mm-1022 Subject: Re: Mean Value Theorem > Two places I¹ve seen the equivalent statement: If f is continuously > differentiable on [0,1], then |f(x)-f(y)|<=sup|f¹(u)||x-y| where the > sup is extended over 0<=u<=1. > I can¹t see why we need f¹ to be continuous. It seems by the MVT we > have > f(x)-f(y)=f¹(c)(x-y) for some c in [0,1], > so that > |f(x)-f(y)| = |f¹(c)||x-y|, for some c in [0,1], > <= sup_u|f¹(u)||x-y|. Maybe they intended f have values in R^n, so MVT is not available. Assuming f¹ exists everywhere and f¹ is integrable, we can still integrate f¹ on [x,y] to get the conclusion. And f¹ bounded will imply f¹ is Riemann integrable, so adding the assumption f¹ is integrable is not needed. So I agree: even in this case, continuity of f¹ is not requried. -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: How fast is the ConÞnued Fraction factorization algorithm? >My impression, based on not very scientiÞc >(or knowledgeable) experiment >is that the Quadratic Sieve (with many polynomials) >is unlikely to factorize number with more than 80 digits >in a reasonable time on a reasonable computer >(say 1 day on my 667MHz laptop). >>What do you mean by unlikely? It makes no sense in this >>context. QS succeeds with virtual certainty in time that depends >>only on the size of n (with some small statistical variation). > That¹s just not true in my experience. > And my comment makes perfect sense. > The quadratic sieve completes in one or two minutes > with some 70-digit numbers, and takes hours with others. > Which is exactly what I would expect. > What reason do you have to suppose the time has only > small statistical variation? > Have you actually tried it? > What if the number is prime? > (I gave random 60-digit numbers for factorisation > as an assignment to a class of 30. > One was prime, and there was a huge variation > in the time taken - using quadratic sieve - > to factorise the others.) This is probably because your implementation runs trial division, the p+-1 and/or a couple of runs of ECM before running the quadratic seive. You can probably test this by factoring a few hundred large numbers and looking a a histogram of the run times. It will be extremely bimodal. The p+-1 and ECM are extremely fast, but probabilistic algoritms (they also have expected run times based on the size of the factors, not on the size of the original number.) The original quadratic seive (if I recall correctly) worked like this: To factor n 1) Choose a factor base of small primes. 2) Choose a random number r between 1 and n-1 3) If the least positive residue of r^2 mod n factors completely using only the primes in the factor base save r and a list of the primes occuring to odd powers. 4) repeat steps 2 and 3 till yo have about as many values of r as there are primes in the factor base. 5) You now have almost enough info to Þnd a pair of numbers such that x^2=y^2 mod n. Improvements have increased the probability of Þnding numbers that factor completely, but the philosopy is the same. The run time will be proportional to the number of trials it takes to Þnd enough numbers that completely factor. === Subject: Re: covering compact set w/squares Oh, I think I did misread. How about this: Given a compact set K in the plane s.t. each pt x is the center of a square Q_x, prove that you can Þnd a subsequence Q_x_i of squares s.t. K is covered by the Q_x_i and sum(over i) Char(Q_x_i) <= 4 Char (union(over i) Q_x_i), where Char(X) is the characteristic (indicator) function of X. >Given a compact set K in the plane s.t. each pt x is the center of a square >Q_x, prove that you can Þnd a subsequence Q_x_i of squares s.t. K is >covered by the Q_x_i and the sum of the areas of each of the Q_x_i is no >more than 4 times the area of the union of the Q_x_i. > Aargh. I think I¹ll just spend the rest of my life posting replies to > this message. I have to say that the two counterexamples I¹ve > given were wrong (I was reading the statement of the problem > Note however that the example showing that it¹s not true that > in a minimal cover no point is in more than four squares was > correct. (I¹m still not sure whether the result itself is true and > I still wonder whether it¹s really what you wanted to prove.) >I think this is a totally geometric thing, and I know it sufÞces to show >that in such a minimal cover no point is contained in more than 4 squares, >but how do i show this? I know how to do this with intervals on the real >line (there the answer is 2 times the area....), but i get lost when going >up a dimension. > ************************ > David C. Ullrich === Subject: Re: how to understand this statement of Second order Necessary Condition? X-AUTHid: rubin > I read the statement of SONC for local minimizer as follows: > Let x* a local minimizer of function f over some constraint set O, > and d a feasible direction at this point x*. If d¹*gradient=0, then > d¹*H*d>=0. > where d¹ is the transpose of d column vector... > I got confused here because this seems to me is a half statement: > If d¹*gradient=0, then d¹*H*d>=0. > but what if d¹*gradient>0 but not =0 at that point x*? > Is this a possible case in the SONC statment? Is that true that when > d¹*gradient>0 but not =0, there is no statement about the SONC? > -Walalal First of all, d having positive inner-product with the gradient is not a problem, since x* is a local minimizer, not a maximizer. I suspect you¹re really interested in the case d¹*gradient < 0. That cannot happen at a local minimizer when d is a feasible direction, since for some small t > 0 x*+td would be feasible and f(x*+td) would be less than f(x*) (by virtue of the gradient condition), contradicting x* being a minimizer. -- Paul ************************************************************* ************ Paul A. Rubin Phone: (517) 432-3509 Department of Management Fax: (517) 432-1111 The Eli Broad Graduate School of Management E-mail: rubin@msu.edu Michigan State University http://www.msu.edu/~rubin/ East Lansing, MI 48824-1122 (USA) ************************************************************* ************ Mathematicians are like Frenchmen: whenever you say something to them, they translate it into their own language, and at once it is something entirely different. J. W. v. GOETHE === Subject: strain softening spring I am a Civil engineer who has been working for the last 20 years in the Þeld as a result of which my math is somewhat basic. I have a good handle of linear systems, but am taking my Þrst baby steps with non-lineear systems. I have read up a few books on non-linear systems looking for a canned solution for a problem I am trying to solve. But while I see a lot of stuff on the DufÞng equation I don¹t see anything that resembles the equation I am trying to solve - a mass-spring-damper system where the spring softens exponentially. i.e. the equation I am trying to solve is as follows: y¹¹ + cy¹ + [ko*e^(-alpha*t)]y = 0 Any help would be most appreciated. sincerely Paul Joseph (pjoseph@excite.com) === Subject: Re: Simple numbers > Hallo; > Well, p(n) is a n-th prime but if n=123456 how do I know what is the > 123456-th prime? I need the correct number (for exampel 1-st prime is -> 2)! >> Hallo, >> I know that every simple number (beside 2 and 3) has its one formula : >> S.N = >> 6*n +/- 1 for some n , but not any n={1,2,3,...+} in formula 6*n +/- 1 >> gives >> the simple number (for example: n=6 => 6*6-1=35 , 35 is not simple , >> 5,7|35) >> By simple number I think you must mean prime. >> so my question is: >> If n={1,2,3,...} what is the formula which gives the simple numbers >> (beside >> 2 and 3) for any n? >> Evidently you don¹t consider p(n) is the n¹th prime to be a formula. >> Any >> particular reason? >> Computational complexity? >> Jon Miller Given any n, there is a simple algorithm that computes p(n). You could spend a lifetime trying to make the algorithm more efÞcient, but the basic algorithm is quite simple. === Subject: Re: Mean Value Theorem >> Two places I¹ve seen the equivalent statement: If f is continuously >> differentiable on [0,1], then |f(x)-f(y)|<=sup|f¹(u)||x-y| where the >> sup is extended over 0<=u<=1. >> I can¹t see why we need f¹ to be continuous. It seems by the MVT we >> have >> >> f(x)-f(y)=f¹(c)(x-y) for some c in [0,1], >> >> so that >> >> |f(x)-f(y)| = |f¹(c)||x-y|, for some c in [0,1], >> <= sup_u|f¹(u)||x-y|. >Maybe they intended f have values in R^n, so MVT is not available. >Assuming f¹ exists everywhere and f¹ is integrable, we can still >integrate f¹ on [x,y] to get the conclusion. And f¹ bounded will >imply f¹ is Riemann integrable, so adding the assumption >f¹ is integrable is not needed. So I agree: even in this >case, continuity of f¹ is not requried. Or you can just say this: If f : R -> R^n is differentiable then |f(x)-f(y)| <= sup_u|f¹(u)||x-y|. Pf: Choose p in R^n with |p| = 1 and = |f(x) - f(y)|. Let g = and apply the MVT to g... QED. Doesn¹t require the fact that f¹ bounded implies that f¹ is Riemann integrable (the proof of which is not obvious to me this second...) ************************ David C. Ullrich === Subject: Re: Œerf¹ function in C > BTW I made implementations in Fortran, HP-41C, Pascal, Modula 2, > Clipper and recently Visual Basic of the Taylor Series expansion of > Phi(x) given in 26.2.11 in A&S since around 1980: > Phi(x) = 1/2 + phi(x)*(x + x^3/3 + x^5/(3*5) + x^7/(3*5*7) + ...) > Here are my VB results against your algorithm (also translated to > Visual Basic). It looks like the VB version loses about 1 digit in > precision against the C version (I¹m not surprised, but it is not too > bad either). > x Marsaglia (my VB Version) > Marsaglia (C Version, from your table) > Klaey (VB) > Maple 20 places > 0.123 0.54894645101643700000 > 0.5489464510164369 > 0.54894645101643700000 > 0.54894645101643675909 > 1.2 0.88493032977829200000 > 0.8849303297782918 > 0.88493032977829200000 > 0.88493032977829173198 > 2.4 0.99180246407540400000 > 0.9918024640754040 > 0.99180246407540400000 > 0.99180246407540387055 > 6.1 0.99999999946965800000 > 0.9999999994696578 > 0.99999999946965800000 > 0.99999999946965767370 > -6.1 0.00000000053034350051 > 0.0000000005303427 > 0.00000000053034221459 > 0.00000000053034232638 > -1.1 0.13566606094638300000 > 0.1356660609463828 > 0.13566606094638200000 > 0.13566606094638267517 > 7.2 0.99999999999969900000 > 0.9999999999996990 > 0.99999999999969900000 > 0.99999999999969893721 > I¹m quite pleased how my algorithm still stands up today :-) > Matthias Kl.8ay > -- > www.kcc.ch Excellent suggestion. While it was not so in the 1960¹s when I developed the method for Fortran via cPhi(x)/phi(x), the ordinary Taylor series about zero for the normal integral, Phi(x)=.5+integral exp(-t^2/2)/sqrt(2*pi), x=0..t = .5+(x-x^3/6+x^5/40-x^7/336+...)/sqrt(2*pi) provides, with current double precision arithmetic, an excellent method for providing normal probabilities arithmetic for the range of x¹s one usually encounters, say to +/- four or Þve sigmas, letting the CPU do the work after being given a few simple instructions. Here is a sample implementation: ----------------------------------------------- #include #include double Phi(double x) { long double z,t=0,s=.3989422804014327L; int i; s*=x; z=s; t=0; for(i=3;s!=t;i+=2) {z*=-x*x*(i-2)/(i*i-i); t=s; s+=z;} return .5+s; } int main(){ double x; while(1){ printf(Enter your x value:); scanf(%lf,&x); printf(Normal Prob(X<%f)=%20.16fn,x,Phi(x)); } } --------------------------------------------- Here are a few results, with the true values provided by setting Digits:=30 in Maple: x Phi(x) .123 .5489464510164368 .5489464510164367590816... true value 2.34 .9903581300546417 .9903581300546416673759... -2.34 .0096418699453583 .0096418699453583326240... 3.45 .9997197067231839 .9997197067231838225884... 4.56 .9999974423189606 .9999974423189605484746... 6.6 .9999999999809187 .9999999999794421109060... -6.6 .0000000000190813 .0000000000205578890939... 6.8 .9999999999950689 .9999999999947690424558... -6.8 .0000000000049310 .0000000000052309575441... 7. 1.0000000000163602 .9999999999987201874561... -7 -.0000000000163602 .0000000000012798125438... Again, I hope to access 80-bit þoating point processors by using long double¹s. With plain double¹s, there is little difference for |x|<4; beyond that accuracy drops to 13 then 12,11,10,... digits until cases such as Phi(6.5), which returns a value >1. Those interested are invited to compare on other platforms the approach using the Taylor series for cPhi(x)/phi(x) with that of the Taylor series for plain Phi(x), using long double¹s then plain double¹s. Either approach may be preferable to the sometimes unavailable and/or mysterious erf. George Marsaglia === Subject: Re: covering compact set w/squares >Oh, I think I did misread. How about this: >Given a compact set K in the plane s.t. each pt x is the center of a square >Q_x, prove that you can Þnd a subsequence Q_x_i of squares s.t. K is >covered by the Q_x_i and >sum(over i) Char(Q_x_i) <= 4 Char (union(over i) Q_x_i), >where Char(X) is the characteristic (indicator) function of X. Not sure. That inequality is just another way to say that no point is contained in more than 4 of the Q_x_i; I¹m not sure whether one can get that, but we know it doesn¹t follow just from taking a minimial cover. When you say Oh, I think I did misread. How about this: it sounds like you¹re still not certain what the statement is. This might be a little simpler if you¹d get the statement staight to begin with. (Um: I¹ve been assuming that this is like a homework problem. If it¹s not, something you¹re trying to do for some other reason, I can tell you a standard result that¹s very much like what you¹re trying to prove, and which you might Þnd equally useful.) >>Given a compact set K in the plane s.t. each pt x is the center of a >square >>Q_x, prove that you can Þnd a subsequence Q_x_i of squares s.t. K is >>covered by the Q_x_i and the sum of the areas of each of the Q_x_i is no >>more than 4 times the area of the union of the Q_x_i. >> Aargh. I think I¹ll just spend the rest of my life posting replies to >> this message. I have to say that the two counterexamples I¹ve >> given were wrong (I was reading the statement of the problem >> Note however that the example showing that it¹s not true that >> in a minimal cover no point is in more than four squares was >> correct. (I¹m still not sure whether the result itself is true and >> I still wonder whether it¹s really what you wanted to prove.) >>I think this is a totally geometric thing, and I know it sufÞces to show >>that in such a minimal cover no point is contained in more than 4 >squares, >>but how do i show this? I know how to do this with intervals on the real >>line (there the answer is 2 times the area....), but i get lost when >going >>up a dimension. >> ************************ >> David C. Ullrich ************************ David C. Ullrich === Subject: Re: puzzle: GCDs of InÞnite Set of Integer Pairs >Playing the physics card isn¹t helpful here. I¹m claiming that it¹s >*mathematically* impossible to randomly choose a real from [0,1]. Not physically >impossible. I¹m asking you to convince me otherwise, use a thought experiment if >you want, or not. >>The canonical example is a spinner which can take on a random angle. >>This gets you [0,2pi) >I think we had this one last time we explored this concept. At some point the >quantum state of the universe takes over. There are a countable number of angles >your spinner can indicate. Not surprising that they all have non-zero >probability, and that one actually happened. Angular momentum is quantized, but angle is not (so far, anyway). In any case, bringing quantum physics to a thought experiment is far less valid than bringing Lebesgue measure to a question about probability. -- Matthew T. Russotto mrussotto@speakeasy.net Extremism in defense of liberty is no vice, and moderation in pursuit of justice is no virtue. But extreme restriction of liberty in pursuit of a modicum of security is a very expensive vice. === Subject: Re: Handling the truth these are *varying* functions that are the factors of (5a_1(x) + 7) > and (5a_2(x) + 7), in the ring of algebraic integers. No. Yes. Here¹s a repost, in case you missed it. You¹ve seen what happens when x=1 and x=2. Let¹s try a different one, where things are simple enough to verify by hand. Suppose we take x = -3. Then we have (5a_1(-3) + 7)(5a_2(-3) + 7) = 7(25(9) + 30(-3) + 2) = 7(137) where the a¹s satisfy a^2 + 4a + 7((-3)^2 + (-3)) = a^2 + 4a + 42. We Þnd that a_1 = -2 + sqrt(-38) and a_2 = -2 - sqrt(-38) It¹s easy enough to verify that the a¹s aren¹t divisible by 7 or by sqrt(7). However, a_1 does share a factor in common with 7, namely w_1 = (1 + 3sqrt(-38))^{1/3} We can verify that 1. w_1^3 divides 7^3, since (1 + 3sqrt(-38))(1 - 3sqrt(-38)) = 343 = 7^3 So w_1 = (1 + 3sqrt(-38))^{1/3} divides 7. 