),gzip(gfe) > Something's bugged me about imaginary exponents ever since learning in > high school that e^ix = cos(x) + isin(x). Now that I am passing on my > alleged wisdom to high school students, I'd like to make sure it is at > least mostly right. After learning that x^n for natural numbers n means x times itself n > times, it seemed like the natural way to extend this to negative > numbers and fractions would be to do it in a way that gets you back > to the natural numbers, where you already know the answer. So if x^(a > +b) = x^a*x^b for natural numbers, then x^(-1) would be defined such > that x^(-1)*x^3 = x^2, which leads you to x^(-1) = 1/x. And x^(1/2) > must be such that x^(1/2)*x^(1/2) = x^(1/2+1/2) = x, so x^(1/2) = +/- > sqrt(x). But that strategy doesn't work for imaginary exponents, since the way > to get back to the integers would be through multiplication, so you > could try to start with (x^i)^i = x^(i*i) = x^(-1) -- but that doesn't > tell you what x^i is. Meanwhile, the book shows how you can start with y = cosx + isinx and > do some implicit differentiation and end up with y = e^ix, but I > couldn't make myself *believe* it. I can follow the math that ends > with y = e^ix, but it seemed just like a neat trick, and what I > couldn't convince myself is that someone else might find an equally > neat trick to prove that, say, e^ix = tan(x) + i*2^x or something > equally exotic. It seems much less convincing than the argument that > x^(-1) = 1/x, which seems to me to forcefully drive home the point > that you could never come up with a trick to show that x^(-1) equals > anything other than 1/x. Is there a cure for my skepticism? :) To that end, here are some > things I would at last like to sort out: 1) Is it true that for the definition of exponentiation a^b, > generalized across all complex numbers for the base and exponent, the > function a^b: > - has a single value if a is 0 or 1 > - has a single value if b is a real integer with no imaginary part > - has multiple values otherwise? 2) If x^(a+bi) has multiple values, and (x^(a+bi))^(c+di) in turn has > multiple values, and meanwhile the expression x^((ac-db) + (bc+ad)i) > is another expression with multiple values (in all cases I mean, or > possibly a single value if the numbers are right), is there always at > least *one* value that appears in both lists? In other words, do even > complex number exponents obey the law (a^b)^c = a^(b*c), if the > expressions on the left and the right have multiple values and you're > allowed to pick the value you want from both lists? 3) If the answer to question #2 is yes, is there an intuitive way to > see that this is true? To me, that would be a beautiful fact, to show > that the exotic function x^i has the property that (x^i)^i = 1/x, for > at least one value chosen from the multiple branches of x^i. As a followup, taking into account the multiple branches, which become apparent when we take the logarithm ln(z) = i ln(x^i) + 2n pi i = (i*i)ln(x) + 2n pi i exponentiating z = e^( -1 ln(x) + 2n pi i ) = x^(-1) You only see the multiple branches when you take logarithm. === Subject: Re: imaginary exponents -- I *want* to believe... posting-account=brOM-AoAAAChaAJEiH5z610-YOTfECd9 Gecko/20070530 Fedora/1.5.0.12-1.fc5 Firefox/1.5.0.12,gzip(gfe),gzip(gfe) > Something's bugged me about imaginary exponents ever since learning in > high school that e^ix = cos(x) + isin(x). Now that I am passing on my > alleged wisdom to high school students, I'd like to make sure it is at > least mostly right. After learning that x^n for natural numbers n means x times itself n > times, it seemed like the natural way to extend this to negative > numbers and fractions would be to do it in a way that gets you back > to the natural numbers, where you already know the answer. So if x^(a > +b) = x^a*x^b for natural numbers, then x^(-1) would be defined such > that x^(-1)*x^3 = x^2, which leads you to x^(-1) = 1/x. And x^(1/2) > must be such that x^(1/2)*x^(1/2) = x^(1/2+1/2) = x, so x^(1/2) = +/- > sqrt(x). But that strategy doesn't work for imaginary exponents, since the way > to get back to the integers would be through multiplication, so you > could try to start with (x^i)^i = x^(i*i) = x^(-1) -- but that doesn't > tell you what x^i is. Meanwhile, the book shows how you can start with y = cosx + isinx and > do some implicit differentiation and end up with y = e^ix, but I > couldn't make myself *believe* it. I can follow the math that ends > with y = e^ix, but it seemed just like a neat trick, and what I > couldn't convince myself is that someone else might find an equally > neat trick to prove that, say, e^ix = tan(x) + i*2^x or something > equally exotic. Hi Bennett, Start with z = r ( cos(a) + i sin(a) ) which is easy to show, as it is simply the trigonometric representation of a vector of length r at an angle a from the positive real axis. Now write the Taylor's series expansion cos(a) = Sum_n=0^infty (-1)^n (a)^(2n)/ (2n)! noting that (-1)^n = ( i*i)^n = (i)^2n cos(a) = Sum_n=0^infty (ia)^2n/(2n)! and i sin(a) = Sum_n=0^infty i(-1)^n(a)^(2n+1)/ (2n+1)! again noting that (-1)^n = (i*i)^n = (i)^2n and that i(-1)^n is thus = (i)^(2n+1) i sin(a) = Sum_n=0^infty (ia)^(2n+1)/(2n+1)! adding the series represenations forming the series representation of cos(a) + i sin(a) = Sum_n=0^infty (ia)^n/n! which is the same as the taylor's series expansion of e^(ia). Having the concept of exponential, it follows that you have the notion of logarithm. z = e^(ia) ln(z) = ia also you have the added item that 1 = e^( i 2 n pi ) z = e^( i a + i 2 n pi) ln(z) = ia + 2n pi i > allowed to pick the value you want from both lists? 3) If the answer to question #2 is yes, is there an intuitive way to > see that this is true? To me, that would be a beautiful fact, to show > that the exotic function x^i has the property that (x^i)^i = 1/x, for > at least one value chosen from the multiple branches of x^i. z = (x^i)^i ln(z) = i ln(x^i) = (i*i)ln(x) = -1 ln(x) z = e^(-1 ln(x)) = e^(ln( x^(-1) = x^(-1) You can see where the multiplication rule of raising a power to a power comes from. -John === Subject: Re: imaginary exponents -- I *want* to believe--- > Something's bugged me about imaginary exponents ever since learning in > high school that e^ix = cos(x) + isin(x). Now that I am passing on my > alleged wisdom to high school students, I'd like to make sure it is at > least mostly right. After learning that x^n for natural numbers n means x times itself n > times, it seemed like the natural way to extend this to negative > numbers and fractions would be to do it in a way that gets you back > to the natural numbers, where you already know the answer. So if x^(a > +b) = x^a*x^b for natural numbers, then x^(-1) would be defined such > that x^(-1)*x^3 = x^2, which leads you to x^(-1) = 1/x. And x^(1/2) > must be such that x^(1/2)*x^(1/2) = x^(1/2+1/2) = x, so x^(1/2) = +/- > sqrt(x). But that strategy doesn't work for imaginary exponents, since the way > to get back to the integers would be through multiplication, so you > could try to start with (x^i)^i = x^(i*i) = x^(-1) -- but that doesn't > tell you what x^i is. Meanwhile, the book shows how you can start with y = cosx + isinx and > do some implicit differentiation and end up with y = e^ix, but I > couldn't make myself *believe* it. I can follow the math that ends > with y = e^ix, but it seemed just like a neat trick, and what I > couldn't convince myself is that someone else might find an equally > neat trick to prove that, say, e^ix = tan(x) + i*2^x or something > equally exotic. It seems much less convincing than the argument that > x^(-1) = 1/x, which seems to me to forcefully drive home the point > that you could never come up with a trick to show that x^(-1) equals > anything other than 1/x. Is there a cure for my skepticism? :) To that end, here are some > things I would at last like to sort out: 1) Is it true that for the definition of exponentiation a^b, > generalized across all complex numbers for the base and exponent, the > function a^b: > - has a single value if a is 0 or 1 > - has a single value if b is a real integer with no imaginary part > - has multiple values otherwise? 2) If x^(a+bi) has multiple values, and (x^(a+bi))^(c+di) in turn has > multiple values, and meanwhile the expression x^((ac-db) + (bc+ad)i) > is another expression with multiple values (in all cases I mean, or > possibly a single value if the numbers are right), is there always at > least *one* value that appears in both lists? In other words, do even > complex number exponents obey the law (a^b)^c = a^(b*c), if the > expressions on the left and the right have multiple values and you're > allowed to pick the value you want from both lists? 3) If the answer to question #2 is yes, is there an intuitive way to > see that this is true? To me, that would be a beautiful fact, to show > that the exotic function x^i has the property that (x^i)^i = 1/x, for > at least one value chosen from the multiple branches of x^i. A simplistic and pragmatic view: (1A) The positive integers form a semigroup without identity element under addition. (1B) Let A be an arbitrary real number different from 1, 0 and -1. Then the powers of A, as defined by repeated multiplication of A with itself, form a semigroup without identity element under multiplication. The semigroups of (1A) and (1B) are isomorphic. This theorem is the mathematical content of one of the easiest parts of secondary-school arithmetic and algebra. (2A) The nonnegative integers form a semigroup with identity element 0 under addition. It is the smallest semigroup with identity that contains the semigroup of (1A). (2B) Let A be a real number as in (1B). Then the powers of A, as defined in (2A), \ together with the number 1 form a semigroup with identity element 1 under multiplication. It is the smallest semigroup with identity that contains the semigroup of (1B). The semigroups of (2A) and (2B) are isomorphic. The important terminology step (linguistic step) one always takes here is to \ define empty sum and empty product. One degree of separation from the tangible concepts of sums consisting of terms and products consisting of factors. The next step consists of extending the semigroups of (2A) and (2B) to groups and to define A to a negative integer power -P as 1/A to the power P. A second degree of separation from good old primary-school multiplication: now we already have negative exponents. The following degrees of separation consist of fractional exponents, irrational exponents and even imaginary exponents. The mathematics is sound: root extraction gives us fractional exponents, continuity and completion yield real exponents and real-valued exponential and logarithmic \ functions. All of this with the isomorphisms of the corresponding addition and multiplication groups in the background. Then there is a great mystery (*): real exponential functions are representable by power series that are BTW everywhere convergent. By plugging in complex numbers in \ these power series we obtain the well-known complex exponential function and its integer, fractional and real powers. By now we are far from the simple facts of primary-school arithmetic, but the algebra behind this all continues to exist by the group isomorphisms and homomorphisms. In short: one does not need an unfounded belief of any kind, but rather the \ brutality of extending the meaning of established terminology, with the risk of literary \ meanings breaking down and turning into nonsense. I leave your specific questions (1), (2) and (3) unanswered. Two hints: (A) Invent, or rather re-invent, the mathematics of complex exponentiation with =complex= base numbers. (B) Take a look at higher programming languages and numerical support packages to see if and how exponentiation (real as well as complex) is defined. Without its mathematics you will get lost, so to say. Some milestones in the history of this subject: Stifel (1487 - 1567) invented negative integer exponents. See http://en.wikipedia.org/wiki/Michael_stifel Cauchy (1789 - 1857) was one of the founding fathers of complex function theory. See http://en.wikipedia.org/wiki/Cauchy Good luck: Johan E. Mebius ----------------------------------------------------------------------------\ ---------- (*) Mystery: WHY this is true I do not know, and I cannot speak meaningfully \ about it. === Subject: Re: pV^gamma=const > pV^gamma=const is a familiar equation from basic thermo. For doing > calculations, it would be very useful to know what the constant is, but > that information is in none of the introductory books I've ever seen. Does > someone know if anybody has ever tried to find that constant, and if so, > where I might read their work? TIA. Google: adiabatic expansion gamma For example: http://www.chem.ufl.edu/~itl/4411L_f00/gamma/gamma.html === Subject: Re: pV^gamma=const posting-account=qKxGxgkAAADAPfYVCc-ZQkIzl0senr2M .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; .NET CLR 1.1.4322; Zune 2.5),gzip(gfe),gzip(gfe) > pV^gamma=const is a familiar equation from basic thermo. æFor doing > calculations, it would be very useful to know what the constant is, but > that information is in none of the introductory books I've ever seen. Does > someone know if anybody has ever tried to find that constant, and if so, > where I might read their work? æTIA. Well, since the units of volume are distance^3, I would expect the units of V^gamma to be distance^(3*gamma). Seriously, the equation p*V^gamma = constant is valid ONLY for adiabatic processes of an ideal gas. The inclusion of the ratio of heat capacitues allows it to hold for poly-atomic gases. Also, the applicability is limited because gamma is NOT a constant. The heat capacities vary with temperature. The heat capacities have contributions from translation (the 'ideal gas' motion), from rotation (due to the spinning of non-spherical molecules and from vibration (internal motions within the molecule), and at higher temperatures from electronic excitations. Each of these motions makes an additive contribution to the heat capacity. The vibrational motion is a particularly complex function of temperature. In adiabatic processes in which temperature does not change very much (so the heat capacities are relatively constant over the range of temperatures of interest) gamma can be treated as a constant, so p*V^gamma = constant is approximately true. Note that the expression also depends on the Tom Davidson Richmond, VA === === Subject: : function is equal almost everywhere to another function = same Lebesgue integral say A is measurable subset of reals Suppose f: A->R is equal to g: A->R almost everwhere. Then f,g both fail to be Lebesgue integrable at same time, or they are both integrable and have the same integral. Correct? === Subject: Re: : function is equal almost everywhere to another function = same Lebesgue integral > say A is measurable subset of reals Suppose f: A->R is equal to g: A->R almost everwhere. Then f,g both fail to be Lebesgue integrable at same time, or they are > both integrable and have the same integral. Correct? Correct. Jose Carlos Santos === Subject: Re: : function is equal almost everywhere to another function = same Lebesgue integral > say A is measurable subset of reals Suppose f: A->R is equal to g: A->R almost everwhere. Then f,g both fail to be Lebesgue integrable at same time, or they are > both integrable and have the same integral. Correct? > What does your textbook say? === Subject: Re: : function is equal almost everywhere to another function = same Lebesgue integral <260220090957331351%anniel@nym.alias.net.invalid> posting-account=2kYL5QoAAAAmH1e_g3WSyZ5Z9zXmRQX3 .NET CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) say A is measurable subset of reals Suppose f: A->R is equal to g: A->R almost everwhere. Then f,g both fail to be Lebesgue integrable at same time, or they are > both integrable and have the same integral. Correct? What does your textbook say? I don't have one. === Subject: alternative solution to a 4 variable system Last night my younger sister came to me with a math problem: u + v = 2 ux + vy = -1 ux^2 + vy^2 = 1 ux^3 + vy^3 = 1 I'm not sure about the constants 2, -1, 1, 1 but I know that their sum was 3. I ended up solving the problem by substituting u and v in the last two equations and it came to a bit much of calculus but the solution was found and the answer was matched against the solution given by the book. Before attempting this ``bruteforce'' attack I tried something that spawned some nice result that I thought I might exploit but failed. I'm curious if I could've gotten anything from that. Here's what I attempted: u(x^3 + x^2 + x + 1) + v(y^3 + y^2 + y + 1) = 3 Now this looks like x^4 - 1 and y^4 - 1 was used to generate the system. But substituting that resulted in: u*(x^4 - 1)/(x - 1) + v(y^4 -1)/(y - 1) = 3 which after some poking around came to a show stopper. Another thing would be to group these as: v(y^4 - x^4 + y^3 - x^4 + y^2 - x^2 + y - x) + 2(x^3 + x^2 + x + 1) = 3 which again was not usefull for me. The equation was found after I substitute v = 2 - u. Can you see any way of doing this better? I'll try and confirm the constants later tonight if needed. -- Everything is simple, we're stupid. gopher://sdf.lonestar.org/1/users/bulibuta === Subject: Re: alternative solution to a 4 variable system posting-account=SvltewoAAAAi7TTYrD3mAaLUHzDiF2d1 Gecko/2009011913 Firefox/3.0.6,gzip(gfe),gzip(gfe) On Feb 26, 10:15æpm, Paul Irofti ux + vy = -1 > ux^2 + vy^2 = 1 > ux^3 + vy^3 = 1 I'm not sure about the constants 2, -1, 1, 1 but I know that their sum > was 3. I ended up solving the problem by substituting u and v in the last two > equations and it came to a bit much of calculus but the solution was > found and the answer was matched against the solution given by the book. Before attempting this ``bruteforce'' attack I tried something that > spawned some nice result that I thought I might exploit but failed. I'm > curious if I could've gotten anything from that. Here's what I attempted: u(x^3 + x^2 + x + 1) + v(y^3 + y^2 + y + 1) = 3 Now this looks like x^4 - 1 and y^4 - 1 was used to generate the system. > But substituting that resulted in: u*(x^4 - 1)/(x - 1) + v(y^4 -1)/(y - 1) = 3 which after some poking around came to a show stopper. Another thing would be to group these as: v(y^4 - x^4 + y^3 - x^4 + y^2 - x^2 + y - x) + 2(x^3 + x^2 + x + 1) = 3 which again was not usefull for me. The equation was found after I > substitute v = 2 - u. Can you see any way of doing this better? I'll try and confirm the > constants later tonight if needed. -- > Everything is simple, we're stupid. > gopher://sdf.lonestar.org/1/users/bulibuta [1] u + v = 2 [2] ux + vy = -1 [3] ux^2 + vy^2 = 1 [4] ux^3 + vy^3 = 1 [1][3] - [2][2] => uv(x-y)^2 = (2)(1)-(-1)^2 = 1 [5] [2][4] - [3][3] => uvxy(x-y)^2 = (-1)(1)-(1)^2 = -2 [6] [1][4] - [2][3] => uv (x^3+y^3-(x^2)y-x(y^2) ) = uv (x+y)(x^2 - xy + y^2 - xy) = uv (x+y)(x-y)^2 = (2)(1)-(-1)(1) = 3 [7] [7]/[5] => x+y = 3 [6]/[5] => xy = -2 You can solve x, y using these 2 relations and substitute back into [1], [2] to get u, v. The trick is to pair up the equations so that if u, v = 0 or x = y, the LHS of the [5-7] contains a common factor uv(x-y)^2 which allow you to extract dependence of x, y out. === Subject: Re: alternative solution to a 4 variable system posting-account=SvltewoAAAAi7TTYrD3mAaLUHzDiF2d1 Gecko/2009011913 Firefox/3.0.6,gzip(gfe),gzip(gfe) > On Feb 26, 10:15æpm, Paul Irofti Last night my younger sister came to me with a math problem: u + v = 2 > ux + vy = -1 > ux^2 + vy^2 = 1 > ux^3 + vy^3 = 1 I'm not sure about the constants 2, -1, 1, 1 but I know that their sum > was 3. I ended up solving the problem by substituting u and v in the last two > equations and it came to a bit much of calculus but the solution was > found and the answer was matched against the solution given by the book. Before attempting this ``bruteforce'' attack I tried something that > spawned some nice result that I thought I might exploit but failed. I'm > curious if I could've gotten anything from that. Here's what I attempted: u(x^3 + x^2 + x + 1) + v(y^3 + y^2 + y + 1) = 3 Now this looks like x^4 - 1 and y^4 - 1 was used to generate the system. > But substituting that resulted in: u*(x^4 - 1)/(x - 1) + v(y^4 -1)/(y - 1) = 3 which after some poking around came to a show stopper. Another thing would be to group these as: v(y^4 - x^4 + y^3 - x^4 + y^2 - x^2 + y - x) + 2(x^3 + x^2 + x + 1) = 3 which again was not usefull for me. The equation was found after I > substitute v = 2 - u. Can you see any way of doing this better? I'll try and confirm the > constants later tonight if needed. -- > Everything is simple, we're stupid. > gopher://sdf.lonestar.org/1/users/bulibuta [1] u + v = 2 > [2] ux + vy = -1 > [3] ux^2 + vy^2 = 1 > [4] ux^3 + vy^3 = 1 [1][3] - [2][2] => uv(x-y)^2 æ = (2)(1)-(-1)^2 = 1 æ æ[5] > [2][4] - [3][3] => uvxy(x-y)^2 = (-1)(1)-(1)^2 = -2 æ æ[6] > [1][4] - [2][3] = æ æuv (x^3+y^3-(x^2)y-x(y^2) ) > æ= uv (x+y)(x^2 - xy + y^2 - xy) > æ= uv (x+y)(x-y)^2 = (2)(1)-(-1)(1) = 3 æ æ æ æ æ æ æ [7] [7]/[5] => x+y = 3 > [6]/[5] => xy æ= -2 You can solve x, y using these 2 relations and substitute > back into [1], [2] to get u, v. The trick is to pair up > the equations so that if u, v = 0 or x = y, the LHS of the > [5-7] contains a common factor uv(x-y)^2 which allow you to > extract dependence of x, y out. Figure out a more generic way to deal with this which can be generalized to higher-dimensions. Look at the right hand side of equations 1-4 in pairs, we get 3 row vectors (2,-1), (-1,1), (1,1). Any 3 vectors in a 2-dim space will be linearly dependent. In particular, 2 (2, -1) + 3 (-1,1) - 1(1,1) = (0,0). 2[Eq1]+3[Eq2]-[Eq3] => u(2+3x-x^2) + v(2+3y-y^2) = 0. 2[Eq2]+3[Eq3]-[Eq4] => ux(2+3x-x^2)+ vy(2+3y-y^2) = 0. or in matrix form [ 1 1 ][ 2+3x-x^2 0 ][ u ] = [ 0 ] [ x y ][ 0 2+3y-y^2][ v ] = [ 0 ] In order to have non-zero solution for (u,v), the matrix in LHS is singular and its determinant vanish. (y-x)(2+3x-x^2)(2+3y-y^2) = 0. So either y = x, or x,y root of the polynomial t^2-3t-2. === Subject: Re: alternative solution to a 4 variable system [-snip-] [1] u + v = 2 > [2] ux + vy = -1 > [3] ux^2 + vy^2 = 1 > [4] ux^3 + vy^3 = 1 [1][3] - [2][2] => uv(x-y)^2 = (2)(1)-(-1)^2 = 1 [5] > [2][4] - [3][3] => uvxy(x-y)^2 = (-1)(1)-(1)^2 = -2 [6] > [1][4] - [2][3] = uv (x^3+y^3-(x^2)y-x(y^2) ) > = uv (x+y)(x^2 - xy + y^2 - xy) > = uv (x+y)(x-y)^2 = (2)(1)-(-1)(1) = 3 [7] Nice find, I did not think about multiplying one equation with itself. [7]/[5] => x+y = 3 > [6]/[5] => xy = -2 This is a very nice solution, just what I was looking for! Can't wait to show this to my sister whne she gets back from hiking. So the way I saw it was a dead end afterall. Meh. -- Everything is simple, we're stupid. gopher://sdf.lonestar.org/1/users/bulibuta === === Subject: Re: Superfunction provides iteration (was Re: Continuous iteration of another quadratic map) Am 25.02.2009 11:26 schrieb mike3: > For example, let's try to find g(2) using this method. Using Newton's > method > we find that G(z) equals 2 at z ~ 0.5064436626559859579387013737 (i.e. > that's the value of G^-1(2).). Computing G(z + 1) yields > ~10.47213595499957939281834734. Sure enough, this is g(2) as a direct > computation of g(2) using g(x) = x^2 + 2phix would show. Nifty, eh? We can compute G(z + 1/2) to get g^(1/2)(2) ~ > 4.280783105324406938219866621. This all just using the series you > posted > plus Newton's method. > :-) Yes, surely. I should have thought of it myself... Looks really nice! Gottfried === Subject: Re: Superfunction provides iteration (was Re: Continuous iteration \ of another quadratic map) <70nmoeFghetmU1@mid.dfncis.de> posting-account=fwSgtAkAAACFnX70ssKwbvm9_oCZVHrx Gecko/2009011913 Firefox/3.0.6,gzip(gfe),gzip(gfe) > Am 25.02.2009 11:26 schrieb mike3: For example, let's try to find g(2) using this method. Using Newton's > method > we find that G(z) equals 2 at z ~ 0.5064436626559859579387013737 (i.e. > that's the value of G^-1(2).). Computing G(z + 1) yields > ~10.47213595499957939281834734. Sure enough, this is g(2) as a direct > computation of g(2) using g(x) = x^2 + 2phix would show. Nifty, eh? We can compute G(z + 1/2) to get g^(1/2)(2) ~ > 4.280783105324406938219866621. This all just using the series you > posted > plus Newton's method. :-) > Yes, surely. I should have thought of it myself... > Looks really nice! > Yes. The only problem is it seems to be difficult to get good initial guesses for the Newton method on the general complex plane that confine to a specific branch one wants. === Subject: Re: Superfunction provides iteration (was Re: Continuous iteration \ of another quadratic map) > Yes. The only problem is it seems to be difficult to get good initial > guesses for the Newton method on the general complex plane that > confine to a specific branch one wants. Have you seen those beautiful shapes on various Newton-method fractals? Many \ of these contain solid areas, which are called 'basins of attraction'. One cannot expect to 'know' in advance which value the method would converge to, because the fractal basin convergence occurs in in the complex plane, is determined 'exactly' by the initial value of z, which either belongs or does not belong \ in a certain basin on the corresponding fractal. Chossing a initial value z_1 which is far away from a previous initial value z_0, may force convergence to move to the fixed points of another basin different from the basin the Newton iteration for z_0 converges to. You probably know of the immense complexity and beauty of the Newton's method fractals. For example, the Newton fractal for the polynomial f(z) = z^3 + 1, has, if memory serves, three major basins of attraction. You can pick up the image from 'The beauty of fractals by Peitgen and Richter. These basins have boundaries between them, where chaotic convergence behavior occurs. The initial value z_0, along with the map: f(z) = z^3 + 1, determines the exact dynamics of convergence. Do you think it would be a trivial task to determine quickly the convergence value, if every fractal and every basin of every fractal has different complex dynamics? I don't know about you, but I don't think so. :-) -- Ioannis === Subject: Integers n such that every group of order n is... I stumbled upon an old sci.math posting, answering the question what are the integers n such that every group of order n is abelian?. http://www.math.niu.edu/~rusin/known-math/97/clt.numbers In particular, at the end of the post, Robin Chapman gives an argument that \ supposedly can be easily extended to find the integers n such that every group of order n is nilpotent. Now, the characterization for integers n such that every group of order n is \ cyclic, or abelian, was given by Gallian in When is Zn the only group of order n? Gallian, J. and Moulton, D. http://tinyurl.com/cuw23d I was wondering if one can come up with such characterizations for 1) nilpotent and 2) solvable groups. So my question is two-fold: a) What integers n are such that every group of order n is nilpotent (and is \ the proof suggested by Robin Chapman a viable one)? b) What integers n are such that every group of order n is solvable (if such \ a characterization exists)? Steve === Subject: Re: Integers n such that every group of order n is... > I stumbled upon an old sci.math posting, answering > the question what are the integers n such that every > group of order n is abelian?. > http://www.math.niu.edu/~rusin/known-math/97/clt.numbers In particular, at the end of the post, Robin Chapman > gives an argument that supposedly can be easily > extended to find the integers n such that every group > of order n is nilpotent. Now, the characterization for integers n such that > every group of order n is cyclic, or abelian, was > given by Gallian in > When is Zn the only group of order n? > Gallian, J. and Moulton, D. > http://tinyurl.com/cuw23d I was wondering if one can come up with such > characterizations for 1) nilpotent and 2) solvable > groups. So my question is two-fold: a) What integers n are such that every group of order > n is nilpotent (and is the proof suggested by Robin > Chapman a viable one)? Just check that no element of prime power order can normalize any subgroup of coprime order (so just use the order of GL(n,p)). There is a nice expression in terms of the Euler phi function. This and some nice generalizations of groups of square-free order are given in: Pazderski, Gerhard. Die Ordnungen, zu denen nur Gruppen mit gegebener Eigenschaft geh.9aren. Arch. Math. (Basel) 10 1959 331-343. http://www.ams.org/mathscinet-getitem?mr=114863 http://dx.doi.org/10.1007/BF01240807 Pakianathan, Jonathan; Shankar, Krishnan. Nilpotent numbers. Amer. Math. Monthly 107 (2000), no. 7, 631-634. http://www.ams.org/mathscinet-getitem?mr=1786236 http://dx.doi.org/10.2307/2589118 > b) What integers n are such that every group of order > n is solvable (if such a characterization exists)? One uses the list of minimal simple groups provided in the series starting with: Thompson, John G. Nonsolvable finite groups all of whose local subgroups are solvable. Bull. Amer. Math. Soc. 74 1968 383-437. http://www.ams.org/mathscinet-getitem?mr=230809 http://dx.doi.org/10.1090/S0002-9904-1968-11953-6 Since S x C_n is a non-solvable group of order |S|*n whenever S is a non-abelian simple group, and the order of a minimal finite non-abelian simple group divides the order of every finite non-abelian simple group and so every finite non-solvable group. === Subject: Re: Integers n such that every group of order n is... > I stumbled upon an old sci.math posting, answering > the question what are the integers n such that > every > group of order n is abelian?. http://www.math.niu.edu/~rusin/known-math/97/clt.numbe > rs In particular, at the end of the post, Robin > Chapman > gives an argument that supposedly can be easily > extended to find the integers n such that every > group > of order n is nilpotent. Now, the characterization for integers n such that > every group of order n is cyclic, or abelian, was > given by Gallian in > When is Zn the only group of order n? > Gallian, J. and Moulton, D. > http://tinyurl.com/cuw23d I was wondering if one can come up with such > characterizations for 1) nilpotent and 2) solvable > groups. So my question is two-fold: a) What integers n are such that every group of > order > n is nilpotent (and is the proof suggested by > Robin > Chapman a viable one)? Just check that no element of prime power order can > normalize any subgroup of coprime order (so just use > the order of GL(n,p)). There is a nice expression > in > terms of the Euler phi function. This and some nice > generalizations of groups of square-free order are > given in: Pazderski, Gerhard. Die Ordnungen, zu denen nur > Gruppen > mit gegebener Eigenschaft geh.9aren. Arch. Math. > (Basel) > 10 1959 331-343. > http://www.ams.org/mathscinet-getitem?mr=114863 > http://dx.doi.org/10.1007/BF01240807 > Pakianathan, Jonathan; Shankar, Krishnan. Nilpotent > numbers. Amer. Math. Monthly 107 (2000), no. 7, > 631-634. > http://www.ams.org/mathscinet-getitem?mr=1786236 > http://dx.doi.org/10.2307/2589118 > > b) What integers n are such that every group of > order > n is solvable (if such a characterization exists)? One uses the list of minimal simple groups provided > in > the series starting with: Thompson, John G. Nonsolvable finite groups all of > whose > local subgroups are solvable. Bull. Amer. Math. Soc. > 74 > 1968 383-437. > http://www.ams.org/mathscinet-getitem?mr=230809 > http://dx.doi.org/10.1090/S0002-9904-1968-11953-6 Since S x C_n is a non-solvable group of order |S|*n > whenever S is a non-abelian simple group, and the > order > of a minimal finite non-abelian simple group divides > the > order of every finite non-abelian simple group and > so > every finite non-solvable group. I wish you didn't make it look so easy. :-) === Subject: Re: Term for a set with no limit points? posting-account=AdyLXQoAAABgRay99CKv1O8Y_7jjivwq InfoPath.1; .NET CLR 2.0.50727; .NET CLR 3.0.04506.30; .NET CLR 3.0.04506.648),gzip(gfe),gzip(gfe) > Exercise 35 on p. 256: (a) Prove Borel's Lemma: > given any sequence whatsoever of real numbers > (a_r) there is a smooth [i.e. C^infinity] function > f:R --> R such that f^(r)(0) = a_r [r'th derivative > at 0 is equal to a_r]. > Astonishing result, which I'd never heard of. (Not too surprising, > as I've never completed a course of study in real analysis. Wish > I had now!) Torture trying to prove it. I haven't had a single > usable idea. (Also not surprising, as today has been an off-day.) > I did, however, come across a reference (in DePree & Swartz): Pugh's book proves a standard result about the rate of growth of the taylor coefficients at a point (as a function of n, where n is the degree of the term) in order for the function to be analytic or non-analytic at a point, and for the problem at hand Pugh gives a hint, which I don't know the details of (I don't have the book with me now) but I think it's something references to this method several times) that amounts to getting the coefficients to grow too quickly at each rational number. Because analytic at a x=b is equivalent to analytic in some neighborhood of x=b, getting non-analyticity at each point in a set dense in the reals will do it. About 7 years ago I posted a length essay (in 2 parts) on C-infinity functions that are nowhere analytic in sci.math. However, the posts were made at Math Forum and, for some reason (as was also the case with many posts of mine from sci.math archive. However, I notice that just a few days before I posted these essays, I announced I'd be doing so Anyway, here are the essays. For some reason certain accent symbols on non-English names/titles are all messed up, although when the posts originally appeared I'm certain they came through at Math Forum's archive pages fine. ESSAY ON NOWHERE ANALYTIC C-INFINITY FUNCTIONS (part 1) http://mathforum.org/kb/message.jspa?messageID=387148 ESSAY ON NOWHERE ANALYTIC C-INFINITY FUNCTIONS (part 2) http://mathforum.org/kb/message.jspa?messageID=387149 Below are the contents, followed by some excerpts from part 1. CONTENTS (for part 1) I. TWO WAYS THAT A C-INFINITY FUNCTION CAN FAIL TO BE ANALYTIC II. EARLY HISTORY OF NOWHERE ANALYTIC C-INFINITY FUNCTIONS III. ZAHORSKI'S 1947 CHARACTERIZATION IV. MOST C-INFINITY FUNCTIONS ARE NOWHERE ANALYTIC V. UNEXPLORED AREAS AND OTHER RESULTS VI. REFERENCES (1-35) CONTENTS (for part 2) I. REFINEMENTS ON THE TWO WAYS THAT A C-INFINITY FUNCTION CAN FAIL TO BE ANALYTIC II. GENERALIZATIONS OF ZAHORSKI'S 1947 CHARACTERIZATION IV. MORE CONCERNING MOST C-INFINITY FUNCTIONS ARE NOWHERE ANALYTIC V. REFERENCES (36-47) SOME EXCERPTS FROM part 1: Every C-infinity function has a formal Taylor series expansion about each point. If this Taylor expansion converges to the original function in neighborhood of x=b, then the function is said to be analytic at x=b. There are two ways this can fail: (C) The Taylor expansion converges in a neighborhood of x=b, but in no neighborhood of x=b does the Taylor expansion converge to the given function. (P) The Taylor expansion fails to converge in every neighborhood of x=b (i.e. the Taylor series at x=b has a zero radius of convergence). Following Zahorski [35], we classify the non-analytic points of a C-infinity function into the following two categories. A point of non-analyticity is called a (C)-point (for Cauchy) if it belongs to (C) above. A point of non-analyticity is called a (P)-point (for Pringsheim) if it belongs to (P) above. III. ZAHORSKI'S 1947 CHARACTERIZATION None of these early examples was shown to have a (C)-point everywhere or to have a (P)-point everywhere. Boas proved in 1935 (see [4] or p. 192 of [5]) that the (P)-points of a nowhere analytic C-infinity function are dense in R. Therefore, it is not possible to have an example in which every point is a (C)-point. However, the possibility that every point could be a (P)-point remained open. The first example of a C-infinity function such that every point is a (P)-point was given by Cartan in 1940 ([6], pp. 20-22). In 1947 Zahorski [35] gave the following characterization for the non-analytic points of a C-infinity function. [An announcement [34] of this result was made in 1946.] THEOREM (Zahorski): A necessary and sufficient condition for two sets of real numbers C and P to be the (C)-points and the (P)-points, respectively, of some C-infinity function is that the following four properties hold: (a) C is a first category F_sigma set. (b) P is a G_delta set. (c) C is disjoint from P. (d) C union P is closed in R. [[ Zahorski died on May 8, 1998 at the age of 84. I believe his work on the singularities of C-infinity functions was part of his Ph.D. research. ]] We mention three corollaries. ** If P is empty in some interval, then the set of (C)-points in that interval is a relatively closed first category subset of that interval. Since every closed first category set is nowhere dense, no such function can be nowhere analytic in that interval. Hence, if a function is nowhere analytic and C-infinity, then every interval must contain some (P)-points, and we get the result that Boas proved in 1935. ** Another corollary is the existence of a C-infinity function having a (P)-point everywhere: Choose C to be empty and P = R. ** Still another corollary is that there exist C-infinity functions belonging to each of the following categories (recall Pringsheim's examples in Section II): (i) C is c-dense in R and P is empty. (ii) C is empty and P is c-dense in R. (iii) Both C and P are c-dense in R. For multivariable versions of Zahorski's theorem, see Bartczak [1], Schmets/Valdivia [27] [28], Siciak [29], and H. Zahorska [33]. IV. MOST C-INFINITY FUNCTIONS ARE NOWHERE ANALYTIC The Baire category theorem can be used to prove the existence of nowhere analytic C-infinity functions, and this has been re-discovered several times. Using the standard metric on C-infinity (more precisely, any metric that generates the topology of uniform convergence for all orders of derivatives on compact sets), Morgenstern [22] (1954) gave a concise proof that the Baire-typical C-infinity function is nowhere analytic. An outline of a proof can be found on pp. 301-302 of Dugundji [12], with Morgenstern's name mentioned. However, Dungundji states that the particular proof he presents is due to Salzmann and Zeller, apparently from a personal communication. For an expanded version in English of Morgenstern's original proof, see pp. 95-97 of Jones [16]. Next up, we have Christensen [9] (1972), who proves the same result, unaware of Morgenstern. Then we have Darst [10] (1973), who was apparently unaware of both Morgenstern's and Christensen's papers. However, Darst followed this up with [11] (1974), where a far stronger Baire-typical result is proved. [See page 26. The result I am alluding to only shows up in the proof of a certain theorem, not in any of his theorem statements.] Darst shows that certain quasi-analytic classes have a Baire-typical set of functions that are nowhere quasi-analytic relative to other quasi-analytic notions. Next, we have Cater [7] (1984), who was aware of both Christensen's and Darst's papers, and probably also of Morgenstern's paper. Cater's paper is well written and his proof has all the details worked out. Siciak [29] (1986), who was aware of Morgenstern's paper, proves a Baire-typical multi-dimensional analog of Morgenstern's result (theorem 10 on p. 144). After this, there is Bernal [2] (1987). Bernal states that a corollary of the main result in his paper is that the Baire-typical C-infinity function is nowhere analytic. [Bernal cites Cater, Christensen, and Darst in his bibliography.] Finally, Ramsamujh [23] (1991) proves that the Baire-typical C-infinity function is nowhere analytic, unaware at that time of any of the preceding papers. Aside from simply being nowhere analytic, can we say anything about the (C)-points and the (P)-points of the Baire-typical C-infinity function? As far as I can tell, it appears that all the sources I mentioned in the previous two paragraphs, except for Bernal [2], Ramsamujh [23], and Siciak [29], prove only that the Baire-typical theorem we know that the set of (P)-points is a G_delta set (actually, this is immediate from the definition), and so we have the interesting observation that the Baire-typical C-infinity function has a Baire-typical set of (P)-points. In fact, Christensen [9] explicitly states his result in this way. However, Bernal [2], Ramsamujh [23], and Siciak [29] manage to prove more. They actually prove that EVERY point is a (P)-point for the Baire-typical C-infinity function. Thus, the Baire-typical C-infinity function has no (C)-points at all. Although this completely settles the matter, the result is unfortunate because it closes the door on the possibility of investigating the Lebesgue measure (or Hausdorff dimension, if measure zero) of the sets of (C)-points and (P)-points of the Baire-typical C-infinity function. Dave L. Renfro === Subject: Re: Inconsistency of the usual axioms of set theory posting-account=yxbZkgkAAABQBvyYeebYQ-PAvi0uT3tG 1.1.4322; .NET CLR 2.0.50727; MS-RTC LM 8; .NET CLR 3.0.4506.2152; .NET CLR 3.5.30729),gzip(gfe),gzip(gfe) > Please let me know how do you discuss the existence of a 1-1 > correspondence between S and > another set, when you did not accept the existence of the set S? The same way we can discuss a single straight horn projecting from the > animal's forehead without accepting the existence of any such animal. I think we need a better example of defined but nonexistent: http://dsc.discovery.com/news/2008/06/11/unicorn-deer-italy.html There's the popular even prime greater than 2, for instance. === Subject: Re: Inconsistency of the usual axioms of set theory <29301058.1235588720025.JavaMail.jakarta@nitrogen.mathforum.org you sure \ are a bunch of cheaters here !! snipping all the relevant parts of your opponents posts ! CHEATERS Keep it up, you're well on the way to JSH levels of ranting. You just need to dress it up in terms of global disaster befalling all the liars who dare to point out errors in your posts. Maybe one day you'll get there. - Tim === Subject: How to prove an infinite symmetric group on positive integer is isomorphic to real number posting-account=9YsO_goAAACoeXJXcnGeSA4nUSdl5-uO 2.0.50727; .NET CLR 1.1.4322; .NET CLR 3.0.04506.30),gzip(gfe),gzip(gfe) A={ f | f : N<---->N } It means A contains all the functions that map N to N, also one-to-one correspondence. And prove that A and R have the same cardinal number. Can anyone help me to solve it? How to construct a mapping from A to R? p.s. N=positive integer R=real number === Subject: Re: How to prove an infinite symmetric group on positive integer is \ isomorphic to real number > A={ f | f : N<---->N } It means A contains all the functions that > map N to N, also one-to-one correspondence. And prove that A and R have the same cardinal number. Can anyone help me to solve it? How to construct a mapping from A to > R? p.s. N=positive integer R=real number You could use continued fractions to get a one-to-one map from A into R. For a map from (0,1) into A: given x in (0,1) with base-2 digits b_1 b_2 b_3 ... consider f that exchanges 2n-1 and 2n if b_n = 1 and leaves them fixed if b_n = 0. -- Robert Israel israel@math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: JSH: One more explanation, factoring solution Originator: tchow@lebesgue.mit.edu.mit.edu (Timothy Chow) >If so, ok, I was wrong. James Harris JSH admits he was wrong? Call Ripley's! -- Tim Chow tchow-at-alum-dot-mit-dot-edu The range of our projectiles---even ... the artillery---however great, will never exceed four of those miles of which as many thousand separate us from the center of the earth. ---Galileo, Dialogues Concerning Two New Sciences === Subject: Re: JSH: One more explanation, factoring solution posting-account=wVv_VwoAAAAVTfUuyxLzug5SzYWCgHj1 Gecko/20081217 Firefox/2.0.0.20,gzip(gfe),gzip(gfe) On Feb 25, 11:09 pm, Tim Smith I'm tired of this entire discussion. A simple solution with some > quadratics has been turned into an endless series of posts from people It's only endless because you refuse to tell us what r(v), s(v), and > t(v) are. -- > --Tim Smith This is likely the end of this episode, and this is as close as we are going to get to JSH ever admitting that this moronic idea is not going to work. As in all previous episodes, his accusations against lying, conspiring, evil, jealous mathematicians were proven unfounded. And as in all previous episodes, the only lesson he will take away from it and remember is that mathematicians were lying, conspiring, evil, and jealous. Marcus. === Subject: Re: JSH: One more explanation, factoring solution It leads to a direct calculation. > It is not possible to do a direct calculation unless you tell us > what the functions r(v), s(v) and t(v) are. Without knowing what > those functions are we cannot do a direct calculation. > Given: x^2 - Dy^2 = 1. Also: y = [+/-2Dv/(f_1 - f_2*v^2 - 2v) +/- 1 -/+(f_1 + f_2*v^2)/(f_1 - > f_2*v^2 - 2v)]/(D-1) and x = +/-(f_1 + f_2*v^2)/(f_1 - f_2*v^2 - 2v) - [+/-2Dv/(f_1 - f_2*v^2 > - > 2v) +/- 1 -/+(f_1 + f_2*v^2)/(f_1 - f_2*v^2 - 2v)]/(D-1) It follows then that (x-1)(x+1) = Dy^2 and you can substitute and multiply out denominators to get: (+/-(f_1 + f_2*v^2) - [+/-2Dv +/- (f_1 - f_2*v^2 - 2v) -/+(f_1 + > f_2*v^2)]/(D-1) - (f_1 - f_2*v^2 - 2v))(+/-(f_1 + f_2*v^2) - [+/-2Dv > +/- (f_1 - f_2*v^2 - 2v) -/+(f_1 + f_2*v^2)]/(D-1) +(f_1 - f_2*v^2 - > 2v)) = D([+/-2Dv +/- (f_1 - f_2*v^2 - 2v) -/+(f_1 + f_2*v^2)]/(D-1) ) > ^2 And then you just differentiate for your two possible cases: d(+/-(f_1 + f_2*v^2) - [+/-2Dv +/- (f_1 - f_2*v^2 - 2v) -/+(f_1 + > f_2*v^2)]/(D-1) - (f_1 - f_2*v^2 - 2v))/dv = 0 and d(+/-(f_1 + f_2*v^2) - [+/-2Dv +/- (f_1 - f_2*v^2 - 2v) -/+(f_1 + > f_2*v^2)]/(D-1) +(f_1 - f_2*v^2 - 2v)) /dv = 0 Solve for v for a min or max. Substitute back, check gcd's with D. Oh, that's wrong. What?! === Subject: Re: JSH: One more explanation, factoring solution On 2009-02-26 00:09:10 -0500, Tim Smith said: > I'm tired of this entire discussion. A simple solution with some > quadratics has been turned into an endless series of posts from people It's only endless because you refuse to tell us what r(v), s(v), and > t(v) are. That's kind of a pattern at this point: JSH posts some hare-brained idea vaguely in the same universe as the factoring problem as the end of all civilisation, sci.math &c ask some really sticky questions, JSH argues that you're all suppressing his brilliant advance, sci.math & co. ask the same really sticky questions and mostly ignore the armchair sociology, JSH gives up on thread N and starts thread S(N), claiming that his ideas from thread N have lead to a new, even deeper crack in the foundations of western civilisation. Not that I mind, really: the only value LEFT in usenet at this point is using trolls and the mentally unsound as exercise equipment for working out your own thoughts and opinions. I certainly do it, even if I do try to post helpful and sincere things to some newsgroups, too. JSH's math skills are kind of remedial, so it's not that hard to find holes in his mathematical arguments, but he puts so much work into complicating them that clearing away the dross is kind of satisfying in the same way that, say, clearing a level in Bejewelled is satisfying. Sure, there's another one just like it around the corner, and ultimately it's just a way to kill time, but it generates a satisfying illusion of achievement, and sometimes you pick up a new pattern. JSH, if you do happen to be reading this, I have some honest advice for you: if you want to treat Usenet as a proving ground for your ideas, for s' sake start actually listening to and processing the criticism your ideas suffer. Remember, your ideas are useless if you can't communicate them. Even if you're right, your hammer is useless if nobody understands it and will destroy absolutely nothing. While I'm at it, could you stop wittering on about crimes and cases and defenses? Internet Lawyer-ing only makes you look like a paranoid schizophrenic. Either bring suit/file charges, in which case your lawyer will advise you not to talk about the case until it's settled in court, or don't, and shut up about it. -o === Subject: Re: JSH: One more explanation, factoring solution > Not that I mind, really: the only value LEFT in usenet at this point is > using trolls and the mentally unsound as exercise equipment for working > out your own thoughts and opinions. This may be (almost) true where mathematics is concerned, but it certainly does not apply generally to usenet. I know of a couple of really useful boards, e.g. comp.lang.fortran and uk.d-i-y === Subject: Re: JSH: One more explanation, factoring solution posting-account=fwSgtAkAAACFnX70ssKwbvm9_oCZVHrx Gecko/2009011913 Firefox/3.0.6,gzip(gfe),gzip(gfe) > Meanwhile, for those of us who want to factor WITHOUT already knowing > non-trivial factors of D, all you've given is this: æ ær, s, and t are functions such that x = r(v)/t(v) and y = s(v)/t(v) > æ æsatisfy x^2 - D y^2 = 1 for rational v, and if you find v that > æ æminimizes something that you won't specify, then GCD( r(v)+t(v), D) > æ æwill be a non-trivial factor of D. Actually, no, there are quite a few other equations and information > also given. And several people have given examples of r(v), s(v), t(v) that fit all > the information you have posted, yet fail to factor. æThe fact remains Not true. æIt's mathematically impossible. > Then provide the r(v), s(v), and t(v) that you have in mind and let's see them work! > you have *never* given an example of r(v), s(v), t(v) that give > non-trivial factors without making knowledge of the factors part of the > construction process. I've merely proven. > And so therefore you have not actually provided an algorithm, and for all we know finding the right functions could require an exponential-time algorithm in which case your method is no better than anything else out there. === Subject: Re: JSH: One more explanation, factoring solution posting-account=fwSgtAkAAACFnX70ssKwbvm9_oCZVHrx Gecko/2009011913 Firefox/3.0.6,gzip(gfe),gzip(gfe) s(v) = 1 > t(v) = (g1 - g2)/2 where g1 g2 = D? OK, you are right--with that definition of r, s, and t, you can factor > D. æToo bad you have to have already factored D in order to find r, s, t. It was an existence proof. > OK, then, but to actually get a working algorithm you'll need a way to reliably determine just what r, s, and t are WITHOUT knowing the factors g1 and g2 of D. So provide the algorithm/formula that determines r, s, and t. > Meanwhile, for those of us who want to factor WITHOUT already knowing > non-trivial factors of D, all you've given is this: æ ær, s, and t are functions such that x = r(v)/t(v) and y = s(v)/t(v) > æ æsatisfy x^2 - D y^2 = 1 for rational v, and if you find v that > æ æminimizes something that you won't specify, then GCD( r(v)+t(v), D) > æ æwill be a non-trivial factor of D. Actually, no, there are quite a few other equations and information > also given. Tell us what r, s, and t are. Rather demanding for someone who has just been called out for > remarkable inconsistencies. > You need those functions for your algorithm to even have a tiny chance at working. If you want to prove that your algorithm works then you must provide enough details to allow someone to actually use it to do something. > But not surprising to me. The story here is not about proof. æI have mathematical proof. æMy > original post quite adeptly steps through an entire mathematical > argument and covers all the necessary bases. > Then why haven't you gone and posted the actual algorithm that actually does the factoring? === Subject: Re: JSH: One more explanation, factoring solution Sorry about the previous partial post; I was away from my machine and using Google. Here's the whole thing (with a few additionaly edits). [JSH] >But how can you deny a proof? Think about it. Imagine you REALLY, >really do not want to accept a proof, but it's a proof! How can you >not? [Gordon Burditt] > You have proved nothing about the SPEED of your method (assuming > it even works, and that it can be converted into an algorithm - [JSH] >It leads to a direct calculation. When you work out all the math, >which is really easy as you have quadratics, you can just calculate >without effort v for each factorization of D-1. Which means that at >most you loop through the combinations of factors of D-1, which is >where you can maybe claim there could be a lot of effort. [Gordon Burditt] > No, you cannot calculate without effort. Even arithmetic takes > time. You have presented nothing to *PROVE* that your method is > faster than existing state-of-the-art factoring methods, or for > that matter faster than trial division. Yes, you actually need to > do things like counting divisions and square roots and determine > how that number scales with the number you are trying to factor. Your proof has an enormous hole in it: you've proved it can factor [marcus_bruckner@yahoo.com] > Yes and no. He claims that it works, and that it is > basically a one-step process: you find the functions, > they are quadratics, and then you can take derivatives and > find where they have minima. This would be very fast - no > searching required. If he were right this would