mm-420 Subject: problems with equationsI need to solve these equations:|x+a|=Sqrt[(y-b)^2+(z-c)^2]|x+d|=Sqrt[(y-e)^2+(z-f)^ 2]|x+g|=Sqrt[(y-h)^2+(z-i)^2]where x,y,z - variablesCould someone, who has Mathematica program or sth else, solve theseequations and send me solution (txt, jpg, ... etc). Please.Solution can be very large. === Subject: Re: problems with equationsWho needs Mathematica?Square all three equations. Take the difference between any one of thesquared equaitons and each of the other two -- producing two linearequaitons! Solve these for x in terms of y and z, then substitute intoone of the original squared equations. Solve the quadratic equation forx. === Subject: Re: problems with equations>I need to solve these equations:>|x+a|=Sqrt[(y-b)^2+(z-c)^2]>|x+d|=Sqrt[(y-e)^2+(z-f )^2]>|x+g|=Sqrt[(y-h)^2+(z-i)^2]>where x,y,z - variables>Could someone, who has Mathematica program or sth else, solve these>equations and send me solution (txt, jpg, ... etc). Please.>Solution can be very large. Square each one, eliminating the square roots and absolute values.(I assume everything is real).Subtract in pairs to get two linear equations in x, y, z.Solve for two of the variables in terms of the third.Then plug everything into one of the three earlier quadratics. The discriminant of the resulting quadratic includes threefactors such as (f - i)^2 + (e - h)^2 - (d - g)^2 .The solution itself is large.-- John Adams served two terms as Vice President and one as President, but lostreelection. Later his son became President despite losing the popular vote.That son lost his reelection attempt badly. Now history is repeating itself.pmontgom@cwi.nl Microsoft Research and CWI Home: San Rafael, California === Subject: hello.....Riemann problem..hello........sirshow that this function is Riemann integrable on [0,1]f(x) = 1/x , x = 1/n (n=1,2,3.....) 0 , otherwise.------------------------um........i think.....let P = {0/n , 1/n , 2/n , ........ , n/n}thusU = sigma Mi * delta_Xi = sigma (1/n)*(1/n) = 1/n i=1~nthus ||p|| -> 0 => U = 0L = 0thus Riemann integrableum........it is right??confirm...please....my doctor... === Subject: Re: hello.....Riemann problem..> hello........sirshow that this function is Riemann integrable on [0,1]f(x) = 1/x , x = 1/n (n=1,2,3.....) 0 , otherwise.Are you sure that this is the right function? I ask this beacause thisfunction is not bounded and therefore cannot be Riemann integrable.I suppose that you mean to write f(x) = x above.> ------------------------um........i think.....let P = {0/n , 1/n , 2/n , ........ , n/n}thusU = sigma Mi * delta_Xi = sigma (1/n)*(1/n) = 1/n> i=1~nAnd this confirms my suspiction that it should be f(x) = xwhen x has the form 1/n. But then the sum would be (1/n).(1/n) + (1/n).0 + (1/n).0 + ... + (1/n).0 = 1/n^2.> thus ||p|| -> 0 => U = 0L = 0thus Riemann integrableum........it is right??fact that ||p|| -> 0. What's important here is that sigma Mi * delta_Xi -> 0 === Subject: Re: hello.....Riemann problem..> And this confirms my suspiction that it should be f(x) = x> when x has the form 1/n. But then the sum would be (1/n).(1/n) + (1/n).0 + (1/n).0 + ... + (1/n).0 = 1/n^2.Just a small correction: the first (1/n).0 should be (1/n).(1/n);therefore the sum is equal to 2/n^2. === Subject: Re: uniform distribution on simplices>>Is there a decent way to let a simple program>>choose uniformly distributed, random points in the space>> { (x1,...,xn) | sum(xi) = 1 and for all i: xi >= 0 }>>?>>(That is: a line-segment for n=2, a triangle for n=3,>>a tetraeder for n=4, etc.)>>What i would like to do is this:>>- choose n random elements of [0,1] (uniformly)>>- order them from small to large: y1, y2, ..., yn>>- define x1 := y1, x(i+1) := y(i+1) - y(i)>>But:>>1. i'm not sure at all whether this will lead>>to a uniform distribution.>>2. perhaps there is a more efficient way to>>simulate a uniform distribution on this set?>>Can anyone help me further?>>Many thanks in advance.>>Herman Jurjus>Let REXP stand for the result of a call for generating an exponentialrandom>variable,>density exp(-y), y>0.>Then:> s=0;> for i to n do {x[i]=REXP; s=s+x[i]; }>will produce a point (x[1]/s, x[2]/s,...,x[n]/s)>that is uniformly distributed over the simplex,>and, aside from a scale factor and n<3, the exponential distribution>is the only one that will do the trick of projecting a random point>(y[1],y[2],...,y[n]) with positive iid coordinates uniformly onto the>simplex by dividing by their sum.>(G. Marsaglia, Uniform distributions over a simplex,>Math. Note 142, Boeing Scientific Research Laboratories , 1961).> Yes - this follows directly from the definition of the Dirichlet> distribution from ratios of gammas (gammata?).>That note shows that>Prob(x[1]>a[1], x[2]>a[2] ,..., x[n]>a[n]) = max[ 0,>1-a[1]-a[2]-a[n] ]^(n-1),>a result that Feller V2, and Knuth V2 attribute to De Finetti in> Giornale Instituto Italian Attuar, 1964.>The fastest way to generate REXP's is probably the ziggurat method of> Marsaglia and Tsang, Journal Statistical Software, v 5, Issue 8, 2000.> http://www.jstatsoft.org/v05/i08/ziggurat.pdf>George Marsaglia> I suspect that Herman's proposed method is more efficient than George's> rejection method. Having looked at your paper, I pose the question of> whether ziggurat method is more efficient than taking logarithms of> uniforms? Taking logarithms of uniforms doesn't come close to the speed of methods such as the ziggurat. Rather than going into details of the try a little C program to take the logarithms of 10^9 uniforms produced by ßoating a fast xorshift RNG(http://www.jstatsoft.org/v08/i14/xorshift.pdf). j^=(j<<13); j^=(j>>17); j^=(j<<5); -log(j*2.328306e-10); Mine took some 93 seconds, while 10^9 calls for REXP's via theziggurat took about 36 seconds, and that included the initial call to create thetables that the ziggurat method requires. While the ziggurat method isparticularly suited for fast generation of normals, RNOR's, it is equally fast forREXP's or most variates with a decreasing or symmetric unimodal density. George Marsaglia === Subject: um.....Riemann problemf(x) = sin (1/x) , x in irrational 0 , otherwiseshow that f(x) is Riemann integrable on [0,1]--------------------------um........i think that f(x) is not impossible.but i don't prove it.i want to use the U, L method.....um.....in my expedient,i know that bouned function on [a,b] is Riemann integrable <=>number of discontinuous is countable.help me....please....... === Subject: re:um.....Riemann problemI learnt the definition of Riemann integral over fifty years ago. Thefunction you are describing would not be Riemann integrable, butwould be Lebesgue integrable. Have the definitions changed? There has developed in recent years the notion of generalized Riemannintegral, which is equivalent to Lebesgue, but does not requiremeasure theory. Is that what you are talking about?----== Posted via Newsfeed.Com - Unlimited-Uncensored-Secure Usenet News==----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: oh...my mistake...sorry.f(x) = sin (1/x) , x in irrational 0 , otherwiseshow that f(x) is not Riemann integrable on [0,1]-------------------------------i want to use U-L method....i want to find partition,esilon such that |U - L| >= ebut i don't use to prove.....um.......help....me....please...... === Subject: Re: um.....Riemann problem> f(x) = sin (1/x) , x in irrational 0 , otherwiseshow that f(x) is Riemann integrable on [0,1]--------------------------um........i think that f(x) is not impossible.but i don't prove it.i want to use the U, L method.....um.....in my expedient,i know that bouned function on [a,b] is Riemann integrable <=>number of discontinuous is countable.No, the theorem is that a bounded function on a compact integrableis Riemann integrable iff the set of dicontinuities has Lebesguemeasure zero. Does your function have any points of continuity?-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the last laugh. Partridge, _Bouncing Back_ (14 times) === Subject: surface - plane> What is the condition that makes the surface a plane ?Choose three points and connect them, choose on two lines a point andconnect to the corners by a line, they cross in a point inside thistriangle.Let a line rotate around this point, so that it alwaystouches the triangle - that will give a plane.Or go from one corner of the triangle in two directions to the othercorners, so you have two arrows, stretch and shrink them, even to theopposite direction and add them and you can get to all points of thisplane. So it's called 2D.Or do the cross-product of the two arrows and you get an arrowperpendicular to the plane, it's called normal and the position of thetail of this normal arrow and it's direction defines a plane too.Three functions -for each direction of space one - dependent on twoindependent variables will define surfaces.Have funHeroP.S.:A square four-legged chair won't wobble on or in a big ball. === Subject: Re: surface - planeIn Euclidean geometry at least, all lines perpendicular to a plane areparallel to each other. === Subject: Re: surface - plane>What is the condition that makes the surface a plane ? If you equation is written in the form f(x,y,z)=c then grad(f) =constant. === Subject: Re: surface - plane> What is the condition that makes the surface a plane ?To be ßat, I suppose... :-) === Subject: Re: surface - plane charset=iso-8859-1> What is the condition that makes the surface a plane ?> To be ßat, I suppose... :-)> Best regards,> Jose Carlos SantosWhich type of stool/chair seems to be always stable with no wobbles? A fourlegged one or a three legged one? === Subject: Re: surface - plane>Which type of stool/chair seems to be always stable with no wobbles? A four>legged one or a three legged one?I can think of both three and four legged chairs that aren't stable atall. How many dimensions does your ßoor have?-- I'm not interested in mathematics that might have anythingto do with reality. -- Easterly, in sci.math === Subject: Re: Partition of positive integers> |> In the following, let us denote > ||> [.] = the integral part> ||> N = the set of positive integers. > ||> For p > 0 define the set A[p]= { [np] ; n in N } .> ||> Find all irrational numbers p , q ,p>1 ,q>1 , such that > ||> (1) A[p] U A[q] = N .> ||> If (1) is satisfied, it's true that 1/p + 1/q = 1 ?> ||> Perhaps, please give a counterexample. Thank you, Alex> |> === ===========> |> The Beatty theorem states that if 1/a + 1/b = 1 a,b irrational then> the sequences [ßoor(na),n>=1] and [ßoor(nb),n>=1] make a partition> of the integers.If 1/a +1/b is too small. then some> integers will be missed.If 1/a + 1/b is too large, then> some integers will be repeated.Since you seem to be interested with only the unions,> it is possible to have the unions equal to N:> take for example a<2 and replace b (defined> by 1/a +1/b =1) by c=b/2, that is also larger> than 1.> For example a=sqrt(2) and c=(2+sqrt(2))/2.> Then A[c] contains A[2c]=A[b] obviously and> N=A[a] union A[b] = A[a] union A[c].Cute, but it raises the obvious question. If the union of these setsand the two sets are disjoint, must it be true that 1/p + 1/q = 1?Achava === Subject: Re: JSH: Short argument, modified Decker example> In an attempt at questioning an important conclusion of mine a Rick> Decker, a professor at Hamilton College, made a post with his own> example of a non-polynomial factorization:(5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) where his a's are roots of a^2 - (x - 1)a + 7(x^2 + x).Making the substitution a_2(x) = b_2(x) - 1 to get(5a_1(x) + 7)(5b_2(x) + 2) = 7(25x^2 + 30x + 2) allows both a_1(0) = 0 and b_2(0) = 0, so let that be a requirement.Therefore, both a_1(x) and b_2(x) have some factor of x itself, which> implies a_1(x) has a factor of 7, so using a_1(x) = 7b_1(x) you have(5(7) b_1(x) + 7)(5b_2(x) + 2) = 7(25x^2 + 30x + 2) and dividing both sides by 7 gives(5b_1(x) + 1)(5b_2(x) + 2) = 25x^2 + 30x + 2.At this point it's all straightforward except that you can actually> check at x=1, you find that doesn't work in the ring of algebraic> integers.> Quite right. Rick Decker however showed that, for x = 1, if you divide both sides by 7 in a different way, ((5 a_1(1) + 7)/sqrt(7)) * ((5 a_2(1) + 7)/sqrt(7)) since a_1(1) = sqrt(-14) and a_2(1) = -sqrt(-14), you get (5 sqrt(-2) + sqrt(7))*(-5 sqrt(-2) + sqrt(7)),which is a product of two algebraic integers, and all thecoefficients are also algebraic integers. So: two arguments that say your factorization is wrong: 1. As you noted, your factorization does not produce algebraic integers, and 2. Rick Decker's factorization does, even though you have said it is impossible. But of course you never quote or acknowledge Rick's factorization. So how would you know? Nora B. === Subject: Tarski's World vs. The language of FOLWould anyone, perhaps, know whether the text/software materialcontained in a book by J. Barwise, J. Etchemendy entitled ÔTarski'sWorld' is *ENTIRELY* incorporated by their later work, i.e. ÔThelanguage of first order logic'? I intend to make a purchase and needto know whether to obtain both or just the latter.I shall be most thankful for any kind (and perhaps reviewing)comments.P.S. I know that there is even a newer book by those authors entitled'Hyperproof', but the accompanying software is for Macintosh only.The homepage of all those books:http://www-csli.stanford.edu/hp/#Hyperproof === Subject: Re: Tarski's World vs. The language of FOL> Would anyone, perhaps, know whether the text/software material> contained in a book by J. Barwise, J. Etchemendy entitled ÔTarski's> World' is *ENTIRELY* incorporated by their later work, i.e. ÔThe> language of first order logic'? I intend to make a purchase and need> to know whether to obtain both or just the latter.I shall be most thankful for any kind (and perhaps reviewing)> comments.P.S. > I know that there is even a newer book by those authors entitled> ÔHyperproof', but the accompanying software is for Macintosh only.The homepage of all those books:http://www-csli.stanford.edu/hp/#HyperproofI don't know about these books, but I took a course in introductorylogic using Barwise and Etchemendy's Language Proof and Logic. Thebook included a bunch of software for doing the exercises, and in factincluded the Tarski's World program.I have mixed feelings about the book. It is intended to be used as aosophical logic book (that is, to spend time teaching methods ofproof and do a bit of meta-theory later) as opposed to a mathematicallogic book (where you skim over proving things in a deductive systemand do a lot of meta-theory). Since its osophical in nature, agood 150 pages are spent trying to convince the reader that theconnectives capture what is intuitively meant by conjunction anddisjunction and so on. It's certainly a useful book, and it doeswhat it intends to do very well. Once you actually start doing formalproofs in FOL, the book excels. There are hundreds of good exercises,some involving Tarski's World (which sort of helps to understandquantificational semantics) and others involving Fitch, a deductivesystem (and proof checking software). At this point, I understand thematerial well enough so that I could jump into just about any logicbook and feel comfortable. So LPL seems a bit basic. But if you're abeginner, I would recommend it highly.(Their webpage is www-csli.stanford.edu) === Subject: Re: Tarski's World vs. The language of FOL> The homepage of all those books:http://www-csli.stanford.edu/hp/#HyperproofI don't know about these books, but I took a course in introductory> logic using Barwise and Etchemendy's Language Proof and Logic. The> book included a bunch of software for doing the exercises, and in fact> included the Tarski's World program.I see. > I have mixed feelings about the book. It is intended to be used as a> osophical logic book (that is, to spend time teaching methods of> proof and do a bit of meta-theory later) as opposed to a mathematical> logic book (where you skim over proving things in a deductive system> and do a lot of meta-theory). Since its osophical in nature, a> good 150 pages are spent trying to convince the reader that the> connectives capture what is intuitively meant by conjunction and> disjunction and so on. It's certainly a useful book, and it does> what it intends to do very well. Once you actually start doing formal> proofs in FOL, the book excels. There are hundreds of good exercises,> some involving Tarski's World (which sort of helps to understand> quantificational semantics) and others involving Fitch, a deductive> system (and proof checking software). At this point, I understand the> material well enough so that I could jump into just about any logic> book and feel comfortable. So LPL seems a bit basic. Very interesting. Thank you.> But if you're a beginner, I would recommend it highly.Thank you for writing. === Subject: Re: Tarski's World vs. The language of FOL> Would anyone, perhaps, know whether the text/software material> contained in a book by J. Barwise, J. Etchemendy entitled ÔTarski's> World' is *ENTIRELY* incorporated by their later work, i.e. ÔThe> language of first order logic'? I intend to make a purchase and need> to know whether to obtain both or just the latter.I have the later book, and it seems to have a complete implementation ofTarski's World.> I shall be most thankful for any kind (and perhaps reviewing)> comments.I like Language, Proof, and Logic a lot. The book includes a log-in code(which I so far haven't used) for submitting answers to the exercises to aserver at Stanford for grading. Cool. I'll probably have my son use that inthe next year or so.