2. w_1^3 divides a_1^3 = 220 - 26sqrt(-38), since (1 + sqrt(-38)(-8 - 2sqrt(-38)) = 220 - 26 sqrt(-38) So w_1 = (1 + 3sqrt(-38))^{1/3} divides a_1 Thus w_1 is a common divisor of a_1 and 7. (In fact, it¹s a gcd of a_1 and 7.) In a similar way, we Þnd that w_2 = (1 - 3sqrt(-38))^{1/3} is a common divisor of a_2 and 7. In this case, with (5a_1 + 7)(5a_2 + 7) = 7(137) We see that the factor 7 on the right splits as w_1 * w_2 = 7 and that f_1 = (5a_1 + 7)/w_1 = 5(-8 - 2sqrt(-38))^{1/3} + (1 - 3sqrt(-38))^{1/3} and f_2 = (5a_1 + 7)/w_1 = 5(-8 + 2sqrt(-38))^{1/3} + (1 + 3sqrt(-38))^{1/3} f_1 * f_2 = 137 exactly as expected. Likewise, it¹s easy to show that w_1, w_2, f_1, and f_2 are all algebraic integers and none of them are units. I hasten to add that this behavior is what happens for almost all values of x, namely that in (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) with the a¹s satisfying a^2 -(x - 1)a + 7(x^2 + x) we will be able to Þnd algebraic integers w_1(x) and w_2(x) with w_1(x) * w_2(x) = 7 and w_i(x) dividing a_i(x) so that with f_i(x) = (5a_1(x) + 7)/w_i(x) we will have f_1(x) * f_2(x) = 25x^2 + 30x + 2 as long as 7 doesn¹t divide 25x^2 + 30x + 2. With more or less difÞculty, one could do the same construction for most integers x, just as I did for x=1 and Keith did for x=2. In most cases we¹ll Þnd that 7 splits into two nonunit factors, distributed between (5a_1+7) and (5a_2+7). In general, as with w_i(-3) = (1 +/- 3sqrt(-38))^{1/3} it won¹t be immediately obvious that w_1(x) and w_2(x) are divisors of 7 and of a_i(x), in the sense that you can¹t look at them and immediately notice that they divide 7 or a_i(x). > Instead you have one of their cohorts Keith Ramsay throwing out a > degree 22 polynomial claiming to have found it some kind of way, yet > if you look at > w_1(x) w_2(x) = 7 > (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) > and > a^2 - (x - 1)a + 7(x^2 + x) > there is no way to derive the w¹s as varying functions of x. > The only thing that actually makes sense is something like > w_1(x) = 7, w_2(x) = 1. No. See above. > I can point out the obvious, like why 7? Why not 13? Why can¹t you > just have something like > w_1(x) w_2(x) = 13 > (5a_1(x) + 13)(5a_2(x) + 13) = 13(25x^2 + 30x + 2) > and > a^2 - (x - 1)a + 13(x^2 + x)? > Mathematically there has to be a reason, right? But you can. Just use (5a_1(x) + 13)(5a_2(x) + 13) = 13(25x^2 + 30x + 3) where the a¹s satisfy a^2 - (x - 2)a + 13(x^2 + x) Then an analogue to what I did above will allow you to Þnd w_1(x) and w_2(x) for most integers x with the w¹s being algebraic integer divisors of 13 and the appropriate a, such that w_1(x) * w_2(x) = 13, just like we saw above. Rick === Subject: Re: Help needed - primes program >Hallo, >I need help with the following : >I made a program which gives all the primes from selected area. I made it in >Pascal. The only problem is that it hasn¹t a very good algorithm which I >need. >Here is the code: >------------------------------------------------------------ --------------- - >--- >Program Primes; >Var > x,y:word; > f:word; > i,n:word; > ok:boolean; >Begin > Write(ŒEnter X and Y -> Œ); > Readln(x,y); > For i:=x to y do > Begin > f:=2; > ok:=true; > While f<=i-1 do > Begin > If i mod f = 0 then > Begin > f:=i; > ok:=false; > End > else > f:=f+1; > End; > If ok=true then Write(i,¹ is a prime. Œ); > End; > Readln; >End. >------------------------------------------------------------ --------------- - >--- >I wanted to give you attachment with the *.EXE Þle but I can¹t so if ypu >need to compile this code download Free Pascal (current version is 1.0.10) >form www.freepascal.org . >Any suggestion and similar is welcome. Why not start at 3 and ignore all the even numbers in the loops, it would double the speed. http://www.nist.gov/dads/HTML/sieve.html including implementations in pascal. === Subject: Re: Help needed - primes program > Hallo, > I need help with the following : > I made a program which gives all the primes from selected area. I made it in > Pascal. The only problem is that it hasn¹t a very good algorithm which I > need. Use the Sieve of Eratosthenes. Google will Þnd several versions and descriptions of it, as will the prime links at http://primepages.org/ Phil -- Unpatched IE vulnerability: XSS in Unparsable XML Files Description: Cross-Site Scripting on any site hosting Þles that can be misrendered in MSXML Reference: http://sec.greymagic.com/adv/gm013-ie/ Exploit: http://sec.greymagic.com/adv/gm013-ie/ === Subject: Re: tensors for tots > I am also familiar with the idea of transforming matrices themselves > to a new basis. If B is the transformation matrix (in my sense), > then the prescription for transforming a matrix A is: > A¹ = BAB^(-1) Observe that this is the Transformation rule for a (1,1)-Tensor. > I¹m uncertain how we wish to transform v* Ok, let there be a Þxed dual vector, say (v*), which is a linear mapping (v*) : V -> R. (V a R-vector space) The value of (v*) at any vector (v) does clearly not depend on the basis, let it be x := <(v*), (v)> (The value of (v*) at (v)) Now we introduce a basis and have coordinate-vectors of (v*) and (v), say v* and v. We write v* and v as a column. We get the value of (v*) at (v) by matrix multiplication of theses two vectors: x = <(v*), (v)> = v*^t v. Now we change the basis. We know the new coordinates of (v), these are v¹ = Bv (your notation :-)) Let v*¹ be the new coordinates of (v*). These are given by v*¹ = Av* with some Matrix A What is A? Look at x = v*^t v = v*¹^t v¹ = (Av*)^t B v = v*^t A^t B v which implies A^t B = identity matrix, A = B^(-1)^t. Here we are. > I have a hunch this has to do with the metric tensor A normal matrix, so to say, is a (1,1)-Tensor. It can be multiplied by a row from the left and by a column from the right, that is by a covariant and a contravariant vector. The metric tensor eats two contravariant vectors, his matrix M represents a different object. It stands for the mapping (v,w) -> v^t M w (here v and w are two columns) The transposing of v is built in. This implies a different behaviour under change of coordinates. === Subject: Re: Sets That Resemble Derivatives Somewhat > 2) [(A-->B)(A¹-->B¹)]¹ = AB¹ U A¹B > 3) [(A¹-->B)(A-->B¹)]¹ = A¹B¹ U AB 2) is just symmetric difference, assuming the existence of some universe === Subject: Re: e is transcendental (was: classes of transcendental numbers ? Why do I even try? Are you even TRYING to learn? Or are you so closed-minded that you can¹t be bothered? >>So? The real part of exp(i pi) is cos(pi), and its imaginary part >>is sin(pi), so all you are saying is that cos(pi) = -1 and >>sin(pi) = 0, and we were already aware of these facts. There is >>NO reason to conclude that exp(i pi) = 0. > I,have a reason , with my due respect. > Panagiotis Stefanides >>Yes, but you DON¹T tell us what your reason is. You can¹t expect us to >>accept your claims without giving support for those claims. So what >>possible reason could you have for expecting us to agree with your claim >>that exp(i pi) = 0? >The reason is simply that exp(ipi0=-1 should be accompanied >by the statement that this is the real part solution. >Is it fair? >>No, that is not a fair comment. exp(i pi), the complex number, is >>equal to -1, the complex number. There is no need to appeal to the >>real part. Also, what does exp(i pi) = -1 is the real part solution >>mean? You look like you are using terminology in a manner not >>recognized in mathematics. >>David McAnally >>-------------- >e^[i*pi] ,as accepted ,is a phasor. >>In most of the relevant Þelds of mathematics, e^[i pi] is a complex >>number. >It is only fair to state that its >polar representation is : >e^[i*pi] = MOD 1 , ARG 180 . >>That is Arg 180 degrees, not just Arg 180. And so what? That does >>not lead to your assertion that exp[i pi] = 0, a result for which >>you have given absolutely no support. Why don¹t you just give up? >>David McAnally >> At the moment, they (the Time Lords) are far from being all-powerful. >> That¹s why it¹s been left up to me and me and me. >> quote by: Patrick Troughton in The Three Doctors >>------- >I, have made myself very explicit. >My original question of the implication >of the imaginary component: e^[ipi]=j*0 (to the related proof) >Complex Notation, chapter 12 ,Electrical Technology 3RD ED. >Edward Hughes Longmans ,page 338: >Stares: > OA*=OB+jOC=OA(COStheta+jSINtheta) >* Symbols representing phasors are printed in bold face italics, while those representing only magnitudes are printed in ordinary italics,.. >Here is very clear the difference between PHASOR and MAGNITUDES I know the difference between a complex number and its modulus (or magnitude). You don¹t have to explain the difference to me. >[which(MAGNITUDES) I, referred to as REAL PART or IMMAGINAR PART, The magnitude of a complex number is generally not equal to either its real or imaginary part. How could you claim that it is? > I should have stated COMPONENTS ]. >>Perhaps you mean that (the imaginary part of e^[i pi]) = 0, in which >>case you are correct, but you have had a lot of problem expressing >>yourself, especially in view of the way that you initially made >>the claim by stating that e^[i pi] = 0, which you described as the >>imaginary part solution, using a terminology that nobody but you >>knows. > e[i*theta]=COStheta+iSINtheta I know that it is true that exp[i theta] = cos(theta)+i sin(theta). It follows that exp[i pi] = -1. Incidentally, e[i theta] = i e theta, unlike what you have written. >thetas could be given and calculations could be performed >for numerical evaluations. And for other results as well. >In books is stated that it is FORMULA >and also terms such as evaluate:(-1+i*sqrt[3])^10 >Are these not solutions to problems? No. Evaluate (-1+i sqrt(3))^10 is a problem for which you can get a solution using exp(i theta) = cos(theta)+i sin(theta). This does not mean that exp(i theta) = cos(theta)+i sin(theta) is itself a solution. You need a problem before you can describe anything as a solution. >>Nobody knows what you mean when you make a statement like exp[i pi] = -1 >>is the real part solution of exp[i pi] = -1, or that exp[i pi] = 0 >>is the imaginary part solution of exp[i pi] = -1. I asked you to >>explain your terminology but you haven¹t bothered. >I, doubt it but ,still I, exlpained it anyway. I did ask you. And you did not explain your bizarre terminology. There was no problem, hence there is no solution, whether real part or imaginary part. >>As I, exlained earlier, but I, am not the kind of not >being bothered This was not a quote from me. Why are you pretending that it was? >>When I take the imaginary part of the equation exp[i pi] = -1, I get >>sin(pi) = 0, a fact which is already known to us. >>Have you thought also of the fact that if you have exp[i pi] = -1 (your >>real part solution) and exp[i pi] = 0 (your imaginary part solution), >>then you could conclude that 0 = exp[i pi] = -1, and get a contradiction? >Well tthere is always the possibility of an answer such as >MULTIVALUED , examplum gratias: It is known that the exponential function is single-valued (in fact, it is known that the exponential function is entire). > e^[2i*pi]=1 > ln[1]=0=2i*pi The logarithm has a branch point at 0. The exponential function does not have a branch point. The analogy is invalid. Because the exponential function is single-valued, my objection here still stands. >I, make use of the comlex notation , but still ,I, have my >natural questions ,and do not accept everything for granted. But you shouldn¹t start making up mathematics to suit yourself. >I, give an example in the form of question: >It is required that COS[-i]+i*COS[-i] be evaluated >so that is its Modulus and Argument >be evaluated (NUMERICALLY]. cos(-i) = cosh(1), so cos(-i)+i cos(-i) = sqrt(2) cosh(1) exp(i p/4). The magnitude is sqrt(2) cosh(1), which is approximately 2.18225, and its argument is pi/4, which is approximately 0.7854. Alternatively, the argument is exactly 45 degrees. David McAnally -------------- === Subject: Re: x^2 + y^4 = z^4 > + y^4 = z^4 has no positive-integer solutions. Is the proof of this > result short enough for some kind soul to post it, or need I make a trip > to the library? (I have citations.) An elementary proof (no elliptic curves) is based on descent. That is, assume a solution exists in positive x, y, z, and show that another one can be found with smaller z. The proof is not particularly difÞcult but too long to type out here (one single-space textbook page). It can be found in many basic number theory books, e.g. Niven and Zuckerman in the chapter on Diophantine equations, p. 107 in the 3rd edition. === Subject: Re: Mean Value Theorem > Two places I¹ve seen the equivalent statement: If f is continuously > differentiable on [0,1], then |f(x)-f(y)|<=sup|f¹(u)||x-y| where the > sup is extended over 0<=u<=1. > I can¹t see why we need f¹ to be continuous. It seems by the MVT we > have > f(x)-f(y)=f¹(c)(x-y) for some c in [0,1], > so that > |f(x)-f(y)| = |f¹(c)||x-y|, for some c in [0,1], > <= sup_u|f¹(u)||x-y|. However, it turns out, later in the discussion, they start talking about rates of convergence, so that there is in fact a need for sup|f¹(u)| to be Þnite--it¹s not just a superþuous nicety. Does anyone know of a case where a textbook purposely included superþuous infomation as part of an exercise? Or how about on a test question? It seems that it would make the problems somewhat more realistic, albeit esthetically unpleasing. It may make students think more about the problem. Students seem to develop the unhealthy assumption that if it is given, then is must be used. === Subject: Re: puzzle: GCDs of InÞnite Set of Integer Pairs > mmmm... in a sense it does. But then again in a sense, most real numbers don¹t > have a proper existence. In terms of proportions, 100% of real numbers are not > thought of, used, speciÞed, and don¹t occur naturally as a measure of > something. I¹m excluding extremely vague usage such as you specify Œall reals > larger than x¹. The set of real numbers that have a Þnite description is countable; this is a superset of the algebraic reals. If your proper existence is equivalent to Þnite description, your assertion is trivial. -- Daniel W. Johnson panoptes@iquest.net http://members.iquest.net/~panoptes/ 039 53 36 N / 086 