be by far > the fastest factoring algorithm out there. Agreed (for the most part). There are algorithms based on JSH's writings (modifications (a) and (c2) near the end of this post) which do work, but which are not one-step algorithms and for which a complexity analysis would therefore be useful. > But he's not right. He thinks he can find one value of > v which minimizes both abs(r(v) - t(v)) and abs(r(v) + t(v)). > In general of course you cannot expect one value to > be a minimum point for two different functions. So this > freaking calculus approach is not going to work. This is a point where his proof is incomplete, but if his proof were right (which it isn't), this would not be a major stumbling block. All you'd need to do to find the right value of v (which I'll call v') is compute the ranges where 0 < abs(r(v) - t(v)) < D and 0 < abs(r(v) + t(v)) < D and find where they overlapped. If the functions exist and are quadratic, this would be easy to do. Take v' to be any rational number in that range. Given that D is not a factor of r(v') - t(v') or r(v') + t(v'), yet (r(v') - t(v')) (r(v') + t(v')) = D s(v')^2, each of r(v') - t(v') and r(v') + t(v') must contain a non-trivial factor of D, which can be extracted by taking gcd(r(v') - t(v'), D) and gcd(r(v') + t(v'), D). Note that the proof depends on r(v) - t(v), r(v) + t(v), and s(v) being integers for all values of v, as JSH originally specified. If you ignore this requirement, you can get lots of false positives that meet the conditions but don't help you factor D. That means you're back to searching for v', which isn't likely to be any better than trial division. It also depends on his proof of the existence of r(v') and t(v') meeting the conditions, which as you note below is flawed. > Then > he confuses himself by nothing that a factorization of > D DOES exist, therefore there must be some value of v > which works. However he overlooks that it can happen > that e.g. r(v) + t(v) has a nontrivial factor in common > with D, but r(v) + t(v) is actually BIGGER than D. Yes. He has ostensively proven (I haven't checked the initial theorem that he bases all this on) the existence of v', r', s', and t' such that 0 < abs(r' + t') < D 0 < abs(r' - t') < D x(v') = r'/t' y(v') = s'/t' but he doesn't seem to realize that r(v') and t(v') are not necessarily the same as r' and t'. In general, r(v') = k r' s(v') = k s' t(v') = k t' where k is an unknown integer, which allows r(v') - t(v') and r(v') + t(v') to be larger than D. This, in my opinion, is the key flaw in his reasoning. > There is the slight additional complication that > his method involves factor D - 1. In real applications > it is going to be a lot easier to factor D - 1 than to > factor D - for one thing, in real (RSA) applications, D > is odd, which means that 2 and (D - 1)/2 are integer > factors of D - 1. He refuses to completely specify r(v), t(v) and s(v), > but this is minor. He write x and y as a function > of v of the form x = r(v)/t(v) and y = s(v)/t(v), > where if v is an integer, then r(v), t(v) and s(v) > are all integers also. You can compute exactly what > these functions are, and they are quadratic in v. I disagree. Strictly you cannot determine what these functions are, since there are no functions that satisfy all the properties that JSH has claimed. Specifically, he has claimed that: (1) r(v) and t(v) are quadratic functions of v and therefore continous and differentiable. (2) r(v), s(v), and t(v) are integer-valued functions. These two requirements combined force the functions r(v) and t(v) to be constant. That would force the function x(v) to be a constant, which is inconsistent with his expression for x(v). Above, you write that if v is an integer, r(v), t(v), and s(v) are integers also. But that's not good enough for JSH's proof. JSH's proof requires that r(v), s(v), and t(v) be integers for all v in their domain (the rationals), or you get the false positives I described above. There are various other ways of dropping one of JSH's specifications: (a) Drop the requirement that the functions be quadratic. By choosing r(v), s(v), and t(v) such that all are integers and gcd(r(v),s(v),t(v))=1, you get a factoring algorithm that works, but is painfully slow since you no longer have nice quadratic functions to minimize. (b) Write r(v), s(v), and t(v) as quadratic functions not of v, but of integers m and n where v = m/n. In this case, the algorithm doesn't work because of the flaw in his existence proof: abs(r(m,n)-t(m,n)) and abs(r(m,n)+t(m,n)) need never be simultaneously smaller than D. (c) If you drop the requirement that r(v), s(v), and t(v) be integer-valued functions, then there are an infinite number of possible choices for r(v), s(v), and t(v), obtainable by multiplying any choice of r(v), s(v), and t(v) by a common rational-valued function of v. One choice of particular interest is (c2) the choice that sets s(v) = 1 for all rational numbers v. This fixes his existence proof, but because the functions are no longer integers, it introduces lots of false positives, and you once again have to search for v'. -- Jim E. Black (domain in headers) How to filter out stupid arguments in 40tude Dialog: [X] Watch/Ignore works on subthreads === Subject: Re: JSH: One more explanation, factoring solution posting-account=DkG3nAkAAAAVuxctYYjlIz6_Yb78PVNd rv:1.8.1.3) Gecko/20070309 Firefox/2.0.0.3,gzip(gfe),gzip(gfe) >It leads to a direct calculation. It is not possible to do a direct calculation unless you tell us > what the functions r(v), s(v) and t(v) are. Without knowing what > those functions are we cannot do a direct calculation. Given: x^2 - Dy^2 = 1. Also: y = [+/-2Dv/(f_1 - f_2*v^2 - 2v) +/- 1 -/+(f_1 + f_2*v^2)/(f_1 - > f_2*v^2 - 2v)]/(D-1) > Let f_1 = (D-1)/f and f_2 = f. Simplify to get y = (+/-)(2v(f - v) / (f^2 - 2fv - (D - 1)v^2) > and x = +/-(f_1 + f_2*v^2)/(f_1 - f_2*v^2 - 2v) - [+/-2Dv/(f_1 - f_2*v^2 > - > 2v) +/- 1 -/+(f_1 + f_2*v^2)/(f_1 - f_2*v^2 - 2v)]/(D-1) > so x = (+/-)(f^2 - 2fv + (D - 1)v^2) / (f^2 - 2fv - (D - 1)v^2) > It follows then that (x-1)(x+1) = Dy^2 and you can substitute and multiply out denominators to get: (+/-(f_1 + f_2*v^2) - [+/-2Dv +/- (f_1 - f_2*v^2 - 2v) -/+(f_1 + > f_2*v^2)]/(D-1) - (f_1 - f_2*v^2 - 2v))(+/-(f_1 + f_2*v^2) - [+/-2Dv > +/- (f_1 - f_2*v^2 - 2v) -/+(f_1 + f_2*v^2)]/(D-1) +(f_1 - f_2*v^2 - > 2v)) = D([+/-2Dv +/- (f_1 - f_2*v^2 - 2v) -/+(f_1 + f_2*v^2)]/(D-1) ) > ^2 Using the + in (+/-) x + 1 = 2(f - v)^2 / (f^2 - 2fv - (D - 1)v^2) x - 1 = 2Dv^2 / (f^2 - 2fv - (D - 1)v^2) Using the - in (+/-) x + 1 = -2Dv^2 / (f^2 - 2fv - (D - 1)v^2) x - 1 = -2(f - v)^2 / (f^2 - 2fv - (D - 1)v^2) so in either case [2(f - v)^2][2Dv^2] = Dy^2 [f^2 - 2fv - (D - 1)v^2] And then you just differentiate for your two possible cases: d(+/-(f_1 + f_2*v^2) - [+/-2Dv +/- (f_1 - f_2*v^2 - 2v) -/+(f_1 + > f_2*v^2)]/(D-1) - (f_1 - f_2*v^2 - 2v))/dv = 0 > d(2(f - v)^2) / dv = 4(f - v)(-1) = 0 when v = f > and d(+/-(f_1 + f_2*v^2) - [+/-2Dv +/- (f_1 - f_2*v^2 - 2v) -/+(f_1 + > f_2*v^2)]/(D-1) +(f_1 - f_2*v^2 - 2v)) /dv = 0 > d(2Dv^2) / dv = 4Dv = 0 when v = 0 so we see the two can't simultaneously be 0. When v = 0 x = +/- 1, y = 0 when v = f, x = +/- 1, y = 0 No factorization help in either case. Rick === Subject: Re: JSH: One more explanation, factoring solution posting-account=fwSgtAkAAACFnX70ssKwbvm9_oCZVHrx Gecko/2009011913 Firefox/3.0.6,gzip(gfe),gzip(gfe) which is really easy as you have quadratics, you can just calculate > without effort v for each factorization of D-1. æWhich means that at > most you loop through the combinations of factors of D-1, which is > where you can maybe claim there could be a lot of effort. I find it interesting that posters like you don't bother finding out > where there might actually be issues, but instead you just work to > cast up doubt--without giving any mathematics worth considering if you > give any at all. > Then why don't you just tell me what the functions r(v), s(v), and t (v) are, exactly? Just write down the formulas: r(v) = ? s(v) = ? t(v) = ? What do I put in the blanks? > Part of the point of this exercise of me replying to your posts isn't > to convince you of anything. It's a puzzle I'm working on, as I ponder after. How do I trust a world that lets people like you operate as you do? Certainly I can just program this thing and remove all the silliness, > but why can't mathematical proof work? > Because you haven't provided any. You haven't given a clear description of the algorithm and from what bits and pieces you have given nothing workable seems to come out! Start off with specifying exactly what the functions r, s, and t are to be, and tell why all the guesses everyone here has made are wrong. === Subject: Re: JSH: One more explanation, factoring solution posting-account=smr6iAoAAACS84DcGY1CagKorl2EXz6F rv:1.9.0.4) Gecko/2008102920 Firefox/3.0.4,gzip(gfe),gzip(gfe) >But how can you deny a proof? Think about it. Imagine you REALLY, >really do not want to accept a proof, but it's a proof! How can you >not? You have proved nothing about the SPEED of your method (assuming > it even works, and that it can be converted into an algorithm - It leads to a direct calculation. When you work out all the math, >which is really easy as you have quadratics, you can just calculate >without effort v for each factorization of D-1. Which means that at >most you loop through the combinations of factors of D-1, which is >where you can maybe claim there could be a lot of effort. No, you cannot calculate without effort. Even arithmetic takes > time. You have presented nothing to *PROVE* that your method is > faster than existing state-of-the-art factoring methods, or for > that matter faster than trial division. Yes, you actually need to > do things like counting divisions and square roots and determine > how that number scales with the number you are trying to factor. Your proof has an enormous hole in it: you've proved it can factor Yes and no. He claims that it works, and that it is > basically a one-step process: you find the functions, > they are quadratics, and then you can take derivatives and > find where they have minima. This would be very fast - no > searching required. If he were right this would be by far > the fastest factoring algorithm out there. Agreed (for the most part). There are algorithms based on JSH's writings (modification (a) near the end of this post) which do work, but which are not one-step algorithms and for which a complexity analysis would therefore be useful. > But he's not right. He thinks he can find one value of > v which minimizes both abs(r(v) - t(v)) and abs(r(v) + t(v)). > In general of courseProxy-Connection: keep-alive Cache-Control: max-age === Subject: Generators in finite groups I feel a bit embarassed asking this as I should know already. Suppose G is a \ group with G=. Therefore G is the set of all products of a and b, so e.g. aaaabbbbabababbababbabbaa is an element. Now suppose G is finite. Is it the case that every element of G has the form \ a^i b^j? Let m = o(a) and n = o(b). If |G| = mn then the result is obviously true; as \ then {a^i b^j; (a,b) e [0,m-1]x[0,n-1]} is a set of mn distinct elements. Do \ there exist any cases where |G| does not equal mn? My intuition says no. I feel like if this was the case, I should be able to \ use Cauchy/Sylow etc to produce an element which contradicts either the \ order hypothesis, or the fact that every element is a product of a and b. But try \ as I might I can't get there! Please help! === Subject: Re: Generators in finite groups >I feel a bit embarassed asking this as I should know already. Suppose G is \ >a group with G=. Therefore G is the set of all products of a and b, so \ >e.g. aaaabbbbabababbababbabbaa is an element. Now suppose G is finite. Is it the case that every element of G has the > form a^i b^j? > no unless G is abelian. Right? abab is not of the form a^i b^j > Let m = o(a) and n = o(b). If |G| = mn then the result is obviously true; \ > as then {a^i b^j; (a,b) e [0,m-1]x[0,n-1]} is a set of mn distinct > elements. Do there exist any cases where |G| does not equal mn? No, G doesn't have to be cyclic. You are thinking G = x which is clearly false in general. My intuition says no. I feel like if this was the case, I should be able > to use Cauchy/Sylow etc to produce an element which contradicts either the \ > order hypothesis, or the fact that every element is a product of a and b. \ > But try as I might I can't get there! Please help! I think you are trying to use examples to prove things in general without realizing that what you are using is only a very small subset of the possibilities. G doesn't have to be abelian or the product of cyclic groups. \ There are many counterexamples to your claims. I think your making the same mistake in both cases. We know that every finite abelian group is a product of cyclic groups. If the group is abelian \ then your first assumption holds because any product can be written as a^i b^j by rearranging the terms. In the second case you are somewhat right, if \ G is abelian then G = A1 x A2... x An and |G| = lcm(|A1|,...,|An|). Ofcourse this doesn't work when G isn't abelian. === Subject: Re: Generators in finite groups posting-account=_l4K0QkAAAC09JhOoK_ZfoJKXOmr_jZf Gecko/2009020911 Ubuntu/8.04 (hardy) Firefox/3.0.6,gzip(gfe),gzip(gfe) >I feel a bit embarassed asking this as I should know already. Suppose G is >a group with G=. Therefore G is the set of all products of a and b, so >e.g. aaaabbbbabababbababbabbaa is an element. Now suppose G is finite. Is it the case that every element of G has the > form a^i b^j? no unless G is abelian. Right? abab is not of the form a^i b^j Abelian is sufficient, but it is not necessary. If either or is normal, then you can rewrite abab. For example, if is normal, then a^{-1}ba = b^k for some k, so abab = a^2(a^{-1}ba)b = a^2 b^k b = a^2 b^(k+1) And if is normal, then bab^{-1} = a^m for some m, so abab = a(bab^{-1})b^2 = a a^m b^2 = a^(m+1)b^2. But even this is not, I think, necessary. All you need is some relation that tells you that ba can be expressed in the form a^i b^j, and unless i=1 or b=1, this relation need not imply normality of either or . -- Arturo Magidin === Subject: Re: Generators in finite groups > Suppose G is a group with G=. .. > Now suppose G is finite. Is it the case that every > element of G has the form a^i b^j? > no unless G is abelian. Right? abab is not of the > form a^i b^j Well it could be. For instance, if G is a group of order 6, then every generating set {a,b} of size 2 has the property that every element of G has the form a^i b^j. > Let m = o(a) and n = o(b). If |G| = mn then the > result is obviously true; as then {a^i b^j; (a,b) > e [0,m-1]x[0,n-1]} is a set of mn distinct > elements. No, G doesn't have to be cyclic. You are thinking G = > x which is clearly false in general. Or more generally, G = , including semidirect products like the non-abelian group of order 6. I should also mention for the OP that generators for my order 12 counterexample are the permutations (1,2,3) and (1,2)(3,4), and for my order 16 counter example, the matrices [ 0, z; -1/z, 0 ], [ 0, 1; -1, 0 ], where z is a primitive 8th root of unity. === Subject: Re: Generators in finite groups > I feel a bit embarassed asking this as I should know > already. Suppose G is a group with G=. Therefore > G is the set of all products of a and b, so e.g. > aaaabbbbabababbababbabbaa is an element. Now suppose G is finite. Is it the case that every > element of G has the form a^i b^j? No, the alternating group of order 12 is a counter example with a of order 3 and b of order 2. > Let m = o(a) and n = o(b). If |G| = mn then the > result is obviously true; as then {a^i b^j; (a,b) e > [0,m-1]x[0,n-1]} is a set of mn distinct elements. This is not true. The quaternion group of order 16 is a counterexample. If a^i*b^j = a^x*b^y, then a^(i-x) = b^(y-j) is in the intersection of and , but need not be the identity. > Do there exist any cases where |G| does not equal mn? Yes, the alternating group of order 12, above. === Subject: Math without Logic posting-account=OWfgwwgAAADQpH2XgMDMe2wuQ7OFPXlE Gecko/2009011913 Firefox/3.0.6,gzip(gfe),gzip(gfe) I would like your comments on the following excerpt from a paper that discusses the pitfalls of teaching advanced mathematics (undergraduate level) without due attention to formal logic. Dan ...In textbooks, we can find careless definitions; for example: f + g is continuous at a point, provided f and g are Of course it is harder to note the extended version of the previous definition: For all functions f , all functions g, and all real numbers a, if f is continuous at a and g is continuous at a, then f + g is continuous at a. It is clear why we teach the former version, but in this case even talented students cannot extend the definitions and theorems noted in short form. According to Selden just only 8.5 percent of his mid-level undergraduate mathematics majors could ñunpackî informally written mathematical statementsinto their logically equivalent formal statements. Some textbook-writers keep quiet about quantifiers, because in their opinion students cannot understand them. Hence if students do not practice this kind of extensions, they have only superficial knowledge about theorems and definitions. With this superficial knowledge students can solve simple problems, but they cannot solve harder ones, e.g. proofs. (p. 2) Source: Mario Bako, Why we need to teach logic and how we can teach it? International Journal for Mathematics Teaching and Learning, October 17, 2002 http://www.cimt.plymouth.ac.uk/journal/bakom.pdf === Subject: Re: Math without Logic > I would like your comments on the following excerpt from a paper that > discusses the pitfalls of teaching advanced mathematics (undergraduate > level) without due attention to formal logic. Dan > ...In textbooks, we can find careless definitions; for example: f + g is continuous at a point, provided f and g are a definition from a theorem and is therefore singularly underequipped to write about either math or logic. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Math without Logic posting-account=T7Gd-QoAAACeQajv7mi_Za6uPu3TpBXy AppleWebKit/525.19 (KHTML, like Gecko) Chrome/1.0.154.48 Safari/525.19,gzip(gfe),gzip(gfe) On Feb 26, 8:07æpm, Dan Christensen discusses the pitfalls of teaching advanced mathematics (undergraduate > level) without due attention to formal logic. Dan ...In textbooks, we can find careless definitions; for example: f + g is continuous at a point, provided f and g are Of course it is harder to note the extended version of the previous > definition: For all functions f , all functions g, and all real numbers a, if f is > continuous at a and g is continuous at a, then f + g is continuous at > a. It is clear why we teach the former version, but in this case even > talented students cannot extend the definitions and theorems noted in > short form. According to Selden just only 8.5 percent of his mid-level > undergraduate mathematics majors could ñunpackî informally written > mathematical statementsinto their logically equivalent formal > statements. Some textbook-writers keep quiet about quantifiers, > because in their opinion students cannot understand them. Hence if > students do not practice this kind of extensions, they have only > superficial knowledge about theorems and definitions. With this > superficial knowledge students can solve simple problems, but they > cannot solve harder ones, e.g. proofs. (p. 2) Source: Mario Bako, Why we need to teach logic and how we can teach > it? International Journal for Mathematics Teaching and Learning, > October 17, 2002http://www.cimt.plymouth.ac.uk/journal/bakom.pdf Well this is just stupid. There is nothing wrong with saying f + g is continuous at a point, provided f and g are Then they formalise it as: For all functions f , all functions g, and all real numbers a, if f is continuous at a and g is continuous at a, then f + g is continuous at a. Who said anything about real numbers? If you're going to be pedantic, you should at least be correct. === Subject: Re: compact set's definition <4772890.1235047517354.JavaMail.jakarta@nitrogen.mathforum.org> > Intuitively compact sets are metric spaces that are complete and bounded. The idea of compactness originated with subsets of R_n. The first exercise is to prove that a complete bounded set is compact. > Counterexample (not in R_n, of course): an infinite discrete space (e.g. the integers) with zero-or-one metric. So easy... === Subject: definition of =?ISO-8859-1?Q?Carath=E9odory_in_measure_the?= =?ISO-8859-1?Q?ory?= Just checking my definition: We say that A has a length by Carath.8eodory if m*(A) + m*(X - A)=b_0 - a_0 where m* is the outer measure and A is a subset of [a_0,b_0) = X . Is this a correct definition? === Subject: Re: definition of Carath?odory in measure theory > Just checking my definition: > We say that A has a length by Carath?odory if > m*(A) + m*(X - A)=b_0 - a_0 > where m* is the outer measure and A is a subset of [a_0,b_0) = X . > Is this a correct definition? The definition according to Caratheodory is that if m* is an outer measure defined on all subsets of a set X, then a set E is m*-measurable if m*(A) = m*(A / E) + m*(A E) for every subset A of X. That is, E serves as a suitable knife for cutting up any other set A into two parts whose outer measures are additive. -- Dave Seaman Third Circuit ignores precedent in Mumia Abu-Jamal ruling. === Subject: Re: definition of =?ISO-8859-1?Q?Carath=E9odory_in_measure_?= =?ISO-8859-1?Q?theory?= > Just checking my definition: > We say that A has a length by Carath.8eodory if > m*(A) + m*(X - A)=b_0 - a_0 where m* is the outer measure and A is a subset of [a_0,b_0) = X . Is this a correct definition? Also, if A satisfies that definition is it said to be m*-measurable? === Subject: Re: Tim Skirvin, Christopher Heckman Re: Small Claims court filing \ on MySpace Re: Lawsuit against Myspace.com (Rupert Murdoch, Tom Anderson, Chris DeWolfe) for slander and not abiding by abuse standards posting-account=9F70KwkAAADloWPH_Z0lOPxF98dtfKh6 Gecko/2008070208 Firefox/3.0.1,gzip(gfe),gzip(gfe) Yes we are. Please send your check for $1000...000 as a retainer. -- > Morris Nyuknyuk > Dewey, Cheetham, & Howe Yours is a joke, but as circumstances change, jokes become easy > reality. Of this Fake MySpace: # > MySpace.com - ~ ArchimedesPlutonium~ - 55 - Female - Holy Catnip ... > MySpace profile for ~.86´ ?ArchimedesPlutonium?.86´~ with pictures, videos, > personal blog, interests, information about me and more. > profile.myspace.com/index.cfm? > fuseaction=user.viewprofile&friendID=118852684 - 63k - It has been up and defaming for several years now. Some of the pictures > trace to Tim Skirvin some trace to Christopher Heckman. [...] Wow. Someone told me that AP was after me again after all this time, but I had no idea he was on the warpath. To set the record straight: 1. I have never had anything to do with Facebook; I already have a website. 2. I have more important things to do than deal with someone who is recognized as a paranoid crank. For instance, currently, I am teaching 300 students this semester. 3. The statement that certain defaming pictures trace back to me is an absolute lie. The only pictures that I have posted are of me, and of my family's dachshunds, and none of these pictures can be considered defaming. Someone may be using my name for nefarious purposes. (I am not ruling out the possibility that AP himself is using my name to do this.) 4. I don't know who this Tim Skirvin is, unless he uses a different name when posting. --- Christopher Heckman === Subject: Re: Tim Skirvin, Christopher Heckman Re: Small Claims court filing \ on MySpace Re: Lawsuit against Myspace.com (Rupert Murdoch, Tom Anderson, Chris DeWolfe) for slander and not abiding by abuse standards posting-account=9F70KwkAAADloWPH_Z0lOPxF98dtfKh6 Gecko/2008070208 Firefox/3.0.1,gzip(gfe),gzip(gfe) > Yes we are. Please send your check for $1000...000 as a retainer. -- > Morris Nyuknyuk > Dewey, Cheetham, & Howe Yours is a joke, but as circumstances change, jokes become easy > reality. Of this Fake MySpace: # > MySpace.com - ~ ArchimedesPlutonium~ - 55 - Female - Holy Catnip ... > MySpace profile for ~.86´ ?ArchimedesPlutonium?.86´~ with pictures, videos, > personal blog, interests, information about me and more. > profile.myspace.com/index.cfm? > fuseaction=user.viewprofile&friendID=118852684 - 63k - It has been up and defaming for several years now. Some of the pictures > trace to Tim Skirvin some trace to Christopher Heckman. [...] Wow. Someone told me that AP was after me again after all this time, > but I had no idea he was on the warpath. To set the record straight: 1. I have never had anything to do with Facebook; Or MySpace or any of the other cookie-cutter webpages which claim to be personal pages, for that matter. > I already have a website. 2. I have more important things to do than deal with someone who is > recognized as a paranoid crank. For instance, currently, I am teaching > 300 students this semester. 3. The statement that certain defaming pictures trace back to me is > an absolute lie. The only pictures that I have posted are of me, and > of my family's dachshunds, and none of these pictures can be > considered defaming. Someone may be using my name for nefarious > purposes. (I am not ruling out the possibility that AP himself is > using my name to do this.) 4. I don't know who this Tim Skirvin is, unless he uses a different > name when posting. [NonBreakingSpace] [NonBreakingSpace] [NonBreakingSpace]--- Christopher Heckman === Subject: Re: Quet-style Prime Grid Game. > Quet-style Prime Grid Game > Played on a 5x5 grid Surely you mean n-by-n? -- Kev === Subject: Re: Quet-style Prime Grid Game. <87wsbfq8d4.fsf@nonospaz.fatphil.org> posting-account=AFsgCgkAAAA3VOfxqn2cTB2LbLN3nbER Gecko/20070319,gzip(gfe),gzip(gfe) > So it was a good idea, early in the game, to aim for the 20 > sector, which has 19 and 17 alongside, so that if one were > a merely passable dart player one could hit one of those and > keep it, being prime. Rather than risk going on and destroying > it down to a 3 or even 2 ! 19 and 17 are alongside 3. 20 has 5 and 1 alongside. Oh damn and blast - quite right. So, I meant, you aim at the 3 and hope not to hit it, but to hit a 17 or 19. (If you hit the 3 despite yourself you still have 2 darts to augment it.) > I'd want to play first (and last). Did you see a noticable bias? In the related game we tried, it didn't seem to matter a lot who played first, but last play was an advantage if there were spare numbers to choose from, but a disadvantage if there weren't. Not much in it though. b === Subject: Question on Complex Made Simple I recently bought Complex Made Simple by Prof. Ullrich, and am having a slightly difficult time comprehending Lemma 3.4. I believe that it is correct, but the proof doesn't convince me. Since I am quite tired, and don't feel like typing up the entire Lemma and proof, I will direct this question to either Prof. Ullrich or other owners of the book: What if K = {1}, rho = 1/2 in the statement of the lemma. The sequence (z_n) I choose is constant: 1/2. Then by definition of K, alpha_n = 1, but the only possible beta_n (-1/2) lies outside the closed disk? It's very possible I am missing something here, but if someone could \ kindly help me to understand, I would be most grateful. Steve === Subject: Re: Question on Complex Made Simple I recently bought Complex Made Simple by Prof. Ullrich, and am having a >slightly difficult time comprehending Lemma 3.4. I believe that it is >correct, but the proof doesn't convince me. Since I am quite tired, and >don't feel like typing up the entire Lemma and proof, I will direct this >question to either Prof. Ullrich or other owners of the book: >What if K = {1}, rho = 1/2 in the statement of the lemma. >The sequence (z_n) I choose is constant: 1/2. Then by definition of K, >alpha_n = 1, but the only possible beta_n (-1/2) lies outside the closed >disk? It's very possible I am missing something here, but if someone could >kindly help me to understand, I would be most grateful. What Rotwang said: D(z, rho) in the proof was a typo for D(z, rho). Oh jesus. I don't think I'm ever going to get this right. Of course what I meant to say was that D(z, rho) was a typo for D(z, rho). > (That's in the proof - it's supposed to be D(z, rho) in the > _statement_ of the lemma.) In fact the proof as presented isn't just wrong, it makes no sense - > D(z, rho) is meanigless since I haven't said what z I'm talking > about. Steve David C. Ullrich Understanding Godel isn't about following his formal proof. > That would make a mockery of everything Godel was up to. > (John Jones, My talk about Godel to the post-grads. > in sci.logic.) -- David C. Ullrich === Subject: Re: Question on Complex Made Simple David C. Ullrich a .8ecrit : > I recently bought Complex Made Simple by Prof. Ullrich, and am having a > slightly difficult time comprehending Lemma 3.4. I believe that it is > correct, but the proof doesn't convince me. Since I am quite tired, and > don't feel like typing up the entire Lemma and proof, I will direct this > question to either Prof. Ullrich or other owners of the book: > What if K = {1}, rho = 1/2 in the statement of the lemma. > The sequence (z_n) I choose is constant: 1/2. Then by definition of K, > alpha_n = 1, but the only possible beta_n (-1/2) lies outside the closed > disk? It's very possible I am missing something here, but if someone could > kindly help me to understand, I would be most grateful. > What Rotwang said: D(z, rho) in the proof was a typo for D(z, rho). Oh jesus. I don't think I'm ever going to get this right. > Of course what I meant to say was that D(z, rho) was a typo > for D(z, rho). I think your z key is stuck (That's in the proof - it's supposed to be D(z, rho) in the > _statement_ of the lemma.) In fact the proof as presented isn't just wrong, it makes no sense - > D(z, rho) is meanigless since I haven't said what z I'm talking > about. Steve > David C. Ullrich Understanding Godel isn't about following his formal proof. > That would make a mockery of everything Godel was up to. > (John Jones, My talk about Godel to the post-grads. > in sci.logic.) > === Subject: Re: Question on Complex Made Simple What Rotwang said: D(z, rho) in the proof was a typo for D(z, rho). Oh jesus. I don't think I'm ever going to get this right. > Of course what I meant to say was that D(z, rho) was a typo > for D(z, rho). Infinite recursion. Again. === Subject: Re: Question on Complex Made Simple > I recently bought Complex Made Simple by Prof. Ullrich That makes two of us. It's for sale in the Netherlands via: http://www.bol.com/nl/p/boeken-engels/complex-made-simple/1001004006484972/i\ ndex.html Han de Bruijn === Subject: Re: Question on Complex Made Simple posting-account=aLpfCwoAAACh4BOs3HOlQBCoxUpEgyxc Gecko/2009011913 Firefox/3.0.6,gzip(gfe),gzip(gfe) > I recently bought Complex Made Simple by Prof. Ullrich, and am having a slightly difficult time comprehending Lemma 3.4. æI believe that it is correct, but the proof doesn't convince me. æSince I am quite tired, and don't feel like typing up the entire Lemma and proof, I \ will direct this question to either Prof. Ullrich or other owners of the book: > What if K = {1}, rho = 1/2 in the statement of the lemma. > The sequence (z n) I choose is constant: 1/2. æThen by definition of K, alpha n = 1, but the only possible beta n (-1/2) lies outside the closed disk? æIt's very possible I am missing something here, but if someone could kindly help me to understand, I would \ be most grateful. I think that every incidence of Dbar(z,rho) in the proof is meant to be Dbar(0,rho). Then the statement for every n there exists alpha n in K and beta n in Dbar(0,rho) such that z n = alpha n + beta n makes sense, since every z n is in Dbar(alpha n,rho) for some alpha n, and z n is in Dbar(alpha n,rho) iff beta n is in Dbar(0,rho), where beta n is defined to be z n - alpha n. === Subject: Re: Question on Complex Made Simple posting-account=aLpfCwoAAACh4BOs3HOlQBCoxUpEgyxc Gecko/2009011913 Firefox/3.0.6,gzip(gfe),gzip(gfe) I recently bought Complex Made Simple by Prof. Ullrich, and am having a slightly difficult time comprehending Lemma 3.4. æI believe that it is correct, but the proof doesn't convince me. æSince I am quite tired, and don't feel like typing up the entire Lemma and proof, I will direct this question to either Prof. Ullrich \ or other owners of the book: > What if K = {1}, rho = 1/2 in the statement of the lemma. > The sequence (z n) I choose is constant: 1/2. æThen by definition of K, alpha n = 1, but the only possible beta n (-1/2) lies outside the closed disk? æIt's very possible I am missing something here, but if someone could kindly help me to understand, I would \ be most grateful. I think that every incidence of Dbar(z,rho) in the proof is meant to > be Dbar(0,rho). Then the statement for every n there exists alpha n > in K and beta n in Dbar(0,rho) such that z n = alpha n + beta n makes > sense, since every z n is in Dbar(alpha n,rho) for some alpha n, and > z n is in Dbar(alpha n,rho) iff beta n is in Dbar(0,rho), where beta n > is defined to be z n - alpha n. P.S. Here is (I think) another way to prove the lemma, which I mention for no reason other than that I think it's quite nice: since K' is obviously bounded, in order to prove it's compact we need only show it's closed. However, the fact that K is compact means that, for every point z in C, there is a point alpha in K such that |z - alpha| = d (z,K) (this is proved by considering a sequence alpha n in K such that |z - alpha n| -> d(z,K) and using sequential compactness). But then it follows that K' is closed, since if z is not in K' then d(z,K) > rho, so that D(z,epsilon) does not intersect K', where epsilon = d(z,K) - rho. === Subject: Re: Question on Complex Made Simple On 26 Feb, 08:25, Steve Dalton > I recently bought Complex Made Simple by Prof. > Ullrich, and am having a slightly difficult time > comprehending Lemma 3.4. æI believe that it is > correct, but the proof doesn't convince me. æSince I > am quite tired, and don't feel like typing up the > entire Lemma and proof, I will direct this question > to either Prof. Ullrich or other owners of the book: > What if K = {1}, rho = 1/2 in the statement of the > lemma. > The sequence (z_n) I choose is constant: 1/2. > æThen by definition of K, alpha_n = 1, but the only > possible beta_n (-1/2) lies outside the closed disk? > æIt's very possible I am missing something here, but > if someone could kindly help me to understand, I > would be most grateful. I think that every incidence of Dbar(z,rho) in the > proof is meant to >be Dbar(0,rho). Aargh. Yes, of course. Major typo, sorry. By all means let us know if you find other problems > like that. > paid > (it won't do you any good, of course). Then the statement for every n there exists alpha_n >in K and beta_n in Dbar(0,rho) such that z_n = > alpha_n + beta_n makes >sense, since every z_n is in Dbar(alpha_n,rho) for > some alpha_n, and >z_n is in Dbar(alpha_n,rho) iff beta_n is in > Dbar(0,rho), where beta_n >is defined to be z_n - alpha_n. David C. Ullrich Understanding Godel isn't about following his formal > proof. > That would make a mockery of everything Godel was up > to. > (John Jones, My talk about Godel to the post-grads. > in sci.logic.) Steve === Subject: Re: Question on Complex Made Simple (And then of course the simplest way to type > overline{D(0,rho} is to simply cut&paste > overline{D(z,rho} from above... aargh.) David C. Ullrich You lazy bum!!!! j/k === Subject: Re: Question on Complex Made Simple <1i2dq4td4l36aubjije6vjl7bbp12g2rmf@4ax.com> posting-account=aLpfCwoAAACh4BOs3HOlQBCoxUpEgyxc Gecko/2009011913 Firefox/3.0.6,gzip(gfe),gzip(gfe) > I recently bought Complex Made Simple by Prof. Ullrich, and am having a slightly difficult time comprehending Lemma 3.4. æI believe that it is correct, but the proof doesn't convince me. æSince I am quite tired, and don't feel like typing up the entire Lemma and proof, I will direct this question to either Prof. Ullrich \ or other owners of the book: > What if K = {1}, rho = 1/2 in the statement of the lemma. > The sequence (z n) I choose is constant: 1/2. æThen by definition of K, alpha n = 1, but the only possible beta n (-1/2) lies outside the closed disk? æIt's very possible I am missing something here, but if someone could kindly help me to understand, I would \ be most grateful. I think that every incidence of Dbar(z,rho) in the proof is meant to > be Dbar(0,rho). Then the statement for every n there exists alpha n > in K and beta n in Dbar(0,rho) such that z n = alpha n + beta n makes > sense, since every z n is in Dbar(alpha n,rho) for some alpha n, and > z n is in Dbar(alpha n,rho) iff beta n is in Dbar(0,rho), where beta n > is defined to be z n - alpha n. P.S. Here is (I think) another way to prove the lemma, which I mention >for no reason other than that I think it's quite nice: since K' is >obviously bounded, in order to prove it's compact we need only show >it's closed. However, the fact that K is compact means that, for every >point z in C, there is a point alpha in K such that |z - alpha| = d >(z,K) (this is proved by considering a sequence alpha n in K such that >|z - alpha n| -> d(z,K) and using sequential compactness). But then it >follows that K' is closed, since if z is not in K' then d(z,K) > rho, >so that D(z,epsilon) does not intersect K', where epsilon = d(z,K) - >rho. Since writing the above I have noticed that, for any metric space and any set K, d(z,K) is a continuous function of z. And so the fact that K' is equal to {z in C | d(z,K) <= rho} implies immediately that K' is closed. Not sure whether this observation serves any purpose. Right. The really right proof is this: If A and B are compact subsets > of the plane then A + B = {a + b : a in A, b in B} is compact. Proof: AxB is compact, and if f(x,y) = x + y then f is continuous > on AxB, so A + B = f(AxB) is compact. QED. Huh. I hadn't thought of that. Decided to give the proof I did because it didn't use anything > about compactness other than the characterization in terms > of convergent subsequences (seemed likely that for some > young readers the convergent subsequence thing will be > the most familiar characterization and/or the one that > makes the most sense). Went for simplicity, or what > seemed to me to be simplicity, over niceness, in many > places in the first half or so. (And then of course the simplest way to type > overline{D(0,rho} is to simply cut&paste > overline{D(z,rho} from above... aargh.) That should be overline{D(z,rho)} (you missed a parenthesis). I demand a refund on this usenet post. === Subject: Re: The scientist as idiot <3f9bq4pok4k5g5shfe17dpc433ls07j875@4ax.com> posting-account=fwSgtAkAAACFnX70ssKwbvm9_oCZVHrx Gecko/2009011913 Firefox/3.0.6,gzip(gfe),gzip(gfe) I recently read Freeman Dyson's The scientist as rebel. æThis is a > collection of book reviews. æI found some of the reviews to be > interesting, but what caught my attention was two things. æMr. Dyson > is pro-religion, and he believes in ESP. æWhile the first thing may be > excused as lots of otherwise intelligent people also fall into this > trap, I was very suprised, to say the least, that he believes in > ESP. [...] My opinion is that YOU are the moron here. [...] Don't be silly. æI also think he's wrong, but there isn't the slightest > doubt that very many otherwise intelligent people fall into the same > trap. æ Who do you think is wrong, exactly? Dyson or the original poster? If it's the latter, does this mean you think scientific research into ESP would be acceptable? (I would, I just don't think that one can jump off and say WOO! It's real!!! Yay!!!!!!!!!!! when you don't have enough evidence. There's a difference between research and blind belief.) === Subject: Re: The scientist as idiot posting-account=fwSgtAkAAACFnX70ssKwbvm9_oCZVHrx Gecko/2009011913 Firefox/3.0.6,gzip(gfe),gzip(gfe) I recently read Freeman Dyson's The scientist as rebel. æThis is a > collection of book reviews. æI found some of the reviews to be > interesting, but what caught my attention was two things. æMr. Dyson > is pro-religion, and he believes in ESP. æWhile the first thing may be > excused as lots of otherwise intelligent people also fall into this > trap, I was very suprised, to say the least, that he believes in > ESP. And how can you be sure that it is not YOU who has fallen into the > there is nothing that exists beyond the visible 3-D world of > materialism trap? æHow can you prove that the tenets of religion > are not some minor reflection of a higher-dimensional much more > advanced and encompassing science? æYou can't prove a negative. To > do so requires YOU to have God-like powers which both you and I assert > NOT to be the case. You must be a fool to believe in religion or ESP. Science is the search for truth, and I believe that there is no place > in science for gods and people with special powers. æWhat is your > opinion? My opinion is that YOU are the moron here. æAll you are saying in > your total ignorance is that you believe there is no place in the > world for any science beyond that which you already understand. Have > you done ANY investigation of ESP? æIf not, then how can you make such > idiot statements about it's existence as a special power? æAre you > so stoopid as to try to assert that all people are born with identical > abilities? æAre you trying to say that ALL science is already well > known by you? Are you saying that the CIA which seems to find ESP > interesting is wasting tax money on stoopidities? æHow about you go do > some investigations on a subject BEFORE you start giving out your > uniformed opinions on it? æAnd while you are at it try to keep those > investigations UNBIASED as science demands. The CIA are a bunch of idiots for wasting the taxpayers money on ESP. The bottom line here is that my opinion is that the idiot is YOU. I hereby declare you an enemy to the cause of truth. Just out of curiosity, what on earth gave you the notion that your > decrees matter to anyone? > Which goes to show that the skeptic can be just as silly, diehard, paranoid and accusatory as the believer. === Subject: Re: The scientist as idiot posting-account=fwSgtAkAAACFnX70ssKwbvm9_oCZVHrx Gecko/2009011913 Firefox/3.0.6,gzip(gfe),gzip(gfe) example, are never proven, though quite frequently they are disproven. > Thus, if you are looking for whether there is *proof* of God or the > paranormal, science would be profoundly mute. I'm curious here. Science never offers proof -- so then why does one talk about a burden of proof to substantiate a claim, but not a burden of disproof to refute a claim, if refutation and not confirmation is the only thing that is possible? === Subject: Re: The scientist as idiot > example, are never proven, though quite frequently they are disproven. > Thus, if you are looking for whether there is *proof* of God or the > paranormal, science would be profoundly mute. > I'm curious here. Science never offers proof -- so then why does one > talk about a burden of proof to substantiate a claim, but not a > burden > of disproof to refute a claim, if refutation and not confirmation is > the only > thing that is possible? Burden of proof is a legal term. Science is not a court of law. It doesn't have to reach a final verdict, and it never does. Anything in science we believe today may be found wrong tomorrow. When physicists talk about theories and the associated experimental evidence, they don't say that an experiment proves a theory, although it may confirm its predictions. What you usually hear is that an experiment has placed bounds on a theory or excluded it. -- Jim E. Black (domain in headers) How to filter out stupid arguments in 40tude Dialog: [X] Watch/Ignore works on subthreads === Subject: Re: The scientist as idiot example, are never proven, though quite frequently they are disproven. > Thus, if you are looking for whether there is *proof* of God or the > paranormal, science would be profoundly mute. > I'm curious here. Science never offers proof -- so then why does one > talk about a burden of proof to substantiate a claim, but not a > burden > of disproof to refute a claim, if refutation and not confirmation is > the only > thing that is possible? Burden of proof is a legal term. Science is not a court of law. It > doesn't have to reach a final verdict, and it never does. Anything in > science we believe today may be found wrong tomorrow. Translation: We can do as we like, just keep funding us. Maybe your grandson will strike it rich someday but in the meantime I'm all right, Jack. With a petty attitude like yours, Black, one can see why empires decline. Science may not be a court of law but we do have them for the purpose of administering justice. === Subject: Re: The scientist as legislator > ------- WOAAHH... AHAHAHAHA... AHAHAHA... ------- [snip crap] We can do as we like, just keep funding us. Maybe your grandson > will strike it rich someday but in the meantime I'm all right, Jack. > With a petty attitude like yours, Black, one can see why empires decline. > Science may not be a court of law but we do have them for the purpose of > administering justice. ::Andro:: Science may not be a court of law but we do have ::Andro:: > them for the purpose of administering justice. Heavy!... Andro, you now have, single handedly, codified > the purpose of socio-physics... How existential!... ahahaha... > None but a fool would claim Nature is just; neither is science. But it works both ways, I can declare Black a cretin and who would he appeal to? === Subject: Re: The scientist as idiot posting-account=sKfmEQkAAAC8kI3Pv6_U_nt9sVsxZ_ou 1.1.4322),gzip(gfe),gzip(gfe) > - Science doesn't offer proof of anything. Scientific theories, for > example, are never proven, though quite frequently they are disproven. > Thus, if you are looking for whether there is *proof* of God or the > paranormal, science would be profoundly mute. I'm curious here. Science never offers proof -- so then why does one > talk about a burden of proof to substantiate a claim, but not a > burden > of disproof to refute a claim, if refutation and not confirmation is > the only > thing that is possible? Probably, the influence of Popper. Continual confirmation of a hypothesis under a limted set of circumstances does not increase the certainty of the hypothesis being true. Newton is always confirmed at everyday speeds. But things go awry at speeds approaching the speed of light. According to Popper a scientific hypothesis is one that can in principle be tested and found to be false. To validate a hypothesis scientists should be attempting to falisfy it rather than looking for easy confirmation. This means that religous claims about god are not scientific as they cannot in priniple be falsofied. God is a supernatural entity that exists outside time and space. It's not very obvious how an hypothesis of this kind could be tested in principle, or in fact what it may mean. So it is puzzling why scientits have religous beliefs. They must be able to strictly compartmentalize the critical evidence-based approach they adopt for their work from the criteria, if any, they used for their personal beliefs. Can an empirical scientist really believe he can reason something into existence by some metaphysical argument or really be convinced that stength of belief is an indicator of correctness of belief. Maths loonies never seem to grasp that , say, dividing 1 by 0 leads to a contradiction, but working scientits surely have reached that level of reasoning power. === Subject: Re: The scientist as idiot sha1:kyLSvI/oskBwwdBSDJ77GZ1rXfU= > So it is puzzling why scientits have religous beliefs. > They must be able to strictly compartmentalize the critical > evidence-based approach they adopt for their work from > the criteria, if any, they used for their personal beliefs. > Can an empirical scientist really believe he can reason > something into existence by some metaphysical argument > or really be convinced that stength of belief is an indicator > of correctness of belief. Most of us have beliefs that are not justified by science. Our beliefs about morality, natural rights, politics, justice and so on are non-scientific beliefs. There is nothing wrong with believing some things which are not empirically refutable, as near as I can figger. Some scientists have chosen to believe in a God with no empirical evidence. Others (scientists and otherwise) find this choice irrational. Perhaps so, but surely not because it is a non-refutable hypothesis. I imagine that you agree that slavery is morally wrong. What experiment or observation do you propose to test this belief? Or do you regard this judgment as a matter of purely subjective opinion, no more factual than the claim that accordion music is sublime? -- Jesse F. Hughes Jesse: Quincy, you should trust me more. Quincy (age 4): Baba, I never trust you. And I've got good reasons. === Subject: Re: The scientist as idiot > - Science doesn't offer proof of anything. Scientific theories, for > example, are never proven, though quite frequently they are disproven. > Thus, if you are looking for whether there is *proof* of God or the > paranormal, science would be profoundly mute. > I'm curious here. Science never offers proof -- so then why does one > talk about a burden of proof to substantiate a claim, but not a > burden > of disproof to refute a claim, if refutation and not confirmation is > the only > thing that is possible? Probably, the influence of Popper. Continual confirmation of a > hypothesis under a limted set of circumstances does > not increase the certainty of the hypothesis being true. Newton > is always confirmed at everyday speeds. But things > go awry at speeds approaching the speed of light. > According to Popper a scientific hypothesis is one that can > in principle be tested and found to be false. To validate a > hypothesis scientists should be attempting to falisfy it rather > than looking for easy confirmation. This means that religous claims > about god are not scientific as they cannot in priniple be > falsofied. God is a supernatural entity that exists outside > time and space. It's not very obvious how an hypothesis > of this kind could be tested in principle, or in fact what it may > mean. > So it is puzzling why scientits have religous beliefs. > They must be able to strictly compartmentalize the critical > evidence-based approach they adopt for their work from > the criteria, if any, they used for their personal beliefs. Compartmentalization is the name of the game. Some people are really amazingly good at it. === Subject: Re: The scientist as idiot According to Popper a scientific hypothesis is one that can >in principle be tested and found to be false. To validate a >hypothesis scientists should be attempting to falisfy it rather >than looking for easy confirmation. This means that religous claims >about god are not scientific as they cannot in priniple be >falsofied. God is a supernatural entity that exists outside >time and space. It's not very obvious how an hypothesis >of this kind could be tested in principle, or in fact what it may >mean. Note that Popper is very careful to say that his falsification > criterion is not a necessary criterion for a proposition to be > meaningful, only for it to be scientific. (It's been a while > since I read about this stuff, but no doubt someone can provide > a reference.) So it is puzzling why scientits have religous beliefs. Not really. What would be (and is) puzzling is when a scientist > has beliefs which are plainly contradicted by empirical evidence, > such as that human bodies can be literally resurrected. There is > nothing very puzzling (to me, at least) in a scientist having > beliefs that cannot be empirically tested, i.e. are not scientific, > but may be meaningful (and even true). But empirical evidence is a big problem in itself. How can I ever check that someone at CERN is speaking the truth, when he says that he/she has the outcome of an expensive experiment which is detecting gravity waves is not a fake ? How can I be sure that people have really landed on the moon ? How can I be certain that the collapse of the buildings at ground zero is not the result of a conspiracy ? The answer is that I can't ! I just have to believe that other people can be trusted. But _can_ they be trusted ? >They must be able to strictly compartmentalize the critical >evidence-based approach they adopt for their work from >the criteria, if any, they used for their personal beliefs. I get the impression that a lot of that kind of doublethink does > go on. Then some people stand up who say that human bodies can be resurrected: http://scriptures.lds.org/john/20 24 But Thomas, one of the twelve, called Didymus, was not with them when Jesus came. 25 The other disciples therefore said unto him, We have seen the Lord. But he said unto them, Except I shall see in his hands the aprint of the nails, and put my finger into the print of the nails, and thrust my hand into his side, I will not believe. Han de Bruijn === Subject: Re: The scientist as idiot <8qtbq4t48d1ml6893cigqvb1asjce14f0s@4ax.com> <82cdd$49a6abae$82a1e228$12260@news1.tudelft.nl> posting-account=suWj4AkAAADE1IvGmj55Nmq3f98qb17e .