> http://www-csli.stanford.edu/hp/#HyperproofYes, quite an interesting Web site. === Subject: Re: Tarski's World vs. The language of FOL> I have the later book, and it seems to have a complete implementation of> Tarski's World.Right. Thank you. That's what I needed to know.> I shall be most thankful for any kind (and perhaps reviewing)> comments.I like Language, Proof, and Logic a lot. I haven't heard of this title. Will look through it. Thank you.> The book includes a log-in code> (which I so far haven't used) for submitting answers to the exercises to a> server at Stanford for grading. Cool. I'll probably have my son use that in> the next year or so.I won't ask you how old he is. :-) > http://www-csli.stanford.edu/hp/#HyperproofYes, quite an interesting Web site.Thank you for writing. === Subject: Re: Sizes of classes|Consider the statement that there exists an ordinal |enumeration of V, the class of all sets. Is this statement |independent of ZFC?Goedel proved that if ZF is consistent, then so is ZF+V=L,and ZF+V=L implies that a certain formula is a well-orderingof the universe. (Goedel's class L has a natural correspondencewith the ordinals.) So your statement is at least consistent.I think, although I don't remember for sure, that it is knownto be independent.Keith Ramsay === Subject: Action Device: Extremely Sorry..> So I have built this Action Device. It is on ßoor of my room. I> pulled the springs to generate enough restoring force.> I marked the position of upper end of bolt. Closing one eye, I pointed> on this mark and I rotated the nut.> The upper end of bolt i.e. point D is moving in upward direction as I> predicted here..> http://www.geocities.com/inertial_propulsion> First Result: It Does Work!> Play this song in full volume...> so gayaa ye jahA.n so gayaa aasamaan> so gayI hai sArI manzile.n> o sArI manzile.n so gayaa hai rastaa> so gayaa ye jahA.n so gayaa aasamaan> Enjoy!> -Abhi.Action Device-Second Test: It does work !!Extremely sorry. This second test proved out to be hoax. -Abhi. === =====================Mohabbat baDe kam ki cheej hai === ========================Subject: Re: Action Device: Extremely Sorry..> Action Device-> Second Test: It does work !!Extremely sorry. This second test proved out to be hoax.ing imbecile wog. One plus one never equals three, not even forvery large values of one. 1) First Law of Thermodynamics: You cannot win. 2) Second Law of Thermodynamics: You can only break even on a verycold day. 3) Third Law of Thermodynamics: It never gets that cold.There is nothing sacred about any aspect of thermodynamics, ignorantstupid wog. You can win with the Casimir effect, you can erect smallsystems in which entropy spontaneously decreases, and negative tempskelvin are trivially created in lasing media, magnetic resonancespectrometries, and deep cryogenic refrigeration strategies.However... jackass...When you tally the bottom line, thermodynamics always wins. An ounceof water has 10^27 molecules. You cannot beat statistics of largenumbers even with 300,00,000 million idiot gods or theBelousov-Zhabotinsky reaction. You are a waste of ßesh and hot air,and you are psychotically determined to stay that way.Crack a book, moron, and become more than the smear that your arenow.GoogleBelousov-Zhabotinsky 5210 hitsHopeless wretched braindead wog. --Uncle Alhttp://www.mazepath.com/uncleal/qz.pdfhttp:// www.mazepath.com/uncleal/eotvos.htm (Do something naughty to physics) === Subject: Re: Action Device: Extremely Sorry..> Action Device-Second Test: It does work !!Extremely sorry. This second test proved out to be hoax.ing imbecile wog. One plus one never equals three, not even for> very large values of one. 1) First Law of Thermodynamics: You cannot win.> 2) Second Law of Thermodynamics: You can only break even on a very> cold day.> 3) Third Law of Thermodynamics: It never gets that cold.There is nothing sacred about any aspect of thermodynamics, ignorant> stupid wog. You can win with the Casimir effect, you can erect small> systems in which entropy spontaneously decreases, and negative temps> kelvin are trivially created in lasing media, magnetic resonance> spectrometries, and deep cryogenic refrigeration strategies.However... jackass...When you tally the bottom line, thermodynamics always wins. An ounce> of water has 10^27 molecules. You cannot beat statistics of large> numbers even with 300,00,000 million idiot gods or the> Belousov-Zhabotinsky reaction. You are a waste of ßesh and hot air,> and you are psychotically determined to stay that way. It's simple to beat the statistics of water. Since water only exists on Earth at STP. So it only exists in pockets. But hydrogen exists everywhere, and anywhere. Since even sand will beat the statistics of a measely 10^90 assorted hydrogen atoms, that's obviously why we send the stupider Psychologists to Russia, Germany, and Korea, and keep the ones who are successful robot people for ourselves. And let the scientists do the sub-aether radar paperwork.Crack a book, moron, and become more than the smear that your are> now.Google> Belousov-Zhabotinsky 5210 hitsHopeless wretched braindead wog. === Subject: Re: Action Device: Extremely Sorry..> Action Device-Second Test: It does work !!> Extremely sorry. This second test proved out to be hoax.ing imbecile wog. One plus one never equals three, not even for> very large values of one. 1) First Law of Thermodynamics: You cannot win.> 2) Second Law of Thermodynamics: You can only break even on a very> cold day.> 3) Third Law of Thermodynamics: It never gets that cold.There is nothing sacred about any aspect of thermodynamics, ignorant> stupid wog. You can win with the Casimir effect, you can erect small> systems in which entropy spontaneously decreases, and negative temps> kelvin are trivially created in lasing media, magnetic resonance> spectrometries, and deep cryogenic refrigeration strategies.However... jackass...When you tally the bottom line, thermodynamics always wins. An ounce> of water has 10^27 molecules. You cannot beat statistics of large> numbers even with 300,00,000 million idiot gods or the> Belousov-Zhabotinsky reaction. You are a waste of ßesh and hot air,> and you are psychotically determined to stay that way. It's simple to beat the statistics of water. > Since water only exists on Earth at STP. So> it only exists in pockets. But hydrogen exists> everywhere, and anywhere. Since even sand will beat> the statistics of a measely 10^90 assorted > hydrogen atoms, that's obviously why we send the stupider > Psychologists to Russia, Germany, and Korea, > and keep the ones who are successful > robot people for ourselves. And let the> scientists do the sub-aether radar paperwork.> All of that follows trivially from Von Neumann's theory of QM, since it's well-know that Von Neumann was a chemist, rather than a photon specialist.> Crack a book, moron, and become more than the smear that your are> now.Google> Belousov-Zhabotinsky 5210 hitsHopeless wretched braindead wog. === Subject: Re: Graph Theory: Adjacency Matrices for Graphs without 3-cycles> One possible matrix for a graph with no 3-cycles is 0 1 0 1 0 1 0 1 ...> 1 0 1 0 1 0 1 0 ...> 0 1 0 1 0 1 0 1 ...> 1 0 1 0 1 0 1 0 ... > etc.This appears to be a sufficient condition for no 3-cycles. Is it also> a necessary condition?A necessary and sufficient condition for a matrix M = (m_ij) to be the> adjacency matrix of some (simple) graph G is that it be a symmetric 0-1> matrix with 0's on the diagonal.A necessary and sufficient condition that G not contain 3-cycles is> that m_ij m_jk m_ki be 0 for all i,j,k, which is equivalent to saying> that M^3 has all 0's on the diagonal, which is in turn equivalent to> saying that the sum of the cubes of the eigenvalues of M is 0. (When> M is a 3x3 matrix, this is in turn equivalent to det M = 0, but I don't> think there is a similarly simple characterization in general.)-Jim FerryI don't think that the matrix of a graph is a true matrix. The 1'sand 0'sdo not represent values; but merely the presence or absence of anedge. The matrix proposed represents the no 3-cycle graphs with themost edges for a given |V|.I do not understand matrix nomenclature and Ôeigenvalues wellenough to comment on your observations. Sorry; J. === Subject: Re: Graph Theory: Adjacency Matrices for Graphs without 3-cycles> One possible matrix for a graph with no 3-cycles is> 0 1 0 1 0 1 0 1 ...> 1 0 1 0 1 0 1 0 ...> 0 1 0 1 0 1 0 1 ...> 1 0 1 0 1 0 1 0 ... > etc.> This appears to be a sufficient condition for no 3-cycles. Is it also> a necessary condition?A necessary and sufficient condition for a matrix M = (m_ij) to be the> adjacency matrix of some (simple) graph G is that it be a symmetric 0-1> matrix with 0's on the diagonal.A necessary and sufficient condition that G not contain 3-cycles is> that m_ij m_jk m_ki be 0 for all i,j,k, which is equivalent to saying> that M^3 has all 0's on the diagonal, which is in turn equivalent to> saying that the sum of the cubes of the eigenvalues of M is 0. (When> M is a 3x3 matrix, this is in turn equivalent to det M = 0, but I don't> think there is a similarly simple characterization in general.)-Jim FerryI don't think that the matrix of a graph is a true matrix. The 1's> and 0's> do not represent values; but merely the presence or absence of an> edge. The matrix proposed represents the no 3-cycle graphs with the> most edges for a given |V|. > I do not understand matrix nomenclature and Ôeigenvalues well> enough to comment on your observations. Sorry; J.I don't know what's not true about it. The adjacency matrix of a graphwith n vertices (numbered 1,2,...,n) is the matrix with entries M_{ij} = 1if is an edge, 0 if it is not. Yes, 1 and 0 represent presence orabsence of an edge. So what? You can still treat this matrix as youwould any other n x n symmetric matrix. Yes, your graph has the most edges of any |V|-vertex graph with no 3-cycles.But not every graph without 3-cycles is a subgraph of it.Consider e.g. the 5-cycle graph 1--2 | | 3| /5--4 (view in fixed-width font)This has the adjacency matrix [ 0 1 0 0 1 ][ 1 0 1 0 0 ][ 0 1 0 1 0 ][ 0 0 1 0 1 ][ 1 0 0 1 0 ]It is not a subgraph of any of your graphs (which don't have any 5-cycles).Department of Mathematics http://www.math.ubc.ca/~israelUniversity of British ColumbiaVancouver, BC, Canada V6T 1Z2 === Subject: Re: Graph Theory: Adjacency Matrices for Graphs without 3-cycles>One possible matrix for a graph with no 3-cycles is> 0 1 0 1 0 1 0 1 ...> 1 0 1 0 1 0 1 0 ...> 0 1 0 1 0 1 0 1 ...> 1 0 1 0 1 0 1 0 ... > etc.>This appears to be a sufficient condition for no 3-cycles. Is it also>a necessary condition?What do you mean by This? You gave one rather special class of > examples. There are many others. > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia > Vancouver, BC, Canada V6T 1Z2This is the general form of the matrix of the graph with the mostedges for a given ÔV'. === Subject: Re: Graph Theory: Adjacency Matrices for Graphs without 3-cycles> One possible matrix for a graph with no 3-cycles is> 0 1 0 1 0 1 0 1 ...> 1 0 1 0 1 0 1 0 ...> 0 1 0 1 0 1 0 1 ...> 1 0 1 0 1 0 1 0 ...> etc.> This appears to be a sufficient condition for no 3-cycles. Is it also> a necessary condition?What do you mean by this? You have indicated an infinite matrix.This is the general form of the matrix. If |V|=5; the proper matrix isthe first five rows and the first five columns. Use the first 20 rowsand 20 columns if |V|= 20.> Do you think this is the only possible adjacency matrix for a graph with no> 3-cycles?The suggested matrix represents a graph with no 3-cycles that has themaximum number of edges.Or maybe you're asking if any graph without 3-cycles is bipartite?> The K3,3 graph is bipartite and has no 3-cycles. > Consider the 5-cycle.The 5-cycle graph does not have the maximum number edges. Its matrixis;0 1 0 0 11 0 1 0 00 1 0 1 00 0 1 0 11 0 0 1 0 The msximum edge matrix for |V| = 5 is;0 1 0 1 01 0 1 0 10 1 0 1 01 0 1 0 10 1 0 1 0 === Subject: Re: Graph Theory: Adjacency Matrices for Graphs without 3-cycles 3QLpj-NoP*NzsIC,boYU]bQ]H'y<#4ga3$21:> Do you think this is the only possible adjacency matrix for a graph with no> 3-cycles?The suggested matrix represents a graph with no 3-cycles that has the> maximum number of edges.You mean Turan's theorem?-- David Eppstein http://www.ics.uci.edu/~eppstein/Univ. of California, Irvine, School of Information & Computer Science === Subject: Re: Graph Theory: Adjacency Matrices for Graphs without 3-cycles> Do you think this is the only possible adjacency matrix for a graph with no> 3-cycles?The suggested matrix represents a graph with no 3-cycles that has the> maximum number of edges.You mean Turan's theorem?I do not understand Turan's theorem. === Subject: Re: Graph Theory: Adjacency Matrices for Graphs without 3-cycles 3QLpj-NoP*NzsIC,boYU]bQ]H'y<#4ga3$21:> The suggested matrix represents a graph with no 3-cycles that has the> maximum number of edges.You mean Turan's theorem?I do not understand Turan's theorem.Turan's theorem tells you the maximum number of edges that can be in a graph with no 3-cycles (or more generally a graph with no complete k-vertex subgraph) and what a graph achieving that maximum has to look like.For the particular case of an n-vertex graph with no 3-cycles, the result is that there can be at most ceiling(n/2)*ßoor(n/2) edges, and that any graph with exactly this many edges is a complete bipartite graph connecting subsets of ceiling(n/2) and ßoor(n/2) vertices.The adjacency matrix of such a graph can be arranged to form the alternating pattern you showed.-- David Eppstein http://www.ics.uci.edu/~eppstein/Univ. of California, Irvine, School of Information & Computer Science === Subject: Examples Of Non-Measurable SetsA colleague of mine recently asked me a question. In math jargon, hewanted to know why probability set functions were not defined for allsubsets of the sample space.So I explained the non-measurable set in one of my analysis texts,Royden I think. This example considers [0, 1) partitioned intoequivalence classes, where two elements of [0, 1) are equivalentiff their difference is a rational number. A set E, defined to bea set consisting of one and only one element from each equivalenceclass, can be shown to be non-measurable. The existence of E followsfrom the axiom of choice.My question: are other examples of a non-measurable set ever invokedin teaching measure theory? What other examples?-- Try http://csf.colorado.edu/pkt/pktauthors/Vienneau.Robert/ Bukharin.html To solve Linear Programs: .../LPSolver.htmlr c A game: .../Keynes.html v s a Whether strength of body or of mind, or wisdom, or i m p virtue, are found in proportion to the power or wealth e a e of a man is a question fit perhaps to be discussed by n e . slaves in the hearing of their masters, but highly @ r c m unbecoming to reasonable and free men in search of d o the truth. -- Rousseau === Subject: Re: Examples Of Non-Measurable Sets>A colleague of mine recently asked me a question. In math jargon, he>wanted to know why probability set functions were not defined for all>subsets of the sample space.>So I explained the non-measurable set in one of my analysis texts,>Royden I think. This example considers [0, 1) partitioned into>equivalence classes, where two elements of [0, 1) are equivalent>iff their difference is a rational number. A set E, defined to be>a set consisting of one and only one element from each equivalence>class, can be shown to be non-measurable. The existence of E follows>from the axiom of choice.>My question: are other examples of a non-measurable set ever invoked>in teaching measure theory? What other examples?There are other examples. One of them, which uses less non-constructive mathematics, is to divide the equivalenceclasses of the irrationals into pairs, such that the sumfrom one member of the pair and the other is rational. This only needs the axiom of choice for two-element sets,which is known to be much weaker than full choice.The Banach-Tarski paradox uses even less choice, havingbeen shown to follow from the Hahn-Banach theorem. -- This address is for information only. I do not claim that these viewsare those of the Statistics Department or of Purdue University.Herman Rubin, Department of Statistics, Purdue Universityhrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: Examples Of Non-Measurable Sets>A colleague of mine recently asked me a question. In math jargon, he>wanted to know why probability set functions were not defined for all>subsets of the sample space.>So I explained the non-measurable set in one of my analysis texts,>Royden I think. This example considers [0, 1) partitioned into>equivalence classes, where two elements of [0, 1) are equivalent>iff their difference is a rational number. A set E, defined to be>a set consisting of one and only one element from each equivalence>class, can be shown to be non-measurable. The existence of E follows>from the axiom of choice.>[...]A tangential response: You and your friend might be interestested in the related question I posed here in August 2002:>If we accept the axiom of choice, then there exists a subset S of the unit interval [0,1] which is not Lebesgue measurable. I propose the following wager to you: We will generate a value X from the uniform probability distribution on [0,1]. If X is in S, I will pay you $1. How much>are you willing to pay to take this wager? What would be the fair>value?-- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: Examples Of Non-Measurable Sets the related question I posed here in August 2002:>If we accept the axiom of choice, then there exists a subset S of the unit interval [0,1] which is not Lebesgue measurable. I propose the following wager to you: We will generate a value X from the uniform probability distribution on [0,1]. If X is in S, I will pay you $1. How much>are you willing to pay to take this wager? What would be the fair>value?This is a psychological question, not a mathematical one. Bydefinition, the fair value of a bet like this depends on theprobability of winning, so the event x in S MUST be measurable forthis to be meaningful. However, there is a sense of fair value whichdepends on the psychology of the individual. This is different fromthe mathematical fair value. Of course the remarks above as to theimpossibility of carrying out this wager in real life are correct. Another psychological use of fair value arises in the St. sburgparadox where we imagine you tossing a coin and receiving 2^n dollarsif you toss n heads before the first tail. The expectation of thewinnings is infinite so there is no fair value but many people wouldbe willing to be the banker for such a bet if they were paid enough. I, for example, would do it for $1,000,000,000,000. That is, you payme one trillion dollars, I'll give you an ordinary quarter to toss andI'll pay you 2^n dollars if you toss n heads in a row before the firsttail. You'd only have to toss 40 heads in a row to bankrupt me. Idoubt you have the cash to make this bet but maybe you could join withthe governments of a few countries to form a syndicate? === Subject: Re: Examples Of Non-Measurable Sets> A colleague of mine recently asked me a question. In math jargon, he> wanted to know why probability set functions were not defined for all> subsets of the sample space.So I explained the non-measurable set in one of my analysis texts,> Royden I think. This example considers [0, 1) partitioned into> equivalence classes, where two elements of [0, 1) are equivalent> iff their difference is a rational number. A set E, defined to be> a set consisting of one and only one element from each equivalence> class, can be shown to be non-measurable. The existence of E follows> from the axiom of choice.> Let ((0,1],B,m) be the probability space with the Borel sigma-field B and the Lebesgue probability measure m. The non-measureable set you refer to, call it V, is not in B. However, one can extend m to a probabilitymeasure m' on the sigma-field generated by the union of B and {V}; that is, B'= sigma(B U {V}).Relative to ((0,1], B', m') the set V is now measureable. Any set not in the sigma-field can be thrown in, in the same manner. But you can't put allof them in. One can show: There is no countably additive probability measure P on the powerset of (0,1] such that P(x)=0 for all xin(0,1]. Since m(x)=0 for all xin(0,1], m can not be extended to a countablyadditive probability measure on the powerset of (0,1]. One can, howeverextend any probability measure to a *finitely* additive probably measure on the collection of all sets. What I'm really trying to say is expressed by D. H. Fremlin in the firstchapter of his Treatise, Measure Theory--he says: Nearly every time I am consulted by a non-specialist who wants to be told a theorem which will solve his problem, I am reminded that pure mathematics in general, and analysis in particular, does not lie in the *theorems* but in the *proofs*. In so far as I have been successful in answering such questions, it has usually been by making a trißing adjustment to a standard argument to produce a non-standard theorem. The ideas are in the details.k.j.h. === Subject: Re: Examples Of Non-Measurable Sets> One can show: There is no countably additive probability measure P on the powerset of > (0,1] such that P(x)=0 for all xin(0,1]. No one can't. Or at least one would become famous if one did. Ulamproved this assuming the continuum hypothesis. But the existence ofsuch (other then identically zero) would mean c is a real-valuedmeasurable cardinal. Most set theorists believe this is consistentwith ZFC (but of course not provable in ZFC).-- http://www.math.ohio-state.edu/~edgar/ === Subject: Re: Examples Of Non-Measurable Sets> My question: are other examples of a non-measurable set ever invoked> in teaching measure theory? What other examples?The Bernstein sets are popular, too. That is a set E such that bothE and its complement meet every uncountable compact set. === Subject: Re: Examples Of Non-Measurable SetsMy question: are other examples of a non-measurable set ever invoked> in teaching measure theory? What other examples?The Bernstein sets are popular, too. That is a set E such that both> E and its complement meet every uncountable compact set.There is also Sierpinski set - nonmeasurable set on the plane such that:no three points of this set are noncolinear,but it meets every closed set of a positive plane measure.You can find details eg. in:Oxtoby Measure and Category (in some Springer series)-- === Subject: Re: graph theory proof>> Problem:>> Use the principle of induction to show that if G = (V,E) is a graph with |V|>> = 2m and G has no 3 -cycles, then |E| <= m^2> I am guessing im supposed to use induction on m. But i dont see how. And>> how does it matter that G has no 3-cycles ?>The typical adjacency matrix of a graph with no 3-cycles looks like>this> 0 1 0 1 0 1> 1 0 1 0 1 0 > 0 1 0 1 0 1 > 1 0 1 0 1 0> 0 1 0 1 0 1 > 1 0 1 0 1 0Are you claiming that a graph with no 3-cycles must be bi-partite? > I didn't realize that I had described a bi-partite graph. If that istrue, then I claim that the largest no 3-cycles graph must bebi-partite.>V even and |E|=(|V|^2)/4-0.25 for V odd.An n cycle with n<>3 has no 3 cycles. The adjacency matrix of the 5> cycle is 0 1 0 0 1> 1 0 1 0 0> 0 1 0 1 0> 0 0 1 0 1> 1 0 0 1 0 > which does not look like yours, and is not bipartite... But my matrix represents a 5-vertex graph with 6 edges and no3-cycles.>Induction on m = 0, 1, 2, ... n is easy.>The number of edges added for each n is = 2*n-1. You mean, the maximum number of edges you can add when going from n-1> vertices to n vertices is 2n-1? How do you prove that?I meant Ôm' not Ôn'. Sorry. Proof of the allegation is predicated onthe validity of the matrix. Jones.PS: I thought you were not going to read any more of my postings. === Subject: Re: graph theory proof Adjunct Assistant Professor at the University of Montana.> Problem:> Use the principle of induction to show that if G = (V,E) is a graph with |V|> = 2m and G has no 3 -cycles, then |E| <= m^2I am guessing im supposed to use induction on m. But i dont see how. And> how does it matter that G has no 3-cycles ?>>The typical adjacency matrix of a graph with no 3-cycles looks like>>this>> 0 1 0 1 0 1>> 1 0 1 0 1 0 >> 0 1 0 1 0 1 >> 1 0 1 0 1 0>> 0 1 0 1 0 1 >> 1 0 1 0 1 0> Are you claiming that a graph with no 3-cycles must be bi-partite? >>I didn't realize that I had described a bi-partite graph. If that is>true, then I claim that the largest no 3-cycles graph must be>bi-partite.You are not being clear, and you are not being precise.Do you mean: Let M(n) be the maximum number of edges among all graphs with n vertices which have no 3-cycles. If G is a graph with n vertices which has no 3-cycles, and G has M(n) edges, then G is bipartite.?In fact, you were claiming much more: you were claiming that G wouldbe bipartite and that the two parts would be as close to equal sizeas possible.If either of this is what you are claiming, then you need to ->prove<-it. And you also need to realize that not every graph which has no3-cycles is a subgraph of a bipartite graph. Talking about typicalgraphs may make sense if you are going to be proving stuff about randomgraphs, but you were not doing so, and even then you need to be veryclear about just what typical means. You were not clear.>>V even and |E|=(|V|^2)/4-0.25 for V odd.> An n cycle with n<>3 has no 3 cycles. The adjacency matrix of the 5>> cycle is> 0 1 0 0 1>> 1 0 1 0 0>> 0 1 0 1 0>> 0 0 1 0 1>> 1 0 0 1 0>> which does not look like yours, and is not bipartite...>But my matrix represents a 5-vertex graph with 6 edges and no>3-cycles.So? The point is that your claims were incorrect as stated. If whatyou ->meant<- was something else, then you should say what youmean. What you ->said<- was false.>PS: I thought you were not going to read any more of my postings.I usually only stick people into kill-files for a limited amount oftime, at least at first.-- === Subject: Re: Modified Decker example, reference> Posters trying to dispute the algebra when presented with the basic> algebra claim that it's the sum (5b_2(x) + 2) that matters, but that> is refuted by noting that the 2 in that expression is a factor of 7(2)> in> 7(25x^2 + 30x + 2)> as the constants are NOT a product of the full sums, but are in fact> products of the constants 7 and 2 within those sums, so by trying to> include variable expressions they are relying on claims insupportable> mathematically.Consider (4/3 + 2/3) (9/2 + 3/2) = (4/3)(9/2) + (4/3)(3/2) + (2/3)(9/2) + (2/3)(3/2)> = 6 + 2 + 3 + 1 = 12> That's ok in the field of algebraic numbers, but doesn't exist in the> ring of algebraic integers.Here we have (a+b)(c+d) = ab + ac + bc + bd = gThe algebra doesn't care what the ring is. and (a+b), (c+d), ab, ac, bc, bd, and g are all integers.> However none of a, b, c or d is an integer.> You sum first for your example, or you're out of the ring.But what about (x+2)/3?Now here's a pop quiz, for algebraic integer x, is (x+2)/3 in the ring> of algebraic integers?Can you *add* first to say that it always is?So the fact that the partial sums as well as full sums are integers,> does not mean that a,b,c, and/or d must be integers.In particular, the fact that the product of the constant terms,> b and d, is an integer does not mean that b and d must be integers. - William HughesThink about (x+2)/3.> High-school teacher: Now class, here is a little pop quiz on integers. Question 1: Is 13 + 1/4 an integer? James: I know teacher! I know! High-school teacher: Yes, James? James: It's not. An integer plus a fraction cannot equal an integer. High-school teacher: Correct. Here's Question 2: is 1/2 + 1/3 an integer? James: I know, teacher! I know! High-school teacher: Yes, James? James: It's not, because it equals 5/6, which isn't a counting number. High-school teacher: Correct, James. Now here's Question 3: Is 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 57/140 an integer? James: I know, teacher! I know! High-school teacher: Yes, James? James: It's not, because all the terms are fractions. They can't possibly add up to an integer! High-school teacher: Thank you, James. Now the next problem involves *algebraic* integers ... === Subject: Re: Modified Decker example, reference> High-school teacher:> Now class, here is a little pop quiz on integers.> Question 1: Is 13 + 1/4 an integer?> James:> I know teacher! I know!> High-school teacher:> Yes, James?> James:> It's not. An integer plus a fraction cannot equal an integer.> High-school teacher:> Correct. Here's Question 2: is 1/2 + 1/3 an integer?> James:> I know, teacher! I know!> High-school teacher:> Yes, James?> James:> It's not, because it equals 5/6, which isn't a counting> number.> High-school teacher:> Correct, James. Now here's Question 3:> Is 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 57/140 an integer?> James:> I know, teacher! I know!> High-school teacher:> Yes, James?> James:> It's not, because all the terms are fractions. They can't> possibly add up to an integer!> High-school teacher:> Thank you, James. Now the next problem involves *algebraic*> integers ...Nora,I think you have nailed a weakness in James' thought processes which drive him to the ridiculous. There are many waysto represent numeric quantities and many tools available to find solutions for algebraic problems.Maybe he has forgotten that the quadratic formula, which provides solutions to quadratic equations, uses radicals--even though the solutions might consist of integers and could be found by factoring. I would guess that in his mind,using radicals sometimes pushes you out of the ring of integers and is therefore an invalid tool for solving somequadratic equations. A similar objection might be made against imaginary numbers which often aid in solving problemsinvolving reals.--There are two things you must never attempt to prove: the unprovable -- and the obvious.--Democracy: The triumph of popularity over principle.--http://www.crbond.com === Subject: Re: Modified Decker example, reference> Posters trying to dispute the algebra when presented with the basic> algebra claim that it's the sum (5b_2(x) + 2) that matters, but that> is refuted by noting that the 2 in that expression is a factor of 7(2)> in> 7(25x^2 + 30x + 2)> as the constants are NOT a product of the full sums, but are in fact> products of the constants 7 and 2 within those sums, so by trying to> include variable expressions they are relying on claims insupportable> mathematically.Consider (4/3 + 2/3) (9/2 + 3/2) = (4/3)(9/2) + (4/3)(3/2) + (2/3)(9/2) + (2/3)(3/2)> = 6 + 2 + 3 + 1 = 12> That's ok in the field of algebraic numbers, but doesn't exist in the> ring of algebraic integers.> Good one. This sheds light on your thinking. Theexpression (4/3 + 2/3)is not an *expression* in the algebraic integers. That is whatyou mean and you are correct. However the *value* of (4/3 + 2/3) clearly, trivially IS an algebraic integer. The way a number is represented symbolically is not theimportant thing. That is essentially cosmetic. You canrepresent a rational number using irrationals; for example, (1/2 - sqrt(3)) + (1/3 + sqrt(3)) = 5/6,even though neither of the terms in parentheses is rational. Mathematics must and do think not about representations ofnumbers, but about their actual values. Thinking that numbersmust be represented in certain restricted forms gets you intoimmediate trouble, as has happened to you here. You are talking about the symbolic representation; the rest of us are talking about the actual value. It goes back to the fact thatyou are really comfortable with math at only the high-schoollevel. That is why you do not trust any factorization thatyou cannot see by inspection, as in high-school algebra problems.That is why you constantly return to simplistic reducible examples,like x^2 + 3*x + 2, etc.: anything deeper than that is just outof reach for you.> Here we have (a+b)(c+d) = ab + ac + bc + bd = gThe algebra doesn't care what the ring is.> and (a+b), (c+d), ab, ac, bc, bd, and g are all integers.> However none of a, b, c or d is an integer.> You sum first for your example, or you're out of the ring.> Sum first or sum last, either way, you end up with thesame numerical answer. The factors are integers, eventhough they are written as sums of non-integers. Again,the representation of a number is not its essence. > But what about (x+2)/3?Now here's a pop quiz, for algebraic integer x, is (x+2)/3 in the ring> of algebraic integers?> Yes, if x = 1, 4, 7, 10, .... In general if x = 3*k + 1where k is any algebraic integer.> Can you *add* first to say that it always is? Add first, then divide, or divide first, then add, youget the same answer! It's an algebraic integer if x is equal to 3*k + 1 where k is an algebraic integer. > So the fact that the partial sums as well as full sums are integers,> does not mean that a,b,c, and/or d must be integers.In particular, the fact that the product of the constant terms,> b and d, is an integer does not mean that b and d must be integers. - William HughesThink about (x+2)/3.> Duh ... OK. See above. Now what? Nora B. === Subject: Re: Modified Decker example, reference> *Successfully* refuting your research - and why don't you at>> least quote Decker's conclusion? Is this a propaganda tactic ->> not stating the opponent's argument because other people might>> actually see that it makes sense ? No.>> Why is it so important to keep noting that he posts from>> Hamilton College? Might it not be more important to quote>> the *substance* of what he said, rather than where he posts>> from ???Why ask why?I just love these examples where you refuse to give meaningful answersto perfectly reasonable questions -- because, of course, you can't.Like when you were blathering nonsense about the Riemann Hypothesis,and you were asked explain what RH *is* and you just said No -- again,because you hadn't a clue how to give a meaningful answer.> Oh yeah, and why don't you **** off as I strongly suggested before?I'm sort of tired of you chasing after my posts like a lost puppy.I also love watching you try to chase away your critics wheneveryour evasions and excuses wear particularly thin. You're hopelesslyoutclassed, and you know it.-- Wayne Brown (HPCC #1104) | When your tail's in a crack, you improvisefwbrown@bellsouth.net | if you're good enough. Otherwise you give | your pelt to the trapper.e^(i*pi) = -1 -- Euler | -- John Myers Myers, Silverlock === Subject: Re: What is JSH's problem?Very very funny. I guess what JSH calls a parrot is someone who believes adefinition and uses it word for word (oh, the mindless little twit). Maybethis is the unfortunate price we pay for the new math : everything isrelative, _including_ defns!Jack> Tim Smith> I'm trying to figure out what the fundamental thing is the JSH doesn't> seem> to grasp, and I *think* it is this, but his argument is so poorly> presented> that I'm not sure. Is this really the trivial point he is missing?