11 55 W === Subject: Re: Handling the truth by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1IFs8R04646; >show that he posts from Hamilton College: You consider mathematicians *incapable* ? Do you or do you not consider mathematicians *incapable* ! Maurice === Subject: intercept length when a random line intersect with ellipsoid shell or parallelepiped shell with Þxed thickness If a random line intersects with a ellipsoid shell or parellelepiped shell with a Þxed thickness, how much is the statistical intercept length? === Subject: Re: covering compact set w/squares No, it¹s not HW...it was an aside in a measure theory course that seemed kind of curious but that I couldn¹t immediately (and at this point just plain couldn¹t) verify. >Oh, I think I did misread. How about this: >Given a compact set K in the plane s.t. each pt x is the center of a square >Q_x, prove that you can Þnd a subsequence Q_x_i of squares s.t. K is >covered by the Q_x_i and >sum(over i) Char(Q_x_i) <= 4 Char (union(over i) Q_x_i), >where Char(X) is the characteristic (indicator) function of X. > Not sure. That inequality is just another way to say that no point > is contained in more than 4 of the Q_x_i; I¹m not sure whether > one can get that, but we know it doesn¹t follow just from taking > a minimial cover. > When you say Oh, I think I did misread. How about this: it > sounds like you¹re still not certain what the statement is. > This might be a little simpler if you¹d get the statement > staight to begin with. > (Um: I¹ve been assuming that this is like a homework > problem. If it¹s not, something you¹re trying to do for > some other reason, I can tell you a standard result that¹s > very much like what you¹re trying to prove, and which you > might Þnd equally useful.) >>Given a compact set K in the plane s.t. each pt x is the center of a >square >>Q_x, prove that you can Þnd a subsequence Q_x_i of squares s.t. K is >>covered by the Q_x_i and the sum of the areas of each of the Q_x_i is no >>more than 4 times the area of the union of the Q_x_i. >> Aargh. I think I¹ll just spend the rest of my life posting replies to >> this message. I have to say that the two counterexamples I¹ve >> given were wrong (I was reading the statement of the problem >> Note however that the example showing that it¹s not true that >> in a minimal cover no point is in more than four squares was >> correct. (I¹m still not sure whether the result itself is true and >> I still wonder whether it¹s really what you wanted to prove.) >>I think this is a totally geometric thing, and I know it sufÞces to show >>that in such a minimal cover no point is contained in more than 4 >squares, >>but how do i show this? I know how to do this with intervals on the real >>line (there the answer is 2 times the area....), but i get lost when >going >>up a dimension. >> ************************ >> David C. Ullrich > ************************ > David C. Ullrich === > ummm...except it¹s slightly a parody site. it¹s not real. Why don¹t you > actually read the whole site before you start condemning it. Moron. > Uh..I think he¹s promoting it. You might want to think hard about that > last word you used. > ;^) Uh...doesn¹t READ like a promotion. I think I¹d rather play dumb. - Nick === Subject: Re: No Set Contains Every Computable Natural Russell Easterly says... >The set of all natural numbers is recursive by Þat. >Why bother to include computability in the deÞnition? >A human must decide if x is a member of this set >before x can be given to the TM. Right. The notion of recursive set is only nontrivial for proper subsets of the naturals. For the complete set, it is recursive by deÞnition. -- Daryl McCullough Ithaca, NY === Subject: Full Beal Conjecture (revised) Sorry for the mistake. My proof can be obtained at === Subject: Quadratics and transformation It seems to me that I should be able to reasonably easily get a general idea of the type(s) of transformations that have been applied to x^2 just by looking at the Þnished quadratic. That is, if I have 3x^2 + 7x + 2, can I easily relate the coefÞcients and constant to the dilations, reþections and translations that this function represents compared to x^2 ? I have tried thinking about it as a combined function, that is, f(x)=x^2, g[f(x)]=3x^2 + 7x + 2 and trying to Þnd g(x). Well, I haven¹t tried with this example but one I tried earlier turned into a horrible mess and didn¹t give me what I wanted - I assumed that g(x) is simply the inverse of g[f(x)] so maybe that¹s why it got so messy ? It also seemed to be a lot of work! I¹m aware that I could determine the turning point and work out the translations and that the roots would indicate what horizontal dilation has been applied. I expect that if I calculate two other points I would also be able to determine reþection and vertical dilation. Of course I could always graph it but I¹m hoping to achieve what I want from inspection of the description of the function (is that the right terminology?). Any and all help appreciated. Ivan. === Subject: re:JSH: Why am I so important? I see what he¹s up to, we¹ll Þll this thread with posts and he¹ll go to somewhere else and say that he¹s important and will have this thread to back him up, and the chain goes on forever. Don¹t reply to him. http://www.newsfeed.com The #1 Newsgroup Service in the World! >100,000 Newsgroups ---= 19 East/West-Coast Specialized Servers - Total Privacy via Encryption =--- === Subject: re:Handling the truth How I hate it when wanna-be mathematicians write out cute identities and call it Œresearch¹ and then make wars about the philosophical implications of some useless notions. Then they have a go about their moral superiority and stuff... Stick to mainstream math and stop f*cking people¹s brains. A megalomaniac cat is no way a Œbetter person¹ than the humble folk. http://www.newsfeed.com The #1 Newsgroup Service in the World! >100,000 Newsgroups ---= 19 East/West-Coast Specialized Servers - Total Privacy via Encryption =--- === Subject: Re: HELL :-) (Was: Re: the anticlassicalist }{ ii: the spectre continues) > *MATHEMATICS IS FULL OF TRICKS* French even allows you the conjugation : ma th.8ematique ta th.8ematique sa th.8ematique notre th.8fme .88 tics votre th.8fme attique leurre, t¹es mat, hic ! > *REDUCING TOPICS TO THE SCIENCE OF NUMBER IS NUMEROLOGY* I¹d call that accounting. === Subject: Re: pentiamond rep-tiles, continued > That leaves these two: > ____ __/ > /____ /___/ > I see no reason that these should not be rep-tilable. Neither do I see > a rep-tiling of either! Does anybody? I now see a reason that the pentiamond on the right should not be rep-tilable. Suppose you have such a rep-tiling. Look at the little pentiamonds along the edge of the big one, starting at the top vertex and following the edges on the right and bottom. You will Þnd that all the boundaries between the little pentiamonds have the same orientation relative to the edge of the big one: _______________ You cannot escape from this pattern, even when you pass a 120-degree angle. When you reach the lower left vertex, you¹re stuck. No solution yet on the I-, or straight, pentiamond. It can¹t be done on a scale of 9x9 or smaller. -:- To what do I owe the honor of this unexpected visit, Lord Ruthven . . . alias Lyford Pemberton! H. C. Artmann, Tom Parker, International Detective -- Col. G. L. Sicherman home: colonel@mail.monmouth.com work: sicherman@att.com web: === Subject: Re: errors in an argument >Let¹s just look at the probability of a protien made up of >100 amino acids forming by chance. All the amino acids >have to line up just right, and the probability of this >happening in any given try is 1/100! or 10^(-158). > There are only 20 different amino acids. > The more subtle error is that nobody says that life required > exactly this protein to be produced by a random permutation. > There are probably a huge number of different proteins that > could do more or less whatever this one does. And its production > would not be a matter of random permutations, but rather of > the gradual accumulation of small improvements. You might > look up genetic algorithms: this is not just a theoretical > construct but a method which is actually used to obtain a > nearly-optimal solution to a difÞcult problem by simulating > the process of evolution. > That said, I think it¹s fair to add: my impression, as an > outsider in this Þeld, is that while evolution itself (for > organisms that already have the usual genetic machinery) has pretty > good mathematical models, we are still far from understanding > at a quantitative level how that genetic machinery might have > arisen in the Þrst place. > Robert Israel israel@math.ubc.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia > Vancouver, BC, Canada V6T 1Z2 Robert, of course, mentioned exactly the errors I was thinking of, with the second being that we don¹t know how many possible proteins there are with 100 amino acids in them. To say that for each one, the probability is essentially 0, and so the collective probability for the set of proteins is 0 is quite a mistake. If I were trying to explain the error to someone, I would probably use the analogy of rolling a 20 sided dice 100 times and recording the result. By the proof above, one can¹t do this, since the probability of any particular sequence is (essentially) 0. I guess what I would really like to know is how poorly I should think of the person who made the argument. Here are some qualiÞers: He boasts of his degree in mathematics. Also, in his talk, he indicated that he would try to teach us how to think critically about what scientists say. On the other hand, he is now in his mid 60¹s, so he is far removed from his schooling. Finally, almost certainly (this is a guess of mine) he did not formulate the proof himself, but it had been presented to him. He liked the argument and missed the þaws. Should I not be too critical of his wishful thinking that this proof actually means something? By the way, I certainly agree completely with Robert¹s last paragraph. John === Subject: Re: Big Rip NY Science Times Is it my imagination, or are this idiot¹s ramblings becoming even *more* delusional and illucid than before? -E === Subject: Re: How to diagonalize a Hermitian matrix >Hello group, > Could somebody please tell me a numerical algorithm to >diagonalize a Hermitian matrix H, so that I end up not only with the >eigenvalues perched along the diagonal, but also with the unitary >matrix U that conjugates with H to diagonalize it. > I should probably say what I really want. I have a >positive-deÞnite Hermitian matrix H, and I need its matrix square >root. The above is simply the approach that occurred to me for >obtaining that square-root. > There are available algorithms to do this diagonalization. > Any good library should have such. > I am assuming that by square root you mean the Hermitian > square root. If you just want a matrix A with H=A*A~, the > conjugate transpose of A, the Cholesky decomposition is the > fast way to do it. > One can use Newton¹s method to obtain the square root of a > matrix; the rate of convergence is not bad. There are other > ways as well. I suggest you ask the question in > sci.math.num-analysis. think that is what I am looking for. I will try on the group you suggest. Achava === Subject: Re: Sets That Resemble Derivatives Somewhat there¹s a general theme that some people have explored that says that the fact that certain sorts of differential operators obey leibniz-like identities is conceptually subordinate to the fact that certain sorts of boundary operators in geometry and topology obey leibniz-like identities, roughly (or sometimes precisely, depending on the particular concept of boundary being used) boundary(x X y) = (boundary(x) X y) + (x X boundary(y)). -- [e-mail address jdolan@math.ucr.edu] === Subject: Big Rip http://qedcorp.com/destiny/ under construction DARK MATTERS SURROUND DARK ENERGY with my comments, additions and physics corrections ÒTwo big stories from the world of physics may portend the arrival of new weapons of mass destruction far more powerful and compact than atomic bombs. In recent years it has been discovered that our universe is being blown apart by a mysterious anti-gravity effect called dark energy. Mainstream physicists are scrambling to explain this mysterious acceleration in the expansion of the universe. Some physicists even believe that the expansion will lead to The Big Rip when all of the matter in the universe is torn asunder - from clusters of galaxies in appears to be made of two unknowns - roughly 23% is dark matter, an invisible source of gravity, and roughly 73% is dark energy, an invisible anti-gravity force. Ordinary matter constitutes perhaps 4 percent of the universe. Recently the British science news journal New Scientist revealed that the American military is pursuing new types of exotic bombs - including a new class of isomeric gamma ray weapons.Ó That was an original idea of mine in 1963 at Cornell and I discussed it with Hans Bethe. That is one of the reasons Ron Bullough invited me to Harwell in 1966. No doubt others thought of it but probably later. I thought of it while at Tech/Ops in Lexington, Mass working for George Parrant Jr. ÒUnlike conventional atomic and hydrogen bombs, the new weapons would trigger the release energy by absorbing radiation, and respond by re-emitting a far more powerful radiation. In this new category of gamma-ray weapons, a nuclear isomer absorbs x-rays and re-emits higher frequency gamma rays. The emitted gamma radiation has been reported to release 60 times the energy of the x-rays that trigger the effect.Ó Gamma-ray weapons could trigger next arms race 19:00 13 August 03 ÒExclusive from New Scientist Print Edition. Subscribe and get 4 free issues. An exotic kind of nuclear explosive being developed by the US Department of Defense could blur the critical distinction between conventional and nuclear weapons. The work has also raised fears that weapons based on this technology could trigger the next arms race. The explosive works by stimulating the release of energy from the nuclei of certain elements but does not involve nuclear Þssion or fusion. The energy, emitted as gamma radiation, is thousands of times greater than that from conventional chemical explosives. The technology has already been included in the Department of Defense¹s Militarily Critical Technologies List, which says: Such extraordinary energy density has the potential to revolutionise all aspects of warfare. Scientists have known for many years that the nuclei of some elements, such as hafnium, can exist in a high-energy state, or nuclear isomer, that slowly decays to a low-energy state by emitting gamma rays. For example, hafnium-178m2, the excited, isomeric form of hafnium-178, has a half-life of 31 years. The possibility that this process could be explosive was discovered when Carl Collins and colleagues at the University of Texas at Dallas demonstrated that they could artiÞcially trigger the decay of the hafnium isomer by bombarding it with low-energy X-rays (New Scientist print edition, 3 July 1999). The experiment released 60 times as much energy as was put in, and in theory a much greater energy release could be achieved.Ó http://www.newscientist.com/news/news.jsp?id=ns99994049 I was thinking in 1963 of a gamma ray laser pumping a nuclear isomeric transition. Bethe at the time said it wouldn¹t work and basically discouraged me working on it. Bekkum continued: ÒIn the summer of 2000 I contacted Nick Cook, the former aviation editor and aerospace consultant to Jane¹s Defence Weekly, the international military affairs journal. Cook had been investigating black budget super-secret research into exotic physics for advanced propulsion technologies.