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; InfoPath.2; .NET CLR 3.5.21022; Tablet PC 2.0; .NET CLR 1.1.4322),gzip(gfe),gzip(gfe) ............................. snip ................................ Not really. æWhat would be (and is) puzzling is when a scientist > has beliefs which are plainly contradicted by empirical evidence, > such as that human bodies can be literally resurrected. æThere is > nothing very puzzling (to me, at least) in a scientist having > beliefs that cannot be empirically tested, i.e. are not scientific, > but may be meaningful (and even true). But empirical evidence is a big problem in itself. How can I ever check > that someone at CERN is speaking the truth, when he says that he/she has > the æoutcome of an expensive experiment which is detecting gravity waves > is not a fake ? How can I be sure that people have really landed on the > moon ? How can I be certain that the collapse of the buildings at ground > zero is not the result of a conspiracy ? The answer is that I can't ! I just have to believe that other people can be trusted. But can they > be trusted ? > ****************************************************************************\ **** A pity you stopped up there: can you trust the people that said they were your parents were ACTUALLY your parents? Can you trust the DNA test, and the lab, that could answer that question? Can you trust books and people that said they knew someone called Hitler and ruled Germany some 70 years ago? Can you trust , and this time it is only books, that there was someone called Napoleon?.......etc......etc......etc Yes, some trust is called for in almost any realm of our lives, and wrt science there's a rather standard procedure (just as there is with mathematics as well): whenever someone stands up and claims something, there must be at least 2-3 independients teams around the world that can check and verify whatever is being claimed. cannot be reproduced and verified several times it won't be accepted as a closed case. Since claims, practically of any kind, regarding gods, spirits, angels, etc. cannot be reproduced and verified, at least not so far, under controlled conditions, religions cannot still claim they proved something in the same way scientist can, sometimes, claim the same thing. Tonio === Subject: Re: The scientist as idiot <8qtbq4t48d1ml6893cigqvb1asjce14f0s@4ax.com> <82cdd$49a6abae$82a1e228$12260@news1.tudelft.nl> posting-account=suWj4AkAAADE1IvGmj55Nmq3f98qb17e .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; InfoPath.2; .NET CLR 3.5.21022; Tablet PC 2.0; .NET CLR 1.1.4322),gzip(gfe),gzip(gfe) > ............................. snip ................................ Not really. æWhat would be (and is) puzzling is when a scientist > has beliefs which are plainly contradicted by empirical evidence, > such as that human bodies can be literally resurrected. æThere is > nothing very puzzling (to me, at least) in a scientist having > beliefs that cannot be empirically tested, i.e. are not scientific, > but may be meaningful (and even true). But empirical evidence is a big problem in itself. How can I ever check > that someone at CERN is speaking the truth, when he says that he/she has > the æoutcome of an expensive experiment which is detecting gravity waves > is not a fake ? How can I be sure that people have really landed on the > moon ? How can I be certain that the collapse of the buildings at ground > zero is not the result of a conspiracy ? The answer is that I can't ! I just have to believe that other people can be trusted. But can they > be trusted ? *************************************************************************** \ ***** A pity you stopped up there: can you trust the people that said they > were your parents were ACTUALLY your parents? Can you trust the DNA > test, and the lab, that could answer that question? Can you trust > books and people that said they knew someone called Hitler and ruled > Germany some 70 years ago? Can you trust , and this time it is only > books, that there was someone called > Napoleon?.......etc......etc......etc Yes, some trust is called for in almost any realm of our lives, and > wrt science there's a rather standard procedure (just as there is with > mathematics as well): whenever someone stands up and claims something, > there must be at least 2-3 independients teams around the world that > can check and verify whatever is being claimed. You'll notice that in a court of law, where presumably there are > higher stakes for the parties involved, there is no requirement for > 2-3 independent investigators to check and verify whatever is claimed. > Furthermore, it is never the expectation that the jury or judges would > be able to go to the scene of the crime and reproduce the gathering of > the evidence for themselves. And yet, the level of certainty in the > determination of truth is at least as high as it is in scientific > investigation. How do you account for that? > ****************************************************************************\ ***** 1) First, I do not believe that a court of law has a higher degree of certainty nor do I trust it better than scientists....all this, generally speaking' 2) Second, I don't think legal requirements have anything to do with scientific ones, and even in that case: I think, and I certainly do hope, that courts of love decide something NOT based on 1 or two eyewitnesses ONLY, though here we get into the witnesses' reliability. 3)I cannot decide whether the level of certainty of determinatiomn of truth, whatever that may be, in in a court of law is at least as high as in scientific investigation. Do you have any data to support this affirmation of yours? I honestly doubt it greatly. 4) The jury system seems to me to be pretty flawed, and I do distrust it. I'd rather have honest, serious judges decide on these things, and when cases are more serious I'd put 3 or more judges to work. Tonio === Subject: Re: The scientist as idiot <8qtbq4t48d1ml6893cigqvb1asjce14f0s@4ax.com> <82cdd$49a6abae$82a1e228$12260@news1.tudelft.nl> posting-account=suWj4AkAAADE1IvGmj55Nmq3f98qb17e .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; InfoPath.2; .NET CLR 3.5.21022; Tablet PC 2.0; .NET CLR 1.1.4322),gzip(gfe),gzip(gfe) > ............................. snip ................................ Not really. æWhat would be (and is) puzzling is when a scientist > has beliefs which are plainly contradicted by empirical evidence, > such as that human bodies can be literally resurrected. æThere is > nothing very puzzling (to me, at least) in a scientist having > beliefs that cannot be empirically tested, i.e. are not scientific, > but may be meaningful (and even true). But empirical evidence is a big problem in itself. How can I ever check > that someone at CERN is speaking the truth, when he says that he/she has > the æoutcome of an expensive experiment which is detecting gravity waves > is not a fake ? How can I be sure that people have really landed on the > moon ? How can I be certain that the collapse of the buildings at ground > zero is not the result of a conspiracy ? The answer is that I can't ! I just have to believe that other people can be trusted. But can they > be trusted ? *************************************************************************** \ ***** A pity you stopped up there: can you trust the people that said they > were your parents were ACTUALLY your parents? Can you trust the DNA > test, and the lab, that could answer that question? Can you trust > books and people that said they knew someone called Hitler and ruled > Germany some 70 years ago? Can you trust , and this time it is only > books, that there was someone called > Napoleon?.......etc......etc......etc Yes, some trust is called for in almost any realm of our lives, and > wrt science there's a rather standard procedure (just as there is with > mathematics as well): whenever someone stands up and claims something, > there must be at least 2-3 independients teams around the world that > can check and verify whatever is being claimed. You'll notice that in a court of law, where presumably there are > higher stakes for the parties involved, there is no requirement for > 2-3 independent investigators to check and verify whatever is claimed. > Furthermore, it is never the expectation that the jury or judges would > be able to go to the scene of the crime and reproduce the gathering of > the evidence for themselves. And yet, the level of certainty in the > determination of truth is at least as high as it is in scientific > investigation. How do you account for that? *************************************************************************** \ ****** 1) First, I do not believe that a court of law has a higher degree of > certainty nor do I trust it better than scientists....all this, > generally speaking' 2) Second, I don't think legal requirements have anything to do with > scientific ones, and even in that case: I think, and I certainly do > hope, that courts of love decide something NOT based on 1 or two > eyewitnesses ONLY, though here we get into the witnesses' reliability. 3)I cannot decide whether the level of certainty of determinatiomn of > truth, whatever that may be, in in a court of law is at least as high > as in scientific investigation. Do you have any data to support this > affirmation of yours? I honestly doubt it greatly. 4) The jury system seems to me to be pretty flawed, and I do distrust > it. I'd rather have honest, serious judges decide on these things, and > when cases are more serious I'd put 3 or more judges to work. Tonio Science and laws are *practices*, and their methodologies are > determined by their practical success. If there is obvious room for > fundamental improvement in methodologies (other than tweaks), then > neither the trial-by-jury system nor the scientific method would not > have endured for the centuries that they have. In the courtroom, the trial-by-jury methodology is used systematically > to decide whether to remove someone's liberty or their life. No such > risk is ever incurred in the scientific method, except by incidental > accident. Therefore, the methodology has to be sufficiently reliable > to support the seriousness of the consequence. By general consensus, > we have agreed that trial-by-jury and the rules of evidence comprise > *the most reliable method* we have on hand for arriving at the truth > in this case. It's worthwhile doing the deeper dive to explore why > this is so, and why the scientific method would be less successful in > *actual practice*. But as a side note, it's easy to notice in your response that you are > expressing your level of trust -- you just trust scientists more than > you trust lawyers, judges, police investigators, and jurors. But > that's a *personal* choice on your part. > *************************************************************************** Of course those are personal choices on my part: are there any other kind of choices? And of course I do trust scientists more than lawyers, police and jurors: do you know anyone that thinks otherwise? And if you do, do you knwo why is that so? And about jury: many countries (I think most of them)don't have that system, which seems to be somehow connected to english law, so it is used ONLY in some countries. As far as I am aware, the scientific method is used ANYWHERE where science is done. Tonio === Subject: Re: The scientist as idiot > Well, I dispute that, too. Many here consider medicine to be a > science. But Asian medicine is much different than Western medicine, > and includes many aspects that are different from (and in fact > repellent to) Western scientific standards. Next you'll be telling us that Western science is a male construct, designed to oppress women. Medicine uses science (physics, chemistry, biology), and is scientific to a considerable extent, but clearly is not a science, any more than engineering is a science. There are many medical practices around the world, but they are by no means all equivalent - some work much better than others, in the scientific sense that outcomes can be predicted. This should be no surprise, after all Western medical practice of 100 years ago was far less effective than the modern version. I expect medicine in all countries to converge on a common set of methods, and it will be much more like the Western version than the Asian version, because it will be based on Western science, which is not really Western - physics in China is not fundamentally different from physics in Denmark. This should not be a surprise either, since experiments conducted in both places will yield the same results. === Subject: Re: The scientist as idiot ............................. snip ................................ > Not really. What would be (and is) puzzling is when a scientist > has beliefs which are plainly contradicted by empirical evidence, > such as that human bodies can be literally resurrected. There is > nothing very puzzling (to me, at least) in a scientist having > beliefs that cannot be empirically tested, i.e. are not scientific, > but may be meaningful (and even true). > But empirical evidence is a big problem in itself. How can I ever check > that someone at CERN is speaking the truth, when he says that he/she has > the outcome of an expensive experiment which is detecting gravity waves > is not a fake ? How can I be sure that people have really landed on the > moon ? How can I be certain that the collapse of the buildings at ground > zero is not the result of a conspiracy ? The answer is that I can't ! > I just have to believe that other people can be trusted. But _can_ they > be trusted ? > *************************************************************************** \ ***** > A pity you stopped up there: can you trust the people that said they > were your parents were ACTUALLY your parents? Can you trust the DNA > test, and the lab, that could answer that question? Can you trust > books and people that said they knew someone called Hitler and ruled > Germany some 70 years ago? Can you trust , and this time it is only > books, that there was someone called > Napoleon?.......etc......etc......etc > Yes, some trust is called for in almost any realm of our lives, and > wrt science there's a rather standard procedure (just as there is with > mathematics as well): whenever someone stands up and claims something, > there must be at least 2-3 independients teams around the world that > can check and verify whatever is being claimed. > You'll notice that in a court of law, where presumably there are > higher stakes for the parties involved, there is no requirement for > 2-3 independent investigators to check and verify whatever is claimed. > Furthermore, it is never the expectation that the jury or judges would > be able to go to the scene of the crime and reproduce the gathering of > the evidence for themselves. And yet, the level of certainty in the > determination of truth is at least as high as it is in scientific > investigation. How do you account for that? > *************************************************************************** \ ****** > 1) First, I do not believe that a court of law has a higher degree of > certainty nor do I trust it better than scientists....all this, > generally speaking' > 2) Second, I don't think legal requirements have anything to do with > scientific ones, and even in that case: I think, and I certainly do > hope, that courts of love decide something NOT based on 1 or two > eyewitnesses ONLY, though here we get into the witnesses' reliability. > 3)I cannot decide whether the level of certainty of determinatiomn of > truth, whatever that may be, in in a court of law is at least as high > as in scientific investigation. Do you have any data to support this > affirmation of yours? I honestly doubt it greatly. > 4) The jury system seems to me to be pretty flawed, and I do distrust > it. I'd rather have honest, serious judges decide on these things, and > when cases are more serious I'd put 3 or more judges to work. > Tonio > Science and laws are *practices*, and their methodologies are > determined by their practical success. If there is obvious room for > fundamental improvement in methodologies (other than tweaks), then > neither the trial-by-jury system nor the scientific method would not > have endured for the centuries that they have. In the courtroom, the trial-by-jury methodology is used systematically > to decide whether to remove someone's liberty or their life. No such > risk is ever incurred in the scientific method, except by incidental > accident. Therefore, the methodology has to be sufficiently reliable > to support the seriousness of the consequence. By general consensus, > we have agreed that trial-by-jury and the rules of evidence comprise > *the most reliable method* we have on hand for arriving at the truth > in this case. It's worthwhile doing the deeper dive to explore why > this is so, and why the scientific method would be less successful in > *actual practice*. But as a side note, it's easy to notice in your response that you are > expressing your level of trust -- you just trust scientists more than > you trust lawyers, judges, police investigators, and jurors. But > that's a *personal* choice on your part. *************************************************************************** Of course those are personal choices on my part: are there any other > kind of choices? > And of course I do trust scientists more than lawyers, police and > jurors: do you know anyone that thinks otherwise? And if you do, do > you knwo why is that so? Hey Tonio, I trust you! ;-) === Subject: Re: The scientist as idiot <8qtbq4t48d1ml6893cigqvb1asjce14f0s@4ax.com> posting-account=sKfmEQkAAAC8kI3Pv6_U_nt9sVsxZ_ou 1.1.4322),gzip(gfe),gzip(gfe) According to Popper a scientific hypothesis is one that can >in principle be tested and found to be false. To validate a >hypothesis scientists should be attempting to falisfy it rather >than looking for easy confirmation. This means that religous claims >about god are not scientific as they cannot in priniple be >falsofied. God is a supernatural entity that exists outside >time and space. It's not very obvious how an hypothesis >of this kind could be tested in principle, or in fact what it may >mean. Note that Popper is very careful to say that his falsification > criterion is not a necessary criterion for a proposition to be > meaningful, only for it to be scientific. æ(It's been a while > since I read about this stuff, but no doubt someone can provide > a reference.) So it is puzzling why scientits have religous beliefs. Not really. æWhat would be (and is) puzzling is when a scientist > has beliefs which are plainly contradicted by empirical evidence, > such as that human bodies can be literally resurrected. æThere is > nothing very puzzling (to me, at least) in a scientist having > beliefs that cannot be empirically tested, i.e. are not scientific, > but may be meaningful (and even true). There is a big difference between a meaningful statement and a meaningful statement about the physical world. murder is illegal, murder is wrong, murder is sinful. Collecting empirical evidence won't tell you whether murder is wrong or whether sqrt(2) is a rational number. But claiming that god exists is a statement about the physical world. A working scientist would surely be aware that empirical claims need empirical evidence to back them up and also that mathematical theorems can be proved rigorously using an approach that is not empirical. But what is then the basis for a belief that cannot be verified empirically or non-empirically ? Belief for belief's sake ? >They must be able to strictly compartmentalize the critical >evidence-based approach they adopt for their work from >the criteria, if any, they used for their personal beliefs. I get the impression that a lot of that kind of doublethink does > go on. -- > Angus Rodgers === Subject: Re: The scientist as idiot Some people want to believe that god has an existence outside of human thought processes, and then the issue of interaction with the physical world arises. On this point I agree with you about the inconsistency of certain kinds of religious belief and science. To the extent that a person believes in the existence of an entity that interacts with the physical world (creates it, for example), that person is showing the limits of his intelligence, in my view. Clearly we all have limits, and our motivations are based on emotions not logic. === Subject: Re: The scientist as idiot According to Popper a scientific hypothesis is one that can > in principle be tested and found to be false. To validate a > hypothesis scientists should be attempting to falisfy it rather > than looking for easy confirmation. This means that religous claims > about god are not scientific as they cannot in priniple be > falsofied. God is a supernatural entity that exists outside > time and space. It's not very obvious how an hypothesis > of this kind could be tested in principle, or in fact what it may > mean. > Note that Popper is very careful to say that his falsification > criterion is not a necessary criterion for a proposition to be > meaningful, only for it to be scientific. (It's been a while > since I read about this stuff, but no doubt someone can provide > a reference.) So it is puzzling why scientits have religous beliefs. > Not really. What would be (and is) puzzling is when a scientist > has beliefs which are plainly contradicted by empirical evidence, > such as that human bodies can be literally resurrected. There is > nothing very puzzling (to me, at least) in a scientist having > beliefs that cannot be empirically tested, i.e. are not scientific, > but may be meaningful (and even true). There is a big difference between a meaningful statement > and a meaningful statement about the physical world. > murder is illegal, murder is wrong, murder is sinful. > Collecting empirical evidence won't tell you whether > murder is wrong or whether sqrt(2) is a rational > number. But claiming that god exists is a statement > about the physical world. Hmmm. One can attach many meanings to god exists - or no meaning at all. The most meaningful interpretation, as far as I'm concerned, is that god exists in the same sense that science fiction exists. They are both creations of the human imagination. === Subject: Re: The scientist as idiot posting-account=fwSgtAkAAACFnX70ssKwbvm9_oCZVHrx Gecko/2009011913 Firefox/3.0.6,gzip(gfe),gzip(gfe) According to Popper a scientific hypothesis is one that can >in principle be tested and found to be false. To validate a >hypothesis scientists should be attempting to falisfy it rather >than looking for easy confirmation. This means that religous claims >about god are not scientific as they cannot in priniple be >falsofied. God is a supernatural entity that exists outside >time and space. It's not very obvious how an hypothesis >of this kind could be tested in principle, or in fact what it may >mean. Note that Popper is very careful to say that his falsification > criterion is not a necessary criterion for a proposition to be > meaningful, only for it to be scientific. æ(It's been a while > since I read about this stuff, but no doubt someone can provide > a reference.) So it is puzzling why scientits have religous beliefs. Not really. æWhat would be (and is) puzzling is when a scientist > has beliefs which are plainly contradicted by empirical evidence, > such as that human bodies can be literally resurrected. æThere is > nothing very puzzling (to me, at least) in a scientist having > beliefs that cannot be empirically tested, i.e. are not scientific, > but may be meaningful (and even true). There is a big difference between a meaningful statement > and a meaningful statement about the physical world. > murder is illegal, murder is wrong, murder is sinful. > Collecting empirical evidence won't tell you whether > murder is wrong or whether sqrt(2) is a rational > number. But claiming that god exists is a statement > about the physical world. I disagree. If you think the word exists is what determines things, > then consider also Beauty exists, traceless invisible souls exist, > my spouse's love exists or even the mathematical so-called 'Monster > Group' exists. Science deals with that which is subject to verification by > measurement. That which does not apply to this criterion is outside > the bounds of science. A working scientist would > surely be aware that empirical claims need empirical evidence > to back them up and also that mathematical theorems > can be proved rigorously using an approach that is not > empirical. But what is then the basis for a belief > that cannot be verified empirically or non-empirically ? > Belief for belief's sake ? This is an example of what I said in my other post. To be blunt, it is > math/science bigotry to say that ONLY knowledge obtained by > mathematical proof or empirical test is classifiable as reliable > knowledge, and remarkably, it is usually a non-scientist that makes > such a claim. And in fact, most cognitive scientists would disagree > with you. There are MANY kinds of knowledge, all obtained with > different methodologies and different complexions of certainty. > So where would religion fit into all of that -- and what implications or lack thereof does this have for the famous question of whether or not God exists that people have debated for hundreds or thousands of years? === Subject: Re: The scientist as idiot > But claiming that god exists is a statement about the physical > world. God exists may mean pretty much anything. It is not at all apparent it is a statement about the physical world without some further elucidation: just what sort of God is at issue? -- Aatu Koskensilta (aatu.koskensilta@uta.fi) Wovon man nicht sprechen kann, dar.9fber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: The scientist as idiot But claiming that god exists is a statement about the >physical world. God exists may mean pretty much anything. It is not at > all apparent it is a statement about the physical world > without some further elucidation: just what sort of God > is at issue? Perhaps God is an inaccessible cardinal. Then claiming God exists would have specific, known consequences, the consistency of the ZFC axioms, at least. I am unclear on how these consequences could be observed, though. However, in that case, God exists would not be a statement about the physical world. Apparently, God is not an inaccessible cardinal. Unfortunate. Jim Burns === Subject: Re: The scientist as idiot <874oyhfqb5.fsf@alatheia.dsl.inet.fi> <49A6B5C3.4050206@osu.edu> posting-account=d-ESTAkAAAAG0l03yI1WJgsTVXx4ebeJ Gecko/20080829 Firefox/2.0.0.17,gzip(gfe),gzip(gfe) > Perhaps God is an inaccessible cardinal. Then claiming > God exists would have specific, known consequences, > the consistency of the ZFC axioms, at least. I am > unclear on how these consequences could be observed, > though. If one is going to play games like this, then you have to be very careful with your definitions and what you mean. For example, what do you mean by exists. At first glance it seems so simple like asking if a tree or a rock exists. That is one interpretation. But one can also ask does the sqrt(2) exist or how about sqrt(-1)? Clearly all math exists which means has an existence as a mental construct. The question of whether mental constructs are things which is to say have some manner of material existence (perhaps extra-dimensional) has not been answered, although there are hints that thoughts are things. On the other hand, it's very clear that the self-consistent mental constructs known as mathematics do not have a material existence in the usual 3D space that we perceive although a great many scientists promote the fiction that they do. Thus it's clear that to ask about the existence of God one must be VERY clear not only about what kind of existence we are talking about, but also just what kind of hypothesis we are using as a model for God. Many populist concepts (old man with a long beard in the sky) are not any more viable than the flat earth theory. More reasonable definitions (The entire universe with it's laws as a sentient super-being) still leaves open many questions including the material one. One can however, narrow the search to more reasonable parameters. We can suggest that if the Universe is a sentient being it not only knows everything (is self-aware) but also may allow communications from it's lesser parts to it (as we feel a pain in the ass) and from it to it's lesser parts (as we move our hands to play the piano). Now we are getting somewhere. We can ask is there any hint that such communications ever take place. And lo, we find that not only are current claims for it all over the place, but history is rife with such claims! However, note the very difficult science here. A person receiving a communication from God would naturally be very impressed with it and would be hard to convince that what they experienced had some other explanation. The rest of the people who were not privy to the communication will remain skeptical. Plus, there is the added factor of how do you set up the experiment to prove that the communication was from what we called God and not from some lesser being such as Gaia (the planet Earth also theorized to be sentient and self-aware) or the sentient Galaxy? These are not simple issues. And all religions appear to have something to add to the discussion based on what they view as their experience. But there are nevertheless no direct simple answers. > However, in that case, God exists would not be a > statement about the physical world. Apparently, > God is not an inaccessible cardinal. Unfortunate. So obviously the first order of business would be to make a determination as to whether God exists refers to a mental or physical existence (possibly both). However, the key point here is that there is NO issue as to whether or not one believes in anything. Belief is not proof and it is not science. A religious person in some experiment might say I experienced thus and so, and that is important data even if they only imagined they experienced it, but belief would in ordinary circumstances not be relevant. HOWEVER, there is a shoe in the soup here. Data and experiments have suggested that in matters of God and ESP etc. that the belief level of the participant produces great modulation upon the achievement of credible results. Even the belief levels of those observing but not participating in the experiments can do this! Well this is an unexpected fly in the scientific ointment, isn't it? Science is used to dealing with primitive and trivial situations where the experiment just lies there to be measured. Once the thing measured starts pushing back, things can get very dicey. Does this mean that science fails? Well hardly. Does it mean that the phenomena doesn't exist and should be just swept under the rug? That isn't science, is it? It just means that the scientist has to grow up and start tackling some SERIOUS questions rather than simply wallowing in childhood simplicity. That is what it means. Got it? === Subject: Re: The scientist as idiot posting-account=fwSgtAkAAACFnX70ssKwbvm9_oCZVHrx Gecko/2009011913 Firefox/3.0.6,gzip(gfe),gzip(gfe) > I recently read Freeman Dyson's The scientist as rebel. æThis is a > collection of book reviews. æI found some of the reviews to be > interesting, but what caught my attention was two things. æMr. Dyson > is pro-religion, and he believes in ESP. æWhile the first thing may be > excused as lots of otherwise intelligent people also fall into this > trap, I was very suprised, to say the least, that he believes in > ESP. Science is the search for truth, and I believe that there is no place > in science for gods and people with special powers. æWhat is your > opinion? Al Lal My opinion is that scientists are people and that their minds are > subject to the very same kind of foibles as everyone else, > *especially* on matters like religion or politics or the paranormal > that are not subject to normal scientific investigation methods. It's > also true that leading scientists are exceptionally curious and this > very often leads them into odd domains, precisely because they are > poorly understood. PD One would hope that scientists, with their powerful intellect, would > be able to seprate fact from fiction. æI believe in truth above all, > and know that there is no proof for god or the paranormal. æOrganized > religion is a poltical tool for controlling people, and should be > understood by all as such. > Let's just toss out politics. === Subject: Re: The scientist as idiot posting-account=fwSgtAkAAACFnX70ssKwbvm9_oCZVHrx Gecko/2009011913 Firefox/3.0.6,gzip(gfe),gzip(gfe) I recently read Freeman Dyson's The scientist as rebel. æThis is a > collection of book reviews. æI found some of the reviews to be > interesting, but what caught my attention was two things. æMr. Dyson > is pro-religion, and he believes in ESP. æWhile the first thing may be > excused as lots of otherwise intelligent people also fall into this > trap, I was very suprised, to say the least, that he believes in > ESP. Science is the search for truth, and I believe that there is no place > in science for gods and people with special powers. æWhat is your > opinion? Al Lal My opinion is that scientists are people and that their minds are > subject to the very same kind of foibles as everyone else, > *especially* on matters like religion or politics or the paranormal > that are not subject to normal scientific investigation methods. Hmm. Since religion is not subject to normal scientific investigation methods, then does this mean it is unreasonable/illogical to ask for a scientific proof or disproof that God exists? === Subject: Re: The scientist as idiot posting-account=fwSgtAkAAACFnX70ssKwbvm9_oCZVHrx Gecko/2009011913 Firefox/3.0.6,gzip(gfe),gzip(gfe) methods, then does this mean it is unreasonable/illogical to ask for > a scientific proof or disproof that God exists? Yes, I believe it is both unreasonable and unnecessary. And of course, > it is unlikely to succeed, because it is not one of those things that > science is good at. > So then how come there is so much focus on that in those atheism vs theism debates, anyway? I guess those debates aren't as logical then as their participants (on either side) like to claim, at least not when they start claiming scientific proofs/disproofs. And most, at least on the Internet, often quickly degenerate into silly personal attacks and other crap too. === Subject: Re: The scientist as idiot > methods, then does this mean it is unreasonable/illogical to ask for > a scientific proof or disproof that God exists? Yes, I believe it is both unreasonable and unnecessary. And of course, > it is unlikely to succeed, because it is not one of those things that > science is good at. > So then how come there is so much focus on that in those atheism vs > theism debates, anyway? I guess those debates aren't as logical > then > as their participants (on either side) like to claim, at least not > when they > start claiming scientific proofs/disproofs. And most, at least on > the > Internet, often quickly degenerate into silly personal attacks and > other > crap too. The explanation is simple: God is a mental virus. Those who are infected, fail to understand that they are infected, hence declare their infection: God exists. Those who are not infected, declare again their non-infection: God does not exist. Those who are infected, cannot understand they are infected, by virtue of the very existence of their own infection. Those who are uninfected, understand fully not only that they are not infected, but that those who are infected have an erroneous view of reality becuase of their infection, hence their argument for the non-existence of God is more valid than the argument for the existence of God, of the infected ones. Hence the result :-) -- Ioannis === Subject: Re: The scientist as idiot > methods, then does this mean it is unreasonable/illogical to ask for > a scientific proof or disproof that God exists? Yes, I believe it is both unreasonable and unnecessary. And of course, > it is unlikely to succeed, because it is not one of those things that > science is good at. So then how come there is so much focus on that in those atheism vs > theism debates, anyway? I guess those debates aren't as logical > then > as their participants (on either side) like to claim, at least not > when they > start claiming scientific proofs/disproofs. And most, at least on > the > Internet, often quickly degenerate into silly personal attacks and > other > crap too. The explanation is simple: God is a mental virus. Those who are infected, > fail to understand that they are infected, hence declare their infection: > God exists. Those who are not infected, declare again their non-infection: God does not > exist. Those who are infected, cannot understand they are infected, by virtue of the > very existence of their own infection. Those who are uninfected, understand fully not only that they are not > infected, but that those who are infected have an erroneous view of reality > becuase of their infection, hence their argument for the non-existence of > God is more valid than the argument for the existence of God, of the > infected ones. Hence the result :-) > -- > Ioannis Hence the position: I am healthy. You are sick, but you don't know > you are sick, because you are sick. The way that I know that I am the > healthy one and you are the sick one is that I am not like you. Who's the healthy one again? The one who uses the least amount of information, according to Occam's Razor. And that's in my view not the one who assumes what is uneccessary (God) to explain reality, since no direct evidence for such a being exists. > PD -- Ioannis === Subject: Re: The scientist as idiot posting-account=d-ESTAkAAAAG0l03yI1WJgsTVXx4ebeJ Gecko/20080829 Firefox/2.0.0.17,gzip(gfe),gzip(gfe) > Mr. Dyson > is pro-religion, and he believes in ESP. While the first thing may be > excused as lots of otherwise intelligent people also fall into this > trap, I was very suprised, to say the least, that he believes in > ESP. CLANG! CLANG! CLANG! Religion alarm going off! Just look at the WORD you are using: believes! That is the terminology of religion NOT science! A scientist would say, I have studied this or so and so has observed and reported that! When we start demanding that one believes in Maxwell science is out the window! No faith-based physics for me, Here's a nice hymn for your religion: There is but one God Allah, And Mohammed is his prophet! There is but one electric field And Maxwell is it's prophet! (J. Slepian) One question: Do you BELIEVE in the ONE ELECTRIC FIELD? Benj === Subject: Re: The scientist as idiot posting-account=d-ESTAkAAAAG0l03yI1WJgsTVXx4ebeJ Gecko/20080829 Firefox/2.0.0.17,gzip(gfe),gzip(gfe) > == That was REALLY funny: you're about to trash him/her because she/he > said he knows there is no proof for god or the paranormal, and you > begin by asking him how he knows that, and whether he...is god! Was SUPPOSED to be funny! :-) > And still his words may not be that exaggerated: under some acceptions > of the word proof, one could argue that there is no known proof of > existence of gods, which is different, imo, from saying (or > understanding) that somene said that there can't be/won't ever be such > a proof. Fact it that when one starts to talk about proof one immediately has to ask just what might be acceptable for such a term. Proof to one person might not be proof at all to another! Before one can discuss such a topic it is necessary to define terms. The term proof being one of the hardest of all to define. But in spite of that, the truth nonetheless remains that anyone attempting to assert a negative statement: This or that does not or worse could neverexist might as well go online right now and order their I am a moron T-shirt! Obviously to prove a negative, one must be totally aware of ALL positive items. This means that you must be omniscient! I do not feel I am too far out of line if I assert that I have never personally observed an omniscient person and I am aware of no reports of such persons as well. ...of course that does not mean I am asserting that no such person exists.... :-) However, there are various people about with great knowledge of very narrow restricted topics. === Subject: Re: The scientist as idiot > But in spite of that, the truth nonetheless remains that anyone > attempting to assert a negative statement: This or that does not or > worse could neverexist might as well go online right now and order > their I am a moron T-shirt! Obviously to prove a negative, one must > be totally aware of ALL positive items. This means that you must be > omniscient! There are no solutions in the positive integers to the equation x^3 + y^3 = z^3. Bart -- Cheerfully resisting change since 1959. === Subject: Re: The scientist as idiot posting-account=d-ESTAkAAAAG0l03yI1WJgsTVXx4ebeJ Gecko/20080829 Firefox/2.0.0.17,gzip(gfe),gzip(gfe) > The bottom line here is that my opinion is that the idiot is YOU. I hereby declare you an enemy to the cause of truth. Al Lal WELL! I didn't expect the Spanish Inquisition! === Subject: Re: The scientist as idiot <3f9bq4pok4k5g5shfe17dpc433ls07j875@4ax.com> posting-account=d-ESTAkAAAAG0l03yI1WJgsTVXx4ebeJ Gecko/20080829 Firefox/2.0.0.17,gzip(gfe),gzip(gfe) >[...] My opinion is that YOU are the moron here. [...] Don't be silly. I also think he's wrong, but there isn't the slightest > doubt that very many otherwise intelligent people fall into the same > trap. My guess is even that the majority of highly intelligent and > well-educated people (at least in the UK, where I live) would agree > with the OP. There is no reason to insult him. I'm sorry, but there IS a reason to insult him! His argument is that by virtue of special knowledge, intelligence, education, the achievements of the best persons in the (science) profession, and their political standing we should blindly accept HIS viewpoint with regard to certain alleged phenomena. And he implies we should do so without ANY proof whatsoever, which spits squarely in the eye of science. The added fact that he clearly has done not even the most cursory investigation of the various phenomena before spouting wild pronouncements that we are expected to swallow without any hesitancy stands a valid proof here in this global forum before the whole world that the entity who signs as Al Lal has no credible knowledge of science or of the scientific method. In short, he is promoting his OWN religion and not science. Furthermore, by asserting that any scientist who does not agree with his dogma is an idiot, he places himself BEYOND merely uniformed, uneducated, and ignorant, but well into propagandist for a given (left wing) political viewpoint. As such, he therefore opens himself up to the use propaganda techniques by others against his political position. I did choose to do this by using the term moron rather the one of the other common techniques such as reference to adjustments of a tinfoil helmet. There. Is that a scholarly enough justification of my post for you? > It is hard not to be > impressed (and intimidated!) by science, and crazy not to be grateful > for the wonderful intellectual freedom it brings (even the freedom to > criticise science itself). Making a religion out of science is one > step too far, but it is very understandable that people do it. Also, > even though I think scientism is insane and narcissistic, I'd far > rather be stuck in a lift with a believer in scientism than with > a religious believer whose mind is still stuck in the Middle Ages and > thinks God planted fossils in order to fool us and test our faith. Personally I'd rather be stuck in an elevator with an open-minded scientist, who makes a religion out of never rejecting ANY observation or erroneous ideas without a credible investigation. I'm talking here about persons who will NOT reject the myths of the Middle Ages or the Greeks or the Bible no matter HOW much they might have been corrupted over time or be poor descriptions due to the lack of general knowledge of people living during the age, without FIRST closely examining them for fragments of truth that nevertheless may have been inadvertently passed on to posterity in spite of the problems with the method of transmission. If you are not willing to seriously ask the question: Is there anything in this myth that just MIGHT be true?, then you are not really ready to study the subject. Hint: They found Troy just where it was supposed to be after dismissing it as a myth for so long. > We'd have a lot more to talk about (while waiting for an engineer > to fix the lift using sound scientific engineering practices) even Actually that refers the 20th century, a previous more civilized age. Today it would be some low paid grunt who really doesn't have an elevator clue who you hope won't kill you all by doing the wrong thing. Unless you can find a terrorist as the one in the movie Brazil who breaks the law by actually fixing things. ;-) > while disagreeing passionately at a fundamental level. I'd even be > a bit worried about anyone who hasn't at least gone through a phase > of believing that science is the only respectable form of knowledge. The problem is not that people think science is the ONLY respectable form of knowledge, but rather that science has been purposely kept quite limited to areas where it doesn't much pose a threat to traditional activities. (Politics, economics, religion, etc.) The lie that is taught is that a phenomena like ESP, because it is transitory and not very repeatable under current understanding, is somehow OUTSIDE the science fence. Note that this is equivalent to saying that lightning can't be studied and hence does not exist because it is not repeatable in a lab. > If it's a mistake (as I firmly believe it is), it's a mistake which > reflects well on the person who makes it. Given that nobody is very > sane, scientism is a form of insanity that has a lot to recommend it. > It at least suggests that the believer is capable of changing his > mind when faced with evidence and argument; and that is a good step > on the way to sanity. Well, exactly! Which is why the Al Lal post is so deserving of criticism. It is NOT scientism. It is asking that views be accepted WITHOUT evidence! It is asking that one blindly accept the views of the speaker based solely upon the fact that the speaker himself is saying them. It's arrogance in the extreme that is wrapped in a tissue of what is purported to be science. It simply doesn't fly. It needs to be called out for what it is. I hope in some small way I started that process. ;-) > /He/ might well think he is stuck in the lift > with a raving idiot, but there is no need to reciprocate. :-) I do believe I sufficiently justified reciprocation above! Benj === Subject: Re: The scientist as idiot One would hope that scientists, with their powerful > intellect, would > be able to separate fact from fiction. Yes, one would hope so. However, sometimes it is a > task that is > extremely difficult, time-consuming and beyond the > powerful > intellect of the scientist. Of course that is no > reason not to try. I believe in truth above all, > and know that there is no proof for god or the > paranormal. You know that there is no such? Please explain > just HOW you know > this? Are you God? Do you have ESP? Ah, I see. You > are an idiot! > Anyway, science does not BEGIN with proof it begins > with > investigations and hints one observes about how > things may work. > As science investigates only then does the actual > proof start to > emerge. You on the other hand, seem to wish to ignore > science and go > straight to religion by making an assertion of > truth without any > proof on your part. And like all religion there may > or may not be a > bit of truth in your assertions. The key is to > investigate them and > find the facts. Without such an investigation, it's > all taken on > faith. I see no reason to take any of your statements > on faith. Well, you right-wing twat. you misunderstand (or, maybe like your idol, you are misunderestimating) the claim being made: that claims are being made for which there is no proof. Until proof can be produced, we have to treat this in a skeptical way. Once reliable evidence is available, it will then be believed. It seems the most reasonable way of going about figuring out. Or, do you have a better method? You almost seem to be arguing that if something cannot be proven, then it is true. Should we then believe in _anything_ just because it cannot be proven? Organized > religion is a poltical tool for controlling people, > and should be > understood by all as such. Ah, yes, Mr. Pinko, the opiate of the people! Isn't > that your > dogma? Hey, a knife can be used to kill or to heal, > but that does not > reflect upon the knife! If you truly believed in > science and a > search for truth you'd get up off your ignorant fat > ass and start > looking into the dark corners of phenomena that > science can't explain. At this moment it cannot explain. It does not mean that it is not _explainable_ by science. And you are missing another distinction: the OP is criticizing here _Organized Religion_, and not religion itself. It is _religion_ that was described as the opium of the people, not organized religion. Whether you like it or not, religion has provided (still does today) a socially-justified excuse for all sorts of abuses. True that atheists have also commited abuses, but not in the name of atheism, so atheism cannot easily be used as an excuse for crimes as religion so often is. > THERE is where advancement of science will be found, > not in filling in > a few blank spaces in handbook tables! If instead of > looking there, > you simply start denying that such unexplained > phenomena can exist, > then clearly it is YOU promoting a religion of > faith-based physics > instead of true science. I take it you've never pondered the statement that > the highest science > (technology) is indistinguishable from magic? Leftist moron. > (Who is not familiar with the excellent research done > into the > paranormal in the USSR during the cold war.) Who's the moron here?. You have no basis for attributing any political motivation for his claims; it is just your sorry inability to face up to life without hiding behind your big man up there that makes you defensive. Maybe some day you will grow up to a level above the fairy-tale stories of the snake and the apple. But it doesn't seem likely. > You running late for your book burnin' meeting with dubya . And go pick up your white sheets for your KKK meeting. === Subject: Re: Pioneering British computer developments posting-account=rsfFvAoAAADqHCqmunbjtmaKjfuP-pwP Mozilla/4.0 (compatible; MSIE 6.0; Windows NT 5.1; SV1) ; SLCC1; .