> Essentially, he seems to think that:> Given a ring R, with subring S, if ab=c, with a, b, c in S, then if> xy=c, x,y in R, x and y must be in S.> Is that it, or is there a more subtle false statement he thinks is true?> See e.g.> http://mathquest.com/discuss/sci.math/a/m/363567/363567> Harris has been doing this sort of thing for eight years.> LH === Subject: Re: Inequality Poser> Let a, b, and c = 1> 27[(1+1)(1+1)(1+1)]^2 >= 64 *1*1*1 (1+1+1)^3> 27* [2*2*2]^2 >= 64 * 27> okay, let a, b, and c = .5> 27 [1]^2 >= 64*.5*.5*.5 * 1.5^3> Dagnabbit.If a=b=c, then LHS = 27 [ (2a)^3 ]^2 = 27 * 64 * a^6 RHS = 64 a a a (3a)^3 = 64 * a^3 * 27 * a^3 = 27 * 64 * a^6The restriction on a,b,c could be loosened a bit; if any of them is zero, then LHS >= 0, and RHS == 0.equality when a=b=c, or abc=0.>> Show that, for real a,b,c > 0, 27[(a + b)(b + c)(c + a)]^2 >= 64abc(a +>> b + c)^3It's possible to rearrange terms to getShow: [((a+b)/2) ((a+c)/2) ((b+c)/2)]^2 >= abc ((a+b+c)/3)^3from here.-- ------------------------Mark Jeffrey Tilfordtilford@ugcs.caltech.edu === Subject: Re: Is this series converge ? ETAtAhQ03w9T9o4mOFm68hbR2tAGdCCgcAIVAI/ XXd7DO2JcHk60ULWqs85gP4cg I can't tell whether it converes or not. I need to know the formula forthe nth term, and in principle a finite set of terms admits an infinitenumber of possibilities for the nth term.So I'm GUESSING that you've rearranged the terms in the series sum(n= 0 to inf.) (1/2^n)and dropped off one term (1/2). If tht is the case, then one arguesthat the above series is absolutely convergent, and that guarantees anyrearrangement will convrege to the same thing. Except here the 1/2 islost so your sum is 1+1/2 instead of 2. === Subject: Re: Is this series converge ?In sci.math, Ignacio Larrosa Ca.96estro:> En el mensaje:hksre1-cli.ln1@lexi2.athghost7038suus.net,> The Ghost In The Machine escribi.97:>> In sci.math, hinoon>> :>1+1/8+1/4+1/32+1/16+1/128+1/ 64+...>I try to check this by detecting n-th term and applying>root test.>But, I don't know what is n-th term.>> It would appear that this series is:>> t_0 = 1>> t_n = 1/2^(n+2) if n is odd>> = 1/2^n if n is even>> Since this series is absolutely convergent (0 <= t_n <= 1/2^n,>> terms of a known absolutely convergent series),>> we can play all sorts of games. The simplest is arguably to>> split it into three parts:>> 1>> 1/8 + 1/32 + ... = 1/8 * (1 + 1/4 + 1/16 + ...) = 1/24>> 1/4 + 1/16 + ... = 1/4 * (1 + 1/4 + 1/16 + ...) = 1/121/8 + 1/32 + ... = 1/8 * (1 + 1/4 + 1/16 + ...) = 1/8(1/(1 - 1/4) = 1/6> 1/4 + 1/16 + ... = 1/4 * (1 + 1/4 + 1/16 + ...) = 1/4(1/(1 - 1/4) = 1/3Whoops...yep, the brain definitely needed more coffee/rest/sex/somethingthere...mea culpa. :-)I will not forget the 1 in a geometric sum.I will not forget the 1 in a geometric sum.I will not forget the 1 in a geometric sum....[46 repetitions snipped]I will not forget the 1 in a geometric sum.:-)>> and therefore the sum is 1 + 1/24 + 1/12 = 1 + 1/8 = 1.125.and therefore the sum is 1 + 1/6 + 1/3 = 1 + 1/2 = 3/2.Alternativally, as the series is absolutely convergent, its sum isSum(1/2^n, n, 0, inf) - 1/2 = 1/(1 - 1/2) - 1/2 = 2 - 1/2 = 3/2> Yeppers.-- #191, ewill3@earthlink.netIt's still legal to go .sigless. === Subject: Re: fundamental period ETAsAhRebDJg2QhfMthryhAfsvdZYEeGBAIUfxKPLme/IHtyUE71EKynVXG+ fTc= The second definitiion is appropriate for sequences where perforce theindependent variable is an integer. The first definition is reall ageneral one.All periods must be multiples of teh fundamental period, so if there isan irrational period then there cannot be a rational period.But you can consider some funcrtions to be superpositions of periodicfunctions with different periods. Search for quasiperiodic funcitons. === Subject: Re: Square root extractionzeroremovethistype@yahoo.com.spamblock (or somebody else of the same name) .> eg (pseudocode)> z=sqrt(y);> if z=int(z) then (y was the perfect square sought) else (it wasn't -> set up next y and test again.)Dodgy! Set this up on any real processor then for sufficiently large integer N that processor will be unable to distinguish sqr(N^2+1) from N. Your if-test will then pass and you would be led to conclude that N^2+1 is a perfect square when by construction it cannot be.A quick test on my HP97 calculator shows it gives the incorrect conclusion for N >= 100000, indicating that 10,000,000,001 is a perfect square.-- Paul TownsendI put it down there, and when I went back to it, there it was GONE!Interchange the alphabetic elements to reply === Subject: Re: perfect square<401ae27e@rutgers.edu> thusly:> Hey, can anyone help me with a proof of why 2,3,7,8 can't be the last> digits of a perfect square? I'm pretty sure that I should start of with> the fact> that any number can be written in the form 10a+ b where 0<=b<=9. But that> means that b can't = 2,3,7, or 8. Here, I get stuck, thanks.-GregThe final digits of perfect 4th powers are even more restricted - they can only be 0, 1, 5 or 6. Any reason why this should be so?On the other hand, a perfect 5th power can end in any digit - always the same digit as the original integer ends in. Again, any reason why this should be so?-- Paul TownsendI put it down there, and when I went back to it, there it was GONE!Interchange the alphabetic elements to reply === Subject: Re: perfect square>On the other hand, a perfect 5th power can end in any digit - always the >same digit as the original integer ends in. Again, any reason why this >should be so? And the same goes for perfect p'th powers in base 2p where p is any odd prime.See Fermat's little theorem.Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: Horror with integrals ETAtAhUAugx0scGmwiFgeK+gI5/ Y3OjRqsECFHLIMnfEUsmIz37WULChTHoQaCXi Consider the area between each hump of the curve abd the x-axis. Youare to prove that as you go along, the humps get successively smallerand they contribute strictly alternating signs to the integral. Thenyou have a standard recipe fo a convergent alternating series. === Subject: Re: Derived groupWe have nonabelian group G of order 24,> and we know that it has> a normal subgroup H of order 8.> Problem: Find derived subgroup G'> (aka [G:G] - commutator subgroup)> Surely it must be subgroup of H,> (|G/H|=3, so G/H is abelian)> but I think it's really smaller,> probably cyclic group of order 2.No. For example, G = SL(2,3), which is the semi-direct product> Q_8 x| C_3 with presentation a^-1, c^3 = 1, a^c = b, b^c = ab>, has derived group [G,G] => which is its normal subgroup H =~ Q_8. (This is clear> since none of the 3 subgroups of order 4 is normal in G, and the> only subgroup of order 2 is , but G/ =~ PSL(2,3) =~> A_4 is non-abelian.)The other counter-example, with |[G,G]| = 4, is the direct> product G = A_4 x C_2.The remaining 2 groups, namely the direct products G = Q_8 x C_3> and G = D_8 x C_3, do both have [G,G] cyclic of order 2. -- === Subject: Re: Derived group> I can't sleep because of one problem:We have nonabelian group G of order 24,> and we know that it has> a normal subgroup H of order 8.> Problem: Find derived subgroup G' > (aka [G:G] - commutator subgroup)Any ideas?> Is it possible to definitely answer > this question at all?> (Of course without checking by hand> all 12 nonabelian groups of order 24 :) )Surely it must be subgroup of H,> (|G/H|=3, so G/H is abelian)> but I think it's really smaller,> probably cyclic group of order 2.Are we able to say something about structure> of such group G ?> Hint: G/H is abelian iff H contains all the commutators of G. Then youonly have to look at groups of order 8.Patrick === Subject: Re: Derived group>> I can't sleep because of one problem:> We have nonabelian group G of order 24,>> and we know that it has>> a normal subgroup H of order 8.>> Problem: Find derived subgroup G' >> (aka [G:G] - commutator subgroup)> Any ideas?>> Is it possible to definitely answer >> this question at all?>> (Of course without checking by hand>> all 12 nonabelian groups of order 24 :) )> Surely it must be subgroup of H,>> (|G/H|=3, so G/H is abelian)>> but I think it's really smaller,>> probably cyclic group of order 2.> Are we able to say something about structure>> of such group G ?>>Hint: G/H is abelian iff H contains all the commutators of G. Then you>only have to look at groups of order 8.It has already been pointed out that this problem does not havea unique answer. You can get two different answers with the samegroup Q_8 of order 8, namely G=SL(2,3) giving G' = Q_8,and G = Q_8 x C_3 giving G' = C_2, so it is not really true to say that you only have to look at groups of order 8.Derek Holt. === Subject: Re: sqrt(-1)=0/0> In sci.logic, ZZBunker> :>> In sci.physics, Earle Jones>> 0/0 is every number, 0 is their placeholder.0/0=undefined, so 0 cannot be a place holder for all x in N, I > or R. > 0> denotes the absence of quantity which is instrinsically a > quantity> (ie. nothing is something).Nothing has no constituent parts, and therefore cannot be > divided by> or multiplied against...*> Do not equate Ônothing' with Ôzero'.earle> *Zero denotes the absence of quantity which is intrisically a quantity,> since osophically, nothing is something.> *>> ÔNothing' is the opposite of Ôsomething'.> ÔZero' is a point on the number line, like 6, -23, pi, and 1,931.> Pedant point: the number line doesn't exist either, at least not>> as a physical object. Then again, the only reason zero needs>> to be specially treated anyway is because it *is* the arithmetic>> identity and therefore X * 0 = 0 for all X in the field.>> No other number in the field has that property.*You are saying that zero is unique.You are right.But then so are all of the other numbers. Every number has something that can be said about it that cannot be said about any other number.Do you remember the proof of this: There are no non-interesting numbers?earle* === Subject: Re: sqrt(-1)=0/0In sci.logic, Earle Jones> In sci.logic, ZZBunker>> :> In sci.physics, Earle Jones> or R. > 0> denotes the absence of quantity which is instrinsically a > quantity> (ie. nothing is something).Nothing has no constituent parts, and therefore cannot be > divided by> or multiplied against...*> Do not equate Ônothing' with Ôzero'.earle> *Zero denotes the absence of quantity which is intrisically a quantity,> since osophically, nothing is something.*> ÔNothing' is the opposite of Ôsomething'.'Zero' is a point on the number line, like 6, -23, pi, and 1,931.Pedant point: the number line doesn't exist either, at least not> as a physical object. Then again, the only reason zero needs> to be specially treated anyway is because it *is* the arithmetic> identity and therefore X * 0 = 0 for all X in the field.> No other number in the field has that property.*> You are saying that zero is unique.You are right.But then so are all of the other numbers. Every number has something > that can be said about it that cannot be said about any other number.Do you remember the proof of this: There are no non-interesting > numbers?No, but I can probably reconstruct many variants of it. :-)Let B_1, B_2, B_3, ... be a set of non-interesting numbers.(B = boring. Surprise.) Order the set such that0 <= abs(B_1) <= abs(B_2) <= ... . Then B_1 becomes aninteresting number, as it's either the closest boring numberto 0, or one of the two.Therefore, the sequence cannot exist and all numbers are interesting.:-)Of course 0 and 1 are unique in that both are identities; 0 isthe arithmetic identity, 1 the multiplicative. Does it meananything? I can't say; I know nothing about it...:-)earle> *-- #191, ewill3@earthlink.netIt's still legal to go .sigless. === Subject: Re: sqrt(-1)=0/0> In sci.logic, ZZBunker> :>> In sci.physics, Earle Jones>> 0/0 is every number, 0 is their placeholder.0/0=undefined, so 0 cannot be a place holder for all x in N, I or R. > 0> denotes the absence of quantity which is instrinsically a quantity> (ie. nothing is something).Nothing has no constituent parts, and therefore cannot be divided by> or multiplied against...*> Do not equate Ônothing' with Ôzero'.earle> *Zero denotes the absence of quantity which is intrisically a quantity,> since osophically, nothing is something.> *>> ÔNothing' is the opposite of Ôsomething'.> ÔZero' is a point on the number line, like 6, -23, pi, and 1,931.> Pedant point: the number line doesn't exist either, at least not>> as a physical object. Then again, the only reason zero needs>> to be specially treated anyway is because it *is* the arithmetic>> identity and therefore X * 0 = 0 for all X in the field.>> No other number in the field has that property. Double pendant point. That's not true, > since zero is not handled in any proof.> Only it's current day dork symbol is > handled. It's handled like gravity. Gently, gently, move it a little bit to the left, > move it a little bit to right, Then move it faster. > Then all of sudden, vailo, it's magic. The result is > a Quantum Chemist Dork gibbering about the holistic > CONSCIOUSNESS of parallel universes, and cold fusion.> You're not making a whole lot of sense here.How are we supposed to handle numbers? I don't know. Since set theory was supposed to answer that question, not me.> How are we supposed to handle zero? I always handled it with a 0, rather than a how. How mathematicans & musicians handle it, depends on piano players and Newton, not people who know things about music and logic. > [rest snipped] === Subject: Re: sqrt(-1)=0/0In sci.logic, ZZBunker:>> In sci.logic, ZZBunker>> :> In sci.physics, Earle Jones> 0> denotes the absence of quantity which is instrinsically a quantity> (ie. nothing is something).Nothing has no constituent parts, and therefore cannot be divided by> or multiplied against...*> Do not equate Ônothing' with Ôzero'.earle> *Zero denotes the absence of quantity which is intrisically a quantity,> since osophically, nothing is something.*> ÔNothing' is the opposite of Ôsomething'.'Zero' is a point on the number line, like 6, -23, pi, and 1,931.Pedant point: the number line doesn't exist either, at least not> as a physical object. Then again, the only reason zero needs> to be specially treated anyway is because it *is* the arithmetic> identity and therefore X * 0 = 0 for all X in the field.> No other number in the field has that property.> Double pendant point. That's not true, >> since zero is not handled in any proof.>> Only it's current day dork symbol is >> handled. It's handled like gravity.> Gently, gently, move it a little bit to the left, >> move it a little bit to right, Then move it faster. >> Then all of sudden, vailo, it's magic. The result is >> a Quantum Chemist Dork gibbering about the holistic >> CONSCIOUSNESS of parallel universes, and cold fusion.>> You're not making a whole lot of sense here.> How are we supposed to handle numbers?> I don't know. Since set theory was supposed to answer that> question, not me.Set theory answers questions about sets, which are a differentbrand of abstract entity from what is commonly termed a number.Of course one can have a set of numbers, or a number (nonnegativeinteger) of sets.Of course one can get bogged down here; if one has 0 sets, onehas the empty set, leading to either a very odd osophicalproblem, or a contradiction.One can construct a series of sets that map 1-1 with thewhole numbers, starting with 0 <=> empty set.>> How are we supposed to handle zero? I always handled it with a 0, rather than a how.> How mathematicans & musicians handle it, depends on piano> players and Newton, not people who know things about > music and logic.That didn't make a whole lot of sense, either. >> [rest snipped]-- #191, ewill3@earthlink.netIt's still legal to go .sigless. === Subject: Re: Simple set/probability proof>Consider {E_k} a collection of decreasing sets, their countable>>intersection is E. Can someone please outline a simple proof for>> oo>>E_n = U (E_kE_(k+1)) U E , for each n.>> k=n>>Now if P is a probability function, how does one prove that>>lim P(E_n)= P(E)>>n-->oo>I assume you know that if A is a decreasing sequence of events, then > lim PA_n = P(int(A)). (This fact is a standard consequence of countable > additivity.) It is clear that (E_n) is a decreasing sequence and E > is a subset of int(E_n). Show that if an outcome is not in E, then > it is not in int(E_n).I think that was what he was trying to prove as a standard consequence> of countable additivity (where he represented each E_n in the> decreasing sequence as a countable union of disjoint sets).It seems he needed some help even on showing that it was a decomposition> into disjoint sets ...You are correct. In fact the question arose in the proof ofcountable additivity implies continuity and your post was veryhelpful in establishing the related set identity, which the text(Chung) simply stated as obvious. It is obvious from a Venndiagram, but, as you demonstrate, not quite that simple to establishformally. Again many thanks for your help and also for Dr.Herschkorn's. === Subject: HELP! stuck on hamilton cycle problemFor a natural number n, define the graph G_n = (V_n, E_n) as* V_n being the set of all n-tuples consisting of 0 and 1. For example: V_2= { (0,0), (0,1), (1,0), (1,1) }* Two vertices are connected with an edge if and only if the correspondingtuples differ in exactly one position. For example there is an edge between(0,0) and (1,0) but not between (0,0) and (1,1)PROBLEM: construct a hamilton circuit in G_nI can construct a hamilton circuit for n = 1, 2 or 3. But do i do it forany n ???? === Subject: Re: HELP! stuck on hamilton cycle problem> For a natural number n, define the graph G_n = (V_n, E_n) as> * V_n being the set of all n-tuples consisting of 0 and 1. For example: V_2> = { (0,0), (0,1), (1,0), (1,1) }> * Two vertices are connected with an edge if and only if the corresponding> tuples differ in exactly one position. For example there is an edge between> (0,0) and (1,0) but not between (0,0) and (1,1)> PROBLEM: construct a hamilton circuit in G_n> I can construct a hamilton circuit for n = 1, 2 or 3. But do i do it for> any n ????How do you do it for n = 1? === Subject: Re: HELP! stuck on hamilton cycle problem> PROBLEM: construct a hamilton circuit in G_nI can construct a hamilton circuit for n = 1, 2 or 3. But do i do it for> any n ????It seems to me that there is one-to-one mapping between Hamiltonian circuitson G_n and n-bit circular gray codes. There are plenty of references on thenet on generating gray codes, or it is easy to discover a simple recursiveway to generate them if you don't feel like Googling.