Ó Uh Oh :) ÒI had been monitoring electronic discussions between various American and Russian scientists theorizing about rectifying the quantum vacuum for advanced space drive. Several groups of scientists, partitioned into various research organizations, were exploring what NASA calls Breakthrough Propulsion Physics - exotic technologies for advanced space travel to traverse the vast distances between stars. Partly inspired by the pulp science Þction stories of their youth, and partly by recent reports of multiple radar tracking tapes of unidentiÞed objects performing impossible maneuvers in the sky, these scientists were on a quest to uncover the most likely new physics for star travel. The NASA program was run by Marc Millis, under the Advanced Space Transportation Program OfÞce (ASTP). Joe Firmage, a Silicon Valley entrepreneur, who at age 28 had found risen to CEO of a three billion dollar internet Þrm, began to fund research in parallel with NASA. He hired a NASA Ames nano-technology scientist, Creon Levit, to run the International Space Sciences Organization, Joe did that because I suggested it. I introduced Creon to Joe. ÒCook was intrigued by the apparent connections between various private investors, defense contractors, NASA, INSCOM (American military intelligence), and the CIA. While researching exotic propulsion technologies Cook had heard rumors of a new kind of weapon, a sub-quantum atomic bomb, being whispered about in the dark halls of defense research.Ó I think that must have come from me regarding J. P. Vigier¹s tight atomic states with experiments in Beograd by Z. Maric and G. Dragic. But how did Cook hear about that? We brought Vigier to ISSO in San Francisco several times along with physicist Gennady Shipov from Moscow. That story with photographs of Vigier and the group is in my autobiography ÒDestiny Matrix.Ó Dragic A, Maric Z, Vigier JP; Phys. Lett. A 265 (2000) 163. New quantum mechanical tight bound states and Œcold fusion¹. Creon Levit and Vigier met with Maric in Budapest in 2000. Bekkum who is one of my Òon line studentsÓ continued: ÒSub-quantum physics is a controversial re-interpretation of quantum theory, based on so-called pilot wave theories, where an information that the predictions of ordinary quantum mechanics could be recast into a pilot wave information theory. Recently Anthony Valentini of the Perimeter Institute has suggested that ordinary quantum theory may be a special case of pilot wave theories, leaving open the possibility of new and exotic non-quantum technologies. Even thought rumors of a sub-quantum bomb may be purely fantasy ÉÓ It¹s not fantasy. It might not work, Maric and Dragic in Beograd, while not ostensibly trying to make a weapon by any means, were trying to test Vigier¹s basic theory of the spatially extended electron, which I think is basically a correct idea and Þts my own ideas including why the electron appears to shrink to less than 10-16 cm under high resolution imaging (i.e. scattering) and how the electric charge distribution is contained by the strongly attractive zero point energy exotic vacuum dark matter core (Abraham-Becker-Lorentz-Poincare stress problem of 100 years ago). This only works in the Bohm hidden or extra variable interpretation. That is, a classical spatially extended electric charge distribution is unstable. It explodes under its own self-repulsion. This is why physicists had to postulate a point electron because they did not understand that the strong gravity attraction of the positive zero point pressure in a possible state of exotic vacuum would hold the charge together. As Herbert Frohlich told me at UCSD in La Jolla in 1966 Òthe basic thing wrong with physics is the idea of the point electron.Ó The bad idea of the point electron gives the inÞnite energy in quantum electrodynamics. Richard Feynman told me in his ofÞce at Cal Tech in 1968 that ÒinÞnite renormalization is a shell game, and it is a scandal in physics that no one could do betterÓ than what he had done. They did not know 100 years ago that 1/3 or so of the universe was this kind of exotic vacuum. For example, there is a huge sphere of exotic vacuum of w = -1 positive pressure that holds our galaxy together preventing our solar system from going off into space on its own. This sphere looks like w = 0 Òcold dark matterÓ from our vantage point. What works on this large scale also works on the small scale of the single electron (and all the charged lepto-quarks). A neutrino has some mass and is simply a micro-geon of pure zero point energy with positive pressure. ÒÉ there is no question that physicists seriously contemplate a phase transition in the quantum vacuum as a real possibility. The quantum vacuum deÞes common sense, because empty space in quantum Þeld theory appear and disappear far too quickly to be detected directly, but their existence has been conÞrmed by experiments that demonstrate their inþuence on ordinary matter.Ó A major component of the physical quantum vacuum consists of virtual electrons frothing and bubbling at the Fermi surface edge of the Dirac negative energy sea. This is because of the Pauli exclusion principle that only none or one electron per quantum state. A virtual electron pops out of the vacuumÕs Fermi surface leaving a hole behind. The hole is the virtual positron. The result is a Òvirtual electron-positron pair.Ó However, the virtual electron and the virtual positron attract because they have opposite charges and they are exchanging virtual photons. Therefore, some of them form a more stable bound state. An enormous number of these virtual pairs Bose-Einstein condense into the same center of mass quantum wave packet forming the ÒVacuum Coherence FieldÓ (AKA ÒInþation FieldÓ). This is a dynamic steady state of detailed balance in which there is a continual inþow and outþow of virtual pairs into and out of this giant quantum or Òmacro-quantumÓ Òsuperþuid.Ó Essentially this is a vacuum phase transition, similar to the BCS transition from a normal metal to an electrical superconductor, from the globally þat micro-quantum electrodynamic vacuum without any gravity at all to the curved macro-quantum electrodynamic vacuum with emergent gravity. EinsteinÕs Þeld equation of general relativity can be derived from the phase wiggles and ripples in the robust stable macroscopically occupied center of mass quantum wave packet of the bound state of the virtual electron-positron pair. The exotic vacuum dark energy and dark matter are simply the amplitude wiggles and ripples of this same virtual pair quantum wave packet. The wave packet spreads over the entire 3D space of the post-inþationary bubble on which our Hubble-horizoned universe is located along with an inÞnity of parallel American. If the world hologram idea is correct, take the surface area of the expanding Hubble sphere that is the causal retarded boundary of 3D space of our past light cone at Earth and divide it by the quantum of area. That gives us the number of Bekenstein-Shannon c-bits and explains the Òarrow of timeÓ (AKA Second Law of Thermodynamics) of increasing thermodynamic entropy in terms of the dynamical expansion of the 3D space of the universe. Lenny Susskind calls this ÒDeSitter Space.Ó Such research should be forbidden! Too late. PandoraÕs Box is open. SchrodingerÕs Cat has jumped out of it. ÒIn the early 1970¹s Soviet physicists were concerned that the vacuum of our universe was in fact only one possible state of empty space. The fundamental state of empty space is called the true vacuum. Our universe was considered to reside in a false vacuum, protected from the true vacuum by the wall of our world. A change from one vacuum state to another is known as a phase transition. This is analogous to the transition between frozen and liquid water. Lev Okun, a Russian physicist and historian recalls Andrei Sakharov, the father of the Soviet hydrogen bomb, expressing his concern about research into the phase transitions of the vacuum. If the wall between the vacuum states was to be breached, calculations showed that an unstoppable expanding bubble would continue to grow until it destroyed our entire universe! Sakharov declared Such research should be forbidden! since there was always the possibility that an experiment might accidentally trigger a vacuum phase transition.Ó British Astronomer Royal, Sir Martin Rees, Master of Trinity College, and Director of the Cambridge University Institute of Theoretical Astronomy on Madingley Road where Stephen Hawking works discusses all this in Chapter 9 of his important book ÒOur Final Hour.Ó ÒCould the wall of our universe be breached from within? The amount of energy required to punch a hole through the wall appeared to be enormous, and no known natural physical phenomena, even the most energetic, had punched through either. A recent report commissioned to examine potential dangers at the Large Hadron Collider, one of the next the best of our existing knowledge. Others are not so certain, however. At least one of the Russian physicists I had corresponded with was said to have been a former associate of Andrei Sakharov. He strongly hinted at new theories the Russians had developed which allow for the manipulation of the fundamental constants of nature, but he never revealed more than a sketch of his ideas. He claimed that a breakthrough was within reach, perhaps within Þve years ÉÓ Who was that? Not George Ryazanov? ÒRecent theoretical explorations may suggest another approach to the physics of the vacuum. The invisible gravitating dark matter could be the other side of the invisible dark energy coin, and that suggests the possibility of manipulating the vacuum for energy release.Ó Now this is my original idea that you got from our communications over the past few years. I am the only physicist in the world today, as far as I know who has suggested this and has already published it in my two books of 2002 so itÕs in the ofÞcial record at the Library of Congress. ÒIf a controllable parameter could be found to mediate the balance between the invisible dark forces, the result would unleash the vacuum energy of creation in all of its awful power and majesty. If it were possible to control the dark sides of the force then spacetime, the arena where everything we know takes place, could be bent and twisted with inÞnitely greater ease than was ever suspected. This would open Pandora¹s box to everything from vacuum energy weapons of mass destruction (capable of destroying the universe!) to spacetime warp drives and time machines.Ó Exactly, the above is the thesis of all my books since 2002 at least. ÒA quick survey of the international electronic archive of physics papers at www.arXiv.org shows that research into the vacuum of spacetime for energy production is alive and well.Ó I do not think that is true. You need to cite speciÞcs here. There are lots of þakey new age papers on Òfree energyÓ on the Web written by people without any real credentials but they are not on www.arXiv.org which is not even allowing competent ÒfringeÓ papers in controversial topics like Òcold fusion.Ó So what exactly are you thinking of here? Indeed, Paul Ginsparg, who controls the archive, does not even allow Carlos Castro to publish conservative competent papers on ÒClifford AlgebrasÓ which are not ÒfringeÓ at all! ÒMost authors are independent researchers struggling with limited funding and resources, yet their theoretical results suggest that somewhere in Nick Cook¹s black world, a major breakthrough has already taken place. Most likely the United States and Russia are in the lead, but China, France, Ukraine, Iran, India and Saudi Arabia all have scientists actively pursuing the fundamental physics that determine the fabric of our reality, and are seeking the theory and the means to access the enormous energies locked inside of the vacuum since the creation of the universe. Even if the black budget world has yet to unleash the enormous potential of vacuum energy, there are signs that those in power may have begun to take notice. Dr. Harold Puthoff, a scientist with strong government connections, who has previously worked on classiÞed projects for the CIA, is a major proponent of vacuum energy physics. Nick Cook¹s book, The Hunt for Zero Point, and his recent stories on zero point energy in Jane¹s Defence Weekly have also brought attention to the dangers and military potential of vacuum research. The American intelligence community Þnanced so-called psychic spies for over twenty years and through four presidential administrations. It is highly unlikely that they would ignore the potential of the quantum vacuum. Dr. George Chapline, of the Lawrence Livermore National Laboratory, and Dr. Jack Sarfatti in San Francisco, knew each other in the sixties in La Jolla, have independently proposed that the quantum vacuum may unstable to the formation of coherent virtual processes. Sarfatti suggests that gravity is an emergent property determined by the physics of the vacuum. His idea is to Þnd a means of directly interacting with the physics of the vacuum that controls the shape of spacetime. Such a possibility would be consistent with the reported success of Evgeny Podkletnov, the Russian scientist who is experimenting with spinning superconducting disks. Podkletnov¹s most recent papers report the appearance of a mysterious coherent beam of gravity like radiation with a measured force of 1000 G. In an interview on BBC radio, Nick Cook pointed out one immediate application of the Podkletnov beam - the destruction of missiles and satellites in þight or in orbit around the earth. Cook showed the BBC internal documents from Boeing, the American aerospace contractor, proving their interest in Podkletnov¹s research.Ó This beam stuff I am suspicious of. Of course if the experiment is good, I have to think more about it. I am not so sure if PodkletnovÕs experiment is any good and has been replicated. ÒThe connections between Podkletnov¹s results, and the kind of vacuum research explored by Sarfatti, beginning in 1999 at the International Space Sciences Organization are the latest threads in a trail that most likely originates in cold war disinformation, a game played by East and West against each other. Glasnost has shifted the balance of partnerships and the positions of the players, but not the stakes of an outcome that would leave the world with even more proliÞc and powerful weapons of mass destruction.