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; InfoPath.1; .NET \ CLR 1.1.4322),gzip(gfe),gzip(gfe) But if truth be told the computer has had a long and honourable history > that stretches back to the closing years of the World War II. > And, say conservations and computer history enthusiasts, Britain played > a big part in the development of the modern computer. At Bletchley Park forerunners of modern computers were built to help the > Allies crack German codes. > Colossus was crucial for D-Day operations Conceived, designed and built by Tommy Flowers, Allen Coombs and Max > Newman, the first Colossus was working in 1943 - three years ahead of > the rival pioneering American machine known as Eniac. The breaking of Nazi Germany codes with large computing machines began in > Poland...but it was not possible to get the computers out of Poland so they > had to be duplicated or perhaps next generation computers were developed. Where the hell did you get that fairy tail? The Poles at that time > still had horses when teh Germans had tanks. Hardy. Tanks for the memory, they might have said. Must admit, never heard about Poles code-cracking, anyway - never mind building computers! Bletchley certainly had a good go at Enigma, yet most of the stuff I have read suggests the answers were leaked to them - or to somebody! On the other hand, the Enigma machine was far too transpositional to be decoded in time for relevant action: they changed codes all the time, we are told! Anyway, when you depend on thousands of tubes, switches, a percentage failing constantly, how can you be sure of anything at all? -- foolsrushin. === Subject: Re: web function > does there exist a function which looks like this http://www.arago.utwente.nl/focusonline/img_dump/79/cosmic_web.jpg more specificly has such a function been studied before ? let f(x) be a taylor series expanded at some finite complex point. let the picture denote a part of the complex plane. let black denote divergeance. and all other colors convergeance. let f(x) have the property that for every y where f(y) converges , > there are at most 3 angles (y origin) where it is connected to the > other converging points. thus f(x) 'connects at most 3 curves' everywhere. and all converging points are connected to all other converging > points. ( i dont have the precise terminology for this in english , sorry ) i call that a ' web function ' and if its given by a taylor series a > ' taylor web function ' i included a pic to make clearer what i mean. examples of ' taylor web functions ' ( like the picture ) are ?? does this relate to fractals ? tommy1729 ps : in analogue to real-analytic , the set of reals is replaced by > the domain of convergeance , the 'web' , so i guess we could say f(x) > has to be ' web-analytic ' :) on the other hand , maybe not because of cauchy-riemann ? There has been some success modeling this galactic distribution by a fractal \ disttribution of D ~= 1.2 which is in good agreement with observation. There's not too much detail freely available on the web unless you have (paid) access to Physical Review Letters. http://prola.aps.org/abstract/PRL/v58/i26/p2818_1 === Subject: Re: web function posting-account=McZ3aQkAAADz6LV-boDe1LcriRhf3lj3 Gecko/2009011913 Firefox/2.0.0.9;MEGAUPLOAD 1.0,gzip(gfe),gzip(gfe) > does there exist a function which looks like this http://www.arago.utwente.nl/focusonline/img dump/79/cosmic web.jpg more specificly has such a function been studied before ? let f(x) be a taylor series expanded at some finite complex point. let the picture denote a part of the complex plane. let black denote divergeance. and all other colors convergeance. let f(x) have the property that for every y where f(y) converges , there are at most 3 angles (y origin) where it is connected to the other \ converging points. thus f(x) 'connects at most 3 curves' everywhere. and all converging points are connected to all other converging points. ( i dont have the precise terminology for this in english , sorry ) i call that a ' web function ' and if its given by a taylor series a ' taylor web function ' i included a pic to make clearer what i mean. examples of ' taylor web functions ' ( like the picture ) are ?? does this relate to fractals ? > tommy1729 ps : in analogue to real-analytic , the set of reals is replaced by the domain of convergeance , the 'web' , so i guess we could say f(x) has to be \ ' web-analytic ' :) on the other hand , maybe not because of cauchy-riemann ? Connectivity of Internet is wery similar to this picture, similarly webs of citations. kunzmilan === Subject: Definition of a bijection posting-account=5cC6XAoAAADEpYLKdk1mwUjYYKkDEgzo AppleWebKit/525.19 (KHTML, like Gecko) Chrome/1.0.154.48 Safari/525.19,gzip(gfe),gzip(gfe) Is the following definition of a bijection sufficient? I like it because it's concise, but it may not be correct. Let X and Y be sets. A function f: X->Y is called a BIJECTION if the preimage of every element of Y with respect to f has cardinality exactly one. Trevor === Subject: Re: Definition of a bijection posting-account=5cC6XAoAAADEpYLKdk1mwUjYYKkDEgzo AppleWebKit/525.19 (KHTML, like Gecko) Chrome/1.0.154.48 Safari/525.19,gzip(gfe),gzip(gfe) To avoid any potential debate about whether we need BIJECTION to define CARDINALITY, let me replace that definition with a new one: Let X and Y be sets. A function f: X->Y is called a BIJECTION if the preimage of every element of Y with respect to f contains exactly one element. Trevor === Subject: Re: Definition of a bijection > To avoid any potential debate about whether we need BIJECTION to > define CARDINALITY, let me replace that definition with a new one: > Let X and Y be sets. A function f: X->Y is called a BIJECTION if the > preimage of every element of Y with respect to f contains exactly one > element. Yes. Moreover, we can say that f is an injection if the preimage of each element of Y has at most one element, and f is a surjection if the preimage of each element of Y has at least one element. -- Dave Seaman Third Circuit ignores precedent in Mumia Abu-Jamal ruling. === Subject: Re: Definition of a bijection posting-account=5cC6XAoAAADEpYLKdk1mwUjYYKkDEgzo AppleWebKit/525.19 (KHTML, like Gecko) Chrome/1.0.154.48 Safari/525.19,gzip(gfe),gzip(gfe) > Yes. æMoreover, we can say that f is an injection if the preimage of each > element of Y has at most one element, and f is a surjection if the > preimage of each element of Y has at least one element. -Trevor === Subject: Re: JSH: When I say proof > What you need is an efficient algorithm to FIND the > factorization. So far you have fumbled around with variations > of the stupid idea that, given two different functions of the > variable v, you can use calculus to find one value of v which > minimizes both. That is trivially false. What is your problem > with understanding this??? I'm sorry to have to tell you Marcus, but you will never be able to understand his problem. The very thing that makes your brain able to understand mathematics makes it unable to understand James's thinking. === Subject: Re: JSH: When I say proof posting-account=wVv_VwoAAAAVTfUuyxLzug5SzYWCgHj1 Gecko/20081217 Firefox/2.0.0.20,gzip(gfe),gzip(gfe) > What you need is an efficient algorithm to FIND the > factorization. So far you have fumbled around with variations > of the stupid idea that, given two different functions of the > variable v, you can use calculus to find one value of v which > minimizes both. That is trivially false. What is your problem > with understanding this??? I'm sorry to have to tell you Marcus, but you will never be able to > understand his problem. The very thing that makes your brain able to > understand mathematics makes it unable to understand James's thinking. (Sniff!) Must you break this news to me so bluntly? (Sob.) We both just need to learn calculus apparently. Anyway for now JSH, sensing somehow that his intense marketing efforts for this latest idea have not gone over well, is crawling back into sullen silence. Yet once again it is reasonable to ask: has he permanently lost touch with reality? Aren't we overdue for (1) prolonged whining about how we destroyed a math journal, (2) his stunning but unappreciated breakthrough in counting primes and solving the Riemann Hypothesis, (3) his insight that Wiles used 'cum hoc, ergo propter hoc', (4) his own ironclad proof of FLT, (5) his brilliant method of protecting CD piracy, or (6) his sweeping and completely rigorous definition of proof ... so many accomplishments. Wow is all I can say. Wow. Bow-wow-wow. Marcus. === Subject: Re: JSH: When I say proof posting-account=fwSgtAkAAACFnX70ssKwbvm9_oCZVHrx Gecko/2009011913 Firefox/3.0.6,gzip(gfe),gzip(gfe) >No, again, the central claim here is that if æ æ æf(v) = abs(r(v) - t(v)) and æ æ æg(v) = abs(r(v) + t(v)) then you can use calculus to find a value v 0 which >minimizes BOTH functions. æYou need this because, by >your logic, you want both f(v) < D and g(v) < D. Maybe James could try a different approach. æFirst calculate the range > of values of v for which f(v) < D. æThen calculate the minimum value > of g(v) within that range. æIf the minimum is less than D his method > has found a factor, if not then no factor has been found. That's better than the other poster's claims as it IS possible to find > a point where abs(r(v) + t(v)) and abs(r(v) - t(v)) are both less than > D with a rational v, when D is a composite which is the central issue. æAs others here have already shown, these two functions >have their minima at different values of v. æAnd of course >this is what you would expect in general. Agreed. And irrelevant even if true as I said in a previous reply. But at least you're on a better track than the other poster! It's not hard math people. æThese functions graph as quadratics. Maybe that would help some of you? Just graph the damn things and quit arguing over trivialities. Why don't you graph them? I'm still not clear on how exactly > your method works. Only you're clear on that so you are in the best > position to do it. Graphing would help people who are confused. æI'm not. And the functions have been found and been proven not to work by > the Quadratic formula. What exactly is your problem with that? You It's not true. æAnd it actually trivially can't be true. All I did was find that for every rational solution to: x^2 - Dy^2 = 1 you have rational solutions for an ellipse or Pythagorean triples > with (D-1)j^2 + (j+/-1)^2 = (x+y)^2 where æj = ((x+Dy) -/+1)/D. So since there ARE rational solutions to y at a value you claim the > quadratic formula proves there are none, you are wrong. > But how do you use these equations up here to factor D? And what are and how do you use those functions r, s, t? > still keep operating under this deluded idea that your still have an > intact hammer despite it having shattered into a thousand pieces! Except the mathematics hasn't changed and I've been explaining it the > same for days. Posters like yourself just make false statements, as if it matters. James Harris === Subject: Re: fixed point set theory posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 Gecko/2009011913 Firefox/3.0.6,gzip(gfe),gzip(gfe) > it clearly counts till w , w times. as you can see on the picture ( = proof ) A picture here is not a proof. But anyway, okay, we do understand the ORDINAL w^2 = w*w where '^' and '*' are for ordinal operations. (So I'd revise my own previous comments to reflect that.) > since card(w^2) = card(w) w^2 is thus simply a way of putting w into subsets. Whatever a way of putting w into subsets means. It is not the case that w is PARTITIONED in some way you've given. Anyway, you haven't proven 2^aleph w = aleph w where '^' stands for CARDINAL exponentiation, if that is what you mean by your claim: > 2^aleph 'oo' = aleph 'oo' MoeBlee === Subject: Re: fixed point set theory posting-account=9QOSvAoAAACEOWJVSDuswW7dB_0wApQO Gecko/2009020410 Fedora/3.0.6-1.fc10 Firefox/3.0.6,gzip(gfe),gzip(gfe) (squid/2.7.STABLE3) since only set theory can hold the attention of the > majority at sci.math ill post about it again , > despite being more intrested in other subjects. you know or you should know that ' fixed points ' > are fundamental in many branches of math. however standard set theory with its large cardinals > is inconsistant with the concept of ' fixed points '. What a strange thing to say. no , as the OP explained ' fixed points ' are fundamental to math. it is thus logical , despite rarely considered. PART I 1 2 3 4 ... 'oo'. 'oo' is a fixpoint of the successor function : x + 1 = S(x) = x If you say so. Exactly what set are you talking > about when you say oo, and exactly what is the > definition of S(x) for any set x? i already explained in the OP. 1 2 3 4 ... oo can only mean the set of the integers or the set of the AP - integers . no other interpretation is meaningfull. I don't expect you'll actually provide us with those > definitions, but until you do you haven't said > anything > meaningful. has only this 'oo' as a solution. one can argue 'oo' = aleph 0 æ( ordinary integers ) 'oo' = aleph 1 ( AP integers and the alike ) but that has no influence on the rest of this thread > , same conclusions follow. remember , only this 'oo' and not e.g. aleph 'oo' , aleph 34 , oo + 2 , 2 oo - 1 , > w^w^...^w since 'oo' is a fixpoint , and btw also otherwise > the set of integers 1 2 3 ... oo is not well defined. ------------- PART II http://upload.wikimedia.org/wikipedia/commons/8/83/Om > ega squared.png this w^2 can clearly be mapped onto w. thus w^2 = w*w = w apart from the way we count it. basicly w^2 is just the ways we count subsets of w. Now you don't have to give your definitions, because > the things you're talking about are standard gizmos. No, w^2 does not count the subsets of w. i didnt say that , i said w^2 is just the way we count subsets of w. i gave a link to a picture of w^2. it clearly counts till w , w times. as you can see on the picture ( = proof ) since card(w^2) = card(w) w^2 is thus simply a way of putting w into subsets. w^n = w all ordinals belong to a cardinal. it is thus clear that we cannot increase the > cardinality of an ordinal by operations that do would > not change the cardinality if applied to cardinality. ( symbolic w^2 = w <=> R^2 = R ) ---------------- PART III since PART I and PART II have shown that applying > the successor function and using ordinals cannot > increase cardinality beyond the way cardinal > functions do , we are only left with cardinal > functions. the simplest function that increases cardinality is > 2^x. in fact , up to isomorphism , its the only one > together with its repetitions. You just love to babble nonsense. and here you just love to get personal again , instead of rational. nonsense ? æok then , lets turn the tables again ! i said : the simplest function that increases cardinality is > 2^x. in fact , up to isomorphism , its the only one > together with its repetitions. give a counterexample !! with proof !! But that is not how it works... If you claim something, it is up to *you* to provide a proof of that claim. That others cannot (and, in many cases, have no wish to) come up with a counterexample to your claims does most certainly not count as support for the veracity of those claims, let alone it is a proof of that veracity. -- m === Subject: Re: fixed point set theory sha1:favGV7lW5Dq9dYM90vD9iycQv8s= > dont underestimate me . -- Scissors and string, scissors and string, When a man's single, he lives like a king. Needles and pins, needles and pins, When a man marries, his trouble begins. --- Mother Goose === Subject: Re: fixed point set theory posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 Gecko/2009011913 Firefox/3.0.6,gzip(gfe),gzip(gfe) > 1 2 3 4 ... 'oo'. 'oo' is a fixpoint of the successor function : x + 1 = S(x) = x has only this 'oo' as a solution. I take that by 'oo' you mean w (omega), the set of natural numbers? Dave Renfro mentioned this, but I'd like to mention it also: w (omega) is NOT a fixed point for the ORDINAL successor operation. Where 'x+1' stands for xu{x}, it is not the case that w+1 = w. But where '+' stands for CARDINAL addition, yes, it is the case that w +1 = w, though this is more usually notated as aleph_0 + 1 = aleph_0. And in that context it is NOT the case that w is the only x such that x = x+1, contrary to your claim only this 'oo' as a solution. When you see the symbol '+' you have to be careful to distinguish what it stands for in a given context. Ideally, in different contexts you should regard '+' as 'ordinal_+' and in other contexts as 'cardinal_ +', and yet other operations in other contexts, as if they are different symbols even. > http://upload.wikimedia.org/wikipedia/commons/8/83/Omega_squared.png this w^2 can clearly be mapped onto w. You linked to a graphic of some kind. If by 'w^2' by mean wXw (the Cartesian product), then, yes, w^2 is 1-1 with w. > thus w^2 = w*w = w Okay, if by '*' you mean cardinal multiplication, the correct formulation is: card(w^2) = w*w = w or, more usually notated: card(w^2) = aleph_0*aleph_0 = aleph_0 (I'll not mention the notational matter of 'w' and 'aleph_0' from now on.) > basicly w^2 is just the ways we count subsets of w. No, it's not. It's the cardinality of the set of ordered pairs of members of w. That is different from the cardinality of the set of subsets of w. > w^n = w No, it's this: card(w^n) = w (where n is a natural number greater than 0) > all ordinals belong to a cardinal. Okay, for every ordinal b there exists a cardinal k such bek. Do we need choice for that? Hmm? > it is thus clear that we cannot increase the cardinality of an ordinal by \ operations that do would not change the cardinality if applied to cardinality. I don't know what you mean by that. > ( symbolic w^2 = w <=> R^2 = R ) Maybe you mean card(w^2) = w <-> card(R^2) = card(R) (where 'R' stands for the set of real numbers). Anyway the above is a theorem of ZFC, since both sides of the biconditional are themselves theorems. > the simplest function that increases cardinality is 2^x. in fact , up to isomorphism , its the only one together with its repetitions. I don't know about that. Would you mention the particular theorem you have in mind? > 2^ aleph_x = aleph_(x + 1) That is the generalized continuum hypothesis. > for finite n this is a simple structure. I don't know what you mean by that. > but now follows the intresting part : conjecture : aleph_'oo' is the largest cardinal. PROOF : ( with the above in mind ) 2^aleph_x = aleph_x+1 If you take the continuum hypothesis as an axiom. > 2^aleph_x = aleph_S(x) where S(x) is again the successor function In this case, it would be the ORDINAL successor. > since 'oo' was the fixed point of S(x) = x we get No, 'w' is NOT a fixed point for ORDINAL sucessor. > 2^aleph_'oo' = aleph_'oo' No. That doesn't follow from ANYTHING here, because it is NOT the case that S(x) = x where 'S' stands for ORDINAL successor. There's more in your post, but I'm going to stop at this point. MoeBlee === Subject: Re: Alternating Vertical and Horizontal Moves In Grid posting-account=ey-PjQoAAAD_Gi3gK0ZWKofsmviYOWhV Gecko/20081217 Firefox/2.0.0.20,gzip(gfe),gzip(gfe) Hi! This is an interesting problem, that I have never seen before. I will make the assumption that at any point a vertical move may be U or D, and a horizontal move may be L or R. In both the start-at-a-corner and general case, a(1) = 1, a(2) = 4, and a(3) = 6, for starters. Now I will focus only on the start-at-a-corner case, and it may as well be the upper left. Obviously n^2 is an upper bound for a(n), and if we try to establish some reasonable lower bounds, by considering a boustrophedonic traversal of the grid, there is a natural division into cases based on the value of n mod 4. In particular, I conjecture that the optimal algorithm for sufficiently large n = 1 mod 4 is to traverse the grid according to the following rules: 1) Start with down 1) Choose up over down whenever at least 4 more moves can be made as a result of doing so. 2) Alternate between choosing right n - 1 times in a row, then left n - 1 times in a row, then right, etc. To see what I mean by this, the 9 x 9 grid would be traversed like: DRURDRURDRURDRDRDLDLULULDLULDLDLDRDRURURDRDRURDRDLDLULDLULDLULDLU. With a little work this establishes: n mod 4 = 1 -> [ n^2 - 2n + 3 <= a(n) ] For the other cases some slight modifications of the algorithm above produce pretty tight lower bounds as well. In the above case, I would expect equality probably holds, but proving it is another story, it seems like it would be tricky. dc === Subject: Re: Alternating Vertical and Horizontal Moves In Grid >What is the sequence {a(k)} that is defined as follows? Start with an n-by-n grid. Move from square to adjacent square by >moving vertically then horizontally then vertically then horizontally, >following each horizontal move with a vertical move, and following >each vertical move with a horizontal move. >No square may be visited more than once. a(n) = the maximum number of squares visitable in this way on an n-by- >n grid. By brute force search, here are the values of a(n) for n = 1, ..., 8: 1, 4, 7, 14, 19, 32, 39, 58 >A variation: You must start in a corner square. Then the first 8 values are: 1, 4, 6, 14, 18, 32, 38, 58 Note that, so far, if n is even, starting in a corner is just as good a starting place as any other location, while if n is odd and n > 1, the value of a(n) when starting in a corner appears to be just 1 less than that for an optimum starting place. >(It does not matter if the first move is vertical or horizontal, as >far as the values of each a(n) are concerned, of course.) I am guessing that this sequence is really easy to calculate, maybe a >(n) is as simple as a quadratic polynomial in terms of n. Or is the >best way to calculate each a(n) is to do a brute-force search of >possible paths through the grid? The function a(n) is definitely not a polynomial in n. Clearly a(n) is positive and bounded above by n^2, hence, if a(n) was a polynomial in n, it would have to be at most of degree 2. But the calculated values of a(n) as posted above contradict that. So what is the nature of a(n)? Consider the sequence b(n) = n^2 - a(n) Clearly b(n) must nonnegative. I suspect it would not be too hard to prove that b(n) must be positive, for n > 2. We can ask whether the sequence b(n) is bounded above? I suspect not, but if, in fact, b(n) is bounded above, then I would guess that the sequence b(n) repeats. Other than brute force search, you might try for recursive relationships, or at least recursive bounds. quasi === Subject: Re: Alternating Vertical and Horizontal Moves In Grid What is the sequence {a(k)} that is defined as follows? Start with an n-by-n grid. Move from square to adjacent square by >moving vertically then horizontally then vertically then horizontally, >following each horizontal move with a vertical move, and following >each vertical move with a horizontal move. >No square may be visited more than once. a(n) = the maximum number of squares visitable in this way on an n-by- >n grid. By brute force search, here are the values of a(n) for n = 1, ..., 8: 1, 4, 7, 14, 19, 32, 39, 58 A variation: You must start in a corner square. Then the first 8 values are: 1, 4, 6, 14, 18, 32, 38, 58 Note that, so far, if n is even, starting in a corner is just as good >a starting place as any other location, while if n is odd and n > 1, >the value of a(n) when starting in a corner appears to be just 1 less >than that for an optimum starting place. (It does not matter if the first move is vertical or horizontal, as >far as the values of each a(n) are concerned, of course.) I am guessing that this sequence is really easy to calculate, maybe a >(n) is as simple as a quadratic polynomial in terms of n. Or is the >best way to calculate each a(n) is to do a brute-force search of >possible paths through the grid? The function a(n) is definitely not a polynomial in n. Clearly a(n) is >positive and bounded above by n^2, hence, if a(n) was a polynomial in >n, it would have to be at most of degree 2. But the calculated values >of a(n) as posted above contradict that. So what is the nature of a(n)? Consider the sequence b(n) = n^2 - a(n) Clearly b(n) must nonnegative. I suspect it would not be too hard to >prove that b(n) must be positive, for n > 2. We can ask whether the sequence b(n) is bounded above? I suspect not, >but if, in fact, b(n) is bounded above, then I would guess that the >sequence b(n) repeats. Other than brute force search, you might try for recursive >relationships, or at least recursive bounds. Ok, there is a simple pattern that I missed. When n is even, it appears that a(n) = n^2 - n + 2 quasi === Subject: Re: Alternating Vertical and Horizontal Moves In Grid >What is the sequence {a(k)} that is defined as follows? Start with an n-by-n grid. Move from square to adjacent square by >moving vertically then horizontally then vertically then horizontally, >following each horizontal move with a vertical move, and following >each vertical move with a horizontal move. >No square may be visited more than once. a(n) = the maximum number of squares visitable in this way on an n-by- >n grid. By brute force search, here are the values of a(n) for n = 1, ..., 8: 1, 4, 7, 14, 19, 32, 39, 58 A variation: You must start in a corner square. Then the first 8 values are: 1, 4, 6, 14, 18, 32, 38, 58 Note that, so far, if n is even, starting in a corner is just as good >a starting place as any other location, while if n is odd and n > 1, >the value of a(n) when starting in a corner appears to be just 1 less >than that for an optimum starting place. (It does not matter if the first move is vertical or horizontal, as >far as the values of each a(n) are concerned, of course.) I am guessing that this sequence is really easy to calculate, maybe a >(n) is as simple as a quadratic polynomial in terms of n. Or is the >best way to calculate each a(n) is to do a brute-force search of >possible paths through the grid? The function a(n) is definitely not a polynomial in n. Clearly a(n) is >positive and bounded above by n^2, hence, if a(n) was a polynomial in >n, it would have to be at most of degree 2. But the calculated values >of a(n) as posted above contradict that. So what is the nature of a(n)? Consider the sequence b(n) = n^2 - a(n) Clearly b(n) must nonnegative. I suspect it would not be too hard to >prove that b(n) must be positive, for n > 2. We can ask whether the sequence b(n) is bounded above? I suspect not, >but if, in fact, b(n) is bounded above, then I would guess that the >sequence b(n) repeats. Other than brute force search, you might try for recursive >relationships, or at least recursive bounds. Ok, there is a simple pattern that I missed. When n is even, it appears that a(n) = n^2 - n + 2 Moreover, it's fairly easy to show that, for even n, a(n) >= n^2 - n + 2 Assuming n is even, a valid path through n^2 - n + 2 vertices can be explicitly constructed as follows ... Start at (0,0). Go right, up, left, up. Repeat until reaching the top. Since n is even, the last 2 moves must have been right, up, ending at (1,n-1). Now go right, down, ending at (2,n-2). We have so far visited 2n + 1 vertices. Now recursively do the same in the n-2 by n-2 square with upper left corner vertex (2,n-2) and lower right corner vertex (n-1,1). The path described above adheres to the restrictions, and it's easily proved, by induction, that it has exactly n^2 - n + 2 vertices. quasi === Subject: Re: Alternating Vertical and Horizontal Moves In Grid What is the sequence {a(k)} that is defined as follows? Start with an n-by-n grid. Move from square to adjacent square by >moving vertically then horizontally then vertically then horizontally, >following each horizontal move with a vertical move, and following >each vertical move with a horizontal move. >No square may be visited more than once. a(n) = the maximum number of squares visitable in this way on an n-by- >n grid. By brute force search, here are the values of a(n) for n = 1, ..., 8: 1, 4, 7, 14, 19, 32, 39, 58 A variation: You must start in a corner square. Then the first 8 values are: 1, 4, 6, 14, 18, 32, 38, 58 Note that, so far, if n is even, starting in a corner is just as good >a starting place as any other location, while if n is odd and n > 1, >the value of a(n) when starting in a corner appears to be just 1 less >than that for an optimum starting place. (It does not matter if the first move is vertical or horizontal, as >far as the values of each a(n) are concerned, of course.) I am guessing that this sequence is really easy to calculate, maybe a >(n) is as simple as a quadratic polynomial in terms of n. Or is the >best way to calculate each a(n) is to do a brute-force search of >possible paths through the grid? The function a(n) is definitely not a polynomial in n. Clearly a(n) is >positive and bounded above by n^2, hence, if a(n) was a polynomial in >n, it would have to be at most of degree 2. But the calculated values >of a(n) as posted above contradict that. So what is the nature of a(n)? Consider the sequence b(n) = n^2 - a(n) Clearly b(n) must nonnegative. I suspect it would not be too hard to >prove that b(n) must be positive, for n > 2. We can ask whether the sequence b(n) is bounded above? I suspect not, >but if, in fact, b(n) is bounded above, then I would guess that the >sequence b(n) repeats. Other than brute force search, you might try for recursive >relationships, or at least recursive bounds. Ok, there is a simple pattern that I missed. When n is even, it appears that a(n) = n^2 - n + 2 Moreover, it's fairly easy to show that, for even n, a(n) >= n^2 - n + 2 Assuming n is even, a valid path through n^2 - n + 2 vertices can be >explicitly constructed as follows ... Start at (0,0). Go right, up, left, up. Repeat until reaching the top. Since n is even, the last 2 moves must have been right, up, ending at >(1,n-1). Now go right, down, ending at (2,n-2). We have so far visited 2n + 1 vertices. Now recursively do the same in the n-2 by n-2 square with upper left >corner vertex (2,n-2) and lower right corner vertex (n-1,1).| To clarify: At each successive lower level in the recursion, switch up and down. Also, when you finish the 2 x 2 level, stop. >The path described above adheres to the restrictions, and it's easily >proved, by induction, that it has exactly n^2 - n + 2 vertices. quasi === Subject: Re: Alternating Vertical and Horizontal Moves In Grid <4utbq4dhhldvts8o5hl3ld180tan2tbdmq@4ax.com> posting-account=ey-PjQoAAAD_Gi3gK0ZWKofsmviYOWhV Gecko/20081217 Firefox/2.0.0.20,gzip(gfe),gzip(gfe) What is the sequence {a(k)} that is defined as follows? Start with an n-by-n grid. Move from square to adjacent square by >moving vertically then horizontally then vertically then horizontally, >following each horizontal move with a vertical move, and following >each vertical move with a horizontal move. >No square may be visited more than once. a(n) = the maximum number of squares visitable in this way on an n-by- >n grid. By brute force search, here are the values of a(n) for n = 1, ..., 8: 1, 4, 7, 14, 19, 32, 39, 58 A variation: You must start in a corner square. Then the first 8 values are: 1, 4, 6, 14, 18, 32, 38, 58 Note that, so far, if n is even, starting in a corner is just as good >a starting place as any other location, while if n is odd and n > 1, >the value of a(n) when starting in a corner appears to be just 1 less >than that for an optimum starting place. (It does not matter if the first move is vertical or horizontal, as >far as the values of each a(n) are concerned, of course.) I am guessing that this sequence is really easy to calculate, maybe a >(n) is as simple as a quadratic polynomial in terms of n. Or is the >best way to calculate each a(n) is to do a brute-force search of >possible paths through the grid? The function a(n) is definitely not a polynomial in n. Clearly a(n) is >positive and bounded above by n^2, hence, if a(n) was a polynomial in >n, it would have to be at most of degree 2. But the calculated values >of a(n) as posted above contradict that. So what is the nature of a(n)? Consider the sequence b(n) = n^2 - a(n) Clearly b(n) must nonnegative. I suspect it would not be too hard to >prove that b(n) must be positive, for n > 2. We can ask whether the sequence b(n) is bounded above? I suspect not, >but if, in fact, b(n) is bounded above, then I would guess that the >sequence b(n) repeats. Other than brute force search, you might try for recursive >relationships, or at least recursive bounds. Ok, there is a simple pattern that I missed. When n is even, it appears that a(n) = n^2 - n + 2 Moreover, it's fairly easy to show that, for even n, a(n) >= n^2 - n + 2 Assuming n is even, a valid path through n^2 - n + 2 vertices can be >explicitly constructed as follows ... Start at (0,0). Go right, up, left, up. Repeat until reaching the top. Since n is even, the last 2 moves must have been right, up, ending at >(1,n-1). Now go right, down, ending at (2,n-2). We have so far visited 2n + 1 vertices. Now recursively do the same in the n-2 by n-2 square with upper left >corner vertex (2,n-2) and lower right corner vertex (n-1,1).| To clarify: At each successive lower level in the recursion, switch up and > down. Also, when you finish the 2 x 2 level, stop. The path described above adheres to the restrictions, and it's easily >proved, by induction, that it has exactly n^2 - n + 2 vertices. quasi this seems legitimate to me. quasi, by the sounds of it, you have some kind of a computer program capable of a brute force test? can you test my conjecture above for n = 9, 13? i'm thinking we are going tol end up here with a quadratic answer, but different quadratics depending on the value of n mod 4. === Subject: Re: Alternating Vertical and Horizontal Moves In Grid What is the sequence {a(k)} that is defined as follows? Start with an n-by-n grid. Move from square to adjacent square by >moving vertically then horizontally then vertically then horizontally, >following each horizontal move with a vertical move, and following >each vertical move with a horizontal move. >No square may be visited more than once. a(n) = the maximum number of squares visitable in this way on an n-by- >n grid. By brute force search, here are the values of a(n) for n = 1, ..., 8: 1, 4, 7, 14, 19, 32, 39, 58 A variation: You must start in a corner square. Then the first 8 values are: 1, 4, 6, 14, 18, 32, 38, 58 Note that, so far, if n is even, starting in a corner is just as good >a starting place as any other location, while if n is odd and n > 1, >the value of a(n) when starting in a corner appears to be just 1 less >than that for an optimum starting place. (It does not matter if the first move is vertical or horizontal, as >far as the values of each a(n) are concerned, of course.) I am guessing that this sequence is really easy to calculate, maybe a >(n) is as simple as a quadratic polynomial in terms of n. Or is the >best way to calculate each a(n) is to do a brute-force search of >possible paths through the grid? The function a(n) is definitely not a polynomial in n. Clearly a(n) is >positive and bounded above by n^2, hence, if a(n) was a polynomial in >n, it would have to be at most of degree 2. But the calculated values >of a(n) as posted above contradict that. So what is the nature of a(n)? Consider the sequence b(n) = n^2 - a(n) Clearly b(n) must nonnegative. I suspect it would not be too hard to >prove that b(n) must be positive, for n > 2. We can ask whether the sequence b(n) is bounded above? I suspect not, >but if, in fact, b(n) is bounded above, then I would guess that the >sequence b(n) repeats. Other than brute force search, you might try for recursive >relationships, or at least recursive bounds. Ok, there is a simple pattern that I missed. When n is even, it appears that a(n) = n^2 - n + 2 Moreover, it's fairly easy to show that, for even n, a(n) >= n^2 - n + 2 Assuming n is even, a valid path through n^2 - n + 2 vertices can be >explicitly constructed as follows ... Start at (0,0). Go right, up, left, up. Repeat until reaching the top. Since n is even, the last 2 moves must have been right, up, ending at >(1,n-1). Now go right, down, ending at (2,n-2). We have so far visited 2n + 1 vertices. Now recursively do the same in the n-2 by n-2 square with upper left >corner vertex (2,n-2) and lower right corner vertex (n-1,1).| To clarify: At each successive lower level in the recursion, switch up and > down. Also, when you finish the 2 x 2 level, stop. The path described above adheres to the restrictions, and it's easily >proved, by induction, that it has exactly n^2 - n + 2 vertices. quasi this seems legitimate to me. quasi, by the sounds of it, you have some >kind of a computer program capable of a brute force test? can you test >my conjecture above for n = 9, 13? i'm thinking we are going tol end >up here with a quadratic answer, but different quadratics depending on >the value of n mod 4. Yes, that seems very plausible -- 3 quadratics, one for even n, one for n = 1 mod 4, and one for n = 3 mod 4. I do have a brute force test, but it wasn't designed for a heavy load. For n > 8, it struggles. I'll let it run for a while -- hopefully, it will at least compute a(9). quasi === Subject: Re: Alternating Vertical and Horizontal Moves In Grid <4utbq4dhhldvts8o5hl3ld180tan2tbdmq@4ax.com> posting-account=ey-PjQoAAAD_Gi3gK0ZWKofsmviYOWhV Gecko/20081217 Firefox/2.0.0.20,gzip(gfe),gzip(gfe) no, i changed my mind about the mod 4 thing, it looks a lot like your formula is right whenever n is even, and my formula is right (by a similar argument) when n is odd. of course, we have only established them as lower bounds with certainty. === Subject: Re: Alternating Vertical and Horizontal Moves In Grid >no, i changed my mind about the mod 4 thing, it looks a lot like your >formula is right whenever n is even, and my formula is right (by a >similar argument) when n is odd. of course, we have only established >them as lower bounds with certainty. Yes, if we require corner starting point, and assuming n > 1, I think you may be right. Thus, for a corner starting point, and n > 1, the conjecture is: a(n) = n^2 - 2n + 3, if n is odd n^2 - n + 2, if n is even quasi PS -- My Maple program is still working on a(9), assuming a required corner starting point. === Subject: Re: Alternating Vertical and Horizontal Moves In Grid no, i changed my mind about the mod 4 thing, it looks a lot like your >formula is right whenever n is even, and my formula is right (by a >similar argument) when n is odd. of course, we have only established >them as lower bounds with certainty. Yes, if we require corner starting point, and assuming n > 1, I think >you may be right. Thus, for a corner starting point, and n > 1, the conjecture is: a(n) = > n^2 - 2n + 3, if n is odd > n^2 - n + 2, if n is even quasi PS -- My Maple program is still working on a(9), assuming a required >corner starting point. Ok, Maple completed a(9) and is now working on a(11). Good news -- a(9) = 66, as expected. Note, for convenience, we are now always making the assumption of a corner starting point. For even n, it apparently doesn't matter. For odd n > 1, allowing a variable starting point appears to make a(n) increase by 1. quasi === Subject: Re: Alternating Vertical and Horizontal Moves In Grid <4utbq4dhhldvts8o5hl3ld180tan2tbdmq@4ax.com> <4p7cq45og61mehq711tc9qm7ir9avhp4gl@4ax.com> posting-account=ey-PjQoAAAD_Gi3gK0ZWKofsmviYOWhV Gecko/20081217 Firefox/2.0.0.20,gzip(gfe),gzip(gfe) no, i changed my mind about the mod 4 thing, it looks a lot like your >formula is right whenever n is even, and my formula is right (by a >similar argument) when n is odd. of course, we have only established >them as lower bounds with certainty. Yes, if we require corner starting point, and assuming n > 1, I think >you may be right. Thus, for a corner starting point, and n > 1, the conjecture is: a(n) = > n^2 - 2n + 3, if n is odd > n^2 - n + 2, if n is even quasi PS -- My Maple program is still working on a(9), assuming a required >corner starting point. Ok, Maple completed a(9) and is now working on a(11). Good news -- a(9) = 66, as expected. Note, for convenience, we are now always making the assumption of a > corner starting point. For even n, it apparently doesn't matter. For > odd n > 1, allowing a variable starting point appears to make a(n) > increase by 1. quasi this makes sense to me, i am starting to work on a proof of the bound for the even case, as i think it will be the easier one. i have some ideas, but nothing concrete enough to comment on yet. === Subject: Re: Newton's Method: Cute trick. posting-account=DSjMzAoAAAAISwlGnTY2c6bnmUVmdFNf AppleWebKit/528.16 (KHTML, like Gecko) Version/4.0 Safari/528.16,gzip(gfe),gzip(gfe) I have never seen this trick explicitly mentioned anywhere. Of course if f (p) < 0, f (q) > 0, f continuous on [p, q] then there is a root of f in (p, q), and if you use _any_ iterative algorithm, then clamping the next iterated result to [p, q] will get you closer to the root. Some things I noted: (1) If you find x0 such that f (x0) is small enough, but x0 is not in [p, q] then clamping x0 to [p, q] might give a new value where f (x) is _not_ small enough anymore, so care must be taken. (2) Clamping can actually only happen on the first Newton-Raphson step if your convergence criterion is actually met, because you proved that each iteration gets closer to the root of F (x). If you are not 100% sure that the convergence criterion is met, clamping might not help. (3) This doesn't solve the problem of finding x0 close enough to the root. Of course it is no big deal if a and b are known. === Subject: Re: Help!How to solve the equation === Subject: Re: nxn matrices of rank k (k Would a few of you please let me know whether Steven Holzner's Differential > Equations for Dummies is a decent book on this subject? I'm looking for > something very, very, basic, with lots of hand-holding. I do have the basics > of calculus and linalg but no real experience with the diffeqs.... I'm not familiar with Holzner's book; but a gentle introduction is Dennis G. Zill, A First Course in Differential Equations with Applications. Ken Pledger. === Subject: Re: fixed point set theory <8675172.1235686249093.JavaMail.jakarta@nitrogen.mathforum.org> posting-account=9QOSvAoAAACEOWJVSDuswW7dB_0wApQO Gecko/2009020410 Fedora/3.0.6-1.fc10 Firefox/3.0.6,gzip(gfe),gzip(gfe) (squid/2.7.STABLE3) On Feb 26, 5:42æpm, amy666 since only set theory can hold the attention of > the > majority at sci.math ill post about it again , > despite being more intrested in other subjects. you know or you should know that ' fixed points > ' > are fundamental in many branches of math. however standard set theory with its large > cardinals > is inconsistant with the concept of ' fixed > points '. What a strange thing to say. no , as the OP explained ' fixed points ' are > fundamental to math. it is thus logical , despite rarely considered. PART I 1 2 3 4 ... 'oo'. 'oo' is a fixpoint of the successor function : x + 1 = S(x) = x If you say so. Exactly what set are you talking > about when you say oo, and exactly what is the > definition of S(x) for any set x? i already explained in the OP. 1 2 3 4 ... oo can only mean the set of the integers or the set of > the AP - integers . no other interpretation is meaningfull. I don't expect you'll actually provide us with > those > definitions, but until you do you haven't said > anything > meaningful. has only this 'oo' as a solution. one can argue 'oo' = aleph 0 æ( ordinary integers ) 'oo' = aleph 1 ( AP integers and the alike ) but that has no influence on the rest of this > thread > , same conclusions follow. remember , only this 'oo' and not e.g. aleph 'oo' , aleph 34 , oo + 2 , 2 oo - 1 , > w^w^...^w since 'oo' is a fixpoint , and btw also > otherwise > the set of integers 1 2 3 ... oo is not well > defined. ------------- PART II http://upload.wikimedia.org/wikipedia/commons/8/83/Om > ega squared.png this w^2 can clearly be mapped onto w. thus w^2 = w*w = w apart from the way we count it. basicly w^2 is just the ways we count subsets of > w. Now you don't have to give your definitions, > because > the things you're talking about are standard > gizmos. No, w^2 does not count the subsets of w. i didnt say that , i said w^2 is just the way we > count subsets of w. i gave a link to a picture of w^2. it clearly counts till w , w times. as you can see on the picture ( = proof ) since card(w^2) = card(w) w^2 is thus simply a way of putting w into subsets. w^n = w all ordinals belong to a cardinal. it is thus clear that we cannot increase the > cardinality of an ordinal by operations that do > would > not change the cardinality if applied to > cardinality. ( symbolic w^2 = w <=> R^2 = R ) ---------------- PART III since PART I and PART II have shown that > applying > the successor function and using ordinals cannot > increase cardinality beyond the way cardinal > functions do , we are only left with cardinal > functions. the simplest function that increases cardinality > is > 2^x. in fact , up to isomorphism , its the only one > together with its repetitions. You just love to babble nonsense. and here you just love to get personal again , > instead of rational. nonsense ? æok then , lets turn the tables again ! i said : the simplest function that increases cardinality > is > 2^x. in fact , up to isomorphism , its the only one > together with its repetitions. give a counterexample !! with proof !! But that is not how it works... yes , that is exactly how it works !! if you call a claim nonsense then you should at least be capable of disproving it. otherwise it might be true and is thus FAR FROM NONSENSE. should be easy too disproof , since its so clearly 'nonsense' ? Nonsense is not easy or hard to disprove: it is simply wrong camp and the particular claim which I quoted is so vague, has so many undefined terms, and it is so poorly expressed that, to be honest, I do not really know what it is supposed to be claiming. While it may be that you have a clear and precise idea in your mind about what the claim claims, you certainly did not manage to communicate it. Since there are many, many things out there which are not nonsense, I simply cannot find any justification in trying to figure out what you mean and on top of that, provide a counterexample to whatever it is that you meant. -- m === Subject: Re: Suggesting a Poll <1234799392.600058@athprx03> \ <87zlgipnni.fsf@alatheia.dsl.inet.fi> posting-account=9F70KwkAAADloWPH_Z0lOPxF98dtfKh6 Gecko/2008072820 Firefox/3.0.1,gzip(gfe),gzip(gfe) > On 17 Feb 2009 03:18:46 -0500, Bill Dubuque 2.1 æ 41/1890 æ Aatu Koskensilta Huh? æNever was there less of a crank-basher (or crank). My news habits have changed over the years. I used to indulge in > crank bashing to a much larger extent some years ago. I've since > come to the view that such an activity is utterly pointless. My high > score may also be partly explained by my interest in the philosophy of > mathematics and mathematical logic. [...] Or if you talk about cranking something up a notch, that'll do it as well. Of course, you realize that you just raised your crank index by a small amount, with this post, right? 8-) Come to think of it, so did I. In fact, I raised my index by just QUOTING your post. Did Bill Dubuque take that into account? If someone quotes a message that has the word crank in it, then provides a flawless proof of the Twin Prime Conjecture, for their first post, then they are listed as 100% crank material. In other crank-related news, Archimedes Plutonium is suing me for a defaming MySpace page that I (along with someone who I had never heard of before) demonstrably had nothing to do with. --- Christopher Heckman === Subject: Re: The modern mathematical concept of infinity is ... > ... > æ> æ> I have shown that there is no actually infinite number of FISONs. > æ æ> Disregarding that you again do use the words actually infinite > without > æ> proper definition, you have not shown that. > æ æ> AI means in this case: The number of natural is not a natural. Yes, indeed, there is no natural number that gives the number of FISONs. And there is no unnatural number either that gives the number of > FISONs. But there is a non-natural number, often called aleph_0, which gives the number of naturals and the number of FISONs, as these numbers are the same. You have *not* shown that there *is* a natural number that gives the number > of FISONs. æSuppose there was such a number n, what would FISON(n + 1) > be? > Suddenly not a FiSON? If the set of all FISONs was a fixed set, then the number of FISONs > was a fixed number. And this number could easily be proved to be not > an unnatural number. I have no notion what restrictions WM would place on unnatural numbers other than their not being natural numbers, but absent any other restrictions, the non-natural number which is often denoted by aleph_0 is the number of members in the set of all FISONs as well as the number of elements in the set of all natural numbers. > æ> This holds for all natural numbers. As there is not a last one, the > æ> set is potentially infinite. But there is no jump to aleph_0. I have no idea what you mean by this. æThe number of elements of the set > N is not a natural number, and so not in N. æThere is no jump, it is only > a definition about the number of elements in an infinite set. æNothing > more, nor less. This definition is wrong - at least the proof that aleph_0 is larger > than every natural number. It is correct everywhere outside of WM's MathUnRealism. Show your interpretation that a set is not fixed. A number of elements of a linear set of unary representations of > naturals > I > II > III > ... > III...I cannot be distinguished unless there are at least as many symbols in > at least one element. > This number of symbols cannot be larger than any natural by definition > of natural. > This number cannot be a fixed natural by induction n ==> n + 1. > There remains only one possibility: N is a non-fixed set. If that means that, even though ordered by successorship, it does not have a last member, fine. If a set is not fixed > how are we able to know that a particular thing is an element of a set? By a test that works on any element and any non-element, and such a test we have. If we know a natural number, then it belongs to N. All natural numbers > that we can know belong to a FISON. It is WM's own thesis that things which cannot be distinguished do not exist. So unless WM can distinguish that particular FISON that all naturals that we can know must belong to from all others, he is merely blowing more hot air. Each natural belongs to some FISON as a last member and each FISON determines a natural as its last member, so set of naturals and the set of FISONs biject. > I think you will agree. But why > should we discuss natural numbers that cannot be known in principle? Because WM has no principles. === Subject: Re: The modern mathematical concept of infinity is ... > ... > æ> æ> I have shown that there is no actually infinite number of FISONs. > æ æ> Disregarding that you again do use the words actually infinite > without > æ> proper definition, you have not shown that. > æ æ> AI means in this case: The number of natural is not a natural. Yes, indeed, there is no natural number that gives the number of FISONs. And there is no unnatural number either that gives the number of > FISONs. But there is a non-natural number, often called aleph_0, which gives the number of naturals and the number of FISONs, as these numbers are the same. You have *not* shown that there *is* a natural number that gives the number > of FISONs. æSuppose there was such a number n, what would FISON(n + 1) > be? > Suddenly not a FiSON? If the set of all FISONs was a fixed set, then the number of FISONs > was a fixed number. And this number could easily be proved to be not > an unnatural number. I have no notion what restrictions WM would place on unnatural numbers other than their not being natural numbers, but absent any other restrictions, the non-natural number which is often denoted by aleph_0 is the number of members in the set of all FISONs as well as the number of elements in the set of all natural numbers. > æ> This holds for all natural numbers. As there is not a last one, the > æ> set is potentially infinite. But there is no jump to aleph_0. I have no idea what you mean by this. æThe number of elements of the set > N is not a natural number, and so not in N. æThere is no jump, it is only > a definition about the number of elements in an infinite set. æNothing > more, nor less. This definition is wrong - at least the proof that aleph_0 is larger > than every natural number. It is correct everywhere outside of WM's MathUnRealism. Show your interpretation that a set is not fixed. A number of elements of a linear set of unary representations of > naturals > I > II > III > ... > III...I cannot be distinguished unless there are at least as many symbols in > at least one element. > This number of symbols cannot be larger than any natural by definition > of natural. > This number cannot be a fixed natural by induction n ==> n + 1. > There remains only one possibility: N is a non-fixed set. If that means that, even though ordered by successorship, it does not have a last member, fine. If a set is not fixed > how are we able to know that a particular thing is an element of a set? By a test that works on any element and any non-element, and such a test we have. If we know a natural number, then it belongs to N. All natural numbers > that we can know belong to a FISON. It is WM's own thesis that things which cannot be distinguished do not exist. So unless WM can distinguish that particular FISON that all naturals that we can know must belong to from all others, he is merely blowing more hot air. Each natural belongs to some FISON as a last member and each FISON determines a natural as its last member, so set of naturals and the set of FISONs biject. > I think you will agree. But why > should we discuss natural numbers that cannot be known in principle? Because WM has no principles. === Subject: Re: The modern mathematical concept of infinity is ... Nntp-Posting-Host: hera.cwi.nl ... > > AI means in this case: The number of natural is not a natural. > > Yes, indeed, there is no natural number that gives the number of FISONs. > > And there is no unnatural number either that gives the number of > FISONs. O. How do you define unnatural number? > You have *not* shown that there *is* a natural number that gives the > number of FISONs. Suppose there was such a number n, what would > FISON(n + 1) be? > Suddenly not a FiSON? > > If the set of all FISONs was a fixed set, then the number of FISONs > was a fixed number. And this number could easily be proved to be not > an unnatural number. How do you define unnatural number? > > I union all FISONs from F_1 to F_n. As there is no last n, I union > > all. > > No, you do not. For each n you unite all F_i for i <= n. As there > is no last n you do not unite them all. > > So if I say forall n in N, then n cannot assume every natural number? Why do you think so? > Where does it stop? Why should I include the last natural number, that > does not exist. All other naturals are included. You do not unite all FISONs, but you unite all FISONs from F_1 to F_n for every n. There is a distinction. In what you do each union has a finite number of elements and a last element. > > No, this set has not a last element that can be fixed. All we know > > about it is that it is not aleph_0. That is as clear as the fact that > > it is not an irrational number. So we know two properties of this last > > element. > > As there is *no* last element any reasoning about it is nonsense. The > reason is that *every* element has a successor. > > Therefore my n assumes every natural number. Yes, for every n you unite all F_1 to F_n. There is *no* n such that you unite *all* F_k. > > This holds for