-- --Tim Smith === Subject: Re: HELP! stuck on hamilton cycle problem> For a natural number n, define the graph G_n = (V_n, E_n) as* V_n being the set of all n-tuples consisting of 0 and 1. For example:> V_2 = { (0,0), (0,1), (1,0), (1,1) }* Two vertices are connected with an edge if and only if the corresponding> tuples differ in exactly one position. For example there is an edge> between (0,0) and (1,0) but not between (0,0) and (1,1)> This gives the graph of the n-cube. Google found the following page:http://www.math.gatech.edu/~trotter/Section4- EulerHam.htmHTH,Tobias-- Just because you're paranoidDon't mean they're not after youreverse my forename for mail! - saibot === Subject: Re: Calculating the arc length of a curve> I have to calc the arc length of the graph of the function> f(x) = (2/3) x^(3/2)> within the interval [0,1].> As a tip I shall describe the function as a parametrized curve.> Okay, I tried to get a parametrized description with the followingapproach:> y(t) is already given with:> (1) y(t) = (2/3) * t^(3/2)> ---------------------> x(t) determined as follows:> y = (2/3) x^(3/2)> <=> (3/2) y = x^(3/2)> <=> ((3/2)y)^(2/3) = y> <=> x = ((3/2)y)^(2/3)> (2) x(t) = ((3/2)t)^(2/3)> Does (1),(2) describe the function as a parametrized curve or have I> misunderstood what I shall do ?> Is this the general way to find a parametrized description of a function ?> Calculating the arc length should then be easy... there is some formula> given...First take a derivative. In this case, f'(x)=sqrt(x)Then the formula is Integral[a,b,sqrt(1+f'(x))dx]Setting up the integral givesIntegral[0,1,sqrt(1+x)dx]u=1+xdu=dxIntegral[1,2,sqrt(u)du ](2/3)u^(3/2)2/3(sqrt(8))-2/32/3(sqrt(8)-1)1.219If you want to parameterize it, you could parameterize it as x=t,y=(2/3)t^(3/2). Personally, I think involving parameters would be overkillas this integral should be pretty basic without parameters.David Moran === Subject: Re: Calculating the arc length of a curve> I have to calc the arc length of the graph of the functionf(x) = (2/3) x^(3/2)within the interval [0,1].As a tip I shall describe the function as a parametrized curve.It's already parameterized as (x, (2/3)x^(3/2)). === Subject: Re: Calculating the arc length of a curve> I have to calc the arc length of the graph of the function> f(x) = (2/3) x^(3/2)> within the interval [0,1].> As a tip I shall describe the function as a parametrized curve.> It's already parameterized as (x, (2/3)x^(3/2)).Ups,okay... to calc the arc length I then computed:dx/dt = 1dy/dt = x^(1/2)Then:sqrt( (dx/dt)^2 + (dy/dt)^2 ) = sqrt( 1^2 + (x^(1/2))^2 = sqrt(1+x)Then:Integral from 0 to 1 over sqrt(1+x)= Integral from 0 to 1 over (1+x)^(1/2)= [ (2/3) * (1+x)^(3/2) ] from 0 to 1= ( (2/3) * 2^(3/2) ) - ( (2/3) * 1^(3/2) )= (4/3) * sqrt(2) - (2/3)This should be the arc length.Right ? === Subject: Re: Calculating the arc length of a curve charset=Windows-1252message> I have to calc the arc length of the graph of the functionf(x) = (2/3) x^(3/2)within the interval [0,1].As a tip I shall describe the function as a parametrized curve.> It's already parameterized as (x, (2/3)x^(3/2)).> Ups,> okay... to calc the arc length I then computed:> dx/dt = 1> dy/dt = x^(1/2)> Then:> sqrt( (dx/dt)^2 + (dy/dt)^2 ) = sqrt( 1^2 + (x^(1/2))^2 = sqrt(1+x)> Then:> Integral from 0 to 1 over sqrt(1+x)> = Integral from 0 to 1 over (1+x)^(1/2)> = [ (2/3) * (1+x)^(3/2) ] from 0 to 1> = ( (2/3) * 2^(3/2) ) - ( (2/3) * 1^(3/2) )> = (4/3) * sqrt(2) - (2/3)> This should be the arc length.> Right ?Yes. === Subject: Re: Calculating the arc length of a curve charset=Windows-1252> messageI have to calc the arc length of the graph of the functionf(x) = (2/3) x^(3/2)within the interval [0,1].As a tip I shall describe the function as a parametrized curve.It's already parameterized as (x, (2/3)x^(3/2)).> Ups,> okay... to calc the arc length I then computed:> dx/dt = 1> dy/dt = x^(1/2)> Then:> sqrt( (dx/dt)^2 + (dy/dt)^2 ) = sqrt( 1^2 + (x^(1/2))^2 = sqrt(1+x)> Then:> Integral from 0 to 1 over sqrt(1+x)> = Integral from 0 to 1 over (1+x)^(1/2)> = [ (2/3) * (1+x)^(3/2) ] from 0 to 1> = ( (2/3) * 2^(3/2) ) - ( (2/3) * 1^(3/2) )> = (4/3) * sqrt(2) - (2/3)> This should be the arc length.> Right ?> Yes.I calculated that the general formula of an arc length for function:Ax^B with limits M and N is:int[ sqrt[ (AB)^2 * x^(2(B - 1)) + 1 ] ]With A = 2/3, B = 3/2, N = 0 and M = 1, that should give you what you hadcalculated. === Subject: How to explain this statistically? charset=iso-8859-1Suppose you have confidence intervall:m = x +- 1.96 s/sqrt(n).Now, if you compute a one-sided confidence intervall, you'll get:m <= x + 1.64 s/sqrt(n) and m >= x - 1.64 s/sqrt(n).Suppose i'd like to set up a new intervall (double-sided) asm = x +- 1.64 s/sqrt(n).That would be a nice idea since the intervall is shorter, even if the security measure is still 0.95. Of course it's not correct wayof computing things but i can't put my finger on why.I have a guess or two but they are rather tame...Any hints?-- KindlyKonrad-------------------------------------------------- -May all spammers die an agonizing death; have no burial places; their souls be chased by demons in Gehenna from one room to another for all eternity and more.Sleep - thing used by ineffective people as a substitute for coffeeAmbition - a poor excuse for not having enough sense to be lazy--------------------------------------------------- === Subject: Re: How to explain this statistically?> Suppose you have confidence intervall:> m = x +- 1.96 s/sqrt(n).... You may get a useful response from the sci.stat.math or sci.stat.edu news group. === Subject: Taking double-integrals with constraints I have a question concerning correctly defining the bounds of integration. I would like to take the double-integral of a function, z=g(x,y), defined as: z=g(x,y)=2*R*cos(y)/x [1]where R is a positive constant. And there is a function that draws x and y together and is defined as follows: f(x,y)=-1(2b)^4*x^3*siny*cosy-1/(2b)^3*x^2*(cosy-siny)+1/(2b)^ 2*x [2]where x is defined in (0, 2*sqrt(2)*b] (b is a positive constant), and the range of y is (3pi/2, 2pi]. Actually, f(x,y) is the joint probability density function of x, and y in the given ranges, and what I want to do next is to find the cumulative distribution function of z, which is a function of x and y. And z is supposed to be inf >= z >0 Although I know the steps of finding the CDF, I get stuck on some math here. the double-integral expression that I write down is the following: |7pi/4 |-2b/sin(y) |2pi |2b/cos(y) CDF(z)=| | | | | | f(x,y)dxdy + | | f(x,y)dxdy [3] | | | | / / / / y=3pi/2 x=2Rcos(y)/z y=7pi/4 x=2Rcos(y)/zHowever, in carrying out the integraion, I realized two things. First, the constraint of x is bounded by 2*sqrt(2)*b >= x >= 2Rcos(y)/z >=0; and second, when z is very small, say, 0.01, given the choice of R and b, there is a possibility that the lower bound of the inner integeral (i.e., 2Rcos(y)/z) in either of the two terms above may be greater than the upper bound. When I followed the normal procedure of intergration to integrate Eq. 3, the expression I obtained for the CDF(z) breaks down for this low x region, giving me a spike exceeding 1 and some very large negative values as z approaches zero (although when t is big, the curve of CDF looks correct). I would certainly appreciate any help you can give me on how to take proper integraion and compensate for the negative effects when z is close to zero. I have a feeling I may have not have defined the correct lower bounds of the inner integral in Eq. 3, but I don't know where I get 3, and the results that came out from it were correct, showing no spikes and negative values in the CDF, and it was monotolically increasing to 1, as expected. So this verification sort of convinced me that I didn't do the integration right. If anybody can give me some help on this and tell me if there's anything wrong in the way I define the lower bound of the integral, I'm -Ed === Subject: Re: what can we say about variance of a linear system output?>Dear all,>I have a linear system y=Ax, where x is input vector, A is system inmatrix>form, y is output vector,>is there anything we can say about var(y) in terms of var(x) and some>characteristic of A, A is given and known...xisting theorem for this?> I assume by variance, you mean the (variance-)covariance matrix:> Var(z) = E[(z-Ez) (z-Ez)'], where z is a colmnn vector and Ô> indicates transpose.> WLOG, we may assume that Ex = 0. Then it easily follows that> Var(y) = E[(Ax) (Ax)'] = A Var(x) A'.> --> Stephen J. Herschkorn herschko@rutcor.rutgers.eduHi Stephen,Thank you for your help! But I mean the deterministic variance, not theprobalistic one.So covariance matrix does not apply here.By terterministic variance I mean the normal definition of variance, noprobability is related at all... === Subject: Re: what can we say about variance of a linear system output?>Thank you for your help! But I mean the deterministic variance, not the>probalistic one.>So covariance matrix does not apply here.>By terterministic variance I mean the normal definition of variance, no>probability is related at all...And what exactly is that? A population variance? The same laws apply. If you mean sample variance, then you need to be more careful.To repeat, what do you mean by deterministic variance?-- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: what can we say about variance of a linear system output? ...> By terterministic variance I mean the normal definition of variance, no> probability is related at all...In that case, the variance of what comes out depends on what goes in andin the characterisr=tics of the system. I don't think you mean exactlythat. Can you clarify what you want to find.Jerry-- Engineering is the art of making what you want from things you can get.[OSlash][OSlash][OSlash][OSlash] === Subject: Re: how does each row, column's variance related with the matrix variance?>Dear all,>Suppose I have a matrix and want to compute its matrix variance, and Iknow>each row and column's variance, what can I say about the matrix variance?>And expression relating all these terms?> I assume you have a random matrix X and you want to determine> Cov(X_(i,j), X_(k,l)) for all i, j, k, l within the dimensions of the> matrix. Off hand, I don't think you can recover these from the> covariance matrices of the rows and columns. Consider, for example, a 2> x 2 matrix. How would you recover Cov(X(1,1), X(2,2))?> --> Stephen J. Herschkorn herschko@rutcor.rutgers.eduOnce again I did not mean the Cov, I mean the non-probabilistic variance... === Subject: Natural Numbers, OrdinalsI'm curious, natural numbers are defined as the first ordinals:0 := {}1 := {0}2 := {0, 1}3 := {0, 1, 2}n := {0, 1, ... , n-1}What, however if we define them as0 := {}1 := {0}2 := {1}3 := {2}n := {n-1}This of course only works for successor ordinals, not limit ordinals,but does it have some kind of nice structure? I mean, for successorordinals, the other stuff is just frivolous repetition. e.g. 3 = {{},{{}}, {{}, {{}}}} when it could just be 3 = {{{{}}}}, much easier towrite, hehehe.I got on this topic from a typo inhttp://mathworld.wolfram.com/Rank.html which calls 2 = {{{}}}Also, do you think it's perfectly alright to define cardinals asnumbers which are associated with a set, whereby two sets have thesame cardinality if they are equipollent? Why does mathworld say thata special axiom must then be used which declares those numbers exist?What is the convention on existence of things, anyway?-Gregarius === Subject: Re: Natural Numbers, OrdinalsGregory Magarshak> I'm curious, natural numbers are defined as the first ordinals:> 0 := {}> 1 := {0}> 2 := {0, 1}> 3 := {0, 1, 2}> n := {0, 1, ... , n-1}> What, however if we define them as> 0 := {}> 1 := {0} ...Peano's axioms for the integers use (or may use) this second def'n.> This of course only works for successor ordinals, not limit ordinals,> but does it have some kind of nice structure? I mean, for successor> ordinals, the other stuff is just frivolous repetition. e.g. 3 = {{},> {{}}, {{}, {{}}}} when it could just be 3 = {{{{}}}}, much easier to> write, hehehe.For Bourbaki, an ordinal is just the type or type specimen of awell-ordered set. He uses a strong form of the axiom of choice, though,which might be the special axiom mentioned at MathWorld. Things like{{{}}} he calls pseudo-ordinals. Then again, maybe the mystery axiom isthe one called the Axiom of Foundation (which Bourbaki's method does notrequire).LH === Subject: Re: Double Summations question>Is the following true?Summation (from i = 0 to infinity) of Summation (from n = 0 toinfinity) of>(u^n i^n)/n!>=>(Summation (from i = 0 to infinity) of u^i) (Summation (from n = 0 to>infinity) of i^n/n!)Hint: u^n i^n = (u i)^n.> That hint is OK.> Then use: e^x = Summation (from n = 0 to infinity) of x^n /n!> to simplify your formulas.> The series converges only when u is> negative (assuming real variables).> The first series always diverge,> second (geometric series in fact) is convergent if |u| < 1/e.Yes, I understand that the sum turns into summation (from i = 0 to infinity)of e^iu...What I want to do however is turn this sum into something of the form(summation (from i = 0 to inifinity) u^i )*(something)Is this possible?> -- > Best regards, Rafal Kucharski. === Subject: Re: Double Summations question>Is the following true?>Summation (from i = 0 to infinity) of Summation (from n = 0 to infinity)of>(u^n i^n)/n!>=>(Summation (from i = 0 to infinity) of u^i) (Summation (from n = 0 to>infinity) of i^n/n!)> Why should it be?> Hint: u^n i^n = (u i)^n. The series converges only when u is> negative (assuming real variables).> -- Yes, I understand that the sum turns into summation (from i = 0 to infinity)of e^iu...What I want to do however is turn this sum into something of the form(summation (from i = 0 to inifinity) u^i )*(something)Is this possible?> Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: Peer Review>In paper-based publication of research should not peer review be>called rivals' review? A referee may read a paper and benefit from>it and then decide that others should not have the same privilege.> You know of any specific examples where this has happened?Yes. There was a Houston physicist called Wu, who works on low-Tsemiconductors. He knew that the info he sent in for publication would bereplicated by the referees. He used the symbol for the element Yetterbium(Yb) in lieu of Terbium (Tb). Then when he got the galley proof JUST beforechanged his mistake. People were outrgaed, to which he replied that it wasan honest mistake, and if nobody was duplicating his expts, what harm hadbeen done as long as he caught it before it went out for publication. ;>)Jack === Subject: Re: Peer Review>>In paper-based publication of research should not peer review be>>called rivals' review? A referee may read a paper and benefit from>>it and then decide that others should not have the same privilege.>> You know of any specific examples where this has happened?>Yes. There was a Houston physicist called Wu, who works on low-T>semiconductors. He knew that the info he sent in for publication would be>replicated by the referees. He used the symbol for the element Yetterbium>(Yb) in lieu of Terbium (Tb). Then when he got the galley proof JUST before>changed his mistake. People were outrgaed, to which he replied that it was>an honest mistake, and if nobody was duplicating his expts, what harm had>been done as long as he caught it before it went out for publication. ;>)Um:(i) are you conesetter?(ii) do you know any examples in mathematics? I don't care about thatother stuff...(iii) This is _not_ an example of a referee rejecting a paper becausehe decides others should not have the same privilege! The paperwas accepted for publication, so it has no relevance to the questionI asked.(It's also not an example of a different bad thing. If Wu has evidencethat referees _did_ appropriate his work then it would be anexample of a Bad Thing (although not the bad thing I askedabout). But there's nothing about that given above - it's justan example of someone _fearing_ that the Bad Thing_could_ happen.(No, the fact that people were outraged at the last-minutecorrection does not prove anything sinister - editorsdon't like last-minute corrections for perfectly legitimatereasons.))>Jack === Subject: Re: Peer Review>>In paper-based publication of research should not peer review be>>called rivals' review? A referee may read a paper and benefit from>>it and then decide that others should not have the same privilege.