Ó That is true, as shown in ÒDestiny Matrix,Ó however you leave out the most important evidence -UFOs! whole business are the connections. Although Nick Cook never revealed the identity of his deep throat contact called Dr. Dan Marckus in the book The Hunt for Zero Point, there was no question that the Podkletnov results had played a major part in Þtting together the pieces of the puzzle. The amount of interest was in Podkletnov¹s reports by NASA, Boeing, and others in the international arena of aerospace and military research communities was evidence that there was more here to explore than the latest musings of the intellectual elite. The truth is that a fundamental theory of gravity at the scales of subatomic nuclear physics does not exist. The fact is that no one understands the nature of the gravitational Þeld at very small scales. In fact gravity has barely been probed much below one millimeter. Every attempt to unify the physical theories of gravity with the well-known standard model physics of electromagnetism, and the strong and weak nuclear forces, has failed. More importantly there has been recent progress in the exotic areas of mainstream research, such as superstring theory, which suggest new kinds of physics, which might support explanations for Podkletnov¹s impulse gravity effect. One of the current fads in theoretical physics involves large extra dimensions of space that allow a much stronger version of gravity to leak off the membrane world of our ordinary three dimensions. The large dimensional picture allows for the well known forces of electromagnetism, and the strong and weak nuclear forces, to be conÞned to a three dimensional brane-world þoating in a higher dimensional space. Gravitons, the are able to slip off of our brane-world, which explains why the gravitational force is so much weaker than the other forces that hold matter together. Gravitons could be exchanged between our brane-world and another brane þoating nearby in the same higher dimensions. The Sarfatti picture offers a more direct interaction with the new physics than the brane world ideas. Sarfatti¹s vision is to Þnd a means of using electromagnetic Þelds in the Josephson effect to couple to the virtual electron-positron pair giant coherent condensate inþation Þeld inside the vacuum that controls the shape of spacetime to the real electron pair giant coherent condensate of a control high temperature superconductor. UCBÕs Ray Chiao has a similar idea using a superconductor to transduce electromagnetic far Þeld waves to gravity waves with high efÞciency conversion. Sarfatti wants to do the same thing with non-propagating electromagnetic and gravity near Þelds. One wonders if the black budget world may have already produced some of the technology needed to explore and test these new realms.Ó Not a chance. They are clueless about the theory. They are still stuck in Hal¹s PV model and Bernie Haisch¹s zero point ideas, which will never, in my opinion þy. They are not asking the right questions and do not have the right idea in their minds. It is my belief, until I see evidence to the contrary, that I am the only physicist on the planet today who is doing real theoretical work directly relevant to the achievement of practical metric engineering anchored in the now observed reality of dark energy. All my work is public. I would love to be proved wrong on this especially by Hal Puthoff, but I am not holding my breath.;-) Extraordinary claims require extraordinary proof. Everything else I have seen is either on the wrong track asking the wrong questions like the work of Puthoff, Haisch, Ibison & Rueda for example, which at least is real physics that has proved itself wrong in Ibison¹s PV cosmology paper, or else the claims are patently obvious nonsense that Feynman called Cargo Cult Science. There is also the Russian torsion work of Akimov and Shipov and I am not prepared to make a deÞnitive statement on that, as the issue is not simple because of several factors some political. One must be careful there to separate Shipov¹s theoretical work from claims made about practical devices including weapons applications. I do, however, agree with you that there is a real issue here as deÞned in Ch. 9 of Martin Rees¹s Our Final Hour. Hal Puthoff coined the term metric engineering for þying saucer technology. Hal has been working on this problem for many decades and has held high USG security clearances and has been privy to reliable information that the saucers are real and are alien. Otherwise he would not be working on the problem. However, Hal¹s theories, both of the zero point energy and of the gravity Þeld will not solve the problem because they are too naively based and do not ask the right questions. The basic physics required for this task is way beyond the depth of PuthoffÕs self-described ÒengineeringÓ approach and can be found in Rovelli¹s new book on quantum gravity. === Subject: Re: No Set Contains Every Computable Natural > Russell Easterly says... >> The usual notion of recursive set is that a set S is >> recursive as a *subset* of the naturals if there is a >> Turing machine T such that for any *natural* number n >> (in unary notation, to be speciÞc) The set of all natural numbers is not a proper subset of >the set of all natural numbers. > But it *is* a subset. A recursive subset. >> T(n) halts and outputs 1 <-> n is an element of S >> T(n) halts and outputs 0 <-> n is not an element of S >> If x is some input that is not a representation of a natural >> number, then there is no constraint on what T(n) does. > >According to this deÞnition, the set of all natural numbers >is recursive because there exists a TM that doesn¹t even >read the input and always outputs a 1. > Right. > The set of all natural numbers is recursive by Þat. No, not by Þat. It is a trivial consequence of a general deÞnition-- the deÞnition that Daryl McCullough has reproduced for you. > Why bother to include computability in the deÞnition? There is no separate deÞnition for the case of all natural numbers. There is one deÞnition that applies to all subsets of the set of natural numbers. The set of all natural numbers (henceforth N) is a subset of N. We apply the deÞntition to N and Þnd out trivially that it is recursive. It is not a trivial deÞnition. It is quite profound. But it is trivial that N is recursive. > A human must decide if x is a member of this set > before x can be given to the TM. You are seeking to create a new deÞnition and theory of computability. Good. You want to consider a larger context of inputs other than just natural numbers. For example, you want to include inÞnite stings as acceptable inputs. Fine. I would agree with you that if we enlarged the universe of acceptable inputs from just N to a larger set containing inÞnite strings, then N would probably not be recursive in any reasonable new deÞnition of the word. You¹re right, a TM would not be able to conclude that an inÞnite string of 1¹s is not a natural number. I would say that in this enlarged context, that N would be recursively enumerable but not recursive. But your problem is that you are using terms with agreed upon deÞnitions by the mathematical community (like recursive) incorrectly. You have your own personal deÞnition that you have failed to state. This is why mathematicians are arguing with you and pointing out your obvious errors. If you would like to re-deÞne recursive by all means do so. But make your new deÞnition known. It would also be good to give it another name to avoid confusion, like Easterly-Recursive. Until you do this, you will probably continue to make false statements that are easily refuted. Remember, mathematics is precise. By the way there are already in existence more generalized recursion theories, which deal with recursion on other domains besides N. -Leonard Blackburn > Russell > - 2 many 2 count === Subject: Re: Quadratics and transformation > It seems to me that I should be able to reasonably easily get a general > idea of the type(s) of transformations that have been applied to x^2 > just by looking at the Þnished quadratic. > That is, if I have 3x^2 + 7x + 2, can I easily relate the coefÞcients > and constant to the dilations, reþections and translations that this > function represents compared to x^2 ? Yes. 3(x^2 + 7/3*x) + 2, complete the square inside the parentheses. So you end up with a(x-b)^2 + c. b is translation along the x-axis, c along the y-axis, a is the dilation factor. If a is negative, that¹s a reþection. Jon Miller === Subject: Re: Understanding taylor expansion for sine Jonathan Christensen -- > I¹ve been trying to understand the taylor exapnsion for the sine > function, > sin(x) = x - x^3/3! + x^5/5! - x^7/7! + ... > I¹ve been told that it uses the fact that e^(i*x) = cos(x) + i*sin(x), > and I can easily work out the taylor series for e^x, but I¹m not sure > how to use these two pieces to Þnd the expansion for sin(x). > Any help would be appreciated. > Jonathan Christensen --- Outgoing mail is certiÞed Virus Free. Checked by AVG anti-virus system (http://www.grisoft.com). === Subject: Re: e is transcendental (was: classes of transcendental numbers ? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1IHZOR13942; >Why do I even try? Are you even TRYING to learn? Or are you >so closed-minded that you can¹t be bothered? >So? The real part of exp(i pi) is cos(pi), and its imaginary part >is sin(pi), so all you are saying is that cos(pi) = -1 and >sin(pi) = 0, and we were already aware of these facts. There is >NO reason to conclude that exp(i pi) = 0. > I,have a reason , with my due respect. >> Panagiotis Stefanides Yes, but you DON¹T tell us what your reason is. You can¹t expect us to >accept your claims without giving support for those claims. So what >possible reason could you have for expecting us to agree with your claim >that exp(i pi) = 0? >The reason is simply that exp(ipi0=-1 should be accompanied >>by the statement that this is the real part solution. >>Is it fair? No, that is not a fair comment. exp(i pi), the complex number, is >equal to -1, the complex number. There is no need to appeal to the >real part. Also, what does exp(i pi) = -1 is the real part solution >mean? You look like you are using terminology in a manner not >recognized in mathematics. David McAnally -------------- >>e^[i*pi] ,as accepted ,is a phasor. In most of the relevant Þelds of mathematics, e^[i pi] is a complex >number. >It is only fair to state that its >>polar representation is : >e^[i*pi] = MOD 1 , ARG 180 . That is Arg 180 degrees, not just Arg 180. And so what? That does >not lead to your assertion that exp[i pi] = 0, a result for which >you have given absolutely no support. Why don¹t you just give up? David McAnally At the moment, they (the Time Lords) are far from being all-powerful. > That¹s why it¹s been left up to me and me and me. > quote by: Patrick Troughton in The Three Doctors ------- >>I, have made myself very explicit. >>My original question of the implication >>of the imaginary component: e^[ipi]=j*0 (to the related proof) >>Complex Notation, chapter 12 ,Electrical Technology 3RD ED. >>Edward Hughes Longmans ,page 338: >>Stares: >> OA*=OB+jOC=OA(COStheta+jSINtheta) >>* Symbols representing phasors are printed in bold face italics, while those representing only magnitudes are printed in ordinary italics,.. >>Here is very clear the difference between PHASOR and MAGNITUDES >I know the difference between a complex number and its modulus >(or magnitude). You don¹t have to explain the difference to >me. >>[which(MAGNITUDES) I, referred to as REAL PART or IMMAGINAR PART, >The magnitude of a complex number is generally not equal to either >its real or imaginary part. How could you claim that it is? I, gave a reference: Complex Notation, chapter 12 ,Electrical Technology 3RD ED. Edward Hughes Longmans ,page 338: States: OA*=OB+jOC=OA(COStheta+jSINtheta) * Symbols representing phasors are printed in bold face italics, while those representing only magnitudes are printed in ordinary italics,.. What are these magnitudes then ? >> I should have stated COMPONENTS ]. >Perhaps you mean that (the imaginary part of e^[i pi]) = 0, in which >case you are correct, but you have had a lot of problem expressing >yourself, especially in view of the way that you initially made >the claim by stating that e^[i pi] = 0, which you described as the >imaginary part solution, using a terminology that nobody but you >knows. >> e[i*theta]=COStheta+iSINtheta >I know that it is true that exp[i theta] = cos(theta)+i sin(theta). >It follows that exp[i pi] = -1. Incidentally, e[i theta] = i e theta, >unlike what you have written. >>thetas could be given and calculations could be performed >>for numerical evaluations. >And for other results as well. >>In books is stated that it is FORMULA >>and also terms such as evaluate:(-1+i*sqrt[3])^10 >>Are these not solutions to problems? >No. Evaluate (-1+i sqrt(3))^10 is a problem for which you can >get a solution using exp(i theta) = cos(theta)+i sin(theta). This >does not mean that exp(i theta) = cos(theta)+i sin(theta) is itself >a solution. You need a problem before you can describe anything as >a solution. Of course ,this crops up when theta is substituted by a given angle. >Nobody knows what you mean when you make a statement like exp[i pi] = -1 >is the real part solution of exp[i pi] = -1, or that exp[i pi] = 0 >is the imaginary part solution of exp[i pi] = -1. I asked you to >explain your terminology but you haven¹t bothered. >>I, doubt it but ,still I, exlpained it anyway. >I did ask you. And you did not explain your bizarre terminology. There >was no problem, hence there is no solution, whether real part or >imaginary part. How else would do You call them,I, said: I should have stated COMPONENTS >As I, exlained earlier, but I, am not the kind of not >>being bothered >This was not a quote from me. Why are you pretending that it was? You are correct, this was my quote. >When I take the imaginary part of the equation exp[i pi] = -1, I get >sin(pi) = 0, a fact which is already known to us. >Have you thought also of the fact that if you have exp[i pi] = -1 (your >real part solution) and exp[i pi] = 0 (your imaginary part solution), >then you could conclude that 0 = exp[i pi] = -1, and get a contradiction? >>Well tthere is always the possibility of an answer such as >>MULTIVALUED , examplum gratias: >It is known that the exponential function is single-valued (in fact, >it is known that the exponential function is entire). >> e^[2i*pi]=1 >> ln[1]=0=2i*pi >The logarithm has a branch point at 0. The exponential function does >not have a branch point. The analogy is invalid. Because the >exponential function is single-valued, my objection here still stands. >>I, make use of the comlex notation , but still ,I, have my >>natural questions ,and do not accept everything for granted. >But you shouldn¹t start making up mathematics to suit yourself. This is not my intention,believe me. >>I, give an example in the form of question: >>It is required that COS[-i]+i*COS[-i] be evaluated >>so that is its Modulus and Argument >>be evaluated (NUMERICALLY]. >cos(-i) = cosh(1), so cos(-i)+i cos(-i) = sqrt(2) cosh(1) exp(i p/4). >The magnitude is sqrt(2) cosh(1), which is approximately 2.18225, >and its argument is pi/4, which is approximately 0.7854. >Alternatively, the argument is exactly 45 degrees. I, thank You. Panagiotis Stefanides >David McAnally >-------------- === Subject: Re: help with solutions for three questions from the past contest > 3. A boat with an ill passenger is 7.5 mi north of a straight > coastline which runs east and west. A hospital on the coast is 60 > miles from the point on shore south of the boat. If the boat starts > toward shore at 15 mph at the same time an ambulance leaves the > hospital at 60 mph and meets the ambulance, what is the total distance > (to the nearest 0.5 mile) traveled by the boatand the ambulance? ans: > 62.5 Choogu > Boat . > | > | > 7.5| z > | > Shore |___________________________________.Hospital > x 60-x > Hours along distance z at 15 mph must equal hours along distance > 60-x at 60 mph. > And at constant speeds, distance = speed * time. I¹ll be curious if that¹s the best solution. (It may be the one expected.) A better solution would be to minimize the time to the hospital. That¹s a standard calculus question (usually phrased in terms of a swimming pool), with the additional proviso that if the ambulance wouldn¹t be to the meeting point at the time the boat gets there, then this solution should be used. Of course, I wouldn¹t be a proper kibbitzer if I actually worked out the details before posting this criticism. Dare we point out that, if it¹s a true emergency, the coast guard sends a helicopter? Jon Miller === Subject: Re: How to choose a matrix P No D!