all natural numbers. As there is not a last one, the > > set is potentially infinite. But there is no jump to aleph_0. > > I have no idea what you mean by this. The number of elements of the > set N is not a natural number, and so not in N. There is no jump, it is > only a definition about the number of elements in an infinite set. > Nothing more, nor less. > > This definition is wrong - at least the proof that aleph_0 is larger > than every natural number. aleph_0 is larger than every finite cardinal number. There is an isomorphism between the finite cardinal numbers > 0 and the natural numbers. That is the strict formulation. In common speech the natural numbers are identified with the finite cardinal numbers > 0, but that does not mean that they are the same. In the same way the integer 2, the rational 2/1, the real 2.000... and the complex number 2.000... + 0.000... * i are identified with each other. They are not the same, but because of isomorphisms they can be regarded as the same. For instance in the definition a rational is a pair of integers with some particular definitions of addition, subrtaction, multiplication and division. As the integer 2 is not a pair of numbers it is not itself a rational number. On the other hand with the equivalence relation: (a, b) == (c, d) iff a * d = b * c we get equivalence classes in the pairs of integers that actually define the rational numbers, and it is easy to sea that there is an isomorphism between the integers i and the equivalence classes represented by (i, 1). Try to read a bit about how the system of numbers is build since about 1920 in mathematics. > Show your interpretation that a set is not fixed. > > A number of elements of a linear set of unary representations of > naturals Sorry, this makes no sense. We were talking about the axiom of extensionality. You had an interpretation that does not makes sets fixed, at least, you stated such. What is your interpretation was the question. So start off with that axiom and show why with that axiom sets are not necessarily fixed. > If a set is not fixed > how are we able to know that a particular thing is an element of a set? > > If we know a natural number, then it belongs to N. All natural numbers > that we can know belong to a FISON. I think you will agree. But why > should we discuss natural numbers that cannot be known in principle? Can you give a *mathematical* definition of know? Consider trunc(10^10^10 * pi) + 1. I know it is a natural number. So it belongs to the set N? I know that the first place where Li(n) and pi(n) switch place is a natural number, so is it in the set N? Moreover, I know that all preceding numbers when I subtract 1 are natural numbers. They all belong to N? I think you are confusing being able to know it with being able to represent it in a finite number of decimals. Now, how did Robert Israels call that? -- dik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: The modern mathematical concept of infinity is ... posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > How do you define æunnatural number? Every number that is not 1 and not n+1 if n is natural. For instance: zero, negative, most rationals, irrationals, complex, larger than every natural. æ> æ> I union all FISONs from F 1 to F n. As there is no last n, I union > æ> æ> all. > æ æ> No, you do not. æFor each n you unite all F i for i <= n. æAs there > æ> is no last n you do not unite them all. > æ æ> So if I say forall n in N, then n cannot assume every natural number? Why do you think so? æ> Where does it stop? Why should I include the last natural number, that > æ> does not exist. All other naturals are included. You do not unite all FISONs, but you unite all FISONs from F 1 to F n for > every n. æThere is a distinction. æIn what you do each union has a finite > number of elements and a last element. Every FISON has a last element. Every union of FISONs has a last element. Here comes the crucial question: Do you believe that more than any natural number of numbers in unary representation can be distinguished by less than aleph 0 strokes per number? æ> æ> No, this set has not a last element that can be fixed. All we know > æ> æ> about it is that it is not aleph 0. That is as clear as the fact that > æ> æ> it is not an irrational number. So we know two properties of this last > æ> æ> element. > æ æ> As there is *no* last element any reasoning about it is nonsense. æThe > æ> reason is that *every* element has a successor. > æ æ> Therefore my n assumes every natural number. Yes, for every n you unite all F 1 to F n. æThere is *no* n such that you > unite *all* F k. Which is missing? > æ> If we know a natural number, then it belongs to N. All natural numbers > æ> that we can know belong to a FISON. I think you will agree. But why > æ> should we discuss natural numbers that cannot be known in principle? Can you give a *mathematical* definition of know? To know is prior to mathematics. It is a primitive notion. >æConsider > trunc(10^10^10 * pi) + 1. æI know it is a natural number. æSo it belongs > to the set N? æI know that the first place where Li(n) and pi(n) switch > place is a natural number, so is it in the set N? æMoreover, I know that > all preceding numbers when I subtract 1 are natural numbers. æThey all > belong to N? That is not sufficient. Each one must be definied so that everybody knows what number is meant. I think you are confusing being able to know it with being able to > represent it in a finite number of decimals. æNow, how did Robert > Israels call that? I don't know how he would call that. All that is necessary to know here is that a number can be identified as an individuum, such that you and me mean the same, when talking about it. Can you understand what I mean? Do you *know* what I mean when I write 1, 2, [pi], |i|? And here it comes again: Do you believe that more than any natural number of numbers in unary representation can be distinguished by less than aleph 0 strokes per number? === Subject: Re: The modern mathematical concept of infinity is ... How do you define æunnatural number? Every number that is not 1 and not n+1 if n is natural. > For instance: zero, negative, most rationals, irrationals, complex, > larger than every natural. Then WM should have no problem with the existence of at least some unnatural numbers. What boggles the mind is that his prejudices on whether an unnatural number can exist seem to have no logical justification. æ> æ> I union all FISONs from F_1 to F_n. As there is no last n, I union > æ> æ> all. > æ æ> No, you do not. æFor each n you unite all F_i for i <= n. æAs there > æ> is no last n you do not unite them all. > æ æ> So if I say forall n in N, then n cannot assume every natural number? Why do you think so? æ> Where does it stop? Why should I include the last natural number, that > æ> does not exist. All other naturals are included. If the last natural number does not exist, why try to do anything with it? You do not unite all FISONs, but you unite all FISONs from F_1 to F_n for > every n. æThere is a distinction. æIn what you do each union has a finite > number of elements and a last element. Every FISON has a last element. The collection of ALL FISONs no more has a largest than the collection of ALL naturals has a largest. > Every union of FISONs has a last > element. Wm has previously claimed that any natual that cannot be distinguished does not exist, and if true the same restriction must apply to FISONs. So unless WM can distinguish that last FISON in the union of all FISONs in the only way FISONs can be distinguished from each other, by naming the last member of that last FISON, no such last FISON exists, and the union of all FISONs is not, itself, a FISON. Here comes the crucial question: > Do you believe that more than any natural number of numbers in unary > representation can be distinguished by less than aleph_0 strokes per > number? Does WWM believe that the FISON which he claims is the union of all FISONs can be distinguished by any method whatsoever? Yes, for every n you unite all F_1 to F_n. æThere is *no* n such that you > unite *all* F_k. Which is missing? The infinitely many successors to the largest n you consider. æ> If we know a natural number, then it belongs to N. All natural numbers > æ> that we can know belong to a FISON. I think you will agree. But why > æ> should we discuss natural numbers that cannot be known in principle? Can you give a *mathematical* definition of know? To know is prior to mathematics. It is a primitive notion. WM has never got past being prior to mathematics to do any actual mathematics. æConsider > trunc(10^10^10 * pi) + 1. æI know it is a natural number. æSo it belongs > to the set N? æI know that the first place where Li(n) and pi(n) switch > place is a natural number, so is it in the set N? æMoreover, I know that > all preceding numbers when I subtract 1 are natural numbers. æThey all > belong to N? That is not sufficient. Each one must be definied so that everybody > knows what number is meant. Including newborns? > And here it comes again: Do you believe that more than any natural > number of numbers in unary representation can be distinguished by less > than aleph_0 strokes per number? I do not believe that WM's distinguishability criteria have any mathematical relevance outside of his MathUnrealism miniworld. === Subject: Re: The modern mathematical concept of infinity is ... As soon as I conceptualize writing a numeral 1 in a unary expression > every other numeral 1 does not exist. I am silly. But what about the > notion of numbers in time? When we introduce a temporal aspect into our mathematical conceptions, > as we do in e.g. the theory of the creative subject and choice > sequences, it is not the gritty-nitty time we actually experience, > measured by clocks, the rotation of celestial bodies, and so on, but > rather an abstraction. A case in point, it makes absolutely no sense > to speak of the creative subject having chosen to make the value of > the free choice sequence alpha to be 10 at index 13427 three in the > afternoon, May 7th 1897, inspired by the juicy cheese-burger he just > ingested etc. It's transcendental time we're dealing with -- the > idea of things happening and coming into being in a temporal > succession -- just as when we speculate about the cogitations of the > idealised mathematicians (a transcendental mathematician) we're not > asking about the psychological make-up of a real mathematician -- > indeed, details about human psychology are on the whole quite > irrelevant, even though intuitionists of certain school like to speak > of mental constructions and what not. It is in this sense we must > read the rather turgid musings of Brouwer on the notion of number > being based on the experience of twoness rooted in the trascendental > experience of time (or the structure of that experience, or however > you like to put it). Can you say anything about it? > The first point to consider is that Isles means natural > natural number notations to be linguistic entities. Contrary to > numbers of today are different from those of yesterday then this is to be > understood in the sense just explained. -- Michael Press === Subject: Re: The modern mathematical concept of infinity is ... > Can you say anything about it? The first point to consider is that Isles means natural number > natural number notations to be linguistic entities. Contrary to > natural numbers of today are different from those of yesterday then > this is to be understood in the sense just explained. I'm not sure if I've read Isles's papers. I'd venture a guess he has in mind some variant of the standard ultra-finitist line, according to which there are non-isomoprhic natural number sequences. For example, we have the naturals as presented in unary notation. This is a weakly infinite series, in that it has no largest member. It is, however, not strongly infinite, in that it has an upper bound, e.g. 2^2^2^2^2^2^2^2^2^2^2^2^2^2 in the serious of naturals expressed with the aid of the exponentiation function. The totality of these sequences is open ended, since we can always introduce new notational devices allowing us to express larger and larger naturals. (Obviously, we must treat is a natural in this or that series as a vague or indeterminate predicate -- like the number of Buffy quotes one must have memorised to count as a true fan -- perhaps thinking of the naturals as fading into non-existence the farther we go, lest a trivial induction argument yield a contradiction.) -- Aatu Koskensilta (aatu.koskensilta@uta.fi) Wovon man nicht sprechen kann, dar.9fber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: The modern mathematical concept of infinity is ... Nntp-Posting-Host: hera.cwi.nl > Because in set theory *every* set is fixed. It is > impossible for a set to not have at one time an element a while it is > an element at some other time. > > Then N is not belonging to set theory or set theory is not capable of > dealing with sets. Apparently you thing different about what sets are than set theory. Not a surprise. But what sets are in your opinion is not a subject of set theory. > I prove by strokes that there are not more than five strokes. Of course > my proof does not hold for actually more than five strokes, but I consider > only at most five strokes. > > There are not more than five different sets with at most five strokes. > > Therefore the concept of more than five n or (F_n) is false. > > Indeed, the concept of more than five FISONs, all of which have less > than six strokes is wrong. Of course. Nope. Your conclusion was that the concept of some number of FISONs *without qualification* was false. To quote you again: > Of course my proof does not hold for an actually infinite number of > strokes, but I consider only finite numbers of strokes. > > Therefore the concept of actual infinity of n or (F_n) is false . To paraphrase: Of course my proof does not hold for six strokes, but I consider only less then five strokes or less. Therefore the concept of six n or (F_n) is false. This is just a plain translation of *your* words by replacing actual infinity by six. You do not consider an actual infinite number of strokes and therefore the concept of actual infinity of n or (F_n) is false. I do not consider a six strokes and therefore the concept of more than five n or (F_n) is false. In what way are the two reasonings different? -- dik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: The modern mathematical concept of infinity is ... posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > æ> Because in set theory *every* set is fixed. æIt is > æ> impossible for a set to not have at one time an element a while it is > æ> an element at some other time. > æ æ> Then N is not belonging to set theory or set theory is not capable of > æ> dealing with sets. Apparently you thing different about what sets are than set theory. æNot a > surprise. æBut what sets are in your opinion is not a subject of set theory. My interest is to show that the number of natural numbers does not surpass any natural number. Nothing else. Casually I call this collection N. æ> I prove by strokes that there are not more than five strokes. æOf course > æ> my proof does not hold for actually more than five strokes, but I consider > æ> only at most five strokes. > æ æ> There are not more than five different sets with at most five strokes. > æ æ> Therefore the concept of more than five n or (F n) is false. > æ æ> Indeed, the concept of more than five FISONs, all of which have less > æ> than six strokes is wrong. Of course. Nope. æYour conclusion was that the concept of some number of FISONs *without > qualification* was false. æTo quote you again: Not necessary. FISON without further qualification implies finite being. æ> Of course my proof does not hold for an actually infinite number of > æ> strokes, but I consider only finite numbers of strokes. > æ æ> Therefore the concept of actual infinity of n or (F n) is false . To paraphrase: > æOf course my proof does not hold for six strokes, but I consider > æ only less then five strokes or less. æTherefore the concept of six n or > æ (F n) is false. > This is just a plain translation of *your* words by replacing actual > infinity by six. > And finity by 5. > You do not consider an actual infinite number of strokes and therefore > the concept of actual infinity of n or (F n) is false. I do not consider a six strokes and therefore the concept of more than > five n or (F n) is false. In what way are the two reasonings different? A FISON is finite by definition. (In your example has not more than 5 strokes.) === Subject: Re: The modern mathematical concept of infinity is ... > æ> Because in set theory *every* set is fixed. æIt is > æ> impossible for a set to not have at one time an element a while it is > æ> an element at some other time. > æ æ> Then N is not belonging to set theory or set theory is not capable of > æ> dealing with sets. Apparently you thing different about what sets are than set theory. æNot a > surprise. æBut what sets are in your opinion is not a subject of set > theory. My interest is to show that the number of natural numbers does not > surpass any natural number. Nothing else. Casually I call this > collection N. In honest mathematics, the number of natural numbers surpasses each and every natural number. æ> I prove by strokes that there are not more than five strokes. æOf > course > æ> my proof does not hold for actually more than five strokes, but I > consider > æ> only at most five strokes. > æ æ> There are not more than five different sets with at most five strokes. > æ æ> Therefore the concept of more than five n or (F_n) is false. > æ æ> Indeed, the concept of more than five FISONs, all of which have less > æ> than six strokes is wrong. Of course. Nope. æYour conclusion was that the concept of some number of FISONs > *without > qualification* was false. æTo quote you again: Not necessary. FISON without further qualification implies finite > being. While each individual FISON is, by definition, finite, the number of FISONs, like the number of naturals, naturally exceeds every natural since for every natural, n, there are at least n+1 FISONs. Until WM can show us a natural n for which there are NOT n+1 naturals and n+1 FISONs, he is doomed to failure. > You do not consider an actual infinite number of strokes and therefore > the concept of actual infinity of n or (F_n) is false. I do not consider a six strokes and therefore the concept of more than > five n or (F_n) is false. In what way are the two reasonings different? A FISON is finite by definition. (In your example has not more than 5 > strokes.) WM seems too stupid, or too stubborn, to distinguish between the finiteness of an individual FISON as a set and the infiniteness of FISONs collectively as members of other sets. Until Wm learns to distinguish between being a member of a set and being a set having members, he will continue to fail at set theory. === Subject: Re: The modern mathematical concept of infinity is ... posting-account=DLD3MQkAAACJxulKt9xouw3DPpXKssCI GTB5; SLCC1; .NET CLR 2.0.50727; .NET CLR 1.1.4322; .NET CLR 3.5.21022; .NET CLR 3.5.30729; .NET CLR 3.0.30618),gzip(gfe),gzip(gfe) æ> Because in set theory *every* set is fixed. æIt is > æ> impossible for a set to not have at one time an element a while it is > æ> an element at some other time. > æ æ> Then N is not belonging to set theory or set theory is not capable of > æ> dealing with sets. Apparently you thing different about what sets are than set theory. æNot a > surprise. æBut what sets are in your opinion is not a subject of set theory. My interest is to show that the number of natural numbers does not > surpass any natural number. Nothing else. Casually I call this > collection N. Is there no element such that the largest element in N (let's call it n) has a successor n+1? If all elements of N have successors, then there is no largest element in N. If there is no largest element in N, then elements of N can be arbitrarily large. If there is an element in N that has no successor, then N is not closed under addition, multiplication, or subtraction. > æ> I prove by strokes that there are not more than five strokes. æOf course > æ> my proof does not hold for actually more than five strokes, but I consider > æ> only at most five strokes. > æ æ> There are not more than five different sets with at most five strokes. > æ æ> Therefore the concept of more than five n or (F n) is false. > æ æ> Indeed, the concept of more than five FISONs, all of which have less > æ> than six strokes is wrong. Of course. Nope. æYour conclusion was that the concept of some number of FISONs *without > qualification* was false. æTo quote you again: Not necessary. FISON without further qualification implies finite > being. Define 'finite being.' > æ> Of course my proof does not hold for an actually infinite number of > æ> strokes, but I consider only finite numbers of strokes. > æ æ> Therefore the concept of actual infinity of n or (F n) is false . To paraphrase: > æOf course my proof does not hold for six strokes, but I consider > æ only less then five strokes or less. æTherefore the concept of six n or > æ (F n) is false. > This is just a plain translation of *your* words by replacing actual > infinity by six. And finity by 5. You do not consider an actual infinite number of strokes and therefore > the concept of actual infinity of n or (F n) is false. I do not consider a six strokes and therefore the concept of more than > five n or (F n) is false. In what way are the two reasonings different? A FISON is finite by definition. (In your example has not more than 5 > strokes.) > === Subject: Re: The modern mathematical concept of infinity is ... posting-account=dO3EpgkAAAA2h6wij2Pt6SWrMMkd-UVg > It is really easy to see: > Three FISONs cannot be distinguished unless at least one FISON with > three elements is amonst them. > Got it? Sure. For example, two of {1, 2}, {2, 3}, and {3, 4} must be equal, > since > none of these sets has more than two elements. FISON means Finite Initial Segment Of Naturals. Ah. Sorry. In that case, let's try <>, <1>, and <1, 2>. That's three distinct FISONs, none of which contains three elements or more. === Subject: Re: The modern mathematical concept of infinity is ... posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) It is really easy to see: > Three FISONs cannot be distinguished unless at least one FISON with > three elements is amonst them. > Got it? Sure. For example, two of {1, 2}, {2, 3}, and {3, 4} must be equal, > since > none of these sets has more than two elements. FISON means Finite Initial Segment Of Naturals. Ah. Sorry. In that case, let's try <>, <1>, and <1, 2>. That's three > distinct FISONs, none of which contains three elements or more. Sorry, I forgot to mention the second condition. A natural number must be represented in unary representation for the sake of this discussion. The smallest one is 1. The smallest FISON is {1}. === Subject: Re: The modern mathematical concept of infinity is ... posting-account=1lE9SQkAAADFrJsDv61dh1YXcJ_ahy5I assuming the existence of the union of all FISONs leads to > an inconsistency, you must 1. assume its existence I do. At this point we expect WM to make a comment not about the union but about the *elements* of the union > I assume that aleph_0 natural numbers exist and that each one > can be represented in unary notation. It is nice to know that we can depend on some things. - William Hughes === Subject: Re: The modern mathematical concept of infinity is ... posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) If you want to show > assuming the existence of the union of all FISONs leads to > an inconsistency, you must æ æ1. assume its existence I do. At this point we expect WM to make a comment not about > the union but about the *elements* of the union I assume that aleph 0 natural numbers exist and that each one > can be represented in unary notation. It is nice to know that we can depend on some things. I believe in the holy set N that cannot be secularized by any proof concerning its imperfect profane elements, not even by a proof about all of them. Better so? === Subject: Re: The modern mathematical concept of infinity is ... posting-account=1lE9SQkAAADFrJsDv61dh1YXcJ_ahy5I > If you want to show > assuming the existence of the union of all FISONs leads to > an inconsistency, you must 1. assume its existence I do. At this point we expect WM to make a comment not about > the union but about the *elements* of the union I assume that aleph_0 natural numbers exist and that each one > can be represented in unary notation. It is nice to know that we can depend on some things. I believe in the holy set N that cannot be secularized by any proof > concerning its imperfect profane elements, not even by a proof about > all of them. Better so? > Well, you clearly do not believe in N. You think that proving something about all elements of N proves something for N. You have admitted that this is false. - William Hughes === Subject: Re: The modern mathematical concept of infinity is ... Because in set theory *every* set is fixed. æIt is > impossible for a set to not have at one time an element a while it is > an element at some other time. Then N is not belonging to set theory or set theory is not capable of > dealing with sets. What transpires, or doesn't, in WM's anti-mathematical version of set theory is peculiar to that version, and is of no importance in any mathematically acceptable set theories. === Subject: Re: The modern mathematical concept of infinity is ... > That only shows that there may be problems trying to exhibit > aleph_0 *numerals* ... Huh? If we consider any (finite) sequence of |s as a numeral denoting > the natural number given by the number of |s in the sequence, > where's the problem? :-o æ æ|, ||, |||, ... Clearly there are aleph_0 such numerals. Actually, if we allow > for an infinite sequence of |s we might even denote aleph_0 by a > numeral, namely by the infinite sequence of |s. (Talking about mathematical objects/sequences here; not about any > physical entities.) To me a numeral is a physical representation of a number, whether > as marks on paper or sound waves of spoken words, or electrically > excited phosphors on a screen, or whatever other form of > representation one can think of. A number is the set of its representations, whether as marks on paper > or sound waves of spoken words, or electrically excited phosphors on > a screen, or whatever other form of representation one can think of. > A number that has no representation does not exist. Does Wm really believe that there can be anything which is no more than the set of its names? And as such, I am quite willing to say that only finitely many of > them can be exhibited. Then we have no divergence, except we differ in what mathematics is. > I am curious to see whether many young students wish to follow your > way, now after it has been clearly exhibited that you cannot > distinguish most of what you call numbers. But anybody may decide it > by himself. My students would not believe that your mathematics is a > science. Math is not a science, at least in the usual sense of science based only on physical observation. But I deny that a number must have some numeral to represent it > before it can be said to exist in some purely > mathematical non-physical sense. Numbers that cannot be identified are non numbers and not part of > mathematics. The inaccessible real numbers cannot be individually identified but are a necessary part of the real number field used by standard analysis. To claim that they re not a part of mathematics is to display a profound ignorance of standard analysis. > Mathematicians who believe in unidentifyable numbers are > not mathematicians. That may be WM's opinion, but we have een how little his opinion is worth , at least re anything mathemtical. There are thousands of mathematicians for whom all sorts of inaccessible numbers and other inaccessible mathematical objects need to exist for mathematics to work properly, and those mathematicians are the ones who, unlike WM, are creating new math, rather than, like WM, vainly try to tear down that which already exists. === Subject: Re: The modern mathematical concept of infinity is ... <1p3fz0dmtcx6e$.n7svlh24r3w$.dlg@40tude.net> posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > A number that has no representation does not exist. Does Wm really æbelieve that there can be anything which is no more than > the set of its names? Of course. Names and numbers and things that do not exist other than by names like most Gods (if not all). And as such, I am quite willing to say that only finitely many of > them can be exhibited. Then we have no divergence, except we differ in what mathematics is. > I am curious to see whether many young students wish to follow your > way, now after it has been clearly exhibited that you cannot > distinguish most of what you call numbers. But anybody may decide it > by himself. My students would not believe that your mathematics is a > science. Math is not a science, at least in the usual sense of science based only > on physical observation. Math is a special branch of physics. Nobody would know anything about math without physical observation. But I deny that a number must have some numeral to represent it > before it can be said to exist in some purely > mathematical non-physical sense. Numbers that cannot be identified are non numbers and not part of > mathematics. The inaccessible real numbers cannot be individually identified but are > a necessary part of the real number field used by standard analysis. To claim that they are not a part of mathematics is to display a profound > ignorance of standard analysis. That does not hinder their non-existence. Mathematicians who believe in unidentifyable numbers are > not mathematicians. That may be WM's opinion, and that of most sober minds in sciences. There are thousands of mathematicians for whom all sorts of inaccessible > numbers and other inaccessible mathematical objects need to exist for > mathematics to work properly, and those mathematicians are the ones who, > unlike WM, are creating new math, rather than, like WM, ævainly try to > tear down that which already exists There are thousands of believers in astrology, clairvoyance, superstition, and set theory. > Do you agree with Virgil that even most natural numbers cannot be > distinguished? > Or do you claim that aleph 0 unary numerals can be distinguished by > less than aleph 0 strokes for each numeral? > False dichotomy. AS usual, as soon as WM posts, he posts a flawed > argument. This is not an argument at all, neither true nor false. What is false in my dichotomy? === Subject: Re: The modern mathematical concept of infinity is ... <1p3fz0dmtcx6e$.n7svlh24r3w$.dlg@40tude.net> posting-account=_-PQygoAAAAciOn_89sZIlnxfb74FzXU Gecko/2009011913 Firefox/3.0.6,gzip(gfe),gzip(gfe) That only shows that there may be problems trying to exhibit > aleph 0 *numerals* ... Huh? If we consider any (finite) sequence of |s as a numeral denoting > the natural number given by the number of |s in the sequence, > where's the problem? :-o æ æ|, ||, |||, ... Clearly there are aleph 0 such numerals. Actually, if we allow > for an infinite sequence of |s we might even denote aleph 0 by a > numeral, namely by the infinite sequence of |s. (Talking about mathematical objects/sequences here; not about any > physical entities.) To me a numeral is a physical representation of a number, whether > as marks on paper or sound waves of spoken words, or electrically > excited phosphors on a screen, or whatever other form of > representation one can think of. A number is the set of its representations, whether as marks on paper > or sound waves of spoken words, or electrically excited phosphors on > a screen, or whatever other form of representation one can think of. > A number that has no representation does not exist. Does Wm really æbelieve that there can be anything which is no more than > the set of its names? And as such, I am quite willing to say that only finitely many of > them can be exhibited. Then we have no divergence, except we differ in what mathematics is. > I am curious to see whether many young students wish to follow your > way, now after it has been clearly exhibited that you cannot > distinguish most of what you call numbers. But anybody may decide it > by himself. My students would not believe that your mathematics is a > science. Math is not a science, at least in the usual sense of science based only > on physical observation. But I deny that a number must have some numeral to represent it > before it can be said to exist in some purely > mathematical non-physical sense. Numbers that cannot be identified are non numbers and not part of > mathematics. The inaccessible real numbers cannot be individually identified but are > a necessary part of the real number field used by standard analysis. To claim that they re not a part of mathematics is to display a profound > ignorance of standard analysis. Mathematicians who believe in unidentifyable numbers are > not mathematicians. That may be WM's opinion, but we have een how little his opinion is > worth , at least re anything mathemtical. There are thousands of mathematicians for whom all sorts of inaccessible > numbers and other inaccessible mathematical objects need to exist for > mathematics to work properly, and those mathematicians are the ones who, > unlike WM, are creating new math, rather than, like WM, ævainly try to > tear down that which already exists. Does that say Eudoxus/Dedekind/Cauchy is insufficient to represent real numbers? Standard analysis would have that no real number doesn't have a label identified as a possibly infinite sequence of digits selected from a finite radix indicating some unique value. (Dedekind/Cauchy is insufficient to represent the dually complete ordered field / naturally totally ordered outward spiral of the continuum.) Most mathematicians creating useful new mathematics are either a) high above the foundations elaborating branch points in constructions, or b) re-interpreting and re-developing divergences from standard foundations. Transfinite cardinals aren't used in applied physics, there's no known use of them, while non-standard analyses (in for example the meromorphic or in terms of non-real functions) are widely applicable and used. EF, for example, is not a real function, yet, a remarkably simple primitive in terms of its description in terms of fundamental objects of (useful) analysis, with surprising properties illustrating natural features of the continuum. EF is a CDF. That standard analysis is blind to some particularly obvious constructions of numbers by their natural properties re-emphasizes its incompleteness, and in some negatory statements its inconsistency. Of course, standard analysis is remarkably perfect and the bedrock for much of the applications of integral calculus among other fields. The numbers are immune to will. Ross F. === Subject: Re: The modern mathematical concept of infinity is ... posting-account=1lE9SQkAAADFrJsDv61dh1YXcJ_ahy5I > Of course we don't know the largest FISON. > It is not fixed However, since it is a FISON, it has a fixed > largest element. > Does the largest fison have a fixed largest element? -William Hughes === Subject: Re: The modern mathematical concept of infinity is ... posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) Of course we don't know the largest FISON. > It is not fixed However, since it is a FISON, it has a fixed > largest element. Does the largest fison have a fixed largest element? Of course. Every FISON has a largest element. It is not fixed, however, what the largest FISON is. Do you agree with Virgil that even most natural numbers cannot be distinguished? Or do you claim that aleph_0 unary numerals can be distinguished by less than aleph_0 strokes within each numeral? === Subject: Re: The modern mathematical concept of infinity is ... posting-account=1lE9SQkAAADFrJsDv61dh1YXcJ_ahy5I > Of course we don't know the largest FISON. > It is not fixed However, since it is a FISON, it has a fixed > largest element. Does the largest fison have a fixed largest element? Of course. Therefore since N does not have a fixed largest element, we know that N is not a FISON. -William Hughes === Subject: Re: The modern mathematical concept of infinity is ... Of course we don't know the largest FISON. > It is not fixed However, since it is a FISON, it has a fixed > largest element. Do you agree with Virgil that even most natural numbers cannot be > distinguished? > Or do you claim that aleph_0 unary numerals can be distinguished by > less than aleph_0 strokes for each numeral? False dichotomy. AS usual, as soon as WM posts, he posts a flawed argument. === Subject: Re: The modern mathematical concept of infinity is ... > I have no quarrel with those who wish to follow an intuitionalist point > of view, but honest intutionalists only chose to restrict themselves > whereas WM claims the right to impose his views on everyone. There is not a claim but a proof: There are not more FISONs than the > largest FISON has elements. If a largest FISON does not exist this is a meaningless > statement. æ No. Every FISON is finite. That is what counts. What counts in claiming There are not more FISONs than the largest FISON has elements is that there is no such thing as a largest FISON available, because as soon as one identifies any FISON, one proves it not to be largest. > You claim that it is inconsistent to assume that > a largest FISON does not exist, however, you have never > shown this. And it is certainly at least equally inconsistent to assume, or claim, that a largest FISON does exist when, by definition, the existence of any FISON implies the existence of a larger one. > If the union of all FISONs that potentially exist > exists as a real set, there is no largest FISON. > Now, we know that you claim such a union does not > exist as a real (actually existing) set, fine. > You further claim that it is inconsistent to claim > that it does exist. æNot fine. æYour proofs only deal with > *elements* of the union of all FISONs. æIf you want to show > assuming the existence of the union of all FISONs leads to > an inconsistency, you must æ æ1. assume its existence I do. I assume that aleph_0 natural numbers exist and that each one > can be represented in unary notation. There is nothing in the mathematics which requires any natural to have a representation. > I conclude that aleph_0 strokes are required for at least one natural. You may want to assume that, but everywhere outside of your MathUnRealism, it is false. The number of naturals is not a natural. > Contradiction. Then it is a contradiction between your assumptions, WM, that does not exist in the mathematics itself. and > æ æ2. prove something about the union, not something > æ æ æ about the *elements* of the union. What is your opinion: > Are there less than aleph_0 natural numbers? Depends on the definition of aleph_0. If it is, as usual, defined as the number of naturals, then no, but if it has some weird MathUnRealistic definition by WM then we must see that definition before expressing an opinion. === Subject: Re: The modern mathematical concept of infinity is ... I do mathematics (cp. my book) Reprising the work of others does not count as mathematics. > That book may contain some of other people's mathematics, but contains > none of your own The footnote on p. VI is new. Then almost certainly wrong. > === Subject: Re: Alternating Vertical and Horizontal Moves In Grid posting-account=ey-PjQoAAAD_Gi3gK0ZWKofsmviYOWhV Gecko/20081217 Firefox/2.0.0.20,gzip(gfe),gzip(gfe) Hi! This is an interesting problem, that I have never seen before. I will make the assumption that at any point a vertical move may be U or D, and a horizontal move may be L or R. In both the start-at-a-corner and general case, a(1) = 1, a(2) = 4, and a(3) = 6, for starters. Now I will focus only on the start-at-a-corner case, and it may as well be the upper left. Obviously n^2 is an upper bound for a(n), and if we try to establish some reasonable lower bounds, by considering a boustrophedonic traversal of the grid, there is a natural division into cases based on the value of n mod 4. In particular, I conjecture that the optimal algorithm for sufficiently large n = 1 mod 4 is to traverse the grid according to the following rules: 1) Start with down 1) Choose up over down whenever at least 4 more moves can be made as a result of doing so. 2) Alternate between choosing right n - 1 times in a row, then left n - 1 times in a row, then right, etc. To see what I mean by this, the 9 x 9 grid would be traversed like: DRURDRURDRURDRDRDLDLULULDLULDLDLDRDRURURDRDRURDRDLDLULDLULDLULDLU. With a little work this establishes: n mod 4 = 1 -> [ n^2 - 2n + 3 <= a(n) ] For the other cases some slight modifications of the algorithm above produce pretty tight lower bounds as well. In the above case, I would expect equality probably holds, but proving it is another story, it seems like it would be tricky. dc === Subject: Re: Alternating Vertical and Horizontal Moves In Grid >What is the sequence {a(k)} that is defined as follows? Start with an n-by-n grid. Move from square to adjacent square by >moving vertically then horizontally then vertically then horizontally, >following each horizontal move with a vertical move, and following >each vertical move with a horizontal move. >No square may be visited more than once. a(n) = the maximum number of squares visitable in this way on an n-by- >n grid. By brute force search, here are the values of a(n) for n = 1, ..., 8: 1, 4, 7, 14, 19, 32, 39, 58 >A variation: You must start in a corner square. Then the first 8 values are: 1, 4, 6, 14, 18, 32, 38, 58 Note that, so far, if n is even, starting in a corner is just as good a starting place as any other location, while if n is odd and n > 1, the value of a(n) when starting in a corner appears to be just 1 less than that for an optimum starting place. >(It does not matter if the first move is vertical or horizontal, as >far as the values of each a(n) are concerned, of course.) I am guessing that this sequence is really easy to calculate, maybe a >(n) is as simple as a quadratic polynomial in terms of n. Or is the >best way to calculate each a(n) is to do a brute-force search of >possible paths through the grid? The function a(n) is definitely not a polynomial in n. Clearly a(n) is positive and bounded above by n^2, hence, if a(n) was a polynomial in n, it would have to be at most of degree 2. But the calculated values of a(n) as posted above contradict that. So what is the nature of a(n)? Consider the sequence b(n) = n^2 - a(n) Clearly b(n) must nonnegative. I suspect it would not be too hard to prove that b(n) must be positive, for n > 2. We can ask whether the sequence b(n) is bounded above? I suspect not, but if, in fact, b(n) is bounded above, then I would guess that the sequence b(n) repeats. Other than brute force search, you might try for recursive relationships, or at least recursive bounds. quasi === Subject: Re: Alternating Vertical and Horizontal Moves In Grid What is the sequence {a(k)} that is defined as follows? Start with an n-by-n grid. Move from square to adjacent square by >moving vertically then horizontally then vertically then horizontally, >following each horizontal move with a vertical move, and following >each vertical move with a horizontal move. >No square may be visited more than once. a(n) = the maximum number of squares visitable in this way on an n-by- >n grid. By brute force search, here are the values of a(n) for n = 1, ..., 8: 1, 4, 7, 14, 19, 32, 39, 58 A variation: You must start in a corner square. Then the first 8 values are: 1, 4, 6, 14, 18, 32, 38, 58 Note that, so far, if n is even, starting in a corner is just as good >a starting place as any other location, while if n is odd and n > 1, >the value of a(n) when starting in a corner appears to be just 1 less >than that for an optimum starting place. (It does not matter if the first move is vertical or horizontal, as >far as the values of each a(n) are concerned, of course.) I am guessing that this sequence is really easy to calculate, maybe a >(n) is as simple as a quadratic polynomial in terms of n. Or is the >best way to calculate each a(n) is to do a brute-force search of >possible paths through the grid? The function a(n) is definitely not a polynomial in n. Clearly a(n) is >positive and bounded above by n^2, hence, if a(n) was a polynomial in >n, it would have to be at most of degree 2. But the calculated values >of a(n) as posted above contradict that. So what is the nature of a(n)? Consider the sequence b(n) = n^2 - a(n) Clearly b(n) must nonnegative. I suspect it would not be too hard to >prove that b(n) must be positive, for n > 2. We can ask whether the sequence b(n) is bounded above? I suspect not, >but if, in fact, b(n) is bounded above, then I would guess that the >sequence b(n) repeats. Other than brute force search, you might try for recursive >relationships, or at least recursive bounds. Ok, there is a simple pattern that I missed. When n is even, it appears that a(n) = n^2 - n + 2 quasi === Subject: Re: Alternating Vertical and Horizontal Moves In Grid >What is the sequence {a(k)} that is defined as follows? Start with an n-by-n grid. Move from square to adjacent square by >moving vertically then horizontally then vertically then horizontally, >following each horizontal move with a vertical move, and following >each vertical move with a horizontal move. >No square may be visited more than once. a(n) = the maximum number of squares visitable in this way on an n-by- >n grid. By brute force search, here are the values of a(n) for n = 1, ..., 8: 1, 4, 7, 14, 19, 32, 39, 58 A variation: You must start in a corner square. Then the first 8 values are: 1, 4, 6, 14, 18, 32, 38, 58 Note that, so far, if n is even, starting in a corner is just as good >a starting place as any other location, while if n is odd and n > 1, >the value of a(n) when starting in a corner appears to be just 1 less >than that for an optimum starting place. (It does not matter if the first move is vertical or horizontal, as >far as the values of each a(n) are concerned, of course.) I am guessing that this sequence is really easy to calculate, maybe a >(n) is as simple as a quadratic polynomial in terms of n. Or is the >best way to calculate each a(n) is to do a brute-force search of >possible paths through the grid? The function a(n) is definitely not a polynomial in n. Clearly a(n) is >positive and bounded above by n^2, hence, if a(n) was a polynomial in >n, it would have to be at most of degree 2. But the calculated values >of a(n) as posted above contradict that. So what is the nature of a(n)? Consider the sequence b(n) = n^2 - a(n) Clearly b(n) must nonnegative. I suspect it would not be too hard to >prove that b(n) must be positive, for n > 2. We can ask whether the sequence b(n) is bounded above? I suspect not, >but if, in fact, b(n) is bounded above, then I would guess that the >sequence b(n) repeats. Other than brute force search, you might try for recursive >relationships, or at least recursive bounds. Ok, there is a simple pattern that I missed. When n is even, it appears that a(n) = n^2 - n + 2 Moreover, it's fairly easy to show that, for even n, a(n) >= n^2 - n + 2 Assuming n is even, a valid path through n^2 - n + 2 vertices can be explicitly constructed as follows ... Start at (0,0). Go right, up, left, up. Repeat until reaching the top. Since n is even, the last 2 moves must have been right, up, ending at (1,n-1). Now go right, down, ending at (2,n-2). We have so far visited 2n + 1 vertices. Now recursively do the same in the n-2 by n-2 square with upper left corner vertex (2,n-2) and lower right corner vertex (n-1,1). The path described above adheres to the restrictions, and it's easily proved, by induction, that it has exactly n^2 - n + 2 vertices. quasi === Subject: Re: Alternating Vertical and Horizontal Moves In Grid What is the sequence {a(k)} that is defined as follows? Start with an n-by-n grid. Move from square to adjacent square by >moving vertically then horizontally then vertically then horizontally, >following each horizontal move with a vertical move, and following >each vertical move with a horizontal move. >No square may be visited more than once. a(n) = the maximum number of squares visitable in this way on an n-by- >n grid. By brute force search, here are the values of a(n) for n = 1, ..., 8: 1, 4, 7, 14, 19, 32, 39, 58 A variation: You must start in a corner square. Then the first 8 values are: 1, 4, 6, 14, 18, 32, 38, 58 Note that, so far, if n is even, starting in a corner is just as good >a starting place as any other location, while if n is odd and n > 1, >the value of a(n) when starting in a corner appears to be just 1 less >than that for an optimum starting place. (It does not matter if the first move is vertical or horizontal, as >far as the values of each a(n) are concerned, of course.) I am guessing that this sequence is really easy to calculate, maybe a >(n) is as simple as a quadratic polynomial in terms of n. Or is the >best way to calculate each a(n) is to do a brute-force search of >possible paths through the grid? The function a(n) is definitely not a polynomial in n. Clearly a(n) is >positive and bounded above by n^2, hence, if a(n) was a polynomial in >n, it would have to be at most of degree 2. But the calculated values >of a(n) as posted above contradict that. So what is the nature of a(n)? Consider the sequence b(n) = n^2 - a(n) Clearly b(n) must nonnegative. I suspect it would not be too hard to >prove that b(n) must be positive, for n > 2. We can ask whether the sequence b(n) is bounded above? I suspect not, >but if, in fact, b(n) is bounded above, then I would guess that the >sequence b(n) repeats. Other than brute force search, you might try for recursive >relationships, or at least recursive bounds. Ok, there is a simple pattern that I missed. When n is even, it appears that a(n) = n^2 - n + 2 Moreover, it's fairly easy to show that, for even n, a(n) >= n^2 - n + 2 Assuming n is even, a valid path through n^2 - n + 2 vertices can be >explicitly constructed as follows ... Start at (0,0). Go right, up, left, up. Repeat until reaching the top. Since n is even, the last 2 moves must have been right, up, ending at >(1,n-1). Now go right, down, ending at (2,n-2). We have so far visited 2n + 1 vertices. Now recursively do the same in the n-2 by n-2 square with upper left >corner vertex (2,n-2) and lower right corner vertex (n-1,1).| To clarify: At each successive lower level in the recursion, switch up and down. Also, when you finish the 2 x 2 level, stop. >The path described above adheres to the restrictions, and it's easily >proved, by induction, that it has exactly n^2 - n + 2 vertices. quasi === Subject: Re: Alternating Vertical and Horizontal Moves In Grid <4utbq4dhhldvts8o5hl3ld180tan2tbdmq@4ax.com> posting-account=ey-PjQoAAAD_Gi3gK0ZWKofsmviYOWhV Gecko/20081217 Firefox/2.0.0.20,gzip(gfe),gzip(gfe) What is the sequence {a(k)} that is defined as follows? Start with an n-by-n grid. Move from square to adjacent square by >moving vertically then horizontally then vertically then horizontally, >following each horizontal move with a vertical move, and following >each vertical move with a horizontal move. >No square may be visited more than once. a(n) = the maximum number of squares visitable in this way on an n-by- >n grid. By brute force search, here are the values of a(n) for n = 1, ..., 8: 1, 4, 7, 14, 19, 32, 39, 58 A variation: You must start in a corner square. Then the first 8 values are: 1, 4, 6, 14, 18, 32, 38, 58 Note that, so far, if n is even, starting in a corner is just as good >a starting place as any other location, while if n is odd and n > 1, >the value of a(n) when starting in a corner appears to be just 1 less >than that for an optimum starting place. (It does not matter if the first move is vertical or horizontal, as >far as the values of each a(n) are concerned, of course.) I am guessing that this sequence is really easy to calculate, maybe a >(n) is as simple as a quadratic polynomial in terms of n. Or is the >best way to calculate each a(n) is to do a brute-force search of >possible paths through the grid? The function a(n) is definitely not a polynomial in n. Clearly a(n) is >positive and bounded above by n^2, hence, if a(n) was a polynomial in >n, it would have to be at most of degree 2. But the calculated values >of a(n) as posted above contradict that. So what is the nature of a(n)? Consider the sequence b(n) = n^2 - a(n) Clearly b(n) must nonnegative. I suspect it would not be too hard to >prove that b(n) must be positive, for n > 2. We can ask whether the sequence b(n) is bounded above? I suspect not, >but if, in fact, b(n) is bounded above, then I would guess that the >sequence b(n) repeats. Other than brute force search, you might try for recursive >relationships, or at least recursive bounds. Ok, there is a simple pattern that I missed. When n is even, it appears that a(n) = n^2 - n + 2 Moreover, it's fairly easy to show that, for even n, a(n) >= n^2 - n + 2 Assuming n is even, a valid path through n^2 - n + 2 vertices can be >explicitly constructed as follows ... Start at (0,0). Go right, up, left, up. Repeat until reaching the top. Since n is even, the last 2 moves must have been right, up, ending at >(1,n-1). Now go right, down, ending at (2,n-2). We have so far visited 2n + 1 vertices. Now recursively do the same in the n-2 by n-2 square with upper left >corner vertex (2,n-2) and lower right corner vertex (n-1,1).