>> You know of any specific examples where this has happened?>Yes. There was a Houston physicist called Wu, who works on low-T>semiconductors. He knew that the info he sent in for publication would be>replicated by the referees. He used the symbol for the element Yetterbium>(Yb) in lieu of Terbium (Tb). Then when he got the galley proof JUSTbefore>changed his mistake. People were outrgaed, to which he replied that itwas>an honest mistake, and if nobody was duplicating his expts, what harm had>been done as long as he caught it before it went out for publication.;>)> Um:> (i) are you conesetter?Defn? Don't know what that means.> (ii) do you know any examples in mathematics? I don't care about that> other stuff...No. But the Wallace thing wasn't about math either. The reason why you don'tsee it in math so much is because a lot of the work has no direct commercialvalue. But, as that changes, and it will (with applications to computerscience, the stock market, etc), you will _definitely_ see this sort ofthing if you work in such a field (which I assume from your startled tone,you don't).> (iii) This is _not_ an example of a referee rejecting a paper because> he decides others should not have the same privilege! The paper> was accepted for publication, so it has no relevance to the question> I asked.True, but it's such a nice story I couldn't resist the temptation ;>). I'msorry it didn't fit into your mental block. But it is nice to liberate andexpand your mind and learn something outside of your own little experiencesin math though, isn't it? Especially, since it could very well be applicableto you down the road.But seriously, if this thing could happen , it's not too hard to imagineproof but knowing people (in science again, I'm sorry), I _know_ it isentirely possible and I'd lay money that it does happen.> (It's also not an example of a different bad thing. If Wu has evidence> that referees _did_ appropriate his work then it would be an> example of a Bad Thing (although not the bad thing I asked> about). But there's nothing about that given above - it's just> an example of someone _fearing_ that the Bad Thing> _could_ happen.see the significance. I'll try to put it in terms you'll understandthough.It's a little like insider trading. Suppose you worked on low-Tsemiconductors. It's an experimental endeavor, so this means you have abunch of grad students baking up all kinds of and then seeing which oneis the best. You get a letter saying that ingredient X gets you the bestlow-T semiconductor before the sci community does. You get all your gradstudents in the lab baking away trying every permutation of chemicalcombinations with X in it to produce something new and better before thepaper cmes out and the other guys get a crack at it. This is unethical. It'svery simple.> (No, the fact that people were outraged at the last-minute> correction does not prove anything sinister - editors> don't like last-minute corrections for perfectly legitimate> reasons.))>Jack> Jack === Subject: Re: Peer Review> In paper-based publication of research should not peer review be> called rivals' review? A referee may read a paper and benefit from> it and then decide that others should not have the same privilege.> (preprints) to his colleagues in the field, and nowadays posts> it on his web page. So if the paper is rejected for reasons such as> those you fear, everyone in the field will still know he was first.> The other side is this: When a bad paper is submitted, then rejected,> it is natural and common for the author to blame the rejection on> anything he can, such as incompetent referees. While there is surely> some incompetent refereeing, that is minuscule compared to the number of> actual bad papers submitted to mathematics journals.In science it is not. Often, a paper will be sent out to a person who isconsidered an expert in the field. But, if he has a competing theory whichconßicts with yours, you'll be gettin' the old rejection letter. I've seenit happen a bunch of times to good people. This fact really makes what thisperson call rival review true.Jack === Subject: Re: Peer Review> Consider also the story of Darwin and Wallace who arrived at the> theory of natural selection at the same time. Wallace published first> and prior publication is supposed to be what counts. But he published> to Darwin and although Darwin gave him some credit the theory today> bears Darwin's name and not Wallace's.describing the thery and a private letter is not a publication.Furthermore, in his truly voluminous private jottings, letters, andconversations, Wallace never expressed anything but pleasure atDarwin's willingness to share at least partial credit (I am quoting === Subject: Re: Peer Review Consider also the story of Darwin and Wallace who arrived at the> theory of natural selection at the same time. Wallace published first> and prior publication is supposed to be what counts. But he published> to Darwin and although Darwin gave him some credit the theory today> bears Darwin's name and not Wallace's.describing the thery and a private letter is not a publication.> Furthermore, in his truly voluminous private jottings, letters, and> conversations, Wallace never expressed anything but pleasure at> Darwin's willingness to share at least partial credit (I am quotingThis meshes with what I heard from an author on BookTV that had writtena bio on Wallace. Darwin had explained his ideas to enough people so that Wallace hadbasically little choice but to acknowledge that Darwin had come up withnatural selection first. Wallace was also apparently a fair andgracious gentleman, so he never tried to claim priority. The author ofthe bio had some additional things to say about Wallace however. Wallace apparently made some important contributions, which I can'tremember.Also, Wallace had something to do with Darwin publishing his book. Friends were able to pressure Darwin to write because the situationwith Wallace raised the possibility that more people would come up withDarwin's ideas and claim priority.Even if Wallace had published first, I don't see why conesetter thinksthat is what is supposed to be what counts. It can be harder to makeyour case if you were first but didn't publish, but publishing date hasnever been, to my knowledge, the final and only arbiter of prioritydisputes. People are interested in who came up with the idea first,not who published it first. Sometimes this is the same thing, but notalways. === Subject: Re: Peer ReviewEven if Wallace had published first, I don't see why conesetter thinks> that is what is supposed to be what counts. It can be harder to make> your case if you were first but didn't publish, but publishing date has> never been, to my knowledge, the final and only arbiter of priority> disputes. People are interested in who came up with the idea first,> not who published it first. Sometimes this is the same thing, but not> always. Perhaps the points raised by others can all be answered by sayingthat I wanted to convey the idea that editors and referees can beconsigned to history now we have the internet. === Subject: Re: Peer Review> Perhaps the points raised by others can all be answered by saying that I> wanted to convey the idea that editors and referees can be consigned to> history now we have the internet.Except the internet doesn't provide a mechanism to separate the worthwhilefrom the crap. -- --Tim Smith === Subject: Re: Peer Review> Perhaps the points raised by others can all be answered by saying that I> wanted to convey the idea that editors and referees can be consigned to> history now we have the internet.Except the internet doesn't provide a mechanism to separate the worthwhile> from the crap. You are happy to let others decide for you which is which? === Subject: Re: Peer Review Perhaps the points raised by others can all be answered by saying> that I wanted to convey the idea that editors and referees can be> consigned to history now we have the internet.[January 30 issue http://chronicle.com/ ] aboutthose advocating an on-line free-to-read author-paid system ofpublication. This push is coming from the biomedical sciences.But it would still be peer-reviewed. Even they realize that thisstep is essential.-- http://www.math.ohio-state.edu/~edgar/ === Subject: Re: Peer Review>[January 30 issue http://chronicle.com/ ] about>those advocating an on-line free-to-read author-paid system of>publication.May I point out the irony of having to pay $82.50 for a yearly-- I'm not interested in mathematics that might have anythingto do with reality. -- Easterly, in sci.math === Subject: Re: Peer ReviewMay I point out the irony of having to pay $82.50 for a yearly-- http://www.math.ohio-state.edu/~edgar/ === Subject: Re: Peer Review <8083d02b.0401312312.55c04654@posting.google.com> Discussion, linux)> Perhaps the points raised by others can all be answered by saying> that I wanted to convey the idea that editors and referees can be> consigned to history now we have the internet.Who wants to sift through *all* of the various crap published on theinternet in order to find a readable and correct paper on a topic ofinterest?Don't get me wrong. The internet is a great research tool, and thetendency of authors to make preprints available[1] is wonderful. Idon't mind if I never go to the library again.But peer-reviewed publications help me know which authors are reliablein the first place. I learn author's names based on their publicationhistory (or some other personal acquaintance with their work). Iciteseer or some similar service and I rely on the fact that citationsin citeseer typically mean a peer-reviewed paper.Get rid of peer review, and you'll end up with all the cruft you findin arXiv (not to say that there's not also quality papers there, butthere are rather more bad papers there than in peer-reviewedjournals).Footnotes: [1] Sadly, not a common practice in osophy -- and whenosophers *do* make their papers available, chances are good it's aWord document. Bleh.-- I thought it relevant to inform that I notified the FBI a couple ofmonths ago about some of the math issues I've brought up here. -- James S. Harris gives Special Agent Fox a new assignment. === Subject: Re: Peer Review> Consider also the story of Darwin and Wallace who arrived at the> theory of natural selection at the same time. Wallace published first> and prior publication is supposed to be what counts. But he published> to Darwin and although Darwin gave him some credit the theory today> bears Darwin's name and not Wallace's.> Do you have a reference for this story.A Google web search for darwin wallacewill give you many links. In particular, appears to be trustworthy.David === Subject: Re: Peer Review Discussion, linux)>> Consider also the story of Darwin and Wallace who arrived at the>> theory of natural selection at the same time. Wallace published first>> and prior publication is supposed to be what counts. But he published>> to Darwin and although Darwin gave him some credit the theory today>> bears Darwin's name and not Wallace's.>> Do you have a reference for this story.> A Google web search for> darwin wallace> will give you many links. In particular,> Does anyone know if there is a Schaum's outline (or different ones),> covering:> - Algebra: Grouptheory (Groups, Rings, Fields)Schaum's Outline of Modern Abstract Algebra> - Discrete maths: Relations, Maps, modular arithmeticsSchaum's Outline of Discrete MathematicsorSchaum's Outline of Set Theory and Related Topics> - Calculus: Sequences (finite, infinite), SeriesSchaum's Outline of Calculus> Which Schaum's cover these topics in detail ?> Do you know similar literature (not for reading the theory, but for> practicinng - with solutions !) ?There are other cram book series for sale in most university bookstores inthe United States (and available through Amazon.com worldwide), but Schaum'sOutline series tends to be dominant in most markets. Using Amazon.com, orfor that matter Amazon.de, is a good guide to what people are buying.Hope this helps! === Subject: Goldberg dualDoes anyone have a name for the dual of a Goldberg polyhedron?Dick Fischbeck === Subject: urgent help neededHican anyone please send me the answer to the following question. soonas possibleLet A be a set of integer. Define an equivalence relation R on AxA.Show that the relation you defined is an equivalence relation. Draw adiagram for your relation. === Subject: Re: urgent help needed> Hi> can anyone please send me the answer to the following question. soon> as possible> Let A be a set of integer. Define an equivalence relation R on AxA.> Show that the relation you defined is an equivalence relation. Draw a> diagram for your relation.Why is it needed so soon? Your homework can't possibly be due untilMonday at the earliest.Doug === Subject: Re: urgent help needed>> Hi>> can anyone please send me the answer to the following question. soon>> as possible>> Let A be a set of integer. Define an equivalence relation R on AxA.>> Show that the relation you defined is an equivalence relation. Draw a>> diagram for your relation.>Why is it needed so soon? Your homework can't possibly be due until>Monday at the earliest.>DougOn the other hand, you have to sympathize with him for wanting help, becausethe wording of the problem is very strange. At first, I thought thatthe first part of the question was asking for a definition of`equivalence realtion', but after reading the next bit, I presumeit is supposed to mean Given an example of an equivalence relationon AxA. But are you intended to give an example for all possible sets A, orcan you choose your own A?Derek Holt. === Subject: Re: urgent help needed>Hi>can anyone please send me the answer to the following question. soon>as possible>Let A be a set of integer. Define an equivalence relation R on AxA.>Show that the relation you defined is an equivalence relation. Draw a>diagram for your relation.>>Why is it needed so soon? Your homework can't possibly be due until>>Monday at the earliest.>>Doug>On the other hand, you have to sympathize with him for wanting help, because>the wording of the problem is very strange. At first, I thought that>the first part of the question was asking for a definition of>`equivalence realtion', but after reading the next bit, I presume>it is supposed to mean Given an example of an equivalence relation>on AxA. But are you intended to give an example for all possible sets A, or>can you choose your own A?If the problem is exactly as he quoted it then it is indeed very strangely phrased. But it happens a lot around here that theperson posting a homework problem gets the statement totallygarbled (presumably because he doesn't understand themeaning of the words he's garbling) so it's hard to knowwhere the difficulty really comes from.>Derek Holt. === Subject: Re: urgent help needed> Hi> can anyone please send me the answer to the following question. soon> as possible> Let A be a set of integer. Define an equivalence relation R on AxA.> Show that the relation you defined is an equivalence relation. Draw a> diagram for your relation.You forgot your professor's e-mail address so we can send it to them.David Moran === Subject: Re: urgent help needed> Hi> can anyone please send me the answer to the following question. soon> as possibleLet A be a set of integer. Define an equivalence relation R on AxA.> Show that the relation you defined is an equivalence relation. Draw a> diagram for your relation.How soon do you need it? When's your homework due? === Subject: Re: definition of asymptote.X-NFilter: 1.2.0>which of these definitions is true of an asymptote and why?>given f(x) and g(x),>g(x) is an asymptote of f(x) if >lim g(x) / f(x) = 1 as x -> infinity>OR>lim g(x) - f(x) = 0 as x -> infinity>(as an example, does x^2 + x converge to x^2 even though they get>further apart in absolute terms, their relative distance gets smaller)>...c8toNeither actually.As you note, the first one does not require the functions to even comeclose to each other. The limit can be one even as the functionsdiverge. I think we can agree that divergent functions don'tconverge.The second limit is a necessary condition but not a sufficient one.Consider the geometric equivalent in the following example: Take theunit line segment from (0,0) to (1,0). Construct the semicircle abovethe axis centered at (1/2,0) with radius 1/2. Length of line is 1;circumference of semicircle is pi/2. Replace the semicircle with twosemicircles, each having half the radius and centers at (1/4,0) and(3/4,0). Semicircles are now closer to line but length and totalcircumference is unchanged. Keep doubling number of semicircles. Thesemicircles become arbitrarily close to line but circumference isalways 57+% greater than length of line. Consequently, they are notasymptotic.With that background, let f(x) = 0 and define g(x) as: sqrt[.5^2 - (x-.5)^2] when 0 <= x <= 1 sqrt[.25^2 - (x-1.25)^2] when 1 < x <= 1.5 sqrt[.25^2 - (x-1.75)^2] when 1.5 < x <= 2 sqrt[.125^2 - (x-2.125)^2] when 2 < x <= 2.25 sqrt[.125^2 - 9x02.375)^2] when 2.25 < x <= 2.5 etc.Basically, the segment between n and n+1 is a curve made up of 2^nsemicircles, each with a radius of 1/2^n and centered at points((2k-1)/2^(n+1),0) for k = 1, 2, 3, ..., 2^(n).Since g(x) gets arbitrarily close to 0 as x increases, the twofunctions approach the same limit. But they are not asymptotic. <> === Subject: Re: definition of asymptote.>which of these definitions is true of an asymptote and why?