=E so P can¹t be an identity matrix >Now I am thinking of this question. >Suppose >Y1=D - P¹EP >Y2=D-E >where D,E,P : nxn matrix >||P||=1 >How to choose P, so that >rank Y1 >= rank Y2 >and >||Y1||>= ||Y2|| > The identity matrix would seem to work... > Maybe you have some more requirements you¹re not telling us. > Robert Israel israel@math.ubc.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia > Vancouver, BC, Canada V6T 1Z2 === Subject: Teaching philosophy I need to write a description of my teaching philosophy. In order to do so accurately, I think I need to write at greater length and in greater detail on this topic than any hiring committee will want to read. Moreover, I need to present my views accurately but, somehow, in such a manner as not to vitiate the consideration of my application. That being the case, can someone please tell me what my teaching philosophy is? Ignorantly, Allan Adler ara@zurich.ai.mit.edu ************************************************************* *************** * * * Disclaimer: I am a guest and *not* a member of the MIT ArtiÞcial * * Intelligence Lab. My actions and comments do not reþect * * in any way on MIT. Moreover, I am nowhere near the Boston * * metropolitan area. * * * ************************************************************* *************** === Subject: Re: proving integrability of binomial numbers Achava Nakhash, the Loving Snake escribi.97: > Ignacio Larrosa Ca.96estro Tartaglia¹s triangle. I only know of Tartaglia as a contributor to > solving the general cubic. Is this because I mostly have access to > American and other English language sources? Google [Tartaglia triangle] or [Tartaglia triangulo], or of course, [Tartaglia triangolo] ... At secondary school, I always saw Tarataglia¹s triangle (well, actually ŒTri.87ngulo de Tartaglia¹). -- Ignacio Larrosa Ca.96estro A Coru.96a (Espa.96a) ilarrosaQUITARMAYUSCULAS@mundo-r.com === Subject: Re: question about periodic function by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1IHh7M14607; >If you have two periodic functions f(t) and g(t) with periods a and b, >then the function f(t) + g(t) is periodic with period c where c is the >least-common-multiple of a and b. Here the Þrst period is 1/.018=500/9 nd >the second period is 1/.02=50. So the least-common-multiple is 500, i.e. >1/.002. That should be the period of the combined function. How to prove the theorem? Any hint is welcome. TIA. === Subject: Re: strain softening spring >i.e. the equation I am trying to solve is as follows: >y¹¹ + cy¹ + [ko*e^(-alpha*t)]y = 0 >Any help would be most appreciated. >sincerely >Paul Joseph >(pjoseph@excite.com) Maple gives a solution in terms of Bessel functions. I posted a Þle called DE.pdf showing the solution at: http://math.asu.edu/~kurtz/de/ --Lynn === Subject: Re: intercept length when a random line intersect with ellipsoid shell or parallelepiped shell with Þxed thickness ChenHS escribi.97: > If a random line intersects with a ellipsoid shell or parellelepiped > shell with a Þxed thickness, how much is the statistical intercept > length? You must deÞne in a precisse way how the random line is choosed. The answer depend on it and it can be very different. -- Ignacio Larrosa Ca.96estro A Coru.96a (Espa.96a) ilarrosaQUITARMAYUSCULAS@mundo-r.com === Subject: Re: CAN ANYONE HELP ME????? > CAN ANYONE HELP ME????? > The Œcaps lock¹ key is on the left-hand side of the keyboard, above > Œshift¹ and below Œtab¹. It should look like this: > +--------+ > | |<--- | <-- tab key > | --->| | > +--------+-+ > | Caps | <-- caps lock key > | Lock | > +--------+-+ > | / | > | || | <-- shift key > +--------+ > Over on the right-hand side of the keyboard, above the numeric keypad, > should be three LEDs, labelled (from left to right) ŒNum Lock¹, ŒCaps > Lock¹ and ŒScroll Lock¹ > Press your Œcaps lock¹ key until the LED marked ŒCaps Lock¹ is no longer > lit. > Now you are ready to post on usenet. > 1.DETERMINE DOMAIN OF THE FUNCTION > f(X)=X3/X2+2X+1 > What happens when x^2 + 2x + 1 = 0? Excellent! I (along with Baghdad Bob and most readers) assumed he meant (x^3/x^2) + 2x + 1 (following the standard rules for order of operations), but you ferreted out the probable real meaning, which for OP¹s clariÞcation should have been written x^3/(x^2 + 2x + 1). > FIND EXTREME FUNCTIONS f AND DRAW ITS GRAPH extreme points? extreme points of the function? > 2.BY NEWTONS METHOD SOLVE EQUATION: > X3-X-2=0 Your book or your notes have Newton¹s method (or Newton-Raphson iteration). Google also gets about 37,000 hits (only 4,780 for Newton-Raphson), but some of these are advanced treatments. You can still get good info, but you have to look for elementary treatments. Jon Miller === Subject: Re: the anticlassicalist }{ iv: from maps to logic > -=-=-=-=-= from maps to logic =-=-=-=-=-=- > Now let us look at interpretting this arrow symbol, such as in process, > categorial inclusion, rule reÞnement, or progressions of state. We want > to be able to evaluate logical relationships between objects. So we look to > deÞne on a recognition map new objects deÞned by any two objects which > allows us to collectively compare two objects logically. The Þrst one of > these, which can be called the least upper bound or disjunction of two > states Œa¹ and Œb¹, written (a / b) which obeys the rules that > a -> (a / b) > b -> (a / b) > forall x in the lattice with a -> x and b -> x, we have (a / b) -> x > where it is important to differentiate when these are derivationally found > on a lattice (objects already deÞned through previous axioms with the > appropriate universal property derivable) and when the existence of new > objects is being deÞned (for each new deÞnition requires new consistency / > completeness / etc. checks). You mean it provides a natural framework for conjectures ? > duality Doesn¹t tell much about that. Duality is negation in boolean algebras, isn¹t it ? === Subject: Re: question about periodic function >>If you have two periodic functions f(t) and g(t) with periods a and b, >>then the function f(t) + g(t) is periodic with period c where c is the >>least-common-multiple of a and b. Here the Þrst period is 1/.018=500/9 nd >>the second period is 1/.02=50. So the least-common-multiple is 500, i.e. >>1/.002. That should be the period of the combined function. >How to prove the theorem? Any hint is welcome. TIA. If f(t) has period a then f(t + na) = f(t) for any integer n, right? Same idea for g: g(t + mb) = g(t) for any integer m. Hmmmm....how could I Þnd an integer that works for both.....? --Lynn === Subject: Re: Teaching philosophy > I need to write a description of my teaching philosophy. In order to > do so accurately, I think I need to write at greater length and in > greater detail on this topic than any hiring committee will want to read. > Moreover, I need to present my views accurately but, somehow, in such > a manner as not to vitiate the consideration of my application. > That being the case, can someone please tell me what my teaching > philosophy is? What kind of philosophy? Metaphysics? Epistemology? Ethics? Aesthetics? Kantian? Pragmatic? Platonic? Aristotelean? Bob Kolker === Subject: Re: Teaching philosophy >I need to write a description of my teaching philosophy. In order to >do so accurately, I think I need to write at greater length and in >greater detail on this topic than any hiring committee will want to read. >Moreover, I need to present my views accurately but, somehow, in such >a manner as not to vitiate the consideration of my application. >That being the case, can someone please tell me what my teaching >philosophy is? >Ignorantly, >Allan Adler >ara@zurich.ai.mit.edu I would be happy to do so, but I only know how to take multiple choice tests. What are the choices? :-) --Lynn === Subject: Re: How many different resistances with n resistors? Originator: tchow@lagrange.mit.edu.mit.edu (Timothy Chow) >> The Puzzle Corner in MIT¹s magazine Technology Review raised the >> question of the number of distinct resistances achievable by connecting >> ten unit resistors. >Tony Bartoletti (seq. A048211) considers only cases that can be reduced by >applying the formula for parallel/series. If you have 6 resistors there is 1 >conÞguration where this is not possible. I suppose we get an additional >resistance value. What¹s the 6-resistor conÞguration you have in mind? And has anyone calculated the answer to the problem if we don¹t assume series/parallel circuits? -- 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: errors in an argument by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1IIGEO17931; >Should I not be too critical of his wishful thinking that this >proof actually means something? His argument is an overly sophisticated version of one of the most basic arguments against evolution, one that I can¹t rule out. How does super evolved, complex, Þne tuned to the millimeter biological structures appear when their evolutionary beneÞt seems to be absolutely nothing even with 90% or 95% of them installed. People used to point out that eyes are one of these mechanisms from the very beginning, but that, I think, is a bad example because any ability to sense this way, even on a very primitive level, must be a huge advantage. My favorite is birds : How can one evolve into þying ? So many things have to happen in order to create functional wings it seems impossible any awkward animal developing in this direction would survive it¹s cost, or any failing þying attempts. The evolutionist arguments sound totally absurd : they ran fast or jumped from tree to tree and got an advantage slowly developing wings. Sounds like absolute voodoo to me. === Subject: Re: the anticlassicalist }{ v: universal truths > And on these lattices we can make more explicit our theories of negation. > On sci.logic, mitch recently gave a very interesting post on Stone algebras. > When we want to model uses of various deÞnitions, we need to show them as > an axiom set. Stone algebras study the axiom set: > ~(a) / ~(~(a)) <--> T Btw, can complex numbers model it ? Let¹s see ~ : r exp(i s) --> r/2 exp(i s/2) a = 1 ~(a) = -1/2 ~(~(a)) = i/4 let¹s say T is 0... Well, then I am sure a Lorentz-Moebius transform can save the situation, I mean, provide a natural interpretation to the same formula graphics. The Moebius group is strictly 3-transitive, and z / w <---> u can pass as denoting -the- moebius transform that swaps z and w while keeping u invariant. Now idea if the thing can mimick a lattice any further, though. Isn¹t the symmetric group attached to the boolean algebra in such a manner that extremely transitive (but non-symmetric) groups shine as interesting analogues ? === Subject: Re: . 53ab2750 is an anonymous user . > AHahahahaha....AHAHAhahahaha...ahahaha.... > ..... you too get three attaboys for your reply ! > Life¹s a bowl of cherries, pits and nuts and all.... ahahahahaha.......ahahahahanson > Yeah. My niece used to laugh like you ... when she was 2 years old ehhhhh, hmmmmmm.........ahahahaha...... Your retort has a pH 12+ for some reason. Is it due to an after effect from ~ ....after chemo, I, Peter Brown was never the same? Cheer up, Peter, *laugh*, that¹s all that¹s there, that¹s all you got left! Life¹s a bowl of cherries, pits and nuts & all....Do you git it? git it git it? hanson, bonvivant and commentator on all walks of life...AHAhhahaha.. Ahahahahahaha..........AHAHAHHAHHAAHAHAHHAHa........ ahahahahaha.. === Subject: Re: Teaching philosophy >I need to write a description of my teaching philosophy. In order to >do so accurately, I think I need to write at greater length and in >greater detail on this topic than any hiring committee will want to read. >Moreover, I need to present my views accurately but, somehow, in such >a manner as not to vitiate the consideration of my application. >That being the case, can someone please tell me what my teaching >philosophy is? >Ignorantly, >Allan Adler A more serious answer: I promise to cheerfully accept students who don¹t have the prerequisites and can¹t add fractions without a calculator or solve linear equations. I will have them work in groups so, hopefully, someone in each group can do the work. I will be their best friend so they don¹t think of me as a teacher. We will have nice pleasant classes where we all discover things together so nobody feels left out or has their self-esteem bruised. Before each exam I will give out a practice exam ****very**** similar to the real exam so they will know what to expect. I will be very generous with part credit for those that still don¹t get it, even giving re-takes if necessary. I **guarantee** a very high success rate in my classes. There will be so many A¹s and B¹s that I will get awesome student evaluations. Students will þock to my classes because I am such a great teacher. The Dean will notice the improvement in my department¹s success rate and will probably increase it¹s funding. Over time, as my methods catch on, even the University administration and maybe even the Regents will hold our department up as a shining example of what is being done to improve higher education. --Lynn === Subject: Re: HELL :-) (Was: Re: the anticlassicalist }{ ii: the spectre continues) > *MATHEMATICS IS FULL OF TRICKS* > French even allows you the conjugation : > ma th.8ematique > ta th.8ematique > sa th.8ematique > notre th.8fme .88 tics > votre th.8fme attique > leurre, t¹es mat, hic ! > *REDUCING TOPICS TO THE SCIENCE OF NUMBER IS NUMEROLOGY* > I¹d call that accounting. lol :-) mitch === Subject: Re: Simple numbers Well what is that algorithm? > Hallo; > Well, p(n) is a n-th prime but if n=123456 how do I know what is the > 123456-th prime? I need the correct number (for exampel 1-st prime is -> 2)! >> Hallo, >> I know that every simple number (beside 2 and 3) has its one formula : >> S.N = >> 6*n +/- 1 for some n , but not any n={1,2,3,...+} in formula 6*n +/- 1 >> gives >> the simple number (for example: n=6 => 6*6-1=35 , 35 is not simple , >> 5,7|35) >> By simple number I think you must mean prime. >> so my question is: >> If n={1,2,3,...} what is the formula which gives the simple numbers >> (beside >> 2 and 3) for any n? >> Evidently you don¹t consider p(n) is the n¹th prime to be a formula. >> Any >> particular reason? >> Computational complexity? >> Jon Miller > Given any n, there is a simple algorithm that computes p(n). You could spend a > lifetime trying to make the algorithm more efÞcient, but the basic algorithm > is quite simple. === Subject: Re: [arccos((sqrt(5)-1)/2 )] / pi is irrational > > I will use this to show that > if cos x = (1 + sqrt(5))/2 then x is transcendental. (Hence, it is > irrational.) > > But that wasn¹t the original question. The original question > was whether x / pi is irrational. A whole nother kettle of Þsh. > > To make this question actually look interesting, note that if > cos(x)=( (1+sqrt(5)) / 4 ), then x/pi is equal to 1/5. > > (Also, the original question has sqrt(5)-1 instead of sqrt(5)+1 . > Again, cos(x) = ( (sqrt(5)-1)/4 ) implies x/pi = 2/5 . The question is > to show that this 2/5 value turns transcendental when that Œ4¹ on the > denominator changes to a Œ2.