| To clarify: At each successive lower level in the recursion, switch up and > down. Also, when you finish the 2 x 2 level, stop. The path described above adheres to the restrictions, and it's easily >proved, by induction, that it has exactly n^2 - n + 2 vertices. quasi this seems legitimate to me. quasi, by the sounds of it, you have some kind of a computer program capable of a brute force test? can you test my conjecture above for n = 9, 13? i'm thinking we are going tol end up here with a quadratic answer, but different quadratics depending on the value of n mod 4. === Subject: Re: Alternating Vertical and Horizontal Moves In Grid What is the sequence {a(k)} that is defined as follows? Start with an n-by-n grid. Move from square to adjacent square by >moving vertically then horizontally then vertically then horizontally, >following each horizontal move with a vertical move, and following >each vertical move with a horizontal move. >No square may be visited more than once. a(n) = the maximum number of squares visitable in this way on an n-by- >n grid. By brute force search, here are the values of a(n) for n = 1, ..., 8: 1, 4, 7, 14, 19, 32, 39, 58 A variation: You must start in a corner square. Then the first 8 values are: 1, 4, 6, 14, 18, 32, 38, 58 Note that, so far, if n is even, starting in a corner is just as good >a starting place as any other location, while if n is odd and n > 1, >the value of a(n) when starting in a corner appears to be just 1 less >than that for an optimum starting place. (It does not matter if the first move is vertical or horizontal, as >far as the values of each a(n) are concerned, of course.) I am guessing that this sequence is really easy to calculate, maybe a >(n) is as simple as a quadratic polynomial in terms of n. Or is the >best way to calculate each a(n) is to do a brute-force search of >possible paths through the grid? The function a(n) is definitely not a polynomial in n. Clearly a(n) is >positive and bounded above by n^2, hence, if a(n) was a polynomial in >n, it would have to be at most of degree 2. But the calculated values >of a(n) as posted above contradict that. So what is the nature of a(n)? Consider the sequence b(n) = n^2 - a(n) Clearly b(n) must nonnegative. I suspect it would not be too hard to >prove that b(n) must be positive, for n > 2. We can ask whether the sequence b(n) is bounded above? I suspect not, >but if, in fact, b(n) is bounded above, then I would guess that the >sequence b(n) repeats. Other than brute force search, you might try for recursive >relationships, or at least recursive bounds. Ok, there is a simple pattern that I missed. When n is even, it appears that a(n) = n^2 - n + 2 Moreover, it's fairly easy to show that, for even n, a(n) >= n^2 - n + 2 Assuming n is even, a valid path through n^2 - n + 2 vertices can be >explicitly constructed as follows ... Start at (0,0). Go right, up, left, up. Repeat until reaching the top. Since n is even, the last 2 moves must have been right, up, ending at >(1,n-1). Now go right, down, ending at (2,n-2). We have so far visited 2n + 1 vertices. Now recursively do the same in the n-2 by n-2 square with upper left >corner vertex (2,n-2) and lower right corner vertex (n-1,1).| To clarify: At each successive lower level in the recursion, switch up and > down. Also, when you finish the 2 x 2 level, stop. The path described above adheres to the restrictions, and it's easily >proved, by induction, that it has exactly n^2 - n + 2 vertices. quasi this seems legitimate to me. quasi, by the sounds of it, you have some >kind of a computer program capable of a brute force test? can you test >my conjecture above for n = 9, 13? i'm thinking we are going tol end >up here with a quadratic answer, but different quadratics depending on >the value of n mod 4. Yes, that seems very plausible -- 3 quadratics, one for even n, one for n = 1 mod 4, and one for n = 3 mod 4. I do have a brute force test, but it wasn't designed for a heavy load. For n > 8, it struggles. I'll let it run for a while -- hopefully, it will at least compute a(9). quasi === Subject: Re: Alternating Vertical and Horizontal Moves In Grid <4utbq4dhhldvts8o5hl3ld180tan2tbdmq@4ax.com> posting-account=ey-PjQoAAAD_Gi3gK0ZWKofsmviYOWhV Gecko/20081217 Firefox/2.0.0.20,gzip(gfe),gzip(gfe) no, i changed my mind about the mod 4 thing, it looks a lot like your formula is right whenever n is even, and my formula is right (by a similar argument) when n is odd. of course, we have only established them as lower bounds with certainty. === Subject: Re: Alternating Vertical and Horizontal Moves In Grid >no, i changed my mind about the mod 4 thing, it looks a lot like your >formula is right whenever n is even, and my formula is right (by a >similar argument) when n is odd. of course, we have only established >them as lower bounds with certainty. Yes, if we require corner starting point, and assuming n > 1, I think you may be right. Thus, for a corner starting point, and n > 1, the conjecture is: a(n) = n^2 - 2n + 3, if n is odd n^2 - n + 2, if n is even quasi PS -- My Maple program is still working on a(9), assuming a required corner starting point. === Subject: Re: Alternating Vertical and Horizontal Moves In Grid no, i changed my mind about the mod 4 thing, it looks a lot like your >formula is right whenever n is even, and my formula is right (by a >similar argument) when n is odd. of course, we have only established >them as lower bounds with certainty. Yes, if we require corner starting point, and assuming n > 1, I think >you may be right. Thus, for a corner starting point, and n > 1, the conjecture is: a(n) = > n^2 - 2n + 3, if n is odd > n^2 - n + 2, if n is even quasi PS -- My Maple program is still working on a(9), assuming a required >corner starting point. Ok, Maple completed a(9) and is now working on a(11). Good news -- a(9) = 66, as expected. Note, for convenience, we are now always making the assumption of a corner starting point. For even n, it apparently doesn't matter. For odd n > 1, allowing a variable starting point appears to make a(n) increase by 1. quasi === Subject: Re: Alternating Vertical and Horizontal Moves In Grid <4utbq4dhhldvts8o5hl3ld180tan2tbdmq@4ax.com> <4p7cq45og61mehq711tc9qm7ir9avhp4gl@4ax.com> posting-account=ey-PjQoAAAD_Gi3gK0ZWKofsmviYOWhV Gecko/20081217 Firefox/2.0.0.20,gzip(gfe),gzip(gfe) no, i changed my mind about the mod 4 thing, it looks a lot like your >formula is right whenever n is even, and my formula is right (by a >similar argument) when n is odd. of course, we have only established >them as lower bounds with certainty. Yes, if we require corner starting point, and assuming n > 1, I think >you may be right. Thus, for a corner starting point, and n > 1, the conjecture is: a(n) = > n^2 - 2n + 3, if n is odd > n^2 - n + 2, if n is even quasi PS -- My Maple program is still working on a(9), assuming a required >corner starting point. Ok, Maple completed a(9) and is now working on a(11). Good news -- a(9) = 66, as expected. Note, for convenience, we are now always making the assumption of a > corner starting point. For even n, it apparently doesn't matter. For > odd n > 1, allowing a variable starting point appears to make a(n) > increase by 1. quasi this makes sense to me, i am starting to work on a proof of the bound for the even case, as i think it will be the easier one. i have some ideas, but nothing concrete enough to comment on yet. === Subject: Re: Newton's Method: Cute trick. posting-account=DSjMzAoAAAAISwlGnTY2c6bnmUVmdFNf AppleWebKit/528.16 (KHTML, like Gecko) Version/4.0 Safari/528.16,gzip(gfe),gzip(gfe) I have never seen this trick explicitly mentioned anywhere. Of course if f (p) < 0, f (q) > 0, f continuous on [p, q] then there is a root of f in (p, q), and if you use _any_ iterative algorithm, then clamping the next iterated result to [p, q] will get you closer to the root. Some things I noted: (1) If you find x0 such that f (x0) is small enough, but x0 is not in [p, q] then clamping x0 to [p, q] might give a new value where f (x) is _not_ small enough anymore, so care must be taken. (2) Clamping can actually only happen on the first Newton-Raphson step if your convergence criterion is actually met, because you proved that each iteration gets closer to the root of F (x). If you are not 100% sure that the convergence criterion is met, clamping might not help. (3) This doesn't solve the problem of finding x0 close enough to the root. Of course it is no big deal if a and b are known. === Subject: Re: Help!How to solve the equation === Subject: Re: Alternating Vertical and Horizontal Moves In Grid posting-account=ey-PjQoAAAD_Gi3gK0ZWKofsmviYOWhV Gecko/20081217 Firefox/2.0.0.20,gzip(gfe),gzip(gfe) Hi! This is an interesting problem, that I have never seen before. I will make the assumption that at any point a vertical move may be U or D, and a horizontal move may be L or R. In both the start-at-a-corner and general case, a(1) = 1, a(2) = 4, and a(3) = 6, for starters. Now I will focus only on the start-at-a-corner case, and it may as well be the upper left. Obviously n^2 is an upper bound for a(n), and if we try to establish some reasonable lower bounds, by considering a boustrophedonic traversal of the grid, there is a natural division into cases based on the value of n mod 4. In particular, I conjecture that the optimal algorithm for sufficiently large n = 1 mod 4 is to traverse the grid according to the following rules: 1) Start with down 1) Choose up over down whenever at least 4 more moves can be made as a result of doing so. 2) Alternate between choosing right n - 1 times in a row, then left n - 1 times in a row, then right, etc. To see what I mean by this, the 9 x 9 grid would be traversed like: DRURDRURDRURDRDRDLDLULULDLULDLDLDRDRURURDRDRURDRDLDLULDLULDLULDLU. With a little work this establishes: n mod 4 = 1 -> [ n^2 - 2n + 3 <= a(n) ] For the other cases some slight modifications of the algorithm above produce pretty tight lower bounds as well. In the above case, I would expect equality probably holds, but proving it is another story, it seems like it would be tricky. dc === Subject: Re: Alternating Vertical and Horizontal Moves In Grid >What is the sequence {a(k)} that is defined as follows? Start with an n-by-n grid. Move from square to adjacent square by >moving vertically then horizontally then vertically then horizontally, >following each horizontal move with a vertical move, and following >each vertical move with a horizontal move. >No square may be visited more than once. a(n) = the maximum number of squares visitable in this way on an n-by- >n grid. By brute force search, here are the values of a(n) for n = 1, ..., 8: 1, 4, 7, 14, 19, 32, 39, 58 >A variation: You must start in a corner square. Then the first 8 values are: 1, 4, 6, 14, 18, 32, 38, 58 Note that, so far, if n is even, starting in a corner is just as good a starting place as any other location, while if n is odd and n > 1, the value of a(n) when starting in a corner appears to be just 1 less than that for an optimum starting place. >(It does not matter if the first move is vertical or horizontal, as >far as the values of each a(n) are concerned, of course.) I am guessing that this sequence is really easy to calculate, maybe a >(n) is as simple as a quadratic polynomial in terms of n. Or is the >best way to calculate each a(n) is to do a brute-force search of >possible paths through the grid? The function a(n) is definitely not a polynomial in n. Clearly a(n) is positive and bounded above by n^2, hence, if a(n) was a polynomial in n, it would have to be at most of degree 2. But the calculated values of a(n) as posted above contradict that. So what is the nature of a(n)? Consider the sequence b(n) = n^2 - a(n) Clearly b(n) must nonnegative. I suspect it would not be too hard to prove that b(n) must be positive, for n > 2. We can ask whether the sequence b(n) is bounded above? I suspect not, but if, in fact, b(n) is bounded above, then I would guess that the sequence b(n) repeats. Other than brute force search, you might try for recursive relationships, or at least recursive bounds. quasi === Subject: Re: Alternating Vertical and Horizontal Moves In Grid What is the sequence {a(k)} that is defined as follows? Start with an n-by-n grid. Move from square to adjacent square by >moving vertically then horizontally then vertically then horizontally, >following each horizontal move with a vertical move, and following >each vertical move with a horizontal move. >No square may be visited more than once. a(n) = the maximum number of squares visitable in this way on an n-by- >n grid. By brute force search, here are the values of a(n) for n = 1, ..., 8: 1, 4, 7, 14, 19, 32, 39, 58 A variation: You must start in a corner square. Then the first 8 values are: 1, 4, 6, 14, 18, 32, 38, 58 Note that, so far, if n is even, starting in a corner is just as good >a starting place as any other location, while if n is odd and n > 1, >the value of a(n) when starting in a corner appears to be just 1 less >than that for an optimum starting place. (It does not matter if the first move is vertical or horizontal, as >far as the values of each a(n) are concerned, of course.) I am guessing that this sequence is really easy to calculate, maybe a >(n) is as simple as a quadratic polynomial in terms of n. Or is the >best way to calculate each a(n) is to do a brute-force search of >possible paths through the grid? The function a(n) is definitely not a polynomial in n. Clearly a(n) is >positive and bounded above by n^2, hence, if a(n) was a polynomial in >n, it would have to be at most of degree 2. But the calculated values >of a(n) as posted above contradict that. So what is the nature of a(n)? Consider the sequence b(n) = n^2 - a(n) Clearly b(n) must nonnegative. I suspect it would not be too hard to >prove that b(n) must be positive, for n > 2. We can ask whether the sequence b(n) is bounded above? I suspect not, >but if, in fact, b(n) is bounded above, then I would guess that the >sequence b(n) repeats. Other than brute force search, you might try for recursive >relationships, or at least recursive bounds. Ok, there is a simple pattern that I missed. When n is even, it appears that a(n) = n^2 - n + 2 quasi === Subject: Re: Alternating Vertical and Horizontal Moves In Grid >What is the sequence {a(k)} that is defined as follows? Start with an n-by-n grid. Move from square to adjacent square by >moving vertically then horizontally then vertically then horizontally, >following each horizontal move with a vertical move, and following >each vertical move with a horizontal move. >No square may be visited more than once. a(n) = the maximum number of squares visitable in this way on an n-by- >n grid. By brute force search, here are the values of a(n) for n = 1, ..., 8: 1, 4, 7, 14, 19, 32, 39, 58 A variation: You must start in a corner square. Then the first 8 values are: 1, 4, 6, 14, 18, 32, 38, 58 Note that, so far, if n is even, starting in a corner is just as good >a starting place as any other location, while if n is odd and n > 1, >the value of a(n) when starting in a corner appears to be just 1 less >than that for an optimum starting place. (It does not matter if the first move is vertical or horizontal, as >far as the values of each a(n) are concerned, of course.) I am guessing that this sequence is really easy to calculate, maybe a >(n) is as simple as a quadratic polynomial in terms of n. Or is the >best way to calculate each a(n) is to do a brute-force search of >possible paths through the grid? The function a(n) is definitely not a polynomial in n. Clearly a(n) is >positive and bounded above by n^2, hence, if a(n) was a polynomial in >n, it would have to be at most of degree 2. But the calculated values >of a(n) as posted above contradict that. So what is the nature of a(n)? Consider the sequence b(n) = n^2 - a(n) Clearly b(n) must nonnegative. I suspect it would not be too hard to >prove that b(n) must be positive, for n > 2. We can ask whether the sequence b(n) is bounded above? I suspect not, >but if, in fact, b(n) is bounded above, then I would guess that the >sequence b(n) repeats. Other than brute force search, you might try for recursive >relationships, or at least recursive bounds. Ok, there is a simple pattern that I missed. When n is even, it appears that a(n) = n^2 - n + 2 Moreover, it's fairly easy to show that, for even n, a(n) >= n^2 - n + 2 Assuming n is even, a valid path through n^2 - n + 2 vertices can be explicitly constructed as follows ... Start at (0,0). Go right, up, left, up. Repeat until reaching the top. Since n is even, the last 2 moves must have been right, up, ending at (1,n-1). Now go right, down, ending at (2,n-2). We have so far visited 2n + 1 vertices. Now recursively do the same in the n-2 by n-2 square with upper left corner vertex (2,n-2) and lower right corner vertex (n-1,1). The path described above adheres to the restrictions, and it's easily proved, by induction, that it has exactly n^2 - n + 2 vertices. quasi === Subject: Re: Alternating Vertical and Horizontal Moves In Grid What is the sequence {a(k)} that is defined as follows? Start with an n-by-n grid. Move from square to adjacent square by >moving vertically then horizontally then vertically then horizontally, >following each horizontal move with a vertical move, and following >each vertical move with a horizontal move. >No square may be visited more than once. a(n) = the maximum number of squares visitable in this way on an n-by- >n grid. By brute force search, here are the values of a(n) for n = 1, ..., 8: 1, 4, 7, 14, 19, 32, 39, 58 A variation: You must start in a corner square. Then the first 8 values are: 1, 4, 6, 14, 18, 32, 38, 58 Note that, so far, if n is even, starting in a corner is just as good >a starting place as any other location, while if n is odd and n > 1, >the value of a(n) when starting in a corner appears to be just 1 less >than that for an optimum starting place. (It does not matter if the first move is vertical or horizontal, as >far as the values of each a(n) are concerned, of course.) I am guessing that this sequence is really easy to calculate, maybe a >(n) is as simple as a quadratic polynomial in terms of n. Or is the >best way to calculate each a(n) is to do a brute-force search of >possible paths through the grid? The function a(n) is definitely not a polynomial in n. Clearly a(n) is >positive and bounded above by n^2, hence, if a(n) was a polynomial in >n, it would have to be at most of degree 2. But the calculated values >of a(n) as posted above contradict that. So what is the nature of a(n)? Consider the sequence b(n) = n^2 - a(n) Clearly b(n) must nonnegative. I suspect it would not be too hard to >prove that b(n) must be positive, for n > 2. We can ask whether the sequence b(n) is bounded above? I suspect not, >but if, in fact, b(n) is bounded above, then I would guess that the >sequence b(n) repeats. Other than brute force search, you might try for recursive >relationships, or at least recursive bounds. Ok, there is a simple pattern that I missed. When n is even, it appears that a(n) = n^2 - n + 2 Moreover, it's fairly easy to show that, for even n, a(n) >= n^2 - n + 2 Assuming n is even, a valid path through n^2 - n + 2 vertices can be >explicitly constructed as follows ... Start at (0,0). Go right, up, left, up. Repeat until reaching the top. Since n is even, the last 2 moves must have been right, up, ending at >(1,n-1). Now go right, down, ending at (2,n-2). We have so far visited 2n + 1 vertices. Now recursively do the same in the n-2 by n-2 square with upper left >corner vertex (2,n-2) and lower right corner vertex (n-1,1).| To clarify: At each successive lower level in the recursion, switch up and down. Also, when you finish the 2 x 2 level, stop. >The path described above adheres to the restrictions, and it's easily >proved, by induction, that it has exactly n^2 - n + 2 vertices. quasi === Subject: Re: Alternating Vertical and Horizontal Moves In Grid <4utbq4dhhldvts8o5hl3ld180tan2tbdmq@4ax.com> posting-account=ey-PjQoAAAD_Gi3gK0ZWKofsmviYOWhV Gecko/20081217 Firefox/2.0.0.20,gzip(gfe),gzip(gfe) What is the sequence {a(k)} that is defined as follows? Start with an n-by-n grid. Move from square to adjacent square by >moving vertically then horizontally then vertically then horizontally, >following each horizontal move with a vertical move, and following >each vertical move with a horizontal move. >No square may be visited more than once. a(n) = the maximum number of squares visitable in this way on an n-by- >n grid. By brute force search, here are the values of a(n) for n = 1, ..., 8: 1, 4, 7, 14, 19, 32, 39, 58 A variation: You must start in a corner square. Then the first 8 values are: 1, 4, 6, 14, 18, 32, 38, 58 Note that, so far, if n is even, starting in a corner is just as good >a starting place as any other location, while if n is odd and n > 1, >the value of a(n) when starting in a corner appears to be just 1 less >than that for an optimum starting place. (It does not matter if the first move is vertical or horizontal, as >far as the values of each a(n) are concerned, of course.) I am guessing that this sequence is really easy to calculate, maybe a >(n) is as simple as a quadratic polynomial in terms of n. Or is the >best way to calculate each a(n) is to do a brute-force search of >possible paths through the grid? The function a(n) is definitely not a polynomial in n. Clearly a(n) is >positive and bounded above by n^2, hence, if a(n) was a polynomial in >n, it would have to be at most of degree 2. But the calculated values >of a(n) as posted above contradict that. So what is the nature of a(n)? Consider the sequence b(n) = n^2 - a(n) Clearly b(n) must nonnegative. I suspect it would not be too hard to >prove that b(n) must be positive, for n > 2. We can ask whether the sequence b(n) is bounded above? I suspect not, >but if, in fact, b(n) is bounded above, then I would guess that the >sequence b(n) repeats. Other than brute force search, you might try for recursive >relationships, or at least recursive bounds. Ok, there is a simple pattern that I missed. When n is even, it appears that a(n) = n^2 - n + 2 Moreover, it's fairly easy to show that, for even n, a(n) >= n^2 - n + 2 Assuming n is even, a valid path through n^2 - n + 2 vertices can be >explicitly constructed as follows ... Start at (0,0). Go right, up, left, up. Repeat until reaching the top. Since n is even, the last 2 moves must have been right, up, ending at >(1,n-1). Now go right, down, ending at (2,n-2). We have so far visited 2n + 1 vertices. Now recursively do the same in the n-2 by n-2 square with upper left >corner vertex (2,n-2) and lower right corner vertex (n-1,1).| To clarify: At each successive lower level in the recursion, switch up and > down. Also, when you finish the 2 x 2 level, stop. The path described above adheres to the restrictions, and it's easily >proved, by induction, that it has exactly n^2 - n + 2 vertices. quasi this seems legitimate to me. quasi, by the sounds of it, you have some kind of a computer program capable of a brute force test? can you test my conjecture above for n = 9, 13? i'm thinking we are going tol end up here with a quadratic answer, but different quadratics depending on the value of n mod 4. === Subject: Re: Alternating Vertical and Horizontal Moves In Grid What is the sequence {a(k)} that is defined as follows? Start with an n-by-n grid. Move from square to adjacent square by >moving vertically then horizontally then vertically then horizontally, >following each horizontal move with a vertical move, and following >each vertical move with a horizontal move. >No square may be visited more than once. a(n) = the maximum number of squares visitable in this way on an n-by- >n grid. By brute force search, here are the values of a(n) for n = 1, ..., 8: 1, 4, 7, 14, 19, 32, 39, 58 A variation: You must start in a corner square. Then the first 8 values are: 1, 4, 6, 14, 18, 32, 38, 58 Note that, so far, if n is even, starting in a corner is just as good >a starting place as any other location, while if n is odd and n > 1, >the value of a(n) when starting in a corner appears to be just 1 less >than that for an optimum starting place. (It does not matter if the first move is vertical or horizontal, as >far as the values of each a(n) are concerned, of course.) I am guessing that this sequence is really easy to calculate, maybe a >(n) is as simple as a quadratic polynomial in terms of n. Or is the >best way to calculate each a(n) is to do a brute-force search of >possible paths through the grid? The function a(n) is definitely not a polynomial in n. Clearly a(n) is >positive and bounded above by n^2, hence, if a(n) was a polynomial in >n, it would have to be at most of degree 2. But the calculated values >of a(n) as posted above contradict that. So what is the nature of a(n)? Consider the sequence b(n) = n^2 - a(n) Clearly b(n) must nonnegative. I suspect it would not be too hard to >prove that b(n) must be positive, for n > 2. We can ask whether the sequence b(n) is bounded above? I suspect not, >but if, in fact, b(n) is bounded above, then I would guess that the >sequence b(n) repeats. Other than brute force search, you might try for recursive >relationships, or at least recursive bounds. Ok, there is a simple pattern that I missed. When n is even, it appears that a(n) = n^2 - n + 2 Moreover, it's fairly easy to show that, for even n, a(n) >= n^2 - n + 2 Assuming n is even, a valid path through n^2 - n + 2 vertices can be >explicitly constructed as follows ... Start at (0,0). Go right, up, left, up. Repeat until reaching the top. Since n is even, the last 2 moves must have been right, up, ending at >(1,n-1). Now go right, down, ending at (2,n-2). We have so far visited 2n + 1 vertices. Now recursively do the same in the n-2 by n-2 square with upper left >corner vertex (2,n-2) and lower right corner vertex (n-1,1).| To clarify: At each successive lower level in the recursion, switch up and > down. Also, when you finish the 2 x 2 level, stop. The path described above adheres to the restrictions, and it's easily >proved, by induction, that it has exactly n^2 - n + 2 vertices. quasi this seems legitimate to me. quasi, by the sounds of it, you have some >kind of a computer program capable of a brute force test? can you test >my conjecture above for n = 9, 13? i'm thinking we are going tol end >up here with a quadratic answer, but different quadratics depending on >the value of n mod 4. Yes, that seems very plausible -- 3 quadratics, one for even n, one for n = 1 mod 4, and one for n = 3 mod 4. I do have a brute force test, but it wasn't designed for a heavy load. For n > 8, it struggles. I'll let it run for a while -- hopefully, it will at least compute a(9). quasi === Subject: Re: Alternating Vertical and Horizontal Moves In Grid <4utbq4dhhldvts8o5hl3ld180tan2tbdmq@4ax.com> posting-account=ey-PjQoAAAD_Gi3gK0ZWKofsmviYOWhV Gecko/20081217 Firefox/2.0.0.20,gzip(gfe),gzip(gfe) no, i changed my mind about the mod 4 thing, it looks a lot like your formula is right whenever n is even, and my formula is right (by a similar argument) when n is odd. of course, we have only established them as lower bounds with certainty. === Subject: Re: Alternating Vertical and Horizontal Moves In Grid >no, i changed my mind about the mod 4 thing, it looks a lot like your >formula is right whenever n is even, and my formula is right (by a >similar argument) when n is odd. of course, we have only established >them as lower bounds with certainty. Yes, if we require corner starting point, and assuming n > 1, I think you may be right. Thus, for a corner starting point, and n > 1, the conjecture is: a(n) = n^2 - 2n + 3, if n is odd n^2 - n + 2, if n is even quasi PS -- My Maple program is still working on a(9), assuming a required corner starting point. === Subject: Re: Alternating Vertical and Horizontal Moves In Grid no, i changed my mind about the mod 4 thing, it looks a lot like your >formula is right whenever n is even, and my formula is right (by a >similar argument) when n is odd. of course, we have only established >them as lower bounds with certainty. Yes, if we require corner starting point, and assuming n > 1, I think >you may be right. Thus, for a corner starting point, and n > 1, the conjecture is: a(n) = > n^2 - 2n + 3, if n is odd > n^2 - n + 2, if n is even quasi PS -- My Maple program is still working on a(9), assuming a required >corner starting point. Ok, Maple completed a(9) and is now working on a(11). Good news -- a(9) = 66, as expected. Note, for convenience, we are now always making the assumption of a corner starting point. For even n, it apparently doesn't matter. For odd n > 1, allowing a variable starting point appears to make a(n) increase by 1. quasi === Subject: Re: Alternating Vertical and Horizontal Moves In Grid <4utbq4dhhldvts8o5hl3ld180tan2tbdmq@4ax.com> <4p7cq45og61mehq711tc9qm7ir9avhp4gl@4ax.com> posting-account=ey-PjQoAAAD_Gi3gK0ZWKofsmviYOWhV Gecko/20081217 Firefox/2.0.0.20,gzip(gfe),gzip(gfe) no, i changed my mind about the mod 4 thing, it looks a lot like your >formula is right whenever n is even, and my formula is right (by a >similar argument) when n is odd. of course, we have only established >them as lower bounds with certainty. Yes, if we require corner starting point, and assuming n > 1, I think >you may be right. Thus, for a corner starting point, and n > 1, the conjecture is: a(n) = > n^2 - 2n + 3, if n is odd > n^2 - n + 2, if n is even quasi PS -- My Maple program is still working on a(9), assuming a required >corner starting point. Ok, Maple completed a(9) and is now working on a(11). Good news -- a(9) = 66, as expected. Note, for convenience, we are now always making the assumption of a > corner starting point. For even n, it apparently doesn't matter. For > odd n > 1, allowing a variable starting point appears to make a(n) > increase by 1. quasi this makes sense to me, i am starting to work on a proof of the bound for the even case, as i think it will be the easier one. i have some ideas, but nothing concrete enough to comment on yet. === Subject: Re: Newton's Method: Cute trick. posting-account=DSjMzAoAAAAISwlGnTY2c6bnmUVmdFNf AppleWebKit/528.16 (KHTML, like Gecko) Version/4.0 Safari/528.16,gzip(gfe),gzip(gfe) I have never seen this trick explicitly mentioned anywhere. Of course if f (p) < 0, f (q) > 0, f continuous on [p, q] then there is a root of f in (p, q), and if you use _any_ iterative algorithm, then clamping the next iterated result to [p, q] will get you closer to the root. Some things I noted: (1) If you find x0 such that f (x0) is small enough, but x0 is not in [p, q] then clamping x0 to [p, q] might give a new value where f (x) is _not_ small enough anymore, so care must be taken. (2) Clamping can actually only happen on the first Newton-Raphson step if your convergence criterion is actually met, because you proved that each iteration gets closer to the root of F (x). If you are not 100% sure that the convergence criterion is met, clamping might not help. (3) This doesn't solve the problem of finding x0 close enough to the root. Of course it is no big deal if a and b are known. === Subject: Re: Help!How to solve the equation === Subject: Matrix A, (m x n), m>=n with Singular Values (s_j) posting-account=Qq_GEwoAAAD0Kc2k70q6xwuZdqx8sCzT 4334.5003; Windows NT 5.1; SV1; .NET CLR 2.0.50727; .NET CLR 3.0.04506.648; .NET CLR 3.5.21022),gzip(gfe),gzip(gfe) spider-mtc-tf12.proxy.aol.com[400C70AC] (Prism/1.2.1), HTTP/1.1 cache-mtc-ab12.proxy.aol.com[400C744C] (Traffic-Server/6.1.5 [uScM]) Matrix A, (m x n), m>=n with singular values (s_j) Show that the singular values of ( I_n A )^T = sqrt[ 1 + (s_j)^2 ] where I_n identity matrix (n x n). i thought someone who speaks and acts positively on the internet knows the direction to go in. A=UEV'. I can see E but not U,V, or A. when i get out of the car i can drink an Icehouse and make some progress (brood over it). We've covered a lot of SVD in class but with very few examples. === Subject: Re: Matrix A, (m x n), m>=n with Singular Values (s_j) posting-account=O9zR9AkAAACmp918j6u5m5plppeILcze Filter 1.2.0.72; GTB5; .NET CLR 1.0.3705; .NET CLR 1.1.4322; Media Center PC 4.0; .NET CLR 2.0.50727; .NET CLR 3.0.04506.648; .NET CLR 3.5.21022; \ .NET CLR 3.0.4506.2152; .NET CLR 3.5.30729),gzip(gfe),gzip(gfe) > Matrix A, (m x n), m>=n with singular values (s j) Show that the singular values of æ( I n æA æ)^T = sqrt[ 1 + (s j)^2 ] where I n identity matrix (n x n). I n A makes sense only if n = m. Dave === Subject: Re: Matrix A, (m x n), m>=n with Singular Values (s_j) posting-account=Qq_GEwoAAAD0Kc2k70q6xwuZdqx8sCzT 4334.5003; Windows NT 5.1; SV1; .NET CLR 2.0.50727; .NET CLR 3.0.04506.648; .NET CLR 3.5.21022),gzip(gfe),gzip(gfe) spider-mtc-ta07.proxy.aol.com[400C7007] (Prism/1.2.1), HTTP/1.1 cache-mtc-aa10.proxy.aol.com[400C740E] (Traffic-Server/6.1.5 [uScM]) Matrix A, (m x n), m>=n with singular values (s j) Show that the singular values of ?( I n ?A ?)^T = sqrt[ 1 + (s j)^2 ] where I n identity matrix (n x n). I n A makes sense only if n = m. Dave A is m x n B = (I n) is (n+m) x n ( A ) B' = (I n A') B'B = I n + A'A V'B'BV = (S b)'S b = V'(I n + A'A)V where S'S is the diag of B singualr values squared. (S b)'S b = I n + (S a)'(S a) === Subject: Re: Matrix A, (m x n), m>=n with Singular Values (s_j) >Matrix A, (m x n), m>=n with singular values (s_j) Show that the singular values of ( I_n A )^T = sqrt[ 1 + (s_j)^2 ] where I_n identity matrix (n x n). >i thought someone who speaks and acts positively on the internet knows >the direction to go in. A=UEV'. I can see E but not U,V, or A. when i get out of the car i can drink an Icehouse and make some >progress (brood over it). We've covered a lot of SVD in class but >with very few examples. homework hint: there is a definition of singualr values of a general matrix B in terms of eigenvalues of B'*B .'=hermitian conjugate hth peter === Subject: Re: || x y^T ||_2 = || x ||_2 * || y ||_2 [2-norm proof] posting-account=Qq_GEwoAAAD0Kc2k70q6xwuZdqx8sCzT 4334.5003; Windows NT 5.1; SV1; .NET CLR 2.0.50727; .NET CLR 3.0.04506.648; .NET CLR 3.5.21022),gzip(gfe),gzip(gfe) spider-dtc-tf07.proxy.aol.com[CDBC70A7] (Prism/1.2.1), HTTP/1.1 cache-dtc-aa03.proxy.aol.com[CDBC7407] (Traffic-Server/6.1.5 [uScM]) On 2009-02-25 16:32:46 -0400, birdhal...@yahoo.com said: x is an (m x 1) vector and y is an (n x 1) vector. The 2-norm relationship to show is: || x y^T || 2 = || x || 2 * || y || 2 i give up. i'd be grateful to know how. birdhaldol(1998-2009) Some hints only ... What is the rank of matrices of form x y^T? What do you know about || M || 2 in terms of singular values? Did you really mean to apply a matrix norm to a nonsquare matrix? > Plausible but not usual. Assuming the || M || 2 is the usual induced > matrix norm for || v || 2 of vectors. How did you come to be asking these questions? They are more typical > of early undergraduate courses. linear algebra. ?yes the 2-norm of a matrix i only know for a square > diagonal one is the maximum diag element. ?we've been taught svd so > maybe that will come into it > - Show quoted text - you know something you are not telling. because you know answers to all undergraduate problems. please show that the help on this ng is plausible. tell me at least something about the |M| 2 norm i don't already know. gordon, we are linked as if we did acid together now. birdhaldol (1998-2009) === Subject: Re: || x y^T ||_2 = || x ||_2 * || y ||_2 [2-norm proof] On 2009-02-25 22:00:05 -0400, birdhaldol@yahoo.com said: On 2009-02-25 16:32:46 -0400, birdhal...@yahoo.com said: x is an (m x 1) vector and y is an (n x 1) vector. The 2-norm relationship to show is: || x y^T ||_2 = || x ||_2 * || y ||_2 i give up. i'd be grateful to know how. birdhaldol(1998-2009) Some hints only ... What is the rank of matrices of form x y^T? What do you know about || M ||_2 in terms of singular values? Did you really mean to apply a matrix norm to a nonsquare matrix? > Plausible but not usual. Assuming the || M ||_2 is the usual induced > matrix norm for || v ||_2 of vectors. How did you come to be asking these questions? They are more typical > of early undergraduate courses. al > linear algebra. ?yes the 2-norm of a matrix i only know for a squ > are > diagonal one is the maximum diag element. ?we've been taught svd > so > maybe that will come into it > - Show quoted text - you know something you are not telling. Yes. That is why you are only getting hints. You do get a bonus points for admitting that it is a homework type question. Hopefully you will remember the method of finding the nswer rather than just some specifiec answer. > because you know answers to > all undergraduate problems. please show that the help on this ng is > plausible. tell me at least something about the |M|_2 norm i don't already know. > gordon, we are linked as if we did acid together now. Go back to the definition of the induced norm. It is the maximum of something. What? Does knowing the singular values help? Try some special cases like 1. - all singular values equal or 2. - only one is nonzero. > birdhaldol > (1998-2009) === Subject: Re: || x y^T ||_2 = || x ||_2 * || y ||_2 [2-norm proof] <2009022522170016807-gsande@worldnetattnet> posting-account=Qq_GEwoAAAD0Kc2k70q6xwuZdqx8sCzT 4334.5003; Windows NT 5.1; SV1; .NET CLR 2.0.50727; .NET CLR 3.0.04506.648; .NET CLR 3.5.21022),gzip(gfe),gzip(gfe) spider-dtc-tf04.proxy.aol.com[CDBC70A4] (Prism/1.2.1), HTTP/1.1 cache-dtc-aa03.proxy.aol.com[CDBC7407] (Traffic-Server/6.1.5 [uScM]) > On 2009-02-25 22:00:05 -0400, birdhal...@yahoo.com said: > On 2009-02-25 16:32:46 -0400, birdhal...@yahoo.com said: x is an (m x 1) vector and y is an (n x 1) vector. The 2-norm relationship to show is: || x y^T || 2 = || x || 2 * || y || 2 i give up. i'd be grateful to know how. birdhaldol(1998-2009) Some hints only ... What is the rank of matrices of form x y^T? What do you know about || M || 2 in terms of singular values? Did you really mean to apply a matrix norm to a nonsquare matrix? > Plausible but not usual. Assuming the || M || 2 is the usual induced > matrix norm for || v || 2 of vectors. How did you come to be asking these questions? They are more typical > of early undergraduate courses. al > linear algebra. yes the 2-norm of a matrix i only know for a squ > are > diagonal one is the maximum diag element. we've been taught svd > so > maybe that will come into it > - Show quoted text - you know something you are not telling. Yes. That is why you are only getting hints. You do get a bonus points > for admitting that it is a homework type question. Hopefully you will > remember the method of finding the nswer rather than just some specifiec > answer. because you know answers to > all undergraduate problems. ?please show that the help on this ng is > plausible. tell me at least something about the |M| 2 norm i don't already know. > gordon, we are linked as if we did acid together now. Go back to the definition of the induced norm. > It is the maximum of something. What? > Does knowing the singular values help? > Try some special cases like 1. - all singular values equal > or 2. - only one is nonzero. birdhaldol > (1998-2009)- Hide quoted text - - Show quoted text -- Hide quoted text - - Show quoted text - i'll look at || M || 2 = max (x != 0) (|| M x || 2) / (|| x || 2) for special x ,look FOR the singular values of (x y^T), i do enjoy solving the problem myself. i don't recommend that you get high somehow to perhaps come up with a viewpoint and attack the problem from that viewpoint. you are more valuable to an (m x n) country (libertarian) than you can know. birdhaldol === Subject: Re: || x y^T ||_2 = || x ||_2 * || y ||_2 [2-norm proof] posting-account=Qq_GEwoAAAD0Kc2k70q6xwuZdqx8sCzT 4334.5003; Windows NT 5.1; SV1; .NET CLR 2.0.50727; .NET CLR 3.0.04506.648; .NET CLR 3.5.21022),gzip(gfe),gzip(gfe) spider-mtc-ta07.proxy.aol.com[400C7007] (Prism/1.2.1), HTTP/1.1 cache-mtc-aa10.proxy.aol.com[400C740E] (Traffic-Server/6.1.5 [uScM]) i don't know if its ok to use y as the maximizing vector but i get a result: || x y' ||_2 = max (y != 0) || x y' y ||_2 / || y ||_2 = || x (|| y || _2)^2 ||_2 = || x ||_2 || (y ||_2)^2 / || y || = || x ||_2 || y ||_2 should i do it with z || x y' ||_2 = max (z != 0) || x y' z ||_2 / || z ||_2 = ? is ( y' z ) = || y ||_2 * || z ||_2 i don't think so. o well distrust anybody that wants to teach you something === Subject: Re: ||xy^T||_2 = ||x||_2*||y||_2 (2-norm problem) >Can you show this for x an (m x 1) vector and y an (n x 1) vector? birdhaldol (1998-2009) >11 yrs old. what if you use 1) the 2-norm of a matrix B is the square root of the maximum eigenvalue of B*B' = square root of the maximum eigenvalue of B*B' 2) now after applying this your are left with a simpler matrix apply again. 3) done hth peter === Subject: Re: Math without Logic posting-account=OWfgwwgAAADQpH2XgMDMe2wuQ7OFPXlE Gecko/2009011913 Firefox/3.0.6,gzip(gfe),gzip(gfe) > On Feb 26, 8:07æpm, Dan Christensen I would like your comments on the following excerpt from a paper that > discusses the pitfalls of teaching advanced mathematics (undergraduate > level) without due attention to formal logic. Dan ...In textbooks, we can find careless definitions; for example: f + g is continuous at a point, provided f and g are Of course it is harder to note the extended version of the previous > definition: For all functions f , all functions g, and all real numbers a, if f is > continuous at a and g is continuous at a, then f + g is continuous at > a. It is clear why we teach the former version, but in this case even > talented students cannot extend the definitions and theorems noted in > short form. According to Selden just only 8.5 percent of his mid-level > undergraduate mathematics majors could ñunpackî informally written > mathematical statementsinto their logically equivalent formal > statements. Some textbook-writers keep quiet about quantifiers, > because in their opinion students cannot understand them. Hence if > students do not practice this kind of extensions, they have only > superficial knowledge about theorems and definitions. With this > superficial knowledge students can solve simple problems, but they > cannot solve harder ones, e.g. proofs. (p. 2) Source: Mario Bako, Why we need to teach logic and how we can teach > it? International Journal for Mathematics Teaching and Learning, > October 17, 2002http://www.cimt.plymouth.ac.uk/journal/bakom.pdf Well this is just stupid. There is nothing wrong with saying æ f + g is continuous at a point, provided f and g are Then they formalise it as: æ For all functions f , all functions g, and all real numbers a, if f > is > æ continuous at a and g is continuous at a, then f + g is continuous > at > æ a. Who said anything about real numbers? If you're going to be pedantic, > you should at least be correct. I don't think the author is saying that we must use only formal definitions in textbooks. The problem isn't the textbooks, but that students lack a sufficient background in formal logic to fully understand informal statements of definitions and theorems that they Only 8.5 percent of his mid-level undergraduate mathematics majors could ñunpackî informally written mathematical statements into their logically equivalent formal statements. I was just wondering if math students and instructors here had noticed the same thing. Dan === Subject: Re: definition of Carath?odory in measure theory > Just checking my definition: > We say that A has a length by Carath?odory if > m*(A) + m*(X - A)=b_0 - a_0 where m* is the outer measure and A is a subset of [a_0,b_0) = X . Is this a correct definition? The definition according to Caratheodory is that if m* is an outer > measure defined on all subsets of a set X, then a set E is m*-measurable > if m*(A) = m*(A / E) + m*(A E) for every subset A of X. That is, E serves as a suitable knife for > cutting up any other set A into two parts whose outer measures are > additive. === Subject: Re: Alternating Vertical and Horizontal Moves In Grid posting-account=ey-PjQoAAAD_Gi3gK0ZWKofsmviYOWhV Gecko/20081217 Firefox/2.0.0.20,gzip(gfe),gzip(gfe) Hi! This is an interesting problem, that I have never seen before. I will make the assumption that at any point a vertical move may be U or D, and a horizontal move may be L or R. In both the start-at-a-corner and general case, a(1) = 1, a(2) = 4, and a(3) = 6, for starters. Now I will focus only on the start-at-a-corner case, and it may as well be the upper left. Obviously n^2 is an upper bound for a(n), and if we try to establish some reasonable lower bounds, by considering a boustrophedonic traversal of the grid, there is a natural division into cases based on the value of n mod 4. In particular, I conjecture that the optimal algorithm for sufficiently large n = 1 mod 4 is to traverse the grid according to the following rules: 1) Start with down 1) Choose up over down whenever at least 4 more moves can be made as a result of doing so. 2) Alternate between choosing right n - 1 times in a row, then left n - 1 times in a row, then right, etc. To see what I mean by this, the 9 x 9 grid would be traversed like: DRURDRURDRURDRDRDLDLULULDLULDLDLDRDRURURDRDRURDRDLDLULDLULDLULDLU. With a little work this establishes: n mod 4 = 1 -> [ n^2 - 2n + 3 <= a(n) ] For the other cases some slight modifications of the algorithm above produce pretty tight lower bounds as well. In the above case, I would expect equality probably holds, but proving it is another story, it seems like it would be tricky. dc === Subject: Re: Alternating Vertical and Horizontal Moves In Grid >What is the sequence {a(k)} that is defined as follows? Start with an n-by-n grid. Move from square to adjacent square by >moving vertically then horizontally then vertically then horizontally, >following each horizontal move with a vertical move, and following >each vertical move with a horizontal move. >No square may be visited more than once. a(n) = the maximum number of squares visitable in this way on an n-by- >n grid. By brute force search, here are the values of a(n) for n = 1, ..., 8: 1, 4, 7, 14, 19, 32, 39, 58 >A variation: You must start in a corner square. Then the first 8 values are: 1, 4, 6, 14, 18, 32, 38, 58 Note that, so far, if n is even, starting in a corner is just as good a starting place as any other location, while if n is odd and n > 1, the value of a(n) when starting in a corner appears to be just 1 less than that for an optimum starting place. >(It does not matter if the first move is vertical or horizontal, as >far as the values of each a(n) are concerned, of course.) I am guessing that this sequence is really easy to calculate, maybe a >(n) is as simple as a quadratic polynomial in terms of n. Or is the >best way to calculate each a(n) is to do a brute-force search of >possible paths through the grid? The function a(n) is definitely not a polynomial in n. Clearly a(n) is positive and bounded above by n^2, hence, if a(n) was a polynomial in n, it would have to be at most of degree 2. But the calculated values of a(n) as posted above contradict that. So what is the nature of a(n)? Consider the sequence b(n) = n^2 - a(n) Clearly b(n) must nonnegative. I suspect it would not be too hard to prove that b(n) must be positive, for n > 2. We can ask whether the sequence b(n) is bounded above? I suspect not, but if, in fact, b(n) is bounded above, then I would guess that the sequence b(n) repeats. Other than brute force search, you might try for recursive relationships, or at least recursive bounds. quasi === Subject: Re: Alternating Vertical and Horizontal Moves In Grid What is the sequence {a(k)} that is defined as follows? Start with an n-by-n grid. Move from square to adjacent square by >moving vertically then horizontally then vertically then horizontally, >following each horizontal move with a vertical move, and following >each vertical move with a horizontal move. >No square may be visited more than once. a(n) = the maximum number of squares visitable in this way on an n-by- >n grid. By brute force search, here are the values of a(n) for n = 1, ..., 8: 1, 4, 7, 14, 19, 32, 39, 58 A variation: You must start in a corner square. Then the first 8 values are: 1, 4, 6, 14, 18, 32, 38, 58 Note that, so far, if n is even, starting in a corner is just as good >a starting place as any other location, while if n is odd and n > 1, >the value of a(n) when starting in a corner appears to be just 1 less >than that for an optimum starting place. (It does not matter if the first move is vertical or horizontal, as >far as the values of each a(n) are concerned, of course.) I am guessing that this sequence is really easy to calculate, maybe a >(n) is as simple as a quadratic polynomial in terms of n. Or is the >best way to calculate each a(n) is to do a brute-force search of >possible paths through the grid? The function a(n) is definitely not a polynomial in n. Clearly a(n) is >positive and bounded above by n^2, hence, if a(n) was a polynomial in >n, it would have to be at most of degree 2. But the calculated values >of a(n) as posted above contradict that. So what is the nature of a(n)? Consider the sequence b(n) = n^2 - a(n) Clearly b(n) must nonnegative. I suspect it would not be too hard to >prove that b(n) must be positive, for n > 2. We can ask whether the sequence b(n) is bounded above? I suspect not, >but if, in fact, b(n) is bounded above, then I would guess that the >sequence b(n) repeats. Other than brute force search, you might try for recursive >relationships, or at least recursive bounds. Ok, there is a simple pattern that I missed. When n is even, it appears that a(n) = n^2 - n + 2 quasi === Subject: Re: Alternating Vertical and Horizontal Moves In Grid >What is the sequence {a(k)} that is defined as follows? Start with an n-by-n grid. Move from square to adjacent square by >moving vertically then horizontally then vertically then horizontally, >following each horizontal move with a vertical move, and following >each vertical move with a horizontal move. >No square may be visited more than once. a(n) = the maximum number of squares visitable in this way on an n-by- >n grid. By brute force search, here are the values of a(n) for n = 1, ..., 8: 1, 4, 7, 14, 19, 32, 39, 58 A variation: You must start in a corner square. Then the first 8 values are: 1, 4, 6, 14, 18, 32, 38, 58 Note that, so far, if n is even, starting in a corner is just as good >a starting place as any other location, while if n is odd and n > 1, >the value of a(n) when starting in a corner appears to be just 1 less >than that for an optimum starting place. (It does not matter if the first move is vertical or horizontal, as >far as the values of each a(n) are concerned, of course.) I am guessing that this sequence is really easy to calculate, maybe a >(n) is as simple as a quadratic polynomial in terms of n. Or is the >best way to calculate each a(n) is to do a brute-force search of >possible paths through the grid? The function a(n) is definitely not a polynomial in n. Clearly a(n) is >positive and bounded above by n^2, hence, if a(n) was a polynomial in >n, it would have to be at most of degree 2. But the calculated values >of a(n) as posted above contradict that. So what is the nature of a(n)? Consider the sequence b(n) = n^2 - a(n) Clearly b(n) must nonnegative. I suspect it would not be too hard to >prove that b(n) must be positive, for n > 2. We can ask whether the sequence b(n) is bounded above? I suspect not, >but if, in fact, b(n) is bounded above, then I would guess that the >sequence b(n) repeats. Other than brute force search, you might try for recursive >relationships, or at least recursive bounds. Ok, there is a simple pattern that I missed. When n is even, it appears that a(n) = n^2 - n + 2 Moreover, it's fairly easy to show that, for even n, a(n) >= n^2 - n + 2 Assuming n is even, a valid path through n^2 - n + 2 vertices can be explicitly constructed as follows ... Start at (0,0). Go right, up, left, up. Repeat until reaching the top. Since n is even, the last 2 moves must have been right, up, ending at (1,n-1). Now go right, down, ending at (2,n-2). We have so far visited 2n + 1 vertices. Now recursively do the same in the n-2 by n-2 square with upper left corner vertex (2,n-2) and lower right corner vertex (n-1,1). The path described above adheres to the restrictions, and it's easily proved, by induction, that it has exactly n^2 - n + 2 vertices. quasi === Subject: Re: Alternating Vertical and Horizontal Moves In Grid What is the sequence {a(k)} that is defined as follows? Start with an n-by-n grid. Move from square to adjacent square by >moving vertically then horizontally then vertically then horizontally, >following each horizontal move with a vertical move, and following >each vertical move with a horizontal move. >No square may be visited more than once. a(n) = the maximum number of squares visitable in this way on an n-by- >n grid. By brute force search, here are the values of a(n) for n = 1, ..., 8: 1, 4, 7, 14, 19, 32, 39, 58 A variation: You must start in a corner square. Then the first 8 values are: 1, 4, 6, 14, 18, 32, 38, 58 Note that, so far, if n is even, starting in a corner is just as good >a starting place as any other location, while if n is odd and n > 1, >the value of a(n) when starting in a corner appears to be just 1 less >than that for an optimum starting place. (It does not matter if the first move is vertical or horizontal, as >far as the values of each a(n) are concerned, of course.) I am guessing that this sequence is really easy to calculate, maybe a >(n) is as simple as a quadratic polynomial in terms of n. Or is the >best way to calculate each a(n) is to do a brute-force search of >possible paths through the grid? The function a(n) is definitely not a polynomial in n. Clearly a(n) is >positive and bounded above by n^2, hence, if a(n) was a polynomial in >n, it would have to be at most of degree 2. But the calculated values >of a(n) as posted above contradict that. So what is the nature of a(n)? Consider the sequence b(n) = n^2 - a(n) Clearly b(n) must nonnegative. I suspect it would not be too hard to >prove that b(n) must be positive, for n > 2. We can ask whether the sequence b(n) is bounded above? I suspect not, >but if, in fact, b(n) is bounded above, then I would guess that the >sequence b(n) repeats. Other than brute force search, you might try for recursive >relationships, or at least recursive bounds. Ok, there is a simple pattern that I missed. When n is even, it appears that a(n) = n^2 - n + 2 Moreover, it's fairly easy to show that, for even n, a(n) >= n^2 - n + 2 Assuming n is even, a valid path through n^2 - n + 2 vertices can be >explicitly constructed as follows ... Start at (0,0). Go right, up, left, up. Repeat until reaching the top. Since n is even, the last 2 moves must have been right, up, ending at >(1,n-1). Now go right, down, ending at (2,n-2). We have so far visited 2n + 1 vertices. Now recursively do the same in the n-2 by n-2 square with upper left >corner vertex (2,n-2) and lower right corner vertex (n-1,1).