>given f(x) and g(x),>g(x) is an asymptote of f(x) if>lim g(x) / f(x) = 1 as x -> infinity>OR>lim g(x) - f(x) = 0 as x -> infinity>(as an example, does x^2 + x converge to x^2 even though they get>further apart in absolute terms, their relative distance gets smaller)>...c8to> Neither actually.> As you note, the first one does not require the functions to even come> close to each other.But it does require that they get closer to each other in a relativesense.> The limit can be one even as the functions diverge.Right. But if that limit is 1, we write f ~ g, saying that f and g are_asymptotic_ as x -> infinity.Two functions can be asymptotic without the graph of one being an asymptoteto the graph of the other. You may consider that to be terminologicallyunfortunate...> The second limit is a necessary condition but not a sufficient one.For what? It is sufficient for the graph of f(x) to be an asymptote to thatof g(x), according to a common definition of asymptote.> Consider the geometric equivalent in the following example: Take the> unit line segment from (0,0) to (1,0). Construct the semicircle above> the axis centered at (1/2,0) with radius 1/2. Length of line is 1;> circumference of semicircle is pi/2. Replace the semicircle with two> semicircles, each having half the radius and centers at (1/4,0) and> (3/4,0). Semicircles are now closer to line but length and total> circumference is unchanged. Keep doubling number of semicircles. The> semicircles become arbitrarily close to line but circumference is> always 57+% greater than length of line. Consequently, they are not> asymptotic.> With that background, let f(x) = 0 and define g(x) as:> sqrt[.5^2 - (x-.5)^2] when 0 <= x <= 1> sqrt[.25^2 - (x-1.25)^2] when 1 < x <= 1.5> sqrt[.25^2 - (x-1.75)^2] when 1.5 < x <= 2> sqrt[.125^2 - (x-2.125)^2] when 2 < x <= 2.25> sqrt[.125^2 - 9x02.375)^2] when 2.25 < x <= 2.5> etc.> Basically, the segment between n and n+1 is a curve made up of 2^n> semicircles, each with a radius of 1/2^n and centered at points> ((2k-1)/2^(n+1),0) for k = 1, 2, 3, ..., 2^(n).> Since g(x) gets arbitrarily close to 0 as x increases, the two> functions approach the same limit. But they are not asymptotic.Hmm. _Two_ functions? I must assume that you intend the other function tobe the constant function f(x) = 0, the graph of which is the x-axis. Thex-axis is an asymptote of the graph of g(x), according to a commondefinition of asymptote. [However, IIRC, there is an alternative definitionwhich requires that an asymptote be the tangent to the graph at infinity.The graph of g(x) has no such tangent, and so, according to such analternative definition, the graph of g(x) has no asymptote.]David Cantrell === Subject: Roman numerals ...........................>But even this seems incorrect because abacus had the idea of zero as a >placeholder. >I also wonder, if a Roman merchant - or whoever - never transferred this >idea from abacus to his book-keeping on paprys or whatever he used to >scribble down the items he had for inventory or taxation purposes. It just >sounds so much more practical to reduce 89 from 1 000 and come up with 911 >instead of messing around with the confusing Roman numerals. I mean, how >could the merchant even calculate with them if he didn't have an abacus at >hand?It is somewhat quicker to compute with Arabic numerals thanRoman, but not that much. There is little difference in doingit by hand and by abacus, and I doubt that someone used to0Roman numerals would find it a factor of 2 slower than someoneof equal ability used to Arabic numerals.-- This address is for information only. I do not claim that these viewsare those of the Statistics Department or of Purdue University.Herman Rubin, Department of Statistics, Purdue Universityhrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: Socrates' Nothing is Everything> [SNIP]> However, I take issue with the idea that a number needs to be> defined in terms which capture only the essense of counting numbers.> Suppose I define a number as any object which is an element of a set> for which there exists a binary operation and which satisfies some set> of axioms. (Vague, obviously, but I think you might get the idea).> Then our problem with 0 goes away.>> ÔNumber' usually means the kinds of things included in dictionary or> encyclopedic definitions eg integral, rational, real, complex,hypercomplex,> modular and other numbers that mathematicians have thought fit toinvent.> Your definition is too wide. For example, it would include sets of> propositions for which first order logic provides both axioms and binary> relations. Propositions are not numbers even if they have values whichare> number-like in the some of their relations.> I'm not so sure about this. In particular, this definition is a> natural one in the context of model theory. The fact that> propositions have number-like relations is just a consequence of the> fact that the Boolean algebra used in sentential logic is isomorphic> to the finite field with two elements.> What is peculiar about numbers proper is that the systems they inhabituse> the operator Ô+'. Without this operator (and its ilk), the uniquecharacter> of number systems is lacking. The peculiarity is that this operatordefies> closure and points the finger at infinity. This property of infinite> extension is abstracted in the Peano definition of number which deploysthe> device of logical substitution into a string. ie Ô0S' is substitutedinto> Ô0X' indefinitely many times to produce, in principle, any possibleinteger.> In particular, the boolean algebra on which propositional logic is> based on the operations + and * defined by:> 1 + 1 = 1> 0 + 0 = 0> 1 + 0 = 1> 0 + 1 = 1> and:> 1*1 = 1> 0*0 = 0> 1*0 = 0> 0*1 = 0> The semantics are simple: + denotes disjunction and * denotes> conjunction. Negation is obvious. Either * with - or + with - forms> a complete set of connectives for the propositional logic.Yes, strong disjunction can be extended indefinitely to apply to many valuedlogics of any size. The analogy with modular arithmetic can then be seen.The Ô+' operator works fine if closed under a modulus and does nor try toescape to infinity.The choice of any sufficiently large modulus would allow practical closurein aritmetic, but division by the modulus is division by zero, so it wouldßy off anyway.The point is that Ô+' unrestricted is only closed by th impossibility ofreaching the (an) infinite domain by a finitistic operation.> This isn't true of course, because the cosmos would be long dead orcrunched> many times before> quite Ôsmall' numbers (eg 1,234!^9,999!) could be realised in thispedantic> way.> Reminds me of the Law of Big Numbers. (Most integers are very, very> large.)> No matter how elaborate or generalised new kinds of numbers may be, the> property of infinite extension is a necessary one.> Wittgenstein spent some time trying to define the meaning of the word> Ôgame', but gave up, concluding that the possible meanings wereincapable of> definite reduction. The same could be said for the word Ônumber' or anyterm> which resists precise definition.> Ah! A Wittgenstein reader. Excellent.> Perhaps you'll appreciate the following then. I believe we're using> the word number in two different contexts. If by number you mean> something along the lines of counting number, then infinity is> obviously not a number. In this context, 0 may not be a number> either, since one usually never counts zero objects. But I think> there's a more appropriate interpretation of the term which allows for> infinity to be considered a number with the same logical status as any> other number. This perspective, I think, is more desireable, since it> allows the term to apply to a larger class of objects which are often> used in this (logical) context.I entirely agree that number should be extended in any way we wish.My original point was that the Cantorian extension did not create a very'useful' class of numbers, because of their unsociability with what isusually regarded as numbers.I think G H Hardy's advice here is relevent.What is essential in mathematics is that its symbols should be capable ofsome interpretation; generally they are capable of many, and then, so far asmathematics is concerned, it does not matter which we adopt.'Provided we can give a clear meaning to the objects and relation of adomain, the creation of new and interesting mathematical objects isrespectable, whether or not they have been run through the set-theoreticsausage-machine to check if we have got a kosher sausage.Tony > By the way, I took this post off of sci.math and sci.physics, since> this is only tangential in the first and irrelevant in the second.Good. === Subject: Finding all possible bit patternsI am looking for a formula to calculate the number of all possibledifferent patterns of bit with the following parameters:- The vector has 13 free spaces.- There are 7 bits to place.For example, here are all the possibilities for a vector with 4 freespaces and 2 bits to place.110010101001011001010011Total = 6 possibilities.Thank you very much.ergagnon@distributel.net === Subject: Re: Finding all possible bit patterns> I am looking for a formula to calculate the number of all possible> different patterns of bit with the following parameters:> - The vector has 13 free spaces.> - There are 7 bits to place.> For example, here are all the possibilities for a vector with 4 free> spaces and 2 bits to place.> 1100> 1010> 1001> 0110> 0101> 0011> Total = 6 possibilities.> Thank you very much.> ergagnon@distributel.netHow about 13!/(6! * 7!) = 1716-- Using M2, Opera's revolutionary e-mail client: http://www.opera.com/m2/ === Subject: Re: Finding all possible bit patternsLooks like it should be the standard combination formula for n things takenm at a time,C(n,m) = n!/[m!(n-m)!]where n! = n * n-1 * n-2 * ... * 1.With n = 13 and m =7C(13,7) = 13!/[7!6!].= 1716It is easy to verify that for your example, C(4,2) = 4!/[2!2!] = 6> I am looking for a formula to calculate the number of all possible> different patterns of bit with the following parameters:> - The vector has 13 free spaces.> - There are 7 bits to place.> For example, here are all the possibilities for a vector with 4 free> spaces and 2 bits to place.> 1100> 1010> 1001> 0110> 0101> 0011> Total = 6 possibilities.> Thank you very much.> ergagnon@distributel.net === Subject: Re: Basic set theory question (indexed sets)|My friend easily concludes then, the|desired set is U, the universal set, but having read (the first third|of) Foundations of Set Theory, I am confused what U would mean in|this context.You said you read the definition of intersection in a topology book.Typically a topology is defined as a family of subsets of a set X,called open sets, that satisfies certain properties. One of them isclosure under arbitrary union. Another is that the empty set and Xare open. The fact that the empty set is open is a consequence ofclosure under arbitrary union, because the union of the emptyfamily is the empty set. So the empty set being open is redundant.Once in awhile it's mentioned that a topology can also be definedin terms of its family of closed sets, the complements in X of theopen sets. When this is done, the corresponding property is closureunder arbitrary intersection of closed subsets of X. The propertycorresponding to the empty set being open is that X is closed.Now, in such a setting the question is whether the correspondingimplication holds, that if a family of subsets of X is closed underarbitrary intersection, then X is a member of it. I would say thatit does, if one has defined intersection of a family of subsets ofX appropriately. There's a definition in category theory of theintersection of a family of subobjects of an object, and when thefamily is empty, the definition gives you the whole object as theintersection. Note that these are not merely sets, but setstogether with the inclusion in X.This is delicate. Unless the intersection is being taken as anintersection of subsets of a given set, the intersection is notdefined. So if you just start talking about arbitrary intersectionsof a family of subsets of X, they might assume that only nonemptyintersections are intended. In order to be sure that the readeris aware that one means to define the intersection of a familyof subsets of a set X with respect to the containing set X, andhence defining the empty intersection to be X, one pretty muchwill have to say so.But in a definition like this one, saying that the empty intersectionis taken to be X, so that X is required to be among the closed sets,is not much different from just saying that X is assumed to be amongthe closed sets. I would assume that in some cases, it simplifiesproofs which involve a lot of intersections of arbitrary families ofsubsets of a set X to start out by defining intersections of families ofsubsets of X the way I'm describing, so that X itself does not needto be treated as a special case, but it only would simplify it a littlebit, after all. And in the definition of topological space, usually oneworks with the family of open sets, where the problem doesn't arise,since an empty _union_ is always the empty set, independent ofwhether it's taken to be an empty family of subsets of X, or just anempty family of sets.Keith Ramsay === Subject: individual movement for 36 players?Originator: gls@news.monmouth.com (Col. G. L. Sicherman)A bridge player has asked me for an individual movement for 36 players.I don't have one, and I suspect that by conventional methods it would take a few years to find one. Any ideas?-:- What kept you this time? grumbled Pierre the Hideous. Nothing to get excited about, Sire, answered Narko smoothly. I was merely playing `Seven-Man Seymour' with the chambermaid. --Len Cool, _American Pie_-- Col. G. L. Sichermancolonel@mail.monmouth.com === Subject: Re: individual movement for 36 players?It might help to define a movement, because I for one am not familiar withthe term and think it's bridge jargon. Is this some sort of arrangementwhich puts the players together in teams or in opposition? === Subject: Re: individual movement for 36 players?Originator: gls@news.monmouth.com (Col. G. L. Sicherman)> It might help to define a movement, because I for one am not familiar with> the term and think it's bridge jargon. Is this some sort of arrangement> which puts the players together in teams or in opposition?Right on both counts. A movement is a way of organizing bridge play intorounds so that no individual (or pair, or team of four) plays against anyother more than once. Ideally each plays against each once, except thatin an individual movement each plays with each as partner once and asopponent twice. It is called a movement because the players move fromplace to place. Again ideally, the movements from place to place are thesame in each round.For example, here is an individual movement for 8 players: 1 7 2 x 4 5 x 6 8 3Player 1 stays put; the others move 2->3->4->5->6->7->8->2. You can verifythat after 7 rounds, every player has played with every other once as partnerand twice as opponent.The number of players need not be a multiple of four. I have worked outindividual movements for 9 players and 13 players, and I'm not the first.Barry Wolk once sent me an individual movement for 28 players in 9 rounds,in which every player plays once with every other, either as partner oropponent.-:- ÔPATAGEOMETRY: The study of those mathematical properties which remain invariant under brain transplants.-- Col. G. L. Sichermancolonel@mail.monmouth.com === Subject: Re: individual movement for 36 players?the people over there would know the answer. I looked in the Encyclopediaof Bridge and although there were several movements listed, none seemed todo what you wanted. Such beasts exist, because I played a round of bridgewith 20 people a short time ago that did what you want. Richard> It might help to define a movement, because I for one am not familiarwith> the term and think it's bridge jargon. Is this some sort of arrangement> which puts the players together in teams or in opposition?> Right on both counts. A movement is a way of organizing bridge play into> rounds so that no individual (or pair, or team of four) plays against any> other more than once. Ideally each plays against each once, except that> in an individual movement each plays with each as partner once and as> opponent twice. It is called a movement because the players move from> place to place. Again ideally, the movements from place to place are the> same in each round.> For example, here is an individual movement for 8 players:> 1 7> 2 x 4 5 x 6> 8 3> Player 1 stays put; the others move 2->3->4->5->6->7->8->2. You canverify> that after 7 rounds, every player has played with every other once aspartner> and twice as opponent.> The number of players need not be a multiple of four. I have worked out> individual movements for 9 players and 13 players, and I'm not the first.> Barry Wolk once sent me an individual movement for 28 players in 9 rounds,> in which every player plays once with every other, either as partner or> opponent.> -:-> ÔPATAGEOMETRY: The study of those mathematical properties> which remain invariant under brain transplants.