¹) > > J > So sorry for the bleeps in my non-solution. > Yes, the original question did involve (sqrt(5)-1)/(2 *p)i not > sqrt(5)+1 and Lindemann¹s lemma and > my proof are only true if x ne 0. Of course e^0 = 1. > I will go back to the drawing board and see if my original proof can > be Þxed. > Sorry for the whole nother kettle of Þsh! > Ray Steiner Well, I Þnally got the solution(hopefully). I don¹t have time to post all the details right now, but here is a brief outline of the steps: 1). Use cos(2*pi/5)= (-1 + sqrt(5))/4 to Þnd all the roots of unity in Q(exp(2*pi*i/5)). (There are 10 of them.) 2). Now consider cos(pi*r)= (-1 + sqrt(5))/2. Then, by the usual formulas of trigonometry, we get cos(2*pi*r)= 2-sqrt(5) and sin(2*pi*r)= 2*sqrt( -2+sqrt(5)). If r were rational, cos(2*pi*r) + i*sin(2*pi*r) (*) would be a root of unity in Q(exp(2*pi*i/5)). 3). Since (*) doesn¹t match any of the answers of step 1, r cannot be rational. Hence, r is irrational. Hope this does the job! Ray Steiner === Subject: Re: Full Beal Conjecture (revised) > Sorry for the mistake. My proof can be obtained at Did you win the dosh? -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: strain softening spring > I am a Civil engineer who has been working for the last 20 years in > the Þeld as a result of which my math is somewhat basic. I have a > good handle of linear systems, but am taking my Þrst baby steps > with non-lineear systems. > I have read up a few books on non-linear systems looking for a > canned solution for a problem I am trying to solve. But while I see > a lot of stuff on the DufÞng equation I don¹t see anything that > resembles the equation I am trying to solve - a mass-spring-damper > system where the spring softens exponentially. > i.e. the equation I am trying to solve is as follows: > y¹¹ + cy¹ + [ko*e^(-alpha*t)]y = 0 > Any help would be most appreciated. Well, if you want an intuitive (albeit approximate) solution, you should start by considering the relative size of the parameters. SpeciÞcally, you have ordinary damping due to the Œc¹ parameter, and then you have the exponential softening, i.e. the Œalpha¹ parameter. These both have units of inverse time and you should start by determining which range of values you are interested in, e.g. c << alpha and/or c >> alpha. In either of these cases, the time scales are very different, and you can therefore make some assumptions. For instance, if alpha is very small, then the softening happens on a very slow time scale and therefore the spring constant is almost, well, constant. You can therefore assume that the solution is an ordinary damped oscillation and simply replace k0 in the solution with k0 * e^(-alpha*t). Another line of approach is to try to eliminate the damping by substituting y(t) = z(t) * e^(-beta*t) for some suitable chosen value of beta. A third idea is to make a vector plot. Introduce z=y¹ and for each point (y,z) draw a vector whose direction is (y¹,z¹). Try it Þrst with alpha = 0, because then the vectors are independent of time. You will see spirals that lead to the point (0,0). When alpha is nonzero, the vector plot is time-dependent, but you might be able to say something qualitative about them anyway, e.g. whether they spiral faster or slower towards (0,0). -Michael. === Subject: A newbie¹s question -- about real number Two real numbers r1 = 0.89, r2 = 0.889999999..... (with inÞnite 9s), === Subject: Re: A newbie¹s question -- about real number lisong@iastate.edu asks this about real numbers: >Two real numbers r1 = 0.89, r2 = 0.889999999..... (with inÞnite 9s), Theoretically NO. In real life measurement or result calculations in which only two signiÞcant Þgures are reasonable, then r1=r2. The context must be appropriate. G C === Subject: DeÞnition of Separable Space (basic topology question) Let A be a subset of (X,T). Then A is dense in X iff for every non-empty open subset U of X, A / U != {}. I have seen two deÞnitions of Œseparable topological space¹: a) (X,T) is separable if there exists A X, where A is countable and dense in X b) (X,T) is separable if there exists A X, where A is countable and dense in X Given def (a), its simple to prove that all countable spaces X are automatically separable since the subset X is countable and non-trivially intersects every non-empty open subset. However, this won¹t work given def (b). I Þrst became suspicious when I saw a proof that all countable spaces were separable that seemed ridiculously complex in comparison to the seemingly obvious 1 step proof above. I later found deÞnitions of separable like def (b). l8r, Mike N. Christoff === Subject: Re: A newbie¹s question -- about real number Adjunct Assistant Professor at the University of Montana. >Two real numbers r1 = 0.89, r2 = 0.889999999..... (with inÞnite 9s), Both expressions are taken to represent the same real number. The Þrst is the real number 8/10 + 9/100. The second is taken to represent the limit of the inÞnity series 8/10 + 8/100 + (9/10^3 + 9/10^4 + 9/10^5 + ....) Since Sum_{i=3}^{inÞnity} 9/10^i = 9/10^3*(Sum_{i=0}^{inÞnity} 1/10^i), and the series (1/1 + 1/10 + 1/100 + 1/1000 + ...) converges to 1/(1-(1/10)) = 1/(9/10) = 10/9, we have that the original inÞnite series 8/10 + 8/100 + (9/10^3 + 9/10^4 + 9/10^5 + ....) converges to 8/10 + 8/100 + 9/10^3*(10/9) = 8/10 + 8/100 + 1/100 = 8/10 + 9/100, which is equal to r1. -- It¹s not denial. I¹m just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Cauchy¹s Interlacing Theorem Hi folks, I¹m looking for a proof of the following... Let A = [a0 a^h ; a A2] >= 0 be a NxN hermitian matrix where a0 is a scalar > 0 and A2 is the lower-right N-1xN-1 (principal) submatrix of A. According to Cauchy¹s interlacing theorem (1829 I guess), the N Eigenvalues gamma_i of A and the N-1 Eigenvalues alpha_i of A2 interlace, i.e., gamma_1 >= alpha_1 >= gamma_2 >= ... Interestingly, having played a little bit with MATLAB, also the Eigenvalues gamma_i of A and the Eigenvalues beta_i of the N-1xN-1 matrix A2-a*a^h/a0 seem to interlace. However, this seems to be true only for nonnegative-deÞnite A. Does anyone have a proof, or reference to a proof for that?? Initially, I wanted to show that if A >= 0 with rank A >= 2, then also A2-a*a^h/a0 >= 0; but, then I discovered Cauchy¹s theorem... ;) (If rank A = 1, then A2-a*a^h/a0 yields the N-1xN-1 zero matrix, and if rank A >= 2, then rank A2-a*a^h/a0 >= 1, at least this is easily seen). Once the above assumption had been proven, A2-a*a^h/a0 >= 0 would directly follow as beta_N-1 >= gamma_N >= 0 (comparing the smallest Eigenvalues) Christian. === Subject: Re: Collatz Conjecture : Symmetry question. > Every time I post on this problem I get beat up so please be kind. I hope that wasn¹t directed at me. I never beat anyone. It may seem that way to Ernst Berg, but I was just trying to help him by offering constructive criticism. Apparently some people can¹t get past the criticism part and get all bent out of shape to the point where they ignore the constructive part. But that¹s Ernst¹s problem. > I have found many properties in the landscape of this problem and the > one that has most of my interest the gross amount of non-regularity > tightly coupled to large a domain of symmetry. I have noticed 2 forms > of symmetry in this problem which have me thoroughly intrigued, but do > not know where to go with these notions at this point. > Symmetry 1; > I have noticed that given any tree pattern of any size, it is exactly > duplicated inÞnitely many times. For example, take the entire tree > starting at A and extend out all branches to a depth of any N. Then > that pattern taken generically is repeated at an offset of A + > n*(constant value) with (n = 1, 2, 3, ... inÞnity), with the constant > value is a function of N. That seems reasonable since any sequence vector evaluates to a simple formula a¹ = (X*a - Z)/Y which means there are always an inÞnte set of integer solutions to every sequence vector. > Symmetry 2; > I have also noticed that all values at N (here I include all > rationales and integers by taking all possible paths from A to N, > keeping NON_INTEGER branches in this process) produces N¹ sets of a > collective 2^(N-1) number of values where the number of items in each > set follows the Binomial CoefÞcients or Pascal¹s Triangle (which ever > you normally call it). If you take each of these N¹ sets of values and > calculate the ordered set difference (N¹¹) that the difference sets > (N¹¹) are constant and equal for all N independent of A. I¹m not sure I follow this, but since there are only two operations (3x+1 and x/2) every number has two possible ancestors, so carried to N levels (and ignoring whether the ancestors are integers) there would certainly be 2^N numbers at distance N. > This means that using only the N¹¹ sets, all sets at N in groups of > N¹, thus all values at N, for any A, can be calculated without > extending the tree from A to N through all the intermediate N¹s. > In other words, all values of absolute stopping distance of N, for any > A, can be calculated directly with out traversing the tree to Þnd > each of those values. But there are vastly more total numbers at a given level than there are integers, so although this is true, I¹m not sure it leads anywhere. For example, I recently found all the integers that are at level 84 (from 1): 179,441,377 If I included all the potential non-integer ancestors, I would have had to sift through 2^84 or 19,342,813,113,834,066,795,298,816 numbers to Þnd which ones are integers. In this case, it is much quicker to traverse the tree from level 1 up to level 84. I have a program that can build each level from the previous one. Since it only generates integer results, it is much more efÞcient than testing each binary permutaion of the two operations (provided you record the integers you collected from the previous level). It would be nice to be able to just start at an arbitrary level and Þnd all the integers. So although it can be done in theory, it can¹t be done in practice for high levels. Unless you have some means of determining which of the 2^84 numbers at level 84 are integers without having to iterate through them all. > Is this interesting to anyone? Yes. > Or better yet, Is this anything of interest for this problem. I¹m not an expert, so I can¹t say. === Subject: Re: A newbie¹s question -- about real number >lisong@iastate.edu asks this about real numbers: >>Two real numbers r1 = 0.89, r2 = 0.889999999..... (with inÞnite 9s), >Theoretically NO. Hence the rationals are uncountable! Rich === Subject: Graph Theory Textbook I¹m an undergraduate, and this summer I will be participating in an REU for discrete math and combinatorics. I am looking for a good graph theory textbook to learn the basics from. I have seen the books that Dover (which of course are very cheap) has to offer on the subject, but unfortunately they seem too elementary. Is the Springer GTM Graph Theory by Reinhard Diestel any good? Any suggestions are highly welcomed. Jack === Subject: Re: errors in an argument > My favorite is birds : How can one evolve into þying ? So many things > have to happen in order to create functional wings it seems impossible any > awkward animal developing in this direction would survive it¹s cost, or > any failing þying attempts. The evolutionist arguments sound totally > absurd : they ran fast or jumped from tree to tree and got an advantage > slowly developing wings. Sounds like absolute voodoo to me. That depends on the size of the animal. If the animal is small enough, the terminal velocity can result in a non-lethal impact with the ground. (I believe the threshold is around the size of a mouse or a cat.) Incremental levels of control over the landing point then have an obvious beneÞt. -- Daniel W. Johnson panoptes@iquest.net http://members.iquest.net/~panoptes/ 039 53 36 N / 086 11 55 W === Subject: Re: 3 Squares Covering 1 Circle > For b, what does a C3v (invariant under 120 deg rotation) > symmetric arrangement yield? (Place the outermost edge of > the squares tangent to the circle.) > > Here¹s a similar situation that does better than the 80.4% > with part (a). Use this 120 deg symmetry, but place the diagonals > of the square along the radii. The optimum placement of the > squares (they can be shifted along the radii) covers 83.0% of > the circle¹s area. This is the best that I¹ve found, but I don¹t > know if it¹s the best possible. > To be a little more precise, the area is 2.60808705... I have an exact > form, but it¹s pretty messy. This happens when the diagonal of each > square extends .13980931 past the center of the circle (along the > diameter). Oh, that¹s a nice conÞguration to consider! I checked your result, and I agree with your area Ac = 2.60808705, although the extension of the square beyond the center of the circle is a little off: it should be rc = 0.13977317. (I would guess that you obtained the value of Ac to single precision, which yields only half the precision for rc). For general r (distance diameter extends beyond the circle¹s center), the area A of the resulting Þgure can be found as follows: Let u = sqrt(2) - r, v = sqrt(2 - u^2), d = (u - v)/2, and p = (1 + u v)/2, then A = 3 (p - sqrt(p) d - (3 + sqrt(3)) r^2/2 + Arcsin(d)). This area is maximized when r equals the critical value rc: rc = (19 - 5 sqrt(3) - sqrt(176 - 86 sqrt(3))) / (26 sqrt(2)). The formula for Ac = A(rc) simpliÞes greatly (as one would expect). Let dc = rc (3 + sqrt(3))/2. Then Ac = 3 (1 - sqrt(2) dc + Arcsin(dc)). To summarize the results so far, the best result posted for case (a) (no overlap) is (a) pi/6 - 1/2 + sqrt(3)/4 + sqrt(55)/8 + Arccos(3/8) = 2.57003584 = 81.8067816% of the circle, and the best result posted for case (b) (overlap allowed) is (b) Ac (see above) = 2.60808705 = 83.0179893% of the circle. -Jim Ferry === Subject: Re: A newbie¹s question -- about real number > Hence the rationals are uncountable! Nice try, but no cigar. The set of decimal sequences that terminate in all 0¹s or all 9¹s is countable. === Subject: Re: A newbie¹s question -- about real number > Theoretically NO. > In real life measurement or result calculations in which only two signiÞcant > Þgures are reasonable, then r1=r2. The context must be appropriate. Theoretically YES, and given that we discuss mathematics here, that is the only meaningful answer. === Subject: Re: DeÞnition of Separable Space (basic topology question) > Let A be a subset of (X,T). Then A is dense in X iff for every non-empty > open subset U of X, A / U != {}. > I have seen two deÞnitions of Œseparable topological space¹: > a) (X,T) is separable if there exists A X, where A is countable and > dense in X > b) (X,T) is separable if there exists A X, where A is > countable and dense in X > Given def (a), its simple to prove that all countable spaces X are > automatically separable since the subset X is countable and non-trivially > intersects every non-empty open subset. However, this won¹t work given def > (b). I Þrst became suspicious when I saw a proof that all countable spaces > were separable that seemed ridiculously complex in comparison to the > seemingly obvious 1 step proof above. I later found deÞnitions of > separable like def (b). > l8r, Mike N. Christoff I have never seen deÞnition (b). I have seen the symbol $subset$ used to mean subset, not proper subset, so maybe that is what you saw. Certainly a one-point space with the discrete topology is to be considered separable, even though it has no dense proper subsets. -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: How many different resistances with n resistors? >> The Puzzle Corner in MIT¹s magazine Technology Review raised the >> question of the number of distinct resistances achievable by connecting >> ten unit resistors. >Tony Bartoletti (seq. A048211) considers only cases that can be reduced by >applying the formula for parallel/series. If you have 6 resistors there is 1 >conÞguration where this is not possible. I suppose we get an additional >resistance value. > What¹s the 6-resistor conÞguration you have in mind? An H with asymmetric legs. I thought that 6 is the smallest number with this property, but obviously a H with 5 resistors gives us r=1 which seems impossible with other combinations of 5 resistors. I did not yet check, whether the 6-H has a resistance not occuring among the (other) 53 possibilities. I will play with this a little tomorrow. I think we would Þnd a reference in encyclopedia if someone had solved the general case. Klaus > And has anyone calculated the answer to the problem if we don¹t assume > series/parallel circuits? > -- > 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: Help needed - primes program Well how about this: ------------------------------------------------------------- --------------- --- Program PB; Var x,y:word; f:word; i,n:word; ok:boolean; Begin Write(Enter X and Y -> Œ); Readln(x,y); For i:=x to y do Begin f:=3; ok:=true; If (i=1) or (i mod 2 = 0) then Begin f:=i+1; ok:=false; End; While f<=sqrt(i) do Begin If i mod f = 0 then Begin f:=i; ok:=false; End else f:=f+2; End; If ok=true then Write(i,¹ is a prime. Œ); End; Readln; End. ------------------------------------------------------------- --------------- --- It ignores all number with form k*2 , k{2,3,4,...