| To clarify: At each successive lower level in the recursion, switch up and down. Also, when you finish the 2 x 2 level, stop. >The path described above adheres to the restrictions, and it's easily >proved, by induction, that it has exactly n^2 - n + 2 vertices. quasi === Subject: Re: Alternating Vertical and Horizontal Moves In Grid <4utbq4dhhldvts8o5hl3ld180tan2tbdmq@4ax.com> posting-account=ey-PjQoAAAD_Gi3gK0ZWKofsmviYOWhV Gecko/20081217 Firefox/2.0.0.20,gzip(gfe),gzip(gfe) What is the sequence {a(k)} that is defined as follows? Start with an n-by-n grid. Move from square to adjacent square by >moving vertically then horizontally then vertically then horizontally, >following each horizontal move with a vertical move, and following >each vertical move with a horizontal move. >No square may be visited more than once. a(n) = the maximum number of squares visitable in this way on an n-by- >n grid. By brute force search, here are the values of a(n) for n = 1, ..., 8: 1, 4, 7, 14, 19, 32, 39, 58 A variation: You must start in a corner square. Then the first 8 values are: 1, 4, 6, 14, 18, 32, 38, 58 Note that, so far, if n is even, starting in a corner is just as good >a starting place as any other location, while if n is odd and n > 1, >the value of a(n) when starting in a corner appears to be just 1 less >than that for an optimum starting place. (It does not matter if the first move is vertical or horizontal, as >far as the values of each a(n) are concerned, of course.) I am guessing that this sequence is really easy to calculate, maybe a >(n) is as simple as a quadratic polynomial in terms of n. Or is the >best way to calculate each a(n) is to do a brute-force search of >possible paths through the grid? The function a(n) is definitely not a polynomial in n. Clearly a(n) is >positive and bounded above by n^2, hence, if a(n) was a polynomial in >n, it would have to be at most of degree 2. But the calculated values >of a(n) as posted above contradict that. So what is the nature of a(n)? Consider the sequence b(n) = n^2 - a(n) Clearly b(n) must nonnegative. I suspect it would not be too hard to >prove that b(n) must be positive, for n > 2. We can ask whether the sequence b(n) is bounded above? I suspect not, >but if, in fact, b(n) is bounded above, then I would guess that the >sequence b(n) repeats. Other than brute force search, you might try for recursive >relationships, or at least recursive bounds. Ok, there is a simple pattern that I missed. When n is even, it appears that a(n) = n^2 - n + 2 Moreover, it's fairly easy to show that, for even n, a(n) >= n^2 - n + 2 Assuming n is even, a valid path through n^2 - n + 2 vertices can be >explicitly constructed as follows ... Start at (0,0). Go right, up, left, up. Repeat until reaching the top. Since n is even, the last 2 moves must have been right, up, ending at >(1,n-1). Now go right, down, ending at (2,n-2). We have so far visited 2n + 1 vertices. Now recursively do the same in the n-2 by n-2 square with upper left >corner vertex (2,n-2) and lower right corner vertex (n-1,1).| To clarify: At each successive lower level in the recursion, switch up and > down. Also, when you finish the 2 x 2 level, stop. The path described above adheres to the restrictions, and it's easily >proved, by induction, that it has exactly n^2 - n + 2 vertices. quasi this seems legitimate to me. quasi, by the sounds of it, you have some kind of a computer program capable of a brute force test? can you test my conjecture above for n = 9, 13? i'm thinking we are going tol end up here with a quadratic answer, but different quadratics depending on the value of n mod 4. === Subject: Re: Alternating Vertical and Horizontal Moves In Grid What is the sequence {a(k)} that is defined as follows? Start with an n-by-n grid. Move from square to adjacent square by >moving vertically then horizontally then vertically then horizontally, >following each horizontal move with a vertical move, and following >each vertical move with a horizontal move. >No square may be visited more than once. a(n) = the maximum number of squares visitable in this way on an n-by- >n grid. By brute force search, here are the values of a(n) for n = 1, ..., 8: 1, 4, 7, 14, 19, 32, 39, 58 A variation: You must start in a corner square. Then the first 8 values are: 1, 4, 6, 14, 18, 32, 38, 58 Note that, so far, if n is even, starting in a corner is just as good >a starting place as any other location, while if n is odd and n > 1, >the value of a(n) when starting in a corner appears to be just 1 less >than that for an optimum starting place. (It does not matter if the first move is vertical or horizontal, as >far as the values of each a(n) are concerned, of course.) I am guessing that this sequence is really easy to calculate, maybe a >(n) is as simple as a quadratic polynomial in terms of n. Or is the >best way to calculate each a(n) is to do a brute-force search of >possible paths through the grid? The function a(n) is definitely not a polynomial in n. Clearly a(n) is >positive and bounded above by n^2, hence, if a(n) was a polynomial in >n, it would have to be at most of degree 2. But the calculated values >of a(n) as posted above contradict that. So what is the nature of a(n)? Consider the sequence b(n) = n^2 - a(n) Clearly b(n) must nonnegative. I suspect it would not be too hard to >prove that b(n) must be positive, for n > 2. We can ask whether the sequence b(n) is bounded above? I suspect not, >but if, in fact, b(n) is bounded above, then I would guess that the >sequence b(n) repeats. Other than brute force search, you might try for recursive >relationships, or at least recursive bounds. Ok, there is a simple pattern that I missed. When n is even, it appears that a(n) = n^2 - n + 2 Moreover, it's fairly easy to show that, for even n, a(n) >= n^2 - n + 2 Assuming n is even, a valid path through n^2 - n + 2 vertices can be >explicitly constructed as follows ... Start at (0,0). Go right, up, left, up. Repeat until reaching the top. Since n is even, the last 2 moves must have been right, up, ending at >(1,n-1). Now go right, down, ending at (2,n-2). We have so far visited 2n + 1 vertices. Now recursively do the same in the n-2 by n-2 square with upper left >corner vertex (2,n-2) and lower right corner vertex (n-1,1).| To clarify: At each successive lower level in the recursion, switch up and > down. Also, when you finish the 2 x 2 level, stop. The path described above adheres to the restrictions, and it's easily >proved, by induction, that it has exactly n^2 - n + 2 vertices. quasi this seems legitimate to me. quasi, by the sounds of it, you have some >kind of a computer program capable of a brute force test? can you test >my conjecture above for n = 9, 13? i'm thinking we are going tol end >up here with a quadratic answer, but different quadratics depending on >the value of n mod 4. Yes, that seems very plausible -- 3 quadratics, one for even n, one for n = 1 mod 4, and one for n = 3 mod 4. I do have a brute force test, but it wasn't designed for a heavy load. For n > 8, it struggles. I'll let it run for a while -- hopefully, it will at least compute a(9). quasi === Subject: Re: Alternating Vertical and Horizontal Moves In Grid <4utbq4dhhldvts8o5hl3ld180tan2tbdmq@4ax.com> posting-account=ey-PjQoAAAD_Gi3gK0ZWKofsmviYOWhV Gecko/20081217 Firefox/2.0.0.20,gzip(gfe),gzip(gfe) no, i changed my mind about the mod 4 thing, it looks a lot like your formula is right whenever n is even, and my formula is right (by a similar argument) when n is odd. of course, we have only established them as lower bounds with certainty. === Subject: Re: Alternating Vertical and Horizontal Moves In Grid >no, i changed my mind about the mod 4 thing, it looks a lot like your >formula is right whenever n is even, and my formula is right (by a >similar argument) when n is odd. of course, we have only established >them as lower bounds with certainty. Yes, if we require corner starting point, and assuming n > 1, I think you may be right. Thus, for a corner starting point, and n > 1, the conjecture is: a(n) = n^2 - 2n + 3, if n is odd n^2 - n + 2, if n is even quasi PS -- My Maple program is still working on a(9), assuming a required corner starting point. === Subject: Re: Alternating Vertical and Horizontal Moves In Grid no, i changed my mind about the mod 4 thing, it looks a lot like your >formula is right whenever n is even, and my formula is right (by a >similar argument) when n is odd. of course, we have only established >them as lower bounds with certainty. Yes, if we require corner starting point, and assuming n > 1, I think >you may be right. Thus, for a corner starting point, and n > 1, the conjecture is: a(n) = > n^2 - 2n + 3, if n is odd > n^2 - n + 2, if n is even quasi PS -- My Maple program is still working on a(9), assuming a required >corner starting point. Ok, Maple completed a(9) and is now working on a(11). Good news -- a(9) = 66, as expected. Note, for convenience, we are now always making the assumption of a corner starting point. For even n, it apparently doesn't matter. For odd n > 1, allowing a variable starting point appears to make a(n) increase by 1. quasi === Subject: Re: Alternating Vertical and Horizontal Moves In Grid <4utbq4dhhldvts8o5hl3ld180tan2tbdmq@4ax.com> <4p7cq45og61mehq711tc9qm7ir9avhp4gl@4ax.com> posting-account=ey-PjQoAAAD_Gi3gK0ZWKofsmviYOWhV Gecko/20081217 Firefox/2.0.0.20,gzip(gfe),gzip(gfe) no, i changed my mind about the mod 4 thing, it looks a lot like your >formula is right whenever n is even, and my formula is right (by a >similar argument) when n is odd. of course, we have only established >them as lower bounds with certainty. Yes, if we require corner starting point, and assuming n > 1, I think >you may be right. Thus, for a corner starting point, and n > 1, the conjecture is: a(n) = > n^2 - 2n + 3, if n is odd > n^2 - n + 2, if n is even quasi PS -- My Maple program is still working on a(9), assuming a required >corner starting point. Ok, Maple completed a(9) and is now working on a(11). Good news -- a(9) = 66, as expected. Note, for convenience, we are now always making the assumption of a > corner starting point. For even n, it apparently doesn't matter. For > odd n > 1, allowing a variable starting point appears to make a(n) > increase by 1. quasi this makes sense to me, i am starting to work on a proof of the bound for the even case, as i think it will be the easier one. i have some ideas, but nothing concrete enough to comment on yet. === Subject: Re: Newton's Method: Cute trick. posting-account=DSjMzAoAAAAISwlGnTY2c6bnmUVmdFNf AppleWebKit/528.16 (KHTML, like Gecko) Version/4.0 Safari/528.16,gzip(gfe),gzip(gfe) I have never seen this trick explicitly mentioned anywhere. Of course if f (p) < 0, f (q) > 0, f continuous on [p, q] then there is a root of f in (p, q), and if you use _any_ iterative algorithm, then clamping the next iterated result to [p, q] will get you closer to the root. Some things I noted: (1) If you find x0 such that f (x0) is small enough, but x0 is not in [p, q] then clamping x0 to [p, q] might give a new value where f (x) is _not_ small enough anymore, so care must be taken. (2) Clamping can actually only happen on the first Newton-Raphson step if your convergence criterion is actually met, because you proved that each iteration gets closer to the root of F (x). If you are not 100% sure that the convergence criterion is met, clamping might not help. (3) This doesn't solve the problem of finding x0 close enough to the root. Of course it is no big deal if a and b are known. === Subject: Re: Help!How to solve the equation === Subject: I just thought this was cool posting-account=Z3AipgkAAABkoMfyNwddSxsYhXHi5CDt MathPlayer 2.10d; .NET CLR 1.1.4322; PeoplePal 3.0),gzip(gfe),gzip(gfe) Hi Group, When I was in high school or early college or something, I was given a problem something like this: Let a_n and b_n be defined via (a_n + b_n*sqrt(M)) = (a + b*sqrt(M)^n Show that a_n / b_n converges to sqrt(M). I guess it had to be high school or the summer after as I discussed differing solutions with a guy that was in a math class I took freshman year. Along the way I used some straightforward algebraic manipulations to show that a_n and b_n satisfied identical second order linear recursions. I was inordinately proud of this minimal result and saved the calculation and the formula for years, hauling it out only when Pell equations came and it applied. Anyway, while I was waiting to have a tooth extracted today, I realized that DeMoivre's theorem allowed us to regard a_n = cos(n*theta) b_n = sin(n*theta) as special cases of this result. I did not have my records with me, so I rederived the applicable part and got cos((n+2)*theta) = 2*cos(theta)*cos((n+1)*theta) - cos(n*theta) and sin((n+2)*theta) = 2*cos(theta)*sin(n+1)*theta) - sin(n*theta) These are easy enough to prove without doing what I did, of course, but I just thought they were kind of cool, so I am presenting them. Achava === Subject: Re: I just thought this was cool > Hi Group, > When I was in high school or early college or something, I was given a > problem something like this: Let a_n and b_n be defined via (a_n + b_n*sqrt(M)) = (a + b*sqrt(M)^n Show that a_n / b_n converges to sqrt(M). I guess it had to be high school or the summer after as I discussed > differing solutions with a guy that was in a math class I took > freshman year. Along the way I used some straightforward algebraic manipulations to > show that a_n and b_n satisfied identical second order linear > recursions. I was inordinately proud of this minimal result and saved > the calculation and the formula for years, hauling it out only when > Pell equations came and it applied. Anyway, while I was waiting to have a tooth extracted today, I > realized that DeMoivre's theorem allowed us to regard a_n = cos(n*theta) > b_n = sin(n*theta) as special cases of this result. > I did not have my records with me, so I rederived the applicable part > and got cos((n+2)*theta) = 2*cos(theta)*cos((n+1)*theta) - cos(n*theta) and sin((n+2)*theta) = 2*cos(theta)*sin(n+1)*theta) - sin(n*theta) These are easy enough to prove without doing what I did, of course, > but I just thought they were kind of cool, so I am presenting them. When I teach linear algebra, I sometimes ask the students to decide whether or not {sin x, sin (x + 1), sin (x + 2)} is a linearly independent set of functions. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: backward paprabolic equation Thread-Topic: backward paprabolic equation Thread-Index: AcmYBHih4rMDB5cwRySvfsYssSi4eQ== Accept-Language: en-GB what do you think about the following assertion. I can not understand is that interesting by itself or could it have an application to anything? Consider a backward parabolic problem of the type u_t=-Delta u+f(t,x,u) tge 0 This problem lives in smooth bounded n-dimensional domain and the zero boundary conditions are applied. The function f is smooth and bounded. The assertion is that this problem has at least one regular solution u (t,x) defined for all nonnegative t and this solution is bounded in H^1_0(M) for all these t. === Subject: Re: Multiplication and addition of ideals Thread-Topic: Multiplication and addition of ideals Thread-Index: AcmYJR6XzyvNS0iBRXiLFwYmXOCP0Q== I asked a friend of mine this question and one of his first instincts was to ask, when is this semiring finitely generated? The answer might be not very often since even the integers don't give an example, but if there are some nontrivial examples, these might be the most tractable to analyze. -- Tim Chow tchow-at-alum-dot-mit-dot-edu The range of our projectiles---even ... the artillery---however great, will never exceed four of those miles of which as many thousand separate us from the center of the earth. ---Galileo, Dialogues Concerning Two New Sciences === Subject: Re: Multiplication and addition of ideals Thread-Topic: Multiplication and addition of ideals Thread-Index: AcmXqSf19S8NtNvgT6WdUkBjAnrcKg== Accept-Language: en-GB well one thing I notice from this conversation is that the set of ideals of a ring with ideal sum and multiplication does form a complete (idempotent) dioid, in particular, the operation *, given as: J* = sum(i in natural numbers){J^i} is well-defined for an ideal J. Such structures have a partial ordering <= given by: J <= K iff J + K = K. Some preliminary results are that for fixed ideals A, B: the affine equation X = AX + B has a least (with respect to <=) solution A*B, and any solution X satisfies X = A*X. It seems like these dioids are well studied in other contexts, but I have never actually seen anything (in the web-lierature anyhow) about dioids of ideals. There is also something called a cost dioid that is well studied, again for motivating reasons totally unrelated to sets of ideals, which requires the additional condition of the existence of roots of elements. Although this condition fails for general ideals, it may be true for some rings, or for some subsets of the collection of ideals of a ring. Maybe we can steal some of this dioid theory and apply it to the ideal context? Comments welcome! dc === Subject: Matrix A, (m x n), m>=n with Singular Values (s_j) posting-account=Qq_GEwoAAAD0Kc2k70q6xwuZdqx8sCzT 4334.5003; Windows NT 5.1; SV1; .NET CLR 2.0.50727; .NET CLR 3.0.04506.648; .NET CLR 3.5.21022),gzip(gfe),gzip(gfe) spider-mtc-tf12.proxy.aol.com[400C70AC] (Prism/1.2.1), HTTP/1.1 cache-mtc-ab12.proxy.aol.com[400C744C] (Traffic-Server/6.1.5 [uScM]) Matrix A, (m x n), m>=n with singular values (s_j) Show that the singular values of ( I_n A )^T = sqrt[ 1 + (s_j)^2 ] where I_n identity matrix (n x n). i thought someone who speaks and acts positively on the internet knows the direction to go in. A=UEV'. I can see E but not U,V, or A. when i get out of the car i can drink an Icehouse and make some progress (brood over it). We've covered a lot of SVD in class but with very few examples. === Subject: Re: Matrix A, (m x n), m>=n with Singular Values (s_j) posting-account=O9zR9AkAAACmp918j6u5m5plppeILcze Filter 1.2.0.72; GTB5; .NET CLR 1.0.3705; .NET CLR 1.1.4322; Media Center PC 4.0; .NET CLR 2.0.50727; .NET CLR 3.0.04506.648; .NET CLR 3.5.21022; \ .NET CLR 3.0.4506.2152; .NET CLR 3.5.30729),gzip(gfe),gzip(gfe) > Matrix A, (m x n), m>=n with singular values (s j) Show that the singular values of æ( I n æA æ)^T = sqrt[ 1 + (s j)^2 ] where I n identity matrix (n x n). I n A makes sense only if n = m. Dave === Subject: Re: Matrix A, (m x n), m>=n with Singular Values (s_j) posting-account=Qq_GEwoAAAD0Kc2k70q6xwuZdqx8sCzT 4334.5003; Windows NT 5.1; SV1; .NET CLR 2.0.50727; .NET CLR 3.0.04506.648; .NET CLR 3.5.21022),gzip(gfe),gzip(gfe) spider-mtc-ta07.proxy.aol.com[400C7007] (Prism/1.2.1), HTTP/1.1 cache-mtc-aa10.proxy.aol.com[400C740E] (Traffic-Server/6.1.5 [uScM]) Matrix A, (m x n), m>=n with singular values (s j) Show that the singular values of ?( I n ?A ?)^T = sqrt[ 1 + (s j)^2 ] where I n identity matrix (n x n). I n A makes sense only if n = m. Dave A is m x n B = (I n) is (n+m) x n ( A ) B' = (I n A') B'B = I n + A'A V'B'BV = (S b)'S b = V'(I n + A'A)V where S'S is the diag of B singualr values squared. (S b)'S b = I n + (S a)'(S a) === Subject: Re: Matrix A, (m x n), m>=n with Singular Values (s_j) >Matrix A, (m x n), m>=n with singular values (s_j) Show that the singular values of ( I_n A )^T = sqrt[ 1 + (s_j)^2 ] where I_n identity matrix (n x n). >i thought someone who speaks and acts positively on the internet knows >the direction to go in. A=UEV'. I can see E but not U,V, or A. when i get out of the car i can drink an Icehouse and make some >progress (brood over it). We've covered a lot of SVD in class but >with very few examples. homework hint: there is a definition of singualr values of a general matrix B in terms of eigenvalues of B'*B .'=hermitian conjugate hth peter === Subject: Re: Nomenclature for various groups Distribution: world > I also know that nZ represents the sub-group of Z that > is obtained by taking every n-th element of Z (starting with the > identity). It is better to think of it as the subgroup obtained by considering >the set of elements of the form n*x for every x in Z. The reason is that, technically, your description does NOT give the >corresponding subgroup, it just gives a subsemigroup consisting of the >nonnegative multiples of n, I didn't say to only go in one direction. :- and in addition it \ presupposes some >ordering on Z, Guilty as charged. > which while it exists is not part of the cyclic group >structure. Which is actually why I didn't just say cyclic group, but specified Z as my starting point. > (It ->is<- part of its structure as an ->ordered<- group, >but you are not specifying it as an ordered group). Matter of fact, I never heard of them before this afternoon. > Does this nomenclature generalize? Remember that for an additive group, we define n*g for g in G and >integer n, inductively as follows [snip elucidation of Abelian and non-Abelian cases] >That said, again: many object to the notation Z_n for the cyclic >group of order n or for the integers under modular addition modulo n; >particularly, those who come from number theory, where Z_p has a >different meaning. Oh, joy. However, as long as my work is clearly stated to be theorems regarding groups, I should be okay, right? > Many recomment/suggest/demand using Z/nZ for the >additive group modulo n Actually, the book that I'm working from uses this notation when first popping out that group, but then says that Z_n is an alternate symbol. > Alas, we cannot make everyone happy. I'll keep myself happy if my notation is consistent with what's done in a non-trivial plurality of cases. > If not, what is the standard naming to use -- or isn't there one? Note that your questions are not really about nomenclature or naming, >but rather about notation... ... or symbolism. >-- >Arturo Magidin, sans .sig Y'know, if you could get a space after the double-dash, you'd have one. -- Michael F. Stemper #include There is three erors in this sentence. === Subject: Re: Nomenclature for various groups Distribution: world posting-account=_l4K0QkAAAC09JhOoK_ZfoJKXOmr_jZf Gecko/2009011913 Firefox/3.0.6,gzip(gfe),gzip(gfe) On Feb 25, 5:08æpm, mstem...@walkabout.empros.com (Michael Stemper) >That said, again: many object to the notation Z n for the cyclic >group of order n or for the integers under modular addition modulo n; >particularly, those who come from number theory, where Z p has a >different meaning. Oh, joy. However, as long as my work is clearly stated to be theorems regarding > groups, I should be okay, right? Yeah, most people will figure it out from context. You may run into trouble if all you ever mention as n are prime numbers, however, because Z p is the absolutely 100% standard notation for the p-adic integers. > æ æMany recomment/suggest/demand using Z/nZ for the >additive group modulo n Actually, the book that I'm working from uses this notation when first > popping out that group, but then says that Z n is an alternate symbol. I learned it with Z n as an undergrad. Then in grad school I took abstract algebra from Paul Vojta, an arithmetic geometry, who objected strongly to using that notation. æAlas, we cannot make everyone happy. I'll keep myself happy if my notation is consistent with what's done > in a non-trivial plurality of cases. I will not say it is a plurality of cases, as these days there are a lot of number theorists. It is a nontrivial number of people however. >-- >ArturoMagidin, sans .sig Y'know, if you could get a space after the double-dash, you'd have one. I try, but Google edist it out no matter what I try; it edited it out of yours, too. I'm still stuck with their interface, as I haven't had time to set up a newsreader and a newsserver.... Arturo Magidin === Subject: Re: || x y^T ||_2 = || x ||_2 * || y ||_2 [2-norm proof] posting-account=Qq_GEwoAAAD0Kc2k70q6xwuZdqx8sCzT 4334.5003; Windows NT 5.1; SV1; .NET CLR 2.0.50727; .NET CLR 3.0.04506.648; .NET CLR 3.5.21022),gzip(gfe),gzip(gfe) spider-dtc-tf07.proxy.aol.com[CDBC70A7] (Prism/1.2.1), HTTP/1.1 cache-dtc-aa03.proxy.aol.com[CDBC7407] (Traffic-Server/6.1.5 [uScM]) On 2009-02-25 16:32:46 -0400, birdhal...@yahoo.com said: x is an (m x 1) vector and y is an (n x 1) vector. The 2-norm relationship to show is: || x y^T || 2 = || x || 2 * || y || 2 i give up. i'd be grateful to know how. birdhaldol(1998-2009) Some hints only ... What is the rank of matrices of form x y^T? What do you know about || M || 2 in terms of singular values? Did you really mean to apply a matrix norm to a nonsquare matrix? > Plausible but not usual. Assuming the || M || 2 is the usual induced > matrix norm for || v || 2 of vectors. How did you come to be asking these questions? They are more typical > of early undergraduate courses. linear algebra. ?yes the 2-norm of a matrix i only know for a square > diagonal one is the maximum diag element. ?we've been taught svd so > maybe that will come into it > - Show quoted text - you know something you are not telling. because you know answers to all undergraduate problems. please show that the help on this ng is plausible. tell me at least something about the |M| 2 norm i don't already know. gordon, we are linked as if we did acid together now. birdhaldol (1998-2009) === Subject: Re: || x y^T ||_2 = || x ||_2 * || y ||_2 [2-norm proof] On 2009-02-25 22:00:05 -0400, birdhaldol@yahoo.com said: On 2009-02-25 16:32:46 -0400, birdhal...@yahoo.com said: x is an (m x 1) vector and y is an (n x 1) vector. The 2-norm relationship to show is: || x y^T ||_2 = || x ||_2 * || y ||_2 i give up. i'd be grateful to know how. birdhaldol(1998-2009) Some hints only ... What is the rank of matrices of form x y^T? What do you know about || M ||_2 in terms of singular values? Did you really mean to apply a matrix norm to a nonsquare matrix? > Plausible but not usual. Assuming the || M ||_2 is the usual induced > matrix norm for || v ||_2 of vectors. How did you come to be asking these questions? They are more typical > of early undergraduate courses. al > linear algebra. ?yes the 2-norm of a matrix i only know for a squ > are > diagonal one is the maximum diag element. ?we've been taught svd > so > maybe that will come into it > - Show quoted text - you know something you are not telling. Yes. That is why you are only getting hints. You do get a bonus points for admitting that it is a homework type question. Hopefully you will remember the method of finding the nswer rather than just some specifiec answer. > because you know answers to > all undergraduate problems. please show that the help on this ng is > plausible. tell me at least something about the |M|_2 norm i don't already know. > gordon, we are linked as if we did acid together now. Go back to the definition of the induced norm. It is the maximum of something. What? Does knowing the singular values help? Try some special cases like 1. - all singular values equal or 2. - only one is nonzero. > birdhaldol > (1998-2009) === Subject: Re: || x y^T ||_2 = || x ||_2 * || y ||_2 [2-norm proof] <2009022522170016807-gsande@worldnetattnet> posting-account=Qq_GEwoAAAD0Kc2k70q6xwuZdqx8sCzT 4334.5003; Windows NT 5.1; SV1; .NET CLR 2.0.50727; .NET CLR 3.0.04506.648; .NET CLR 3.5.21022),gzip(gfe),gzip(gfe) spider-dtc-tf04.proxy.aol.com[CDBC70A4] (Prism/1.2.1), HTTP/1.1 cache-dtc-aa03.proxy.aol.com[CDBC7407] (Traffic-Server/6.1.5 [uScM]) > On 2009-02-25 22:00:05 -0400, birdhal...@yahoo.com said: > On 2009-02-25 16:32:46 -0400, birdhal...@yahoo.com said: x is an (m x 1) vector and y is an (n x 1) vector. The 2-norm relationship to show is: || x y^T || 2 = || x || 2 * || y || 2 i give up. i'd be grateful to know how. birdhaldol(1998-2009) Some hints only ... What is the rank of matrices of form x y^T? What do you know about || M || 2 in terms of singular values? Did you really mean to apply a matrix norm to a nonsquare matrix? > Plausible but not usual. Assuming the || M || 2 is the usual induced > matrix norm for || v || 2 of vectors. How did you come to be asking these questions? They are more typical > of early undergraduate courses. al > linear algebra. yes the 2-norm of a matrix i only know for a squ > are > diagonal one is the maximum diag element. we've been taught svd > so > maybe that will come into it > - Show quoted text - you know something you are not telling. Yes. That is why you are only getting hints. You do get a bonus points > for admitting that it is a homework type question. Hopefully you will > remember the method of finding the nswer rather than just some specifiec > answer. because you know answers to > all undergraduate problems. ?please show that the help on this ng is > plausible. tell me at least something about the |M| 2 norm i don't already know. > gordon, we are linked as if we did acid together now. Go back to the definition of the induced norm. > It is the maximum of something. What? > Does knowing the singular values help? > Try some special cases like 1. - all singular values equal > or 2. - only one is nonzero. birdhaldol > (1998-2009)- Hide quoted text - - Show quoted text -- Hide quoted text - - Show quoted text - i'll look at || M || 2 = max (x != 0) (|| M x || 2) / (|| x || 2) for special x ,look FOR the singular values of (x y^T), i do enjoy solving the problem myself. i don't recommend that you get high somehow to perhaps come up with a viewpoint and attack the problem from that viewpoint. you are more valuable to an (m x n) country (libertarian) than you can know. birdhaldol === Subject: Re: || x y^T ||_2 = || x ||_2 * || y ||_2 [2-norm proof] posting-account=Qq_GEwoAAAD0Kc2k70q6xwuZdqx8sCzT 4334.5003; Windows NT 5.1; SV1; .NET CLR 2.0.50727; .NET CLR 3.0.04506.648; .NET CLR 3.5.21022),gzip(gfe),gzip(gfe) spider-mtc-ta07.proxy.aol.com[400C7007] (Prism/1.2.1), HTTP/1.1 cache-mtc-aa10.proxy.aol.com[400C740E] (Traffic-Server/6.1.5 [uScM]) i don't know if its ok to use y as the maximizing vector but i get a result: || x y' ||_2 = max (y != 0) || x y' y ||_2 / || y ||_2 = || x (|| y || _2)^2 ||_2 = || x ||_2 || (y ||_2)^2 / || y || = || x ||_2 || y ||_2 should i do it with z || x y' ||_2 = max (z != 0) || x y' z ||_2 / || z ||_2 = ? is ( y' z ) = || y ||_2 * || z ||_2 i don't think so. o well distrust anybody that wants to teach you something === Subject: Re: ||xy^T||_2 = ||x||_2*||y||_2 (2-norm problem) >Can you show this for x an (m x 1) vector and y an (n x 1) vector? birdhaldol (1998-2009) >11 yrs old. what if you use 1) the 2-norm of a matrix B is the square root of the maximum eigenvalue of B*B' = square root of the maximum eigenvalue of B*B' 2) now after applying this your are left with a simpler matrix apply again. 3) done hth peter === Subject: Re: incomplete magic square posting-account=ogMREwkAAAC5xUr8sg7heGtsvzzF18LA Gecko/2009011913 Firefox/3.0.6,gzip(gfe),gzip(gfe) > I have a simple ( and incomplete ) magic ( maybe not so magic ) square where > the only element we are given is the one at the middle 35 and 4 more > integers forming a cross around 35. Clockwise from the top the numbers are > 32, 46, 38 and 24. we can see that 2*35=(32+38) and 2*35=(24+46). > Is there a general way to solve this if the only number given is the one in > the middle? > So you have: a 32 b 24 35 46 c 38 d The magic sum is 105. There are six more rows to complete, and only four missing values. Thus, you can easily make a set of 6 equations with the four unknowns: a + b = 73 a + c = 81 a + d = 70 b + c = 70 b + d = 59 c + d = 67 By the nature of these equations, you can choose any three involving a specific three variables and solve them. Then use any other equation to get the last. In this case, you will find a, b, c, d = 42, 31, 39, 28 is the unique solution. If the ONLY number given is the one in the middle, then there is not enough information to get a unique solution, so almost any magic- square generating method will give a possible answer - just take any magic square and add or subtract the same number from every element to get the one you need to line up. === Subject: Re: can someone test this please still hoping a good friendly guy will have a guy that reply to this :( at least say hi ;/ === Subject: Re: Solutions manual to Separation Process Principles, 2nd Ed.,by Seader, Henley posting-account=4CpmTgoAAAA-reuye4uyyf_yOmR8lxMl Gecko/2009011913 Firefox/3.0.6,gzip(gfe),gzip(gfe) > solutions manual (To search click in keyboard Ctrl+F) > Solutions Manuals in Electronic (PDF)Format! Just contact with , > sendsolutions (at) hotmail.com (my email address), these are parts > of our solutions, if the solution you want is on the list, please > email to me. > NOTICE: if the solutions manual that in my list ,please note it in > your email . > Instructor Solutions manual to : > Solutions manual to A First Course in Probability, (7th), By > Sheldon Ross Solutions manua to Financial Accounting 6e by horngren Harrison > Solutions manual to Advanced Accounting, 9th edition by Hoyle, > Schaefer, & Doupnik > Solutions manual to Complex Variables with Applications (Pie) 3rd > by A.David Wunsch > Solutions manual to Computer Design Fundamentals4E by Mano and Kime. > 4th > Solutions manual to Computer Networks Systems Approach 3ed by davie > peterson > Solutions manual to COMPUTER ORGANIZATION AND ARCHITECTURE DESIGNING > Solutions manual to Cost Accounting, 13/e 13e by Horngren > Solutions manua to Data and Computer Communications, 7th Edition By > Stallings > Solutions manual to Differential Equations and Linear Algebra by > Penney and Edwards, 2nd > Solutions manual to Elementary Differential Equations and Boundary > Value Problems, 8th by Boyce and Diprima > Solutions manual to Elements of engineering electromagnetics (6/ > e) by N.N.RAO > Solutions manual to Engineering electromagnetics (7/ > e) by HAYT > Solutions manual to Engineering Fluid Mechanics, 7th, By Clayton > T. 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Farmer > Solutions manual to Linear Algebra by Jim Hefferon > Solutions manual to Introductory Econometrics for Finance (Chris > Brooks 2002) > Solutions manual to INTRODUCTION TO LINEAR ALGEBRA: Solutions to > Exercises (Gilbert Strang) > Solutions manual to Operational Risk Quantification: Mathematical > Solutions for Analyzing Loss Data > Solutions manual to Fundamentals of Financial Management With Infotrac > Concise 4th by Eugene Brigham > 5th. > Solutions manual to computer system architecture 3rd by M.Morris > Mano > Solutions manual to Design Analysis in Rock Mechanics By William G. > Pariseau > Solutions manual to Digital Design 4th by M. Morris Mano, Michael D. > Ciletti Solutions manual to Engineering and Chemical Thermodynamics by Milo D. > Koretsky > Solutions manual to Fundamentals of Clas Sical Thermodynamics 6th > edition by Van Wylen > Solutions manual to Mechanical Vibrations, 3rd Edition, by Singiresu > S. Rao > Solutions manual to Mechanical Vibrations, Third Edition, by Singiresu > S. Rao > Solutions manual to Microelectronic Circuit Analysis and Design, 3ed. > by Neamen (2006) - [Solutions Manual Only] by Donald A. Neamen > Solutions manual to Modern Control Systems 11th by Richard C Dorf and > Robert H. Bishop > Solutions manual to Modern Organic Synthesis: An Introduction by > Michael H. Nantz, Hasan Palandoken, George S. Zweifel > Solutions manual to Power System Analysis By John J. Grainger, William > D. Stevenson Jr. hi guys i need solution manual Transport Processes and separation process principles =( could u send to my e-mail erdinc8479@hotmail.com ty for your help === Subject: Re: Are the integers as normed vector space complete? > Yes. Every discrete metric space is complete. Careful. Every uniformly discrete metric space is complete, but not every metric space that is discrete (as a topological space) is complete. That is: a metric space (X,d) is (topologically) discrete if for each x in X there is epsilon > 0 such that d(y,x) < epsilon implies y=x. It is uniformly discrete if there is a single epsilon > 0 such that for all x and y, d(y,x) < epsilon implies y = x. If the metric space is a group with a metric invariant under the group operations, the same epsilon that works for one x works for all of them, so in this case discreteness does imply completeness. -- Robert Israel israel@math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Are the integers as normed vector space complete? posting-account=_l4K0QkAAAC09JhOoK_ZfoJKXOmr_jZf Gecko/2009020911 Ubuntu/8.04 (hardy) Firefox/3.0.6,gzip(gfe),gzip(gfe) On Feb 25, 3:40æpm, Robert Israel > Yes. Every discrete metric space is complete. Careful. æEvery uniformly discrete metric space is complete, but not > every metric space that is discrete (as a topological space) is complete. -- Arturo Magidin, sans .sig === Subject: Re: How to prove this <27107722.1235573020209.JavaMail.jakarta@nitrogen.mathforum.org>, > Certainly the thing being summed, which in my version is > ---> arctan(1/sqrt(k)) > and in your version is > ---> arcsin(1/sqrt(k)) > could likely be shown to be irrational. > I would even bet it could be shown to be an > irrational multiple of Pi. Well, not always - for k = 1, they're both rational multiples of pi. arctan(1 / sqrt 3) and arcsin(1 / sqrt 4) are also rational multiples of pi. > But neither of these (if shown) would imply that > the *sum* of the arctans or arcsins could not be > a multiple of pi. What he said. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Alain's T-approximation does not work (was Re: Continuous iteration of another quadratic map) posting-account=fwSgtAkAAACFnX70ssKwbvm9_oCZVHrx Gecko/2009011913 Firefox/3.0.6,gzip(gfe),gzip(gfe) > On Feb 23, 2:50æam, alainvergh...@gmail.com > Am 22.02.2009 23:20 schrieb mike3: What I'd be more curious about though is how you got that > series you posted in the message you gave in terms of the > height. As I'd be curious about a similar case for f(0), i.e. > g(-phi). (phi = golden ratio) Then I could see what that period > 2 orbit looks like when done continuously. > First you build the matrix-operator for the function. The function is æ æ g(x) = a*x + 1*x^2 æ æ æ æ æ æ æ æ where æ æ a = (1+sqrt(5)) = 2*phi For iteration you need g(x) and its powers g(x)^2, g(x)^3,... To compute these powers it suffices to operate on the coefficients only > (square,cube,... the formal powerseries (which is here only a polynomial)) For iteration you combine these coefficients in the same way as you would > do a matrix-multiplication. So write the coefficients of the powers > of g(x) into a matrix: æ g(x)^0 æ = 1 æ+ æ 0*x æ+ æ 0*x^2 + æ0*x^3 + 0*x^4 + ... > æ g(x)^1 æ = 0 æ+ æ a*x æ+ æ 1*x^2 + æ0*x^3 + 0*x^4 + ... > æ g(x)^2 æ = 0 æ+ æ 0*x æ+ a^2*x^2 + 2a*x^3 + 1*x^4 + ... > æ ... > up to a reasonable dimension (to get well approximated results > for fractional iterates) The matrix G has then only the coefficients of the formal powerseries > (and I write it in transposed form) æG = æ 1 æ. æ . æ æ. æ . > æ æ æ æ. æa æ . æ æ. æ . > æ æ æ æ. æ1 æa^2 æ . æ . > æ æ æ æ. æ. 2a æ æa^3 æ. > æ æ æ æ.... ---------------------------------------------- > Second, the use of the matrix G Then a vector V(x) = [1,x,x^2,x^3,...] up to the selected dimension gives > using æ V(x) * G = [ 1, g(x), g(x)^2, g(x)^3 ,... ] In Pari/GP write (for a given dimension n) gp> æV(x) = vector(n,c,x^(c-1)) gp> æY = V(1) * G > æ then Y = [1, g(1) , g(1)^2, g(1)^3,...] correctly up to a certain column > æ The remaining columns are wrong due to truncation to finite size. Since the form of the result-vector is again like V(), a vector of consecutive > powers, the iteration can be performed by the formal relation æ V(x) æ æ* G = V(g(x)) > æ V(g(x)) * G = V(g(g(x)) > æ ... > and so on. However: this is exactly valid only in the limit, where we > assume infinite dimension. For integer iterates k this can be written as æV(gÁk(x)) = V(x) * G^k æ æ æ(theoretically) gp> æ æ æ æY = V(x) * G^k æ æ æ(practically) > æ where Y ~ V(gÁk(x)) æ æ æ æ æonly approximately which is true for the case of infinite dimension, and only approximate > with finite dimension. Just try this in Pari/GP. ----------------------------------------------------------- > 3) fractional iterates Next step is fractional iterate. It is an obvious idea to use > fractional powers of G for this. Fractional powers of G can either be determined by binomial-expansion > with G and I as arguments or by diagonalization. > The general formula for diagonalization of a matrix is æ æG æ = W * D * W^-1 æ æ æ æ æ æ æ// D is diagonal > and > æ æG^h = W * D^h * W^-1 So to find W and D in Pari/GP there is the function mateigen, which > gives a version of W first. gp> æ W = mateigen(G) Then compute WInv as inverse of W gp> æ WInv = W^-1 Then compute D as gp> æ D = WInv * G * W Then you can define a pari-function for arbitrary powers of D: gp> æ ædpow(x) = matrix(n,n,r,c,if(r==c,D[r,c]^x)) and then write gp> æ æGPow = W * dpow(0.5) * WInv to store G^0.5 in a constant matrix. Then you use gp> æ æ æY = V(x) * GPow æ æand Y is approximately ~ V(gÁh(x)) or just define the scalar function, using the relevant coefficients of GPow gp> æ gpow(x) = sum(k=0,n-1, x^k * GPow[1+k,2]) ----------------------------------------- > 4) Remarks: However, the approximation may be poor. I use dimension n=64 for > standard discussion and n=128 if the approximation is difficult. > The mateigen-procedure in Pari/GP may perform poorly because it > doesn't exploit the easiness of the solution for triangular matrices, > for which we can even get exact solutions. I've a version for > triangular matrices of the special form here (matrix-operators on V(x)) > implemented, which gives the eigensystem-decomposition of n=64x64 > or 128x128 triangular-matrices in a second or two and even symbolically > if the matrix-dimension is small. > I could mail it to you, but I think I've already posted it in the > tetration-forum in the matrix-method-thread. Also I use Euler-summation for convergence-acceleration. But this > needs some more explanations - try this first to get used to it. Gottfried Bonjour, I do believe you're using heavy tools. > What about approximating x^2-1 by a conjugate of > a known iterate :Ex (x^2-1) = f-1(f(x))^2 - 2) ,T(x)=x^2-2 > or æ (x^2-1) =~f^-1 o T o f (x) > and (x^2-1)^[r] =~f^-1 o T^r o f (x) The trouble is it only seems to work when x is large. What about small > x? There's even more trouble. The approximation does not work at all! :( Try r = 2, x = 4. You get ~ 222.9955156499789347277396687. Which is close to 223 -- and that's (4^2 - 1)^2 - 2 not (4^2 - 1)^2 - 1. It's even worse for close to 49727 -- and *that* is just ((4^2 - 1)^2 - 2)^2 - 2 not ((4^2 - 1)^2 - 1)^2 - 1. It is not approaching f^n(x) where f(x) = x^2 - 1, it's approaching f^(n-1)(x^2 - 1) where f(x) = x^2 - 2! === Subject: Re: Alain's T-approximation does not work (was Re: Continuous iteration of another quadratic map) posting-account=06BQLAoAAADoC7Y4z9FWcUwGvMa7xMG9 7.4),gzip(gfe),gzip(gfe) On Feb 23, 2:50æam, alainvergh...@gmail.com > Am 22.02.2009 23:20 schrieb mike3: What I'd be more curious about though is how you got that > series you posted in the message you gave in terms of the > height. As I'd be curious about a similar case for f(0), i.e. > g(-phi). (phi = golden ratio) Then I could see what that period > 2 orbit looks like when done continuously. > First you build the matrix-operator for the function. The function is æ æ g(x) = a*x + 1*x^2 æ æ æ æ æ æ æ æ where æ æ a = (1+sqrt(5)) = 2*phi For iteration you need g(x) and its powers g(x)^2, g(x)^3,... To compute these powers it suffices to operate on the coefficients only > (square,cube,... the formal powerseries (which is here only a polynomial)) For iteration you combine these coefficients in the same way as you would > do a matrix-multiplication. So write the coefficients of the powers > of g(x) into a matrix: æ g(x)^0 æ = 1 æ+ æ 0*x æ+ æ 0*x^2 + æ0*x^3 + 0*x^4 + ... > æ g(x)^1 æ = 0 æ+ æ a*x æ+ æ 1*x^2 + æ0*x^3 + 0*x^4 + ... > æ g(x)^2 æ = 0 æ+ æ 0*x æ+ a^2*x^2 + 2a*x^3 + 1*x^4 + ... > æ ... > up to a reasonable dimension (to get well approximated results > for fractional iterates) The matrix G has then only the coefficients of the formal powerseries > (and I write it in transposed form) æG = æ 1 æ. æ . æ æ. æ . > æ æ æ æ. æa æ . æ æ. æ . > æ æ æ æ. æ1 æa^2 æ . æ . > æ æ æ æ. æ. 2a æ æa^3 æ. > æ æ æ æ.... ---------------------------------------------- > Second, the use of the matrix G Then a vector V(x) = [1,x,x^2,x^3,...] up to the selected dimension gives > using æ V(x) * G = [ 1, g(x), g(x)^2, g(x)^3 ,... ] In Pari/GP write (for a given dimension n) gp> æV(x) = vector(n,c,x^(c-1)) gp> æY = V(1) * G > æ then Y = [1, g(1) , g(1)^2, g(1)^3,...] correctly up to a certain column > æ The remaining columns are wrong due to truncation to finite size. Since the form of the result-vector is again like V(), a vector of consecutive > powers, the iteration can be performed by the formal relation æ V(x) æ æ* G = V(g(x)) > æ V(g(x)) * G = V(g(g(x)) > æ ... > and so on. However: this is exactly valid only in the limit, where we > assume infinite dimension. For integer iterates k this can be written as æV(gÁk(x)) = V(x) * G^k æ æ æ(theoretically) gp> æ æ æ æY = V(x) * G^k æ æ æ(practically) > æ where Y ~ V(gÁk(x)) æ æ æ æ æonly approximately which is true for the case of infinite dimension, and only approximate > with finite dimension. Just try this in Pari/GP. ----------------------------------------------------------- > 3) fractional iterates Next step is fractional iterate. It is an obvious idea to use > fractional powers of G for this. Fractional powers of G can either be determined by binomial-expansion > with G and I as arguments or by diagonalization. > The general formula for diagonalization of a matrix is æ æG æ = W * D * W^-1 æ æ æ æ æ æ æ// D is diagonal > and > æ æG^h = W * D^h * W^-1 So to find W and D in Pari/GP there is the function mateigen, which > gives a version of W first. gp> æ W = mateigen(G) Then compute WInv as inverse of W gp> æ WInv = W^-1 Then compute D as gp> æ D = WInv * G * W Then you can define a pari-function for arbitrary powers of D: gp> æ ædpow(x) = matrix(n,n,r,c,if(r==c,D[r,c]^x)) and then write gp> æ æGPow = W * dpow(0.5) * WInv to store G^0.5 in a constant matrix. Then you use gp> æ æ æY = V(x) * GPow æ æand Y is approximately ~ V(gÁh(x)) or just define the scalar function, using the relevant coefficients of GPow gp> æ gpow(x) = sum(k=0,n-1, x^k * GPow[1+k,2]) ----------------------------------------- > 4) Remarks: However, the approximation may be poor. I use dimension n=64 for > standard discussion and n=128 if the approximation is difficult. > The mateigen-procedure in Pari/GP may perform poorly because it > doesn't exploit the easiness of the solution for triangular matrices, > for which we can even get exact solutions. I've a version for > triangular matrices of the special form here (matrix-operators on V(x)) > implemented, which gives the eigensystem-decomposition of n=64x64 > or 128x128 triangular-matrices in a second or two and even symbolically > if the matrix-dimension is small. > I could mail it to you, but I think I've already posted it in the > tetration-forum in the matrix-method-thread. Also I use Euler-summation for convergence-acceleration. But this > needs some more explanations - try this first to get used to it. Gottfried Bonjour, I do believe you're using heavy tools. > What about approximating x^2-1 by a conjugate of > a known iterate :Ex (x^2-1) = f-1(f(x))^2 - 2) ,T(x)=x^2-2 > or æ (x^2-1) =~f^-1 o T o f (x) > and (x^2-1)^[r] =~f^-1 o T^r o f (x) The trouble is it only seems to work when x is large. What about small > x? There's even more trouble. The approximation does not work at all! :( Try r = 2, x = 4. You get ~ 222.9955156499789347277396687. Which > is close to 223 -- and that's (4^2 - 1)^2 - 2 not (4^2 - 1)^2 - 1. > It's even worse for > close to 49727 -- and *that* is just ((4^2 - 1)^2 - 2)^2 - 2 not > ((4^2 - 1)^2 - 1)^2 - 1. It is not approaching f^n(x) where f(x) = x^2 > - 1, > it's approaching f^(n-1)(x^2 - 1) where f(x) = x^2 - 2!- Masquer le texte \ des messages pr.8ec.8edents - - Afficher le texte des messages pr.8ec.8edents - Bonjour Mike, I am not sure I've made my idea very clear : t1(x)=x^2-1 which has to be iterated, t2(x)=x^2-2 continuous iterate supposed known, We may search a conjugaison: t1 = h^[-1] o t2 o h or x^2-1 = h^[-1] ( h(x)^2-2) = x^2-2 I try h(x) = x+1/(2*x) as a first approximation , going far away from zero ... Alain === Subject: Re: Alain's T-approximation does not work (was Re: Continuous iteration of another quadratic map) posting-account=fwSgtAkAAACFnX70ssKwbvm9_oCZVHrx Gecko/2009011913 Firefox/3.0.6,gzip(gfe),gzip(gfe) On Feb 26, 3:25æam, alainvergh...@gmail.com > On Feb 23, 2:50æam, alainvergh...@gmail.com > Am 22.02.2009 23:20 schrieb mike3: What I'd be more curious about though is how you got that > series you posted in the message you gave in terms of the > height. As I'd be curious about a similar case for f(0), i.e. > g(-phi). (phi = golden ratio) Then I could see what that period > 2 orbit looks like when done continuously. > First you build the matrix-operator for the function. The function is æ æ g(x) = a*x + 1*x^2 æ æ æ æ æ æ æ æ where æ æ a = (1+sqrt(5)) = 2*phi For iteration you need g(x) and its powers g(x)^2, g(x)^3,... To compute these powers it suffices to operate on the coefficients only > (square,cube,... the formal powerseries (which is here only a polynomial)) For iteration you combine these coefficients in the same way as you would > do a matrix-multiplication. So write the coefficients of the powers > of g(x) into a matrix: æ g(x)^0 æ = 1 æ+ æ 0*x æ+ æ 0*x^2 + æ0*x^3 + 0*x^4 + ... > æ g(x)^1 æ = 0 æ+ æ a*x æ+ æ 1*x^2 + æ0*x^3 + 0*x^4 + ... > æ g(x)^2 æ = 0 æ+ æ 0*x æ+ a^2*x^2 + 2a*x^3 + 1*x^4 + ... > æ ... > up to a reasonable dimension (to get well approximated results > for fractional iterates) The matrix G has then only the coefficients of the formal powerseries > (and I write it in transposed form) æG = æ 1 æ. æ . æ æ. æ . > æ æ æ æ. æa æ . æ æ. æ . > æ æ æ æ. æ1 æa^2 æ . æ . > æ æ æ æ. æ. 2a æ æa^3 æ. > æ æ æ æ.... ---------------------------------------------- > Second, the use of the matrix G Then a vector V(x) = [1,x,x^2,x^3,...] up to the selected dimension gives > using æ V(x) * G = [ 1, g(x), g(x)^2, g(x)^3 ,... ] In Pari/GP write (for a given dimension n) gp> æV(x) = vector(n,c,x^(c-1)) gp> æY = V(1) * G > æ then Y = [1, g(1) , g(1)^2, g(1)^3,...] correctly up to a certain column > æ The remaining columns are wrong due to truncation to finite size. Since the form of the result-vector is again like V(), a vector of consecutive > powers, the iteration can be performed by the formal relation æ V(x) æ æ* G = V(g(x)) > æ V(g(x)) * G = V(g(g(x)) > æ ... > and so on. However: this is exactly valid only in the limit, where we > assume infinite dimension. For integer iterates k this can be written as æV(gÁk(x)) = V(x) * G^k æ æ æ(theoretically) gp> æ æ æ æY = V(x) * G^k æ æ æ(practically) > æ where Y ~ V(gÁk(x)) æ æ æ æ æonly approximately which is true for the case of infinite dimension, and only approximate > with finite dimension. Just try this in Pari/GP. ----------------------------------------------------------- > 3) fractional iterates Next step is fractional iterate. It is an obvious idea to use > fractional powers of G for this. Fractional powers of G can either be determined by binomial-expansion > with G and I as arguments or by diagonalization. > The general formula for diagonalization of a matrix is æ æG æ = W * D * W^-1 æ æ æ æ æ æ æ// D is diagonal > and > æ æG^h = W * D^h * W^-1 So to find W and D in Pari/GP there is the function mateigen, which > gives a version of W first. gp> æ W = mateigen(G) Then compute WInv as inverse of W gp> æ WInv = W^-1 Then compute D as gp> æ D = WInv * G * W Then you can define a pari-function for arbitrary powers of D: gp> æ ædpow(x) = matrix(n,n,r,c,if(r==c,D[r,c]^x)) and then write gp> æ æGPow = W * dpow(0.5) * WInv to store G^0.5 in a constant matrix. Then you use gp> æ æ æY = V(x) * GPow æ æand Y is approximately ~ V(gÁh(x)) or just define the scalar function, using the relevant coefficients of GPow gp> æ gpow(x) = sum(k=0,n-1, x^k * GPow[1+k,2]) ----------------------------------------- > 4) Remarks: However, the approximation may be poor. I use dimension n=64 for > standard discussion and n=128 if the approximation is difficult. > The mateigen-procedure in Pari/GP may perform poorly because it > doesn't exploit the easiness of the solution for triangular matrices, > for which we can even get exact solutions. I've a version for > triangular matrices of the special form here (matrix-operators on V(x)) > implemented, which gives the eigensystem-decomposition of n=64x64 > or 128x128 triangular-matrices in a second or two and even symbolically > if the matrix-dimension is small. > I could mail it to you, but I think I've already posted it in the > tetration-forum in the matrix-method-thread. Also I use Euler-summation for convergence-acceleration. But this > needs some more explanations - try this first to get used to it. Gottfried Bonjour, I do believe you're using heavy tools. > What about approximating x^2-1 by a conjugate of > a known iterate :Ex (x^2-1) = f-1(f(x))^2 - 2) ,T(x)=x^2-2 > or æ (x^2-1) =~f^-1 o T o f (x) > and (x^2-1)^[r] =~f^-1 o T^r o f (x) The trouble is it only seems to work when x is large. What about small > x? There's even more trouble. The approximation does not work at all! :( Try r = 2, x = 4. You get ~ 222.9955156499789347277396687. Which > is close to 223 -- and that's (4^2 - 1)^2 - 2 not (4^2 - 1)^2 - 1. > It's even worse for > close to 49727 -- and *that* is just ((4^2 - 1)^2 - 2)^2 - 2 not > ((4^2 - 1)^2 - 1)^2 - 1. It is not approaching f^n(x) where f(x) = x^2 > - 1, > it's approaching f^(n-1)(x^2 - 1) where f(x) = x^2 - 2!