> --> Col. G. L. Sicherman> colonel@mail.monmouth.com === Subject: Logarithmic spiral pole findingI would like to receive some help. My problem is as follows:I have the position of 3 points in relation to a coordinate system (mycoordinate system). I know that the 3 points belongs to a logarithmicspiral, whose parameters are known. I need to know how calculate theposition of the pole of the logarithmic spiral in relation to mycoordinate system.Miguel === Subject: the harmonic seriescan anyone offer any insights on finding the sum to n terms of the harmonicseries? i've tried everything, but every expression i try boils down to theoriginal series. i even ended up considering the logarithm of a variablebase, which is something i've never even seen before in a maths problem buteven this approach comes to nothing. i looked on mathworld and they gave aformula but it used a function i've heard of but know nothing about, thegamma function and a constant. i know the answer is already known, but thatsnot very satisifying, i'd rather derive it myself. By i way, i'm not atuniversity yet so i don't have access to the literature, thanks in advancefor any help. (harmonic series is sigma 1/n) === Subject: Re: the harmonic series>can anyone offer any insights on finding the sum to n terms of the harmonic>series? i've tried everything, but every expression i try boils down to the>original series. i even ended up considering the logarithm of a variable>base, which is something i've never even seen before in a maths problem but>even this approach comes to nothing. i looked on mathworld and they gave a>formula but it used a function i've heard of but know nothing about, the>gamma function and a constant. i know the answer is already known, but thats>not very satisifying, i'd rather derive it myself. By i way, i'm not at>university yet so i don't have access to the literature, thanks in advance>for any help. (harmonic series is sigma 1/n)If you want to understand why the partial sum of the harmonic series isthe Euler-Mascheroni constant plus the digamma function, see http://www.whim.org/nebula/math/partharm.htmlThe Euler-Mascheroni constant is approximately .5772156649... If youwant to see one method of computing this constant, see http://www.whim.org/nebula/math/eulergamma.htmlI have used this algorithm to compute over 10000 digits of gamma.If all you need is an approximation of the partial harmonic sum, youcan use the Euler-Maclaurin Sum Formula to get the asymptotic series n --- 1 > - --- k k=1 = log(n) + gamma 1 1 1 1 1 1 + -- - ----- + ------ - ------ + ------ - ------- + ... 2n 12n^2 120n^4 252n^6 240n^8 132n^10The error in this approximation should be less than 1/(47n^12).For more about the Euler-Maclaurin Sum Formula see http://mathworld.wolfram.com/ Euler-MaclaurinIntegrationFormulas.htmlor http://www.whim.org/nebula/math/eulermac.htmlRob Johnson take out the trash before replying === Subject: Re: the harmonic seriesgamma + ln nwhere gamma is the Euler-Mascheroni constanthttp://mathworld.wolfram.com/HarmonicSeries.htmlhttp:/ /mathworld.wolfram.com/Euler-MascheroniConstant.html === Subject : Re: the harmonic series> can anyone offer any insights on finding the sum to n terms of the harmonic> series? i've tried everything, but every expression i try boils down to the> original series. i even ended up considering the logarithm of a variable> base, which is something i've never even seen before in a maths problem but> even this approach comes to nothing. i looked on mathworld and they gave a> formula but it used a function i've heard of but know nothing about, the> gamma function and a constant. i know the answer is already known, but thats> not very satisifying, i'd rather derive it myself. By i way, i'm not at> university yet so i don't have access to the literature, thanks in advance> for any help. (harmonic series is sigma 1/n)Maple reports this in terms of the digamma function Psi...> sum(1/k, k = 1..n); Psi(n + 1) + gammaHere, gamma is Euler's constant, and the digamma functionPsi(x) = Gamma'(x)/Gamma(x) is the logarithmic derivativeof the Gamma function.-- http://www.math.ohio-state.edu/~edgar/ === Subject: Re: the harmonic series> can anyone offer any insights on finding the sum to n terms of the harmonic> series? i've tried everything, but every expression i try boils down to the> original series. i even ended up considering the logarithm of a variable> base, which is something i've never even seen before in a maths problem but> even this approach comes to nothing. i looked on mathworld and they gave a> formula but it used a function i've heard of but know nothing about, the> gamma function and a constant. i know the answer is already known, but thats> not very satisifying, i'd rather derive it myself. By i way, i'm not at> university yet so i don't have access to the literature, thanks in advance> for any help. (harmonic series is sigma 1/n)The harmonic series is 1 + 1/2 + 1/3 + 1/4 +...This sum diverges, becoming larger than any fixed value when enough terms are added, so that the sum of all terms does not exist.One can, obviously, find a value for any finite partial sum, but, unfortunately, there are no nice formulae for the values of such partial sums. === Subject: Relationship of inertial and gravitational inertiaX-PNG: alt.sci.physicsX-NewsOnePostHost: 68.118.203.42X-NewsOnePostAddr: 68.118.203.42X-NOTrace: FMKKBLHLCLDCJBIBMOBLOLNJELKDFCOBCNEMDMIKof Transportation's Bridge Design Unit, we tried to make them complete, briefand easy to understand: I think the following statement defining mass is whatwe might have written if we had been asked to do so: Mass can be defined in terms of either (a) the mutual resistance of twoand/ or passing throught the exact same place and/or (b) the mutualresistance of the penetration of a body resting on a planet's terra firmasurface. Where the bodies physically exert mutual force on each other andare accelerated - their velocities forcibly changed - inversely to theirweight, and/or massiveness. One body cannot exert more (mutual) force on the other's nor can one exertgreater impulse's than the other's. The impenetrability of matter, is theprinciple that relates these two seemingly unrelated phenomea, and makes thisa scientific theory. ----- Posted via NewsOne.Net: Free (anonymous) Usenet News via the Web ----- http://newsone.net/ -- Free reading and anonymous posting to 60,000+ groups NewsOne.Net prohibits users from posting spam. If this or other postsmade through NewsOne.Net violate posting guidelines, email abuse@newsone.net === Subject: Stupid QuestionsIs the empty set open or closed?I'm sure this depends on context.Does it ever make sense to talk about an open empty set?Let x_n be a countable sequence that contains everyrational number in (-1,1).Define a new sequence x'_n = x_n + r.If r is an algebraic irrational, like sqrt(2), does x'_nonly contain algebraic irrationals?If r is a transcendental irrational does x'_n onlycontain transcendentals? === Subject: Re: Stupid Questions> Is the empty set open or closed?> I'm sure this depends on context.> Does it ever make sense to talk about an open empty set?Every element of an empty set does indeed satisfy the requirement to beopen. The empty set does that a lot - it trivially satisfies criteria by notever testing them.> Let x_n be a countable sequence that contains every> rational number in (-1,1).> Define a new sequence x'_n = x_n + r.> If r is an algebraic irrational, like sqrt(2), does x'_n> only contain algebraic irrationals?Well, if x' = x + r, where r is an algebraic irrational, you can find apolynomial R(y) such that y=r is the root of R. R(y-x) can still be seen asa polynomial in y and has root x + r.Since, if I remember correctly, the algebraic irrationals are defined asroots of polynomials with integral coefficients, you can transform R intosuch a polynomial by multiplying R by q when x = p/q.> If r is a transcendental irrational does x'_n only> contain transcendentals?Same thing here, I believe. Were you able to find R(y-x) such that x + r wasthe root of this equation (i.e. it is algebraic), then you would also haveR(y) with r as a root, implying that r is not transcendental. === Subject: Re: Stupid Questions> Is the empty set open or closed?> I'm sure this depends on context.> Does it ever make sense to talk about an open empty set?> Every element of an empty set does indeed satisfy the requirement to be> open. The empty set does that a lot - it trivially satisfies criteria bynot> ever testing them.Sorry, the phrase I'm looking for here is vacuously true. === Subject: Re: Stupid Questions>> Is the empty set open or closed?>> I'm sure this depends on context.>> Does it ever make sense to talk about an open empty set?>> Every element of an empty set does indeed satisfy the requirement to be>> open. The empty set does that a lot - it trivially satisfies criteria by>not>> ever testing them.>Sorry, the phrase I'm looking for here is vacuously true.Aha! Stupid question deserves stupid answer!Derek Holt. === Subject: Re: Stupid Questions>> Is the empty set open or closed?>> I'm sure this depends on context.>> Does it ever make sense to talk about an open empty set?>> Every element of an empty set does indeed satisfy the requirement to be>> open. The empty set does that a lot - it trivially satisfies criteriaby>not>> ever testing them.>Sorry, the phrase I'm looking for here is vacuously true.> Aha! Stupid question deserves stupid answer!> Derek Holt.Not sure what you mean here. Did I mislead the OP? === Subject: Re: Stupid Questions> Is the empty set open or closed?For any topology on any non-empty set S, both S and {} are both open and closed.I'm sure this depends on context.> Does it ever make sense to talk about an open empty set?See above. Let x_n be a countable sequence that contains every> rational number in (-1,1).> Define a new sequence x'_n = x_n + r.In the answers to the following questions, the answers are the same for an arbitrary sequence, x_n, of rationalsSequences are, as sets, countable by definition, possibly finite, since they are defined as images of N for some function.If r is an algebraic irrational, like sqrt(2), does x'_n> only contain algebraic irrationals?Yes, only but not all.If r is a transcendental irrational does x'_n only> contain transcendentals?Yes. === Subject: Request for comments on attempts of proofsIn the ever lasting quest to learn, i hereby request those who are willing ,to comment on the following proofs i constructed. If i errored, im happy toreceive corrected versions :)(1) if gf is surjection, show that g is a surjectionIf gf is surjection then we have gf(x) = z. Let f(x) = y. We then havegf(x) = g(y) = z and thus g is a surjection(2) if gf is injection, show that f is a injectionOm gf is injection we havegf(x) = gf(x') --> f(x) = f(x')Let f(x) = y, Let f(x') = y' and we getg(y) = g(y') --> y = y' thus g is an injectionNow suppose f IS NOT an injection, then there are x != x such that f(x) =f(x'), g(f(x)) = g(f(x')), gf(x) = gf(x') --> gf is not an injection, whichis a contradiction.(3) Show that if gf and g are both bijections then f is also a bijectionIf gf och g are both .8ar bijections we have:gf has an inverse (gf)^-1g has an en inverse (g)^-1Suppose f is a bijection, then it has an inverse f^-1 such that (gf)^-1 =f^-1g^-1 since gf(f^-1g^-1) = gg^-1 = i. Now suppose f is NOT a bijectionthen f has no inverse -> gf has no inverse -> gf is not a bijection, whichis a contradiction. === Subject: Re: Request for comments on attempts of proofs Adjunct Assistant Professor at the University of Montana.>In the ever lasting quest to learn, i hereby request those who are willing ,>to comment on the following proofs i constructed. If i errored, im happy to>receive corrected versions :)>(1) if gf is surjection, show that g is a surjection>If gf is surjection then we have gf(x) = z. Let f(x) = y. We then have>gf(x) = g(y) = z and thus g is a surjectionThe idea is right, but if I were grading this I would deduct pointsfor being unclear. Here is how I would write it in solutions handedout to students (which means it is much more verbose than it reallyneeds to be; I will enclose in brackets things that probably don'thave to be there, but make it for easier reading if they are, IMHO):Say f:X->Y, and g:Y->Z. To show that g is a surjection, let z in Z. [Wewant to show that there exists y in Y such that g(y)=z.] Since gf is asurjection, there exists x in X such that gf(x)=z. Then f(x)=y is inY, and g(y)=g(f(x))=z, so g is a surjection.>(2) if gf is injection, show that f is a injection>Om gf is injection we have>gf(x) = gf(x') --> f(x) = f(x')>Let f(x) = y, Let f(x') = y' and we get>g(y) = g(y') --> y = y' thus g is an injectionThis is false. Consider X={1}, Y = {2,3}, Z = {4}, f sends 1 to 2; andg sends 2 and 3 to 4. Then gf is an injection, but g is not aninjection.Your error is that the only thing you can prove is that g ->RESTRICTEDTO THE IMAGE OF f<- is an injection. But since f is not necessarilysurjective, that does not get you all the way to g being a surjection.>Now suppose f IS NOT an injection, then there are x != x such that f(x) =>f(x'), g(f(x)) = g(f(x')), gf(x) = gf(x') --> gf is not an injection, which>is a contradiction.This is really all you have to do. Note that you are not using that gis an injection.If you want a direct proof:Let x and x' be elements of X such that f(x)=f(x'). We want to showthat x = x'. Since f(x)=f(x'), we must have gf(x) = g(f(x))=g(f(x')) =gf(x'). Since gf is injective, this implies that x=x', as desired.Or, using the other definition of injective:Let x and x' be different elements of X. Then gf(x) is different fromgf(x'), this because gf is injective. Thus, g(f(x)) is different fromg(f(x')), and therefore, f(x) must be different from f(x'). Thisproves f is injective.>(3) Show that if gf and g are both bijections then f is also a bijection>If gf och g are both .8ar bijections we have:>gf has an inverse (gf)^-1>g has an en inverse (g)^-1>Suppose f is a bijection,Hey! You are supposed to PROVE that. Do not suppose it!> then it has an inverse f^-1 such that (gf)^-1 =>f^-1g^-1 since gf(f^-1g^-1) = gg^-1 = i.True, but not really relevant. You assumed what you wanted to prove.> Now suppose f is NOT a bijection>then f has no inverse -> gf has no inverseThis is false. In the example I gave above, f is not a bijection,but gf has an inverse, since it IS a bijection.> -> gf is not a bijection, which>is a contradiction.Try using what you proved above: since gf is a bijection, you alreadyknow that it is an injection, and therefore that f must beinjective. So all you need to do is prove that it is surjective. Let yin Y. Then there exists an x in X such that gf(x) = g(y), because gfis surjective; since g is injective, this implies that f(x)=y, so f issurjective. Since it is both injective and surjective, it isbijective.If you don't have this equivalence for set functions, then you werealmost on the right track: you want to show that f has an inverse. Youknow that there is a function h such that h(gf) = identity_X, and(gf)h = identity_Z. And you know there is a function k such that kg =identity_Y, and gk = identity_Z. You want to find an inverse to f.If f had an inverse, then you know that you would have h = (gf)^-1 =f^{-1}g^{-1}, so hg = f^{-1}g^{-1}g = f^{-1}. This tells you what youmight want to try as an inverse to f: namely, hg.First, note that hg is indeed a function from Y to X, as it should beif it is to be the inverse of f. Next, we need to show that (hg)f =id_X, and that f(hg) = id_Y.Let x in X. Then (hg)f(x) = h(gf(x)) = x, because h(gf)=id_X. So thiscomposition is the identity.And let y in Y. Then we want to show that f(hg(y)) = y. This is theone which is a bit more difficult; clearly we need to use k at somepoint. We do so here. We can compose with the identity of Y withoutchanging the result, so we have:f(hg(y)) = id_Y(f(hg(y))) = (kg)(f(hg(y))) = k(gf(h(g(y))))But gf(h) is the identity on Z, and g(y) is in Z, so this is equal to: = k(g(y)) = kg(y) = ybecause kg is the identity on Y. This proves that f has a two sidedinverse, namely hg, so f is bijective.-- === Subject: Re: Request for comments on attempts of proofsFirst let me extend a sincere thanks for taking the time to comment on myhumble proof attempts. One thing which puzzles me when trying to provesomething in general, be it graph theory or something else, is WHERE tostart. Is there some general thought procedure on how to start out a proof ?>In the ever lasting quest to learn, i hereby request those who arewilling ,>to comment on the following proofs i constructed. If i errored, im happyto>receive corrected versions :)>(1) if gf is surjection, show that g is a surjection>If gf is surjection then we have gf(x) = z. Let f(x) = y. We then have>gf(x) = g(y) = z and thus g is a surjection> The idea is right, but if I were grading this I would deduct points> for being unclear. Here is how I would write it in solutions handed> out to students (which means it is much more verbose than it really> needs to be; I will enclose in brackets things that probably don't> have to be there, but make it for easier reading if they are, IMHO):> Say f:X->Y, and g:Y->Z. To show that g is a surjection, let z in Z. [We> want to show that there exists y in Y such that g(y)=z.] Since gf is a> surjection, there exists x in X such that gf(x)=z. Then f(x)=y is in> Y, and g(y)=g(f(x))=z, so g is a surjection.>(2) if gf is injection, show that f is a injection>Om gf is injection we have>gf(x) = gf(x') --> f(x) = f(x')>Let f(x) = y, Let f(x') = y' and we get>g(y) = g(y') --> y = y' thus g is an injection> This is false. Consider X={1}, Y = {2,3}, Z = {4}, f sends 1 to 2; and> g sends 2 and 3 to 4. Then gf is an injection, but g is not an> injection.> Your error is that the only thing you can prove is that g ->RESTRICTED> TO THE IMAGE OF f<- is