} and it goes till sqrt(i) insted of i. Is this faste or slower than sieve of eEatosthenes? Why? === Subject: Re: A newbie¹s question -- about real number > Two real numbers r1 = 0.89, r2 = 0.889999999..... (with inÞnite 9s), > can we say r1 = r2 ? Yes. The problem is that we present inÞnite decimal expansions in school before their meaning can be properly understood. Once this meaning is made clear, proving r1 = r2 is a very simple exercise. === Subject: Re: Teaching philosophy > I promise to cheerfully accept students who don¹t have the > prerequisites and can¹t add fractions without a calculator or solve > linear equations. I will have them work in groups so, hopefully, > someone in each group can do the work. I will be their best friend so > they don¹t think of me as a teacher. We will have nice pleasant > classes where we all discover things together so nobody feels left out > or has their self-esteem bruised. Before each exam I will give out a > practice exam ****very**** similar to the real exam so they will know > what to expect. I will be very generous with part credit for those > that still don¹t get it, even giving re-takes if necessary. I > **guarantee** a very high success rate in my classes. There will be so > many A¹s and B¹s that I will get awesome student evaluations. Students > will þock to my classes because I am such a great teacher. The Dean > will notice the improvement in my department¹s success rate and will > probably increase it¹s funding. Over time, as my methods catch on, > even the University administration and maybe even the Regents will > hold our department up as a shining example of what is being done to > improve higher education. Hah, pretty good! === Subject: Re: Math Too Advanced For Mainstream Economists Robert Vienneau threatening to hold his breath until he dies once again, > The pitiful child who posts below is teaches economics at a leading > second-tier department of economics in the U.S.A. Hard to believe. Maybe you can get your big brother to beat him up. === Subject: Re: help with solutions for three questions from the past contest > 3. A boat with an ill passenger is 7.5 mi north of a straight > coastline which runs east and west. A hospital on the coast is 60 > miles from the point on shore south of the boat. If the boat starts > toward shore at 15 mph at the same time an ambulance leaves the > hospital at 60 mph and meets the ambulance, what is the total distance > (to the nearest 0.5 mile) traveled by the boatand the ambulance? ans: > 62.5 Choogu > Boat . > | > | > 7.5| z > | > Shore |___________________________________.Hospital > x 60-x > Hours along distance z at 15 mph must equal hours along distance > 60-x at 60 mph. > And at constant speeds, distance = speed * time. > I¹ll be curious if that¹s the best solution. (It may be the one expected.) > A better solution would be to minimize the time to the hospital. This does minimize the time to the hospital! === Subject: Re: question about periodic function >If you have two periodic functions f(t) and g(t) with periods a and b, >then the function f(t) + g(t) is periodic with period c where c is the >least-common-multiple of a and b. Here the Þrst period is 1/.018=500/9 nd >the second period is 1/.02=50. So the least-common-multiple is 500, i.e. >1/.002. That should be the period of the combined function. > How to prove the theorem? Any hint is welcome. TIA. The sum will only be periodic if the ratio of periods is rational. If that ratio is irrational f(t) + g(t) is not periodic. === Subject: Re: Simple numbers > Well what is that algorithm? It¹s called the Sieve of Eratosthenes. -- Dave Seaman Judge Yohn¹s mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Anyone knows a formula for this? (combinatorics) Hi Let n be the number of cards drawn from an ordinary deck (52 cards). For example according to the book, if n=5 i.e if we pick 5 cards randomly from a deck of 52 cards The probability of getting a) A full house is: [C(4,2)*C(4,3)*2*C(13,2)]/C(52,5) b) Exactly two pairs is: [C(4,2)*C(4,2)*C(44,1)*C(13,2)]/C(52,5) So i was wondering if there¹s a general result (a formula)? for a) and b) where n is the number of cards drawn. === Subject: Re: errors in an argument > If I were trying to explain the error to someone, I would probably > use the analogy of rolling a 20 sided dice 100 times and recording > the result. By the proof above, one can¹t do this, since the > probability of any particular sequence is (essentially) 0. What about the probability of any particular sperm actually being the one to fertilize an egg? That would seem to preclude the possibility of sexual reproduction altogether. -- Joe Bramblett, KD5NRH === Subject: Re: I¹d like to join this group! >> Does the Equation below really prove that 2=1 ?? Y N >> 1) X=Y ; Given >> 2) X^2=XY ; Multiply both sides by X >> 3) X^2-Y^2=XY-Y^2 ; Subtract Y^2 from both sides >> 4) (X+Y)(X-Y)=Y(X-Y) ; Factor >> 5) X+Y=Y ; Cancel out (X-Y) term >> 6) 2Y=Y ; Substitute X for Y, by equation 1 >> 7) 2=1 ; Divide both sides by Y > Go back to step 4. What it really proves is 2x0=1x0. > Now try this: > 1) I am nothing without her. (Shakespeare) > 2) ME - SHE = 0 (Algebraic representation) > 3) ME = SHE (Added SHE to both sides) > 4) M = SH (Divided by E) > 5) MIT = SHIT (Multiplied by IT) > So there you have the mathematical proof of what Shakespeare > thought of the future American techincal education. :-) You Guys are too sharp. That was actually printed in OMNI magazine Nov 79 , as proof that 2=1 . ;-) === Subject: Re: Dik Winter¹s claims revisited, dependency issue > James Harris: > |> They¹ve gone so far that Keith Ramsay even claimed to have posted a > |> solution for the w¹s, but how did he pick w¹s from inÞnity? > I¹m not sure what you mean here by pick w¹s from inÞnity. > | He didn¹t. There is a well-deÞned algorithm for computing > |the w¹s, starting with the number (1 + sqrt(-167))/2. If you > |had replaced 7 with 17, the starting number would have been > |(-1 + sqrt(-407))/2 and the result would be different. > Let¹s recall what I posted. > |It appears that Q(sqrt(-167)) has a class group of order 11. > |Let r stand for (1+sqrt(-167))/2 (which is an algebraic integer, > |in spite of the 2 in the denominator). > |r^11 = (-592764018-86559857*r) > | = (44555-222*r) (-12882-2017*r) > |and > |7^11 = (44555-222*r)(44555+222*r). > As an intermediate step, I found the GCD of ((1+sqrt(-167))/2)^11 > and 7^11, which is an 11-th power of the GCD of (1+sqrt(-167))/2 > and 7. > Harris to check the result using a calculator or something like that. > |Also (-12882-2017*r)(-12882-2017*(1-r)) = 2^11 * 3^11, which is > |relatively prime to 7^11. > |The factor 44555-222*r = 44444 - 111 sqrt(-167) and its conjugate > |44444 + 111 sqrt(-167) are roots of x^2-88888*x+7^11=0. One GCD of > |r with 7 is then (44444-111 sqrt(-167))^(1/11), which is a root of > |x^22 - 88888 * x^11 + 7^11 = 0. All those numbers are hard to parse through, so I just went back to considering that a_1(x) + a_2(x) = (x-1) so assuming factors f_1(x), and f_2(x), with your w_1(2) = (44444-111 sqrt(-167))^(1/11), and w_2(x) = (44444+111 sqrt(-167))^(1/11), I have w_1(2) f_1(2) + w_2(2) f_2(2) = 1 and assuming the w¹s are symmetric by sign, I have that f_1(2) must equal f_2(2). But f_1(2)f_2(2) = 6 as the problem I see is that you¹re stuck, if the w¹s are symmetric by sign, so are you saying they are not? James Harris === Subject: Re: How many ways to put 5 balls into 500 ordered cups? > You are given 500 numbered cups and Þve identical balls. Any cup can > hold up to Þve balls. How many ways can you put the Þve balls into > the 500 cups? Three; you can use your left hand, your right hand, or both. Of course, you could use some other sort of mechanical aid, but that would be a question for the engineering groups. -- Joe Bramblett, KD5NRH === Subject: Re: A newbie¹s question -- about real number >> Hence the rationals are uncountable! >Nice try, but no cigar. The set of decimal sequences that terminate in all >0¹s or all 9¹s is countable. No cigar? Bummer. But I *can* Þnd an uncountable number of non-terminating representations of 89/100 and Þgured that since r1<>r2 the conclusion followed. If that is not true then clearly 89/100 is irrational! Even better. Rich === Subject: Re: No Set Contains Every Computable Natural > Russell Easterly says... > The usual notion of recursive set is that a set S is >> recursive as a *subset* of the naturals if there is a >> Turing machine T such that for any *natural* number n >> (in unary notation, to be speciÞc) The set of all natural numbers is not a proper subset of >the set of all natural numbers. But it *is* a subset. A recursive subset. > T(n) halts and outputs 1 <-> n is an element of S >> T(n) halts and outputs 0 <-> n is not an element of S >> If x is some input that is not a representation of a natural >> number, then there is no constraint on what T(n) does. > >According to this deÞnition, the set of all natural numbers >is recursive because there exists a TM that doesn¹t even >read the input and always outputs a 1. Right. > The set of all natural numbers is recursive by Þat. > No, not by Þat. It is a trivial consequence of a general deÞnition-- > the deÞnition that Daryl McCullough has reproduced for you. Using Daryl McCullough¹s deÞnition, I can claim the set of all natural numbers is (1,2,3). Since the string 1111 does not represent a natural number in my system, it is too large to be the input to a Turing machine. > Why bother to include computability in the deÞnition? > There is no separate deÞnition for the case of all natural numbers. > There is one deÞnition that applies to all subsets of the set of natural > numbers. The set of all natural numbers (henceforth N) is a subset of > N. We apply the deÞntition to N and Þnd out trivially that it is > recursive. > It is not a trivial deÞnition. It is quite profound. But it is > trivial that N is recursive. I Þnd the idea that N=(1,2,3) quite profound. > A human must decide if x is a member of this set > before x can be given to the TM. > You are seeking to create a new deÞnition and theory of computability. > Good. You want to consider a larger context of inputs other than > just natural numbers. For example, you want to include inÞnite > stings as acceptable inputs. Who decides the string is Þnite? A TM will always think the input is Þnite. I have given proofs of this. > Fine. I would agree with you that > if we enlarged the universe of acceptable inputs from just N to a larger > set containing inÞnite strings, then N would probably not be > recursive in any reasonable new deÞnition of the word. You¹re right, > a TM would not be able to conclude that an inÞnite string of 1¹s is > not a natural number. A TM will say that any string of 1¹s represents a natural number. > I would say that in this enlarged context, that > N would be recursively enumerable but not recursive. > But your problem is that you are using terms with agreed upon deÞnitions > by the mathematical community (like recursive) incorrectly. You have > your own personal deÞnition that you have failed to state. I used a standard deÞnition of decidable language. The standard deÞnition states that a human must decide that a string is Þnite before giving it to the TM. Ultimately, a human being must decide if a string, x, represents a natural number. Why pretend there is an algorithm that can do this? > This is why > mathematicians are arguing with you and pointing out your obvious > errors. If you would like to re-deÞne recursive by all means do so. I am not redeÞning recursive. I am pointing out that if the deÞnition of decidable languange says there is a TM that can decide if string, x, is a member of the language, then the language of all unary representations of the natural numbers is NOT a decidable language. If we include the requirement that a human being decides whether x is a Þnite string, then I agree N is recursive. This additional requirement allows me to choose N to be any set I want it to be. > But make your new deÞnition known. It would also be good to give > it another name to avoid confusion, like Easterly-Recursive. > Until you do this, you will probably continue to make false statements > that are easily refuted. Are you refuting my claim that N=(1,2,3)? > Remember, mathematics is precise. N is precisely equal to (1,2,3). > By the way there are already in existence more generalized recursion theories, > which deal with recursion on other domains besides N. Can you give references? Obviously, I need to learn more about this stuff. Russell - 2 many 2 count === Subject: Re: Anyone knows a formula for this? (combinatorics) Adjunct Assistant Professor at the University of Montana. >Let n be the number of cards drawn from an ordinary deck (52 cards). >For example according to the book, if n=5 >i.e if we pick 5 cards randomly from a deck of 52 cards >The probability of getting >a) A full house is: >[C(4,2)*C(4,3)*2*C(13,2)]/C(52,5) >b) Exactly two pairs is: >[C(4,2)*C(4,2)*C(44,1)*C(13,2)]/C(52,5) >So i was wondering if there¹s a general result (a formula)? for a) and b) >where n is the number of cards drawn. You should try to understand what the numbers mean. For example, with full house, you have C(52,5): that¹s the number of different ways in which you can randomly pick 5 cards from a deck of 52 cards. It is the total space. If you are randomly pickin k cards from a deck of 52, then that number should obviously be replaced by C(52,k). What do the other numbers mean? I.e., C(4,2)*C(4,3)*2*C(13,2) should be the number of different 5 card hands which are a full house. How is this accomplished? Why does this count that? Well, a full house is determined by stating the following things: (1) The rank of the 3-of-a-kind (i.e., is it 3 aces, 3 jacks, 3 tens, etc); and (2) The rank of the pair; and (3) The suits involved in the 3-of-a-kind; and (4) The suits involved in the pair. That completely determines the hand, since there are no cards leftover in a 5-card hand. Now, C(4,2) means how many ways