- Masquer le texte des messages pr.8ec.8edents - - Afficher le texte des messages pr.8ec.8edents - Bonjour Mike, I am not sure I've made my idea very clear : t1(x)=x^2-1 æwhich has to be iterated, > t2(x)=x^2-2 æcontinuous iterate supposed known, > We may search a conjugaison: > æ æt1 = h^[-1] o t2 o h > or x^2-1 = h^[-1] ( h(x)^2-2) = x^2-2 > I try æh(x) = x+1/(2*x) as a first approximation , > going far away from zero ... Alain Hmm. This seems to work better, although the superfunction identity (see my post) does not seem to transform it correctly to get iteration on smaller x-values. === Subject: sum of fibonaccis: = -1 ? The sum of the fibonacci-numbers is infinite; the partial sums diverge. However if I use the geometric series-formula I get sum = - 1 It fits also the pattern, which emerges, if I sum with some integer m>1 sum k=1,inf fib(k)/m^k = m/(m^2 - m -1) Is this appropriate? Gottfried Helms === Subject: Re: sum of fibonaccis: = -1 ? Am 25.02.2009 23:40 schrieb Gottfried Helms: > Is this appropriate? Gottfried Helms Hi - that I can defend that assignment of -1 to the sum by refering to the analytic continuation of the powerseries 1 + x + x^2 + ... and the closed-form-formula for the geometric series. (is the same way appropriate as...) However, Bill's invention makes me lough... :-) Very nice! Well, I took the result from the two geometric series, which occur, if I chose the Lucas-representation for the fibonacci numbers, and the sum of two analytic continued terms should be acceptable as closed-form formula too. In another context I had the *infinite* sum (though converging) of terms which all result from analytic continued geometric series - but that seems inappropriate: a concurring method gave different result which was better founded. Can we pinpoint the reason for an error/the error in the case of infinitely many values taken by analytic-continuation, or possibly have a means to quantify it/describe it by a function? Gottfried === Subject: Re: sum of fibonaccis: = -1 ? posting-account=AFsgCgkAAAA3VOfxqn2cTB2LbLN3nbER Gecko/20070319,gzip(gfe),gzip(gfe) One can also get it this way... Sumfib = 1 + 3 + 8 + 21 + 55 + ... : 1 + 2 + 5 + 13 + 34 + ... thus... : Sumfib +1 =(1+1)+ 3 + 8 + 21 + 55 + ... : (1 + 2)+ 5 + 13 + 34 + ... : : = (2 + 3) + 8 + 21 + 55 + ... : + (3 + 5) + 13 + 34 + ... : : = 0 + (5 + 8) + 21 + 55 + ... : 0 + (8 + 13) + 34 + ... : : = 0 + 0 +(13 + 21) + 55 + ... : 0 + 0 + (21 + 34) + ... : : = 0 + 0 + 0 + (34 + 55) + ... : 0 + 0 + 0 + (55 + ... : : = 0 + 0 = 0 : thus Sumfib = -1 . This is OC a totally Illegal Telescoping Cancellation (Infinite), (ITSI), but this itsi bitsi sort of calculation can be made rigorous by most of the summability methods Robert Israel alluded to (& Ben Goddard used); details in Hardy's Divergent Series. -- Wittering William * What an intellectual! * - Magdalen College Oxford, King's College Cambridge, * Imperial College London, Trinity College Dublin, * the Sorbonne, Harvard, MIT - he's heard of them all! === Subject: Re: sum of fibonaccis: = -1 ? The sum of the fibonacci-numbers is infinite; the partial sums > diverge. However if I use the geometric series-formula I get > sum = - 1 It fits also the pattern, which emerges, if I sum with some > integer m>1 sum k=1,inf fib(k)/m^k = m/(m^2 - m -1) Is this appropriate? It's appropriate as long as you make it clear that this is not literally a sum, but rather the result of some specified summability method. -- Robert Israel israel@math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: sum of fibonaccis: = -1 ? @mid.dfncis.de: The sum of the fibonacci-numbers is infinite; the partial sums > diverge. However if I use the geometric series-formula I get > sum = - 1 It fits also the pattern, which emerges, if I sum with some > integer m>1 sum k=1,inf fib(k)/m^k = m/(m^2 - m -1) Is this appropriate? It's as appropriate as sum_{k=0}^{inf} 2^k = 1/(1-2) = -1. B. -- Cheerfully resisting change since 1959. === Subject: Re: sum of fibonaccis: = -1 ? posting-account=Z3AipgkAAABkoMfyNwddSxsYhXHi5CDt MathPlayer 2.10d; .NET CLR 1.1.4322; PeoplePal 3.0),gzip(gfe),gzip(gfe) > @mid.dfncis.de: The sum of the fibonacci-numbers is infinite; the partial sums > diverge. However if I use the geometric series-formula I get > æsum = æ- 1 It fits also the pattern, which emerges, if I sum with some > integer m>1 æ sum k=1,inf fib(k)/m^k æ= m/(m^2 - m -1) Is this appropriate? It's as appropriate as sum {k=0}^{inf} æ2^k = 1/(1-2) = -1. B. But that sum is completely appropriate! It is a true infinite sum converging to the limit you give. At least it does if you are in the filed of 2-adic numbers. I can't think of a similar justification for the sum of the Fibonacci numbers, since we are raising units to powers and so there is no obvious way to create a justification. Achava === Subject: Re: sum of fibonaccis: = -1 ? Am 26.02.2009 06:19 schrieb Achava Nakhash, the Loving Snake: > (...) > I can't think of a similar justification for the sum of the Fibonacci > numbers, since we are raising units to powers and so there is no > obvious way to create a justification. Hi Achava - I don't understand this. raising units to powers - where do we this here? Would you mind to explain further? Btw, in my other post I said, that I got the sum by considering the lucas form fib(n) = (phi^n - (1-phi)^n)/sqrt(5) I just took the geometric series fibsum1 = sum {n>0} phi^n = 1/(1 - phi) // by anal.cont. of geometric series = - phi fibsum2 = sum {n>0} (1-phi)^n = 1/(1 - (1-phi)) // by anal.cont. of geometric series = 1/phi = phi-1 and fibsum = (fibsum1 - fibsum2)/sqrt(5) = -(2 phi - 1)/sqrt(5) = -1 Gottfried === Subject: Re: JSH: Factorization and the global economy Bull. Orders of Magnitude. You have no idea at all. >Ah. And you know this how??? >If the algorithms that you claim are 'far more secure' are known to >the >public, then they become available to all. e.g. Baton. See the NSA's >website. Look up their type 1 crypto algorithms. >If, however, they are secret, and you are privy to them, then your >revealing their existence and their security levels is a criminal >security violation. I know clearances and responcabilities very well, thank you. Just do a lot \ of reading and searching and you can run into several of the systems, public \ with some interesting info, watered down, not specific, to what is really there. And how would anyone know that ? NSA website; Cool they have newly declassified crypto from 1938. a lot of other good reading for sci.crypt for background they do have a declassified TS on cold war, interesting they would post such on the internet, known generic stuff, but why show a real declassified? ln 1968 a Soviet Golf-class nuclear submarine on patrol in the Pacific mysteriously went to the bottom with all hands. The Soviets could not locate the wreck, but the U.S. Navy could, and the U.S. began to study the feasibility of capturing it. (all rest white-ed out) They have some code patches for lenux too. If you factored some large #s that is quite impressive, and I wont bother you. I know of another person Brittish who was doing it, was obsessed with it and went off the deep end for good about 5 years ago, but he had factored \ some really big ones, not the blind search stuff either. === Subject: Re: JSH: Factorization and the global economy > Another moron. Do you know who you are talking to?? æHave you actually > done any crypto work for the NSA or DOD?? > Other methods are NOT 'far more secure'. > And the only method that is impossible to break is > a true OTP. æAnd key establishment problems make it > totally impractical to implement. æOne needs a secondary > communication channel for key establishment. Yes, but. æ What about quantum key distribution? It has not yet practical except over short distances. >(a few 10's of miles). Maintaining coherence over >longer distances has been a problem. I expect that these >difficulties will be solved in time. But it is not yet >ready for 'prime time'. > James, you said that you knew next to nothing about cryptography, so I will amplify some of the points Bob makes here. >However, this isn't the only problem. The key must be >truly random, not just pseudo-random, No, you cannot just use rand(). You need TRUE random data which is difficult to generate. A great deal of effort has been expended on this problem - true random data is definitely a problem. Google Yarrow or Fortuna for a couple of software approaches to the problem. For a hardware approach have a look at http://www.av8n.com/turbid/paper/turbid.htm >and it must be as long as the text it is encrypting. That means you need lots and lots of true random data. You do not just have a problem of data quality, you also have a problem of data quantity. >Furthermore, it can only be used ONCE. This is the big one. Once means once. A two time pad is as secure as transmitting the plaintext - not secure at all. Read up on the Venona break of Soviet cyphers: http://en.wikipedia.org/wiki/Venona_project Used only once can impose a lot of restrictions. For example, if your computer operating system can write a running program to a swapfile on disk then you might have to go as far as terminating your hard disk with extreme prejudice (and a sledgehemmer) and dissolving the result in acid. Once means once and once only. These issues are nothing to do with QKD, but are inherent in the use of a One Time Pad. The mathematical proof of its unbreakability depends critically on having TRUE random data used ONCE. If either condition is not met then the mathematical proof is not applicable and the resulting system is breakable. rossum >Thus, if you are encrypting a 10MB >message, you need a random 10MB that must be generated, >communicated, used ONCE, then thrown away. This makes >key management one royal pain in the *ss. Apparently, you, Pubkeybreaker, are known to be The > Real Thing to others in sci.crypt. It seems like too > good an opportunity to pass up, to find out whether > quantum key distribution is The Real Thing too. It is real. It works. But not yet on a large enough scale to >be useful. It can't yet be used if one wants to (say) >encrypt a message with OTP going from LA to (say) Miami. > Apparently, you also enjoy a good flame. I don't mind receiving flames. I have an asbestos suit :-) >Flame away. > === Subject: Re: JSH: Factorization and the global economy James, you said that you knew next to nothing about cryptography, so I > will amplify some of the points Bob makes here. However, this isn't the only problem. The key must be >truly random, not just pseudo-random, No, you cannot just use rand(). You need TRUE random data which is > difficult to generate. A great deal of effort has been expended on > this problem - true random data is definitely a problem. Google > Yarrow or Fortuna for a couple of software approaches to the > problem. For a hardware approach have a look at > http://www.av8n.com/turbid/paper/turbid.htm and it must be as long as the text it is encrypting. That means you need lots and lots of true random data. You do not > just have a problem of data quality, you also have a problem of data > quantity. But you're generating all these lovely quantum measurements anyway. Can't the measurements that are part of the quantum bit-passing also be the source of the one time pad? >Furthermore, it can only be used ONCE. This is the big one. Once means once. A two time pad is as secure as > transmitting the plaintext - not secure at all. Read up on the Venona > break of Soviet cyphers: http://en.wikipedia.org/wiki/Venona_project Used only once can impose a lot of restrictions. For example, if your > computer operating system can write a running program to a swapfile on > disk then you might have to go as far as terminating your hard disk > with extreme prejudice (and a sledgehemmer) and dissolving the result > in acid. Once means once and once only. These issues are nothing to do with QKD, but are inherent in the use > of a One Time Pad. The mathematical proof of its unbreakability > depends critically on having TRUE random data used ONCE. If either > condition is not met then the mathematical proof is not applicable and > the resulting system is breakable. I have a vague appreciation of the importance of using a One Time Pad only once (meaning really, really, really once -- no fooling). It's about on the level of More information good. I can't begin to imagine what I would do with, for example, two messages that I suspected were XORed with the same OTP. I don't understand why I might need a sledge hammer to be certain some files were not used again. Assuming all this stuff is software-mediated (the only way I can imagine me using the stuff), surely, the software is not going to re-use an old OTP accidentally, is it? It's possible someone could break into my computer and get the old OTPs laying around unerased and unsmashed, but in that case, they should just go ahead and read my mail. Jim Burns === Subject: Re: JSH: Factorization and the global economy >I have a vague appreciation of the importance of using a >One Time Pad only once (meaning really, really, really once >-- no fooling). It's about on the level of More information >good. I can't begin to imagine what I would do with, for example, >two messages that I suspected were XORed with the same OTP. Tim has answered this, you XOR the two cyphertexts together which removes all trace of the key, just leaving the XOR of the plaintexts. That is easily decryptable as an English plaintext is going to contain substrings like the , and etc. Using them as trial keys in all possible positions will start to bring out chunks of the other plaintext in the appropriate positions. Other methods are possible. >I don't understand why I might need a sledge hammer to be >certain some files were not used again. Assuming all this >stuff is software-mediated (the only way I can imagine me >using the stuff), surely, the software is not going to >re-use an old OTP accidentally, is it? If the adversary gets hold of your OTP after you have used it ... I have seen stories of the CIA doing exactly that with some of their old mainframe disc packs. When they failed they were terminated rather than let them out of the building for repair. As with all security, you need to judge the level of security you use to be commensurate with the resources available to your potential attackers. rossum === Subject: Re: The Real Purpose Of Math <19796970.1235316588589.JavaMail.jakarta@nitrogen.mathforum.org> <8763j2ffhf.fsf@phiwumbda.org> posting-account=KroL8woAAAACDGpRprxyFi_gYw4Un8Xt 2.0.50727; yplus 5.1.05b),gzip(gfe),gzip(gfe) > The idea is clearly religious, so why not take it to a æforum for religious discussion? What is totally religious about a system of Governing Equations? Yes, what *is* religious about the following inference? æ Governing Equation > existence of a Governor > existence of God > æ (topologically identified with the Governor) That's just topology, that is. æAnyone who denies it is simply > prejudiced against simple mathematics. You've convinced me, I tell you what. > -- > But what if I'm right [and have solved the factoring problem]? æAnd > what if I can't contain the problem and, oh, in six months from now, > you are desperately trying to find food as you run from humans who > have turned to cannibalism to survive? æ-- James S. Harris (12/21/09) A preliminary conceptualization of the Governing Equation (actually an inter-locking system of partial differential equations) is that it would contain the equations associated with the Standard Model and the best and most elegant description of gravitation as special cases. This set of equations possibly will utilize a generalization of complex variables to higher dimensional commutative algebras. http://www.intelrap.com/lt1.html Complex Variables span only the plane but in order to fully understand a higher dimensional brane we need a more powerful math tool. But if we do not believe that a Creator does exist who created many of the things for which we wish then we will never be motivated to search for the rule... the rule of the highest law of the Universe which may be metaphorically described as a United Verse that beckons all those with a genuine thirst to seek an understanding and to seek it first. If a man has a genuine creative wonder then he understands that he stands under a power much greater than himself. Then he may be lifted upon the shoulders of giants to learn the wonders of math and science and the knowledge that when lies are deleted only the truth is left. Newton stood on the shoulders of giants who now seek more new wondering clients to reveal more mysteries of a magnificent creation in which man plays a principal role in the observation. www.intelrap.com === Subject: Re: Proof of Fermat's Last Theorem posting-account=brOM-AoAAAChaAJEiH5z610-YOTfECd9 Gecko/20070530 Fedora/1.5.0.12-1.fc5 Firefox/1.5.0.12,gzip(gfe),gzip(gfe) > According to Fermat's Last Theorem, a cube can not be divided into > two smaller cubes, a fourth power can not be divided into two smaller > fourth powers, and so on. In other words, if a,b,c, and n are integers and a^n+b^n=c^n and > abcn <> 0, then the set of possible solutions for n is {-2, -1, 1, 2}. What is the proof for this? It's pretty elegant, but there isn't enough space here to write it. Michael === Subject: Re: number theory for children The problem is that number theory makes assertions requiring little background to understand, but proving things, which after all is what number theorists do, requires a whole armamentarium of stuff ranging widely over mathematics. OTOH, sure, show children some of the fascinating conclusions. Just be prepared to run for cover if the child asks How do they know that? === Subject: Re: Congrunces and perfect squares This number, when factorised, gives (10^(k+1) - 1)^2 which is a > perfect square. However, is it possible to show that the original number is a perfect > square using congruences from number theory and not by factorising it? Perhaps you mean 10^(2k+2) - 2(10^(k+1)) + 1 > instead of 10^(2k+1) - 2(10^(k+1)) + 1 Oops - I missed that. But yes, as you indicate, the first term was clearly meant to be (10^(2k+1))^2 or equivalently, 10^(2k+2) For the OP: Using the simple substitution x = 10^(k+1) your expression becomes x^2 + 2x + 1 which clearly factors as (x + 1)^2 The substitution makes the factorization more obvious. quasi === Subject: Re: Congrunces and perfect squares > This number, when factorised, gives (10^(k+1) - 1)^2 which is a > perfect square. However, is it possible to show that the original number is a perfect > square using congruences from number theory and not by factorising it? Perhaps you mean 10^(2k+2) - 2(10^(k+1)) + 1 > instead of 10^(2k+1) - 2(10^(k+1)) + 1 Oops - I missed that. But yes, as you indicate, the first term was clearly meant to be (10^(2k+1))^2 or equivalently, 10^(2k+2) For the OP: Using the simple substitution x = 10^(k+1) your expression becomes x^2 + 2x + 1 The above should be x^2 - 2x + 1 >which clearly factors as (x + 1)^2 And hence the above should be (x - 1)^2 >The substitution makes the factorization more obvious. quasi === Subject: Re: Congrunces and perfect squares This number, when factorised, gives (10^(k+1) - 1)^2 which is a > perfect square. However, is it possible to show that the original number is a perfect > square using congruences from number theory and not by factorising it? Perhaps you mean 10^(2k+2) - 2(10^(k+1)) + 1 > instead of 10^(2k+1) - 2(10^(k+1)) + 1 Oops - I missed that. But yes, as you indicate, the first term was clearly meant to be (10^(2k+1))^2 Ugh -- how careless I am. The above should be 10^(k+1) >or equivalently, 10^(2k+2) For the OP: Using the simple substitution x = 10^(k+1) your expression becomes x^2 + 2x + 1 The above should be x^2 - 2x + 1 which clearly factors as (x + 1)^2 And hence the above should be (x - 1)^2 The substitution makes the factorization more obvious. To the OP: I made a mess of things in my last 2 replies, but with the current corrections, it should now be ok. quasi === Subject: Re: Congrunces and perfect squares > This number, when factorised, gives (10^(k+1) - 1)^2 which is a > perfect square. However, is it possible to show that the original number is a perfect > square using congruences from number theory and not by factorising it? Perhaps you mean 10^(2k+2) - 2(10^(k+1)) + 1 > instead of 10^(2k+1) - 2(10^(k+1)) + 1 Oops - I missed that. But yes, as you indicate, the first term was clearly meant to be (10^(2k+1))^2 Ugh -- how careless I am. The above should be 10^(k+1) Ugh again -- I meant (10^(k+1))^2 >or equivalently, 10^(2k+2) For the OP: Using the simple substitution x = 10^(k+1) your expression becomes x^2 + 2x + 1 The above should be x^2 - 2x + 1 which clearly factors as (x + 1)^2 And hence the above should be (x - 1)^2 The substitution makes the factorization more obvious. To the OP: I made a mess of things in my last 2 replies, but with the current >corrections, it should now be ok. My last 3 replies -- all with careless errors. But now it's OK. quasi === Subject: Re: (m-1)x^2-2x+1-m=0 has two real solutions... posting-account=DSjMzAoAAAAISwlGnTY2c6bnmUVmdFNf AppleWebKit/528.16 (KHTML, like Gecko) Version/4.0 Safari/528.16,gzip(gfe),gzip(gfe) > Hy, I'm having troubles with the following problem: Prove that the following equation has two real solutions, for any m, > if m is not 1: > (m-1)x^2-2x+1-m=0 so, to find out how many solutions I apply the quadratic formula: ( 2 +/- sqr(4 - 4 * (m-1) * (1-m) ) / 2 (m-1) And take out the part that matters in this case: 16 - 4 * (m-1) * (1-m) And, since here I'm at a lost. I tried solving this last part, so: > 4 - 4 * (m-1)*(1-m) = 4 - 4 *(m-m^2-1+m)=4-4*(-m^2+2m-1)=4+4m^2-8m > +4=4m^2-8m+8 > This solves in my calculator to 1+1i or 1-1i. Which I have no use for > (to my knowledge). Now the book's answer is: > 4+4(m-1)^2 > 0 > witch I understand proves the point, althouhgt I don't know how I'm > supposed to get from (m-1)*(1-m) to (m-1)^2 (even if I'm feeling it > should be something rather obvious....) The discriminant b^2 - 4ac equals 4 - 4 (m-1)(1-m) which equals 4 + 4 (m-1)(m-1) (what is -(1-m) ? ). So you have a positive number plus a square; the sum is positive, therefore two solutions. But there is the division by 2a = 2 (m-1) which doesn't work out when m = 1. In that case the complete equation is -2x = 0 with solution x = 0 only. === Subject: Re: In need of a closed form for this constant! >How about using an root finding algorithm which >converges faster? >For example, if one start with a close enough initial >value, >Newton-Raphson method ( >http://en.wikipedia.org/wiki/Newton%27s_method > ) >usually converges quadratically (ie. double your >precision in >every iteration). Funny you should ask because I just discovered > a faster convergence by modifying my original > algorithm. The original which produces a slow convergence of (n)-- (3 - (log((sqrt(pi+1))*pi^2))) + pi = n > (3 - (log((sqrt(n+1))*n^2))) + n =n_1 > (3 - (log((sqrt(n_1+1))*n_1^2))) + n_1 = n_2 > (3 - (log((sqrt(n_2+1))*n_2^2))) + n_2 = n_3 > (3 - (log((sqrt(n_3+1))*n_3^2))) + n_3 = n_4 > etc. Modifying for a much faster convergence of (n) (3 - (log((sqrt(pi+1))*pi^2))) *((1/pi)+1) + pi = n > (3 - (log((sqrt(n+1))*n^2))) *((1/n)+1) + n = n_1 > (3 - (log((sqrt(n_1+1))*n_1^2))) *((1/n_1)+1) + n_1 = n_2 > (3 - (log((sqrt(n_2+1))*n_2^2))) *((1/n_2)+1) + n_2 = n_3 > (3 - (log((sqrt(n_3+1))*n_3^2))) *((1/n_3)+1) + n_3 = n_4 > etc. Probably more than twice as fast as my original version. Dan Your iteration converges linearly. I think you need around > 350 iterations to get a 1000 digit precision. In contrast, Newton-Raphson on f(x) = x^5 + x^4 - e^6 need > only about 9 iteration to achieve same accuracy ( with the > additional benefit you don't need to deal with the slower > multi-precision log() function ). BTW, does anyone know what is the relative complexity of > computing multiple precison log vs multiply and divide???? Denote by AGM the aritmethic-geometric mean. All elementary transcendental function can be computed with bit complexity O_B(M(n).log(n)) where M(n) is the bit complexity of multiplication. Borwein and Borwein, Pi_and_the_AGM. Denote by K(k) the complete elliptic integral of the first kind. K(k') (1) pi.----- = log(-) K(k) {q) K(k) 1 ----- = - [theta_3(q)]^2. pi 2 [theta_2(q)]^2 k = ---------------- [theta_3(q)]^2 theta_3(q) = sum_{-oo < n < oo} q^{n^2} theta_2(q) = sum_{-oo < n < oo} q^{(n + 1/2)^2} Fix q, 0 < a < q < b < 1. Step 1: Calculate K(k)/pi Step 2: Calculate k. Step 3: Calculate K(k') from the AGM. Step 4: Calculate log(1/q). -- Michael Press === Subject: Re: In need of a closed form for this constant! <28197189.1235495408224.JavaMail.jakarta@nitrogen.mathforum.org> posting-account=SvltewoAAAAi7TTYrD3mAaLUHzDiF2d1 Gecko/2009011913 Firefox/3.0.6,gzip(gfe),gzip(gfe) BTW, does anyone know what is the relative complexity of > computing multiple precison log vs multiply and divide???? Denote by AGM the aritmethic-geometric mean. All elementary > transcendental function can be computed with bit complexity æ æ O B(M(n).log(n)) where M(n) is the bit complexity of multiplication. æ > Borwein and Borwein, Pi and the AGM. Michael Press Cool, one more item on my bed-side to-read list (growing very fast in past 3 weeks since I start to re-read sci.math ;-p) === Subject: countable collection of dense open subsets of reals with intersection having 0 measure posting-account=2kYL5QoAAAAmH1e_g3WSyZ5Z9zXmRQX3 .NET CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) How can I construct a sequence of dense open subsets of IR such that the intersection is Lesbesque null set? === Subject: Re: countable collection of dense open subsets of reals with intersection having 0 measure > How can I construct a sequence of dense open subsets of IR such that > the intersection is Lesbesque null set? Am I reading this right? Isn't any dense, open subset of the reals (and hence a countable intersection) cocountable? === Subject: Re: countable collection of dense open subsets of reals with intersection having 0 measure posting-account=AdyLXQoAAABgRay99CKv1O8Y_7jjivwq InfoPath.1; .NET CLR 2.0.50727; .NET CLR 3.0.04506.30; .NET CLR 3.0.04506.648),gzip(gfe),gzip(gfe) > Am I reading this right? Isn't any dense, open subset of > the reals (and hence a countable intersection) cocountable? If C is a closed nowhere dense subset of the reals, then the complement of C is a dense open set. Now note that C can be uncountable, for example when C is a Cantor set. Dave L. Renfro === Subject: Re: countable collection of dense open subsets of reals with intersection having 0 measure > How can I construct a sequence of dense open subsets of IR such that > the intersection is Lesbesque null set? Cover the rationals by small open sets. === Subject: Re: countable collection of dense open subsets of reals with intersection having 0 measure posting-account=2kYL5QoAAAAmH1e_g3WSyZ5Z9zXmRQX3 .NET CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) How can I construct a sequence of dense open subsets of IR such that > the intersection is Lesbesque null set? Cover the rationals by small open sets. How can I construct a sequence of dense open subsets of IR such that > the intersection is Lesbesque null set? Cover the rationals by small open sets. Ok, here is my attempt (your hint was very good). Enumerate the rationals Q = {r k} for k=1,2,... Let G n = / ( r k - 1/(n*2*2^k) , r k + 1/(n*2*2^k) ) over n=1,2,... now by subadditivity, m(G n) <= SUM over k of( 1/n*2^k ) = 1/n (1/2 + 1/4 + 1/8 + ...) = 1/n Define G := /G n over n = 1,2,... Now, clearly G 1 superset of G 2 superset of G 3 ... Thus m(G) = lim{n->inf} m(G n) <= lim{n->inf} 1/n = 0 So m(G) = 0. === Subject: Coogi Men's Jeans - discount posting-account=uz1cQgoAAACe_NoiQ9omKJf2BvOAUW4x Mozilla/4.0 (compatible; MSIE 6.0; Windows NT 5.1; SV1) ; Embedded \ Web Browser from: http://bsalsa.com/),gzip(gfe),gzip(gfe) Where can go to find Coogi Men's Jeans? now We have a good collection of Coogi Men's Jeans in different styles and different colors. Please see them : http://www.luxury-fashion.org/static/Apparels/Coogi-Mens-Jeans-61.html http://www.luxury-fashion.org/static/Apparels/Coogi-Mens-Jeans-60.html Our Coogi Men's Jeans are fine quality And find more new fashion cloth,shoes and apparels please view : http://www.luxury-fashion.org Welcome check our other pages or feel free contact us. You can find what do you want here! === Subject: Re: Alternating Vertical and Horizontal Moves In Grid posting-account=ey-PjQoAAAD_Gi3gK0ZWKofsmviYOWhV Gecko/20081217 Firefox/2.0.0.20,gzip(gfe),gzip(gfe) Hi! This is an interesting problem, that I have never seen before. I will make the assumption that at any point a vertical move may be U or D, and a horizontal move may be L or R. In both the start-at-a-corner and general case, a(1) = 1, a(2) = 4, and a(3) = 6, for starters. Now I will focus only on the start-at-a-corner case, and it may as well be the upper left. Obviously n^2 is an upper bound for a(n), and if we try to establish some reasonable lower bounds, by considering a boustrophedonic traversal of the grid, there is a natural division into cases based on the value of n mod 4. In particular, I conjecture that the optimal algorithm for sufficiently large n = 1 mod 4 is to traverse the grid according to the following rules: 1) Start with down 1) Choose up over down whenever at least 4 more moves can be made as a result of doing so. 2) Alternate between choosing right n - 1 times in a row, then left n - 1 times in a row, then right, etc. To see what I mean by this, the 9 x 9 grid would be traversed like: DRURDRURDRURDRDRDLDLULULDLULDLDLDRDRURURDRDRURDRDLDLULDLULDLULDLU. With a little work this establishes: n mod 4 = 1 -> [ n^2 - 2n + 3 <= a(n) ] For the other cases some slight modifications of the algorithm above produce pretty tight lower bounds as well. In the above case, I would expect equality probably holds, but proving it is another story, it seems like it would be tricky. dc === Subject: Re: Alternating Vertical and Horizontal Moves In Grid >What is the sequence {a(k)} that is defined as follows? Start with an n-by-n grid. Move from square to adjacent square by >moving vertically then horizontally then vertically then horizontally, >following each horizontal move with a vertical move, and following >each vertical move with a horizontal move. >No square may be visited more than once. a(n) = the maximum number of squares visitable in this way on an n-by- >n grid. By brute force search, here are the values of a(n) for n = 1, ..., 8: 1, 4, 7, 14, 19, 32, 39, 58 >A variation: You must start in a corner square. Then the first 8 values are: 1, 4, 6, 14, 18, 32, 38, 58 Note that, so far, if n is even, starting in a corner is just as good a starting place as any other location, while if n is odd and n > 1, the value of a(n) when starting in a corner appears to be just 1 less than that for an optimum starting place. >(It does not matter if the first move is vertical or horizontal, as >far as the values of each a(n) are concerned, of course.) I am guessing that this sequence is really easy to calculate, maybe a >(n) is as simple as a quadratic polynomial in terms of n. Or is the >best way to calculate each a(n) is to do a brute-force search of >possible paths through the grid? The function a(n) is definitely not a polynomial in n. Clearly a(n) is positive and bounded above by n^2, hence, if a(n) was a polynomial in n, it would have to be at most of degree 2. But the calculated values of a(n) as posted above contradict that. So what is the nature of a(n)? Consider the sequence b(n) = n^2 - a(n) Clearly b(n) must nonnegative. I suspect it would not be too hard to prove that b(n) must be positive, for n > 2. We can ask whether the sequence b(n) is bounded above? I suspect not, but if, in fact, b(n) is bounded above, then I would guess that the sequence b(n) repeats. Other than brute force search, you might try for recursive relationships, or at least recursive bounds. quasi === Subject: Re: Alternating Vertical and Horizontal Moves In Grid What is the sequence {a(k)} that is defined as follows? Start with an n-by-n grid. Move from square to adjacent square by >moving vertically then horizontally then vertically then horizontally, >following each horizontal move with a vertical move, and following >each vertical move with a horizontal move. >No square may be visited more than once. a(n) = the maximum number of squares visitable in this way on an n-by- >n grid. By brute force search, here are the values of a(n) for n = 1, ..., 8: 1, 4, 7, 14, 19, 32, 39, 58 A variation: You must start in a corner square. Then the first 8 values are: 1, 4, 6, 14, 18, 32, 38, 58 Note that, so far, if n is even, starting in a corner is just as good >a starting place as any other location, while if n is odd and n > 1, >the value of a(n) when starting in a corner appears to be just 1 less >than that for an optimum starting place. (It does not matter if the first move is vertical or horizontal, as >far as the values of each a(n) are concerned, of course.) I am guessing that this sequence is really easy to calculate, maybe a >(n) is as simple as a quadratic polynomial in terms of n. Or is the >best way to calculate each a(n) is to do a brute-force search of >possible paths through the grid? The function a(n) is definitely not a polynomial in n. Clearly a(n) is >positive and bounded above by n^2, hence, if a(n) was a polynomial in >n, it would have to be at most of degree 2. But the calculated values >of a(n) as posted above contradict that. So what is the nature of a(n)? Consider the sequence b(n) = n^2 - a(n) Clearly b(n) must nonnegative. I suspect it would not be too hard to >prove that b(n) must be positive, for n > 2. We can ask whether the sequence b(n) is bounded above? I suspect not, >but if, in fact, b(n) is bounded above, then I would guess that the >sequence b(n) repeats. Other than brute force search, you might try for recursive >relationships, or at least recursive bounds. Ok, there is a simple pattern that I missed. When n is even, it appears that a(n) = n^2 - n + 2 quasi === Subject: Re: Alternating Vertical and Horizontal Moves In Grid >What is the sequence {a(k)} that is defined as follows? Start with an n-by-n grid. Move from square to adjacent square by >moving vertically then horizontally then vertically then horizontally, >following each horizontal move with a vertical move, and following >each vertical move with a horizontal move. >No square may be visited more than once. a(n) = the maximum number of squares visitable in this way on an n-by- >n grid. By brute force search, here are the values of a(n) for n = 1, ..., 8: 1, 4, 7, 14, 19, 32, 39, 58 A variation: You must start in a corner square. Then the first 8 values are: 1, 4, 6, 14, 18, 32, 38, 58 Note that, so far, if n is even, starting in a corner is just as good >a starting place as any other location, while if n is odd and n > 1, >the value of a(n) when starting in a corner appears to be just 1 less >than that for an optimum starting place. (It does not matter if the first move is vertical or horizontal, as >far as the values of each a(n) are concerned, of course.) I am guessing that this sequence is really easy to calculate, maybe a >(n) is as simple as a quadratic polynomial in terms of n. Or is the >best way to calculate each a(n) is to do a brute-force search of >possible paths through the grid? The function a(n) is definitely not a polynomial in n. Clearly a(n) is >positive and bounded above by n^2, hence, if a(n) was a polynomial in >n, it would have to be at most of degree 2. But the calculated values >of a(n) as posted above contradict that. So what is the nature of a(n)? Consider the sequence b(n) = n^2 - a(n) Clearly b(n) must nonnegative. I suspect it would not be too hard to >prove that b(n) must be positive, for n > 2. We can ask whether the sequence b(n) is bounded above? I suspect not, >but if, in fact, b(n) is bounded above, then I would guess that the >sequence b(n) repeats. Other than brute force search, you might try for recursive >relationships, or at least recursive bounds. Ok, there is a simple pattern that I missed. When n is even, it appears that a(n) = n^2 - n + 2 Moreover, it's fairly easy to show that, for even n, a(n) >= n^2 - n + 2 Assuming n is even, a valid path through n^2 - n + 2 vertices can be explicitly constructed as follows ... Start at (0,0). Go right, up, left, up. Repeat until reaching the top. Since n is even, the last 2 moves must have been right, up, ending at (1,n-1). Now go right, down, ending at (2,n-2). We have so far visited 2n + 1 vertices. Now recursively do the same in the n-2 by n-2 square with upper left corner vertex (2,n-2) and lower right corner vertex (n-1,1). The path described above adheres to the restrictions, and it's easily proved, by induction, that it has exactly n^2 - n + 2 vertices. quasi === Subject: Re: Alternating Vertical and Horizontal Moves In Grid What is the sequence {a(k)} that is defined as follows? Start with an n-by-n grid. Move from square to adjacent square by >moving vertically then horizontally then vertically then horizontally, >following each horizontal move with a vertical move, and following >each vertical move with a horizontal move. >No square may be visited more than once. a(n) = the maximum number of squares visitable in this way on an n-by- >n grid. By brute force search, here are the values of a(n) for n = 1, ..., 8: 1, 4, 7, 14, 19, 32, 39, 58 A variation: You must start in a corner square. Then the first 8 values are: 1, 4, 6, 14, 18, 32, 38, 58 Note that, so far, if n is even, starting in a corner is just as good >a starting place as any other location, while if n is odd and n > 1, >the value of a(n) when starting in a corner appears to be just 1 less >than that for an optimum starting place. (It does not matter if the first move is vertical or horizontal, as >far as the values of each a(n) are concerned, of course.) I am guessing that this sequence is really easy to calculate, maybe a >(n) is as simple as a quadratic polynomial in terms of n. Or is the >best way to calculate each a(n) is to do a brute-force search of >possible paths through the grid? The function a(n) is definitely not a polynomial in n. Clearly a(n) is >positive and bounded above by n^2, hence, if a(n) was a polynomial in >n, it would have to be at most of degree 2. But the calculated values >of a(n) as posted above contradict that. So what is the nature of a(n)? Consider the sequence b(n) = n^2 - a(n) Clearly b(n) must nonnegative. I suspect it would not be too hard to >prove that b(n) must be positive, for n > 2. We can ask whether the sequence b(n) is bounded above? I suspect not, >but if, in fact, b(n) is bounded above, then I would guess that the >sequence b(n) repeats. Other than brute force search, you might try for recursive >relationships, or at least recursive bounds. Ok, there is a simple pattern that I missed. When n is even, it appears that a(n) = n^2 - n + 2 Moreover, it's fairly easy to show that, for even n, a(n) >= n^2 - n + 2 Assuming n is even, a valid path through n^2 - n + 2 vertices can be >explicitly constructed as follows ... Start at (0,0). Go right, up, left, up. Repeat until reaching the top. Since n is even, the last 2 moves must have been right, up, ending at >(1,n-1). Now go right, down, ending at (2,n-2). We have so far visited 2n + 1 vertices. Now recursively do the same in the n-2 by n-2 square with upper left >corner vertex (2,n-2) and lower right corner vertex (n-1,1).| To clarify: At each successive lower level in the recursion, switch up and down. Also, when you finish the 2 x 2 level, stop. >The path described above adheres to the restrictions, and it's easily >proved, by induction, that it has exactly n^2 - n + 2 vertices. quasi === Subject: Re: Alternating Vertical and Horizontal Moves In Grid <4utbq4dhhldvts8o5hl3ld180tan2tbdmq@4ax.com> posting-account=ey-PjQoAAAD_Gi3gK0ZWKofsmviYOWhV Gecko/20081217 Firefox/2.0.0.20,gzip(gfe),gzip(gfe) What is the sequence {a(k)} that is defined as follows? Start with an n-by-n grid. Move from square to adjacent square by >moving vertically then horizontally then vertically then horizontally, >following each horizontal move with a vertical move, and following >each vertical move with a horizontal move. >No square may be visited more than once. a(n) = the maximum number of squares visitable in this way on an n-by- >n grid. By brute force search, here are the values of a(n) for n = 1, ..., 8: 1, 4, 7, 14, 19, 32, 39, 58 A variation: You must start in a corner square. Then the first 8 values are: 1, 4, 6, 14, 18, 32, 38, 58 Note that, so far, if n is even, starting in a corner is just as good >a starting place as any other location, while if n is odd and n > 1, >the value of a(n) when starting in a corner appears to be just 1 less >than that for an optimum starting place. (It does not matter if the first move is vertical or horizontal, as >far as the values of each a(n) are concerned, of course.) I am guessing that this sequence is really easy to calculate, maybe a >(n) is as simple as a quadratic polynomial in terms of n. Or is the >best way to calculate each a(n) is to do a brute-force search of >possible paths through the grid? The function a(n) is definitely not a polynomial in n. Clearly a(n) is >positive and bounded above by n^2, hence, if a(n) was a polynomial in >n, it would have to be at most of degree 2. But the calculated values >of a(n) as posted above contradict that. So what is the nature of a(n)? Consider the sequence b(n) = n^2 - a(n) Clearly b(n) must nonnegative. I suspect it would not be too hard to >prove that b(n) must be positive, for n > 2. We can ask whether the sequence b(n) is bounded above? I suspect not, >but if, in fact, b(n) is bounded above, then I would guess that the >sequence b(n) repeats. Other than brute force search, you might try for recursive >relationships, or at least recursive bounds. Ok, there is a simple pattern that I missed. When n is even, it appears that a(n) = n^2 - n + 2 Moreover, it's fairly easy to show that, for even n, a(n) >= n^2 - n + 2 Assuming n is even, a valid path through n^2 - n + 2 vertices can be >explicitly constructed as follows ... Start at (0,0). Go right, up, left, up. Repeat until reaching the top. Since n is even, the last 2 moves must have been right, up, ending at >(1,n-1). Now go right, down, ending at (2,n-2). We have so far visited 2n + 1 vertices. Now recursively do the same in the n-2 by n-2 square with upper left >corner vertex (2,n-2) and lower right corner vertex (n-1,1).| To clarify: At each successive lower level in the recursion, switch up and > down. Also, when you finish the 2 x 2 level, stop. The path described above adheres to the restrictions, and it's easily >proved, by induction, that it has exactly n^2 - n + 2 vertices. quasi this seems legitimate to me. quasi, by the sounds of it, you have some kind of a computer program capable of a brute force test? can you test my conjecture above for n = 9, 13? i'm thinking we are going tol end up here with a quadratic answer, but different quadratics depending on the value of n mod 4. === Subject: Re: Alternating Vertical and Horizontal Moves In Grid What is the sequence {a(k)} that is defined as follows? Start with an n-by-n grid. Move from square to adjacent square by >moving vertically then horizontally then vertically then horizontally, >following each horizontal move with a vertical move, and following >each vertical move with a horizontal move. >No square may be visited more than once. a(n) = the maximum number of squares visitable in this way on an n-by- >n grid. By brute force search, here are the values of a(n) for n = 1, ..., 8: 1, 4, 7, 14, 19, 32, 39, 58 A variation: You must start in a corner square. Then the first 8 values are: 1, 4, 6, 14, 18, 32, 38, 58 Note that, so far, if n is even, starting in a corner is just as good >a starting place as any other location, while if n is odd and n > 1, >the value of a(n) when starting in a corner appears to be just 1 less >than that for an optimum starting place. (It does not matter if the first move is vertical or horizontal, as >far as the values of each a(n) are concerned, of course.) I am guessing that this sequence is really easy to calculate, maybe a >(n) is as simple as a quadratic polynomial in terms of n. Or is the >best way to calculate each a(n) is to do a brute-force search of >possible paths through the grid? The function a(n) is definitely not a polynomial in n. Clearly a(n) is >positive and bounded above by n^2, hence, if a(n) was a polynomial in >n, it would have to be at most of degree 2. But the calculated values >of a(n) as posted above contradict that. So what is the nature of a(n)? Consider the sequence b(n) = n^2 - a(n) Clearly b(n) must nonnegative. I suspect it would not be too hard to >prove that b(n) must be positive, for n > 2. We can ask whether the sequence b(n) is bounded above? I suspect not, >but if, in fact, b(n) is bounded above, then I would guess that the >sequence b(n) repeats. Other than brute force search, you might try for recursive >relationships, or at least recursive bounds. Ok, there is a simple pattern that I missed. When n is even, it appears that a(n) = n^2 - n + 2 Moreover, it's fairly easy to show that, for even n, a(n) >= n^2 - n + 2 Assuming n is even, a valid path through n^2 - n + 2 vertices can be >explicitly constructed as follows ... Start at (0,0). Go right, up, left, up. Repeat until reaching the top. Since n is even, the last 2 moves must have been right, up, ending at >(1,n-1). Now go right, down, ending at (2,n-2). We have so far visited 2n + 1 vertices. Now recursively do the same in the n-2 by n-2 square with upper left >corner vertex (2,n-2) and lower right corner vertex (n-1,1).| To clarify: At each successive lower level in the recursion, switch up and > down. Also, when you finish the 2 x 2 level, stop. The path described above adheres to the restrictions, and it's easily >proved, by induction, that it has exactly n^2 - n + 2 vertices. quasi this seems legitimate to me. quasi, by the sounds of it, you have some >kind of a computer program capable of a brute force test? can you test >my conjecture above for n = 9, 13? i'm thinking we are going tol end >up here with a quadratic answer, but different quadratics depending on >the value of n mod 4. Yes, that seems very plausible -- 3 quadratics, one for even n, one for n = 1 mod 4, and one for n = 3 mod 4. I do have a brute force test, but it wasn't designed for a heavy load. For n > 8, it struggles. I'll let it run for a while -- hopefully, it will at least compute a(9). quasi === Subject: Re: Alternating Vertical and Horizontal Moves In Grid <4utbq4dhhldvts8o5hl3ld180tan2tbdmq@4ax.com> posting-account=ey-PjQoAAAD_Gi3gK0ZWKofsmviYOWhV Gecko/20081217 Firefox/2.0.0.20,gzip(gfe),gzip(gfe) no, i changed my mind about the mod 4 thing, it looks a lot like your formula is right whenever n is even, and my formula is right (by a similar argument) when n is odd. of course, we have only established them as lower bounds with certainty. === Subject: Re: Alternating Vertical and Horizontal Moves In Grid >no, i changed my mind about the mod 4 thing, it looks a lot like your >formula is right whenever n is even, and my formula is right (by a >similar argument) when n is odd. of course, we have only established >them as lower bounds with certainty. Yes, if we require corner starting point, and assuming n > 1, I think you may be right. Thus, for a corner starting point, and n > 1, the conjecture is: a(n) = n^2 - 2n + 3, if n is odd n^2 - n + 2, if n is even quasi PS -- My Maple program is still working on a(9), assuming a required corner starting point. === Subject: Re: Alternating Vertical and Horizontal Moves In Grid no, i changed my mind about the mod 4 thing, it looks a lot like your >formula is right whenever n is even, and my formula is right (by a >similar argument) when n is odd. of course, we have only established >them as lower bounds with certainty. Yes, if we require corner starting point, and assuming n > 1, I think >you may be right. Thus, for a corner starting point, and n > 1, the conjecture is: a(n) = > n^2 - 2n + 3, if n is odd > n^2 - n + 2, if n is even quasi PS -- My Maple program is still working on a(9), assuming a required >corner starting point. Ok, Maple completed a(9) and is now working on a(11). Good news -- a(9) = 66, as expected. Note, for convenience, we are now always making the assumption of a corner starting point. For even n, it apparently doesn't matter. For odd n > 1, allowing a variable starting point appears to make a(n) increase by 1. quasi === Subject: Re: Alternating Vertical and Horizontal Moves In Grid <4utbq4dhhldvts8o5hl3ld180tan2tbdmq@4ax.com> <4p7cq45og61mehq711tc9qm7ir9avhp4gl@4ax.com> posting-account=ey-PjQoAAAD_Gi3gK0ZWKofsmviYOWhV Gecko/20081217 Firefox/2.0.0.20,gzip(gfe),gzip(gfe) no, i changed my mind about the mod 4 thing, it looks a lot like your >formula is right whenever n is even, and my formula is right (by a >similar argument) when n is odd. of course, we have only established >them as lower bounds with certainty. Yes, if we require corner starting point, and assuming n > 1, I think >you may be right. Thus, for a corner starting point, and n > 1, the conjecture is: a(n) = > n^2 - 2n + 3, if n is odd > n^2 - n + 2, if n is even quasi PS -- My Maple program is still working on a(9), assuming a required >corner starting point. Ok, Maple completed a(9) and is now working on a(11). Good news -- a(9) = 66, as expected. Note, for convenience, we are now always making the assumption of a > corner starting point. For even n, it apparently doesn't matter. For > odd n > 1, allowing a variable starting point appears to make a(n) > increase by 1. quasi this makes sense to me, i am starting to work on a proof of the bound for the even case, as i think it will be the easier one. i have some ideas, but nothing concrete enough to comment on yet. === Subject: Re: Newton's Method: Cute trick. posting-account=DSjMzAoAAAAISwlGnTY2c6bnmUVmdFNf AppleWebKit/528.16 (KHTML, like Gecko) Version/4.0 Safari/528.16,gzip(gfe),gzip(gfe) I have never seen this trick explicitly mentioned anywhere. Of course if f (p) < 0, f (q) > 0, f continuous on [p, q] then there is a root of f in (p, q), and if you use _any_ iterative algorithm, then clamping the next iterated result to [p, q] will get you closer to the root. Some things I noted: (1) If you find x0 such that f (x0) is small enough, but x0 is not in [p, q] then clamping x0 to [p, q] might give a new value where f (x) is _not_ small enough anymore, so care must be taken. (2) Clamping can actually only happen on the first Newton-Raphson step if your convergence criterion is actually met, because you proved that each iteration gets closer to the root of F (x). If you are not 100% sure that the convergence criterion is met, clamping might not help. (3) This doesn't solve the problem of finding x0 close enough to the root. Of course it is no big deal if a and b are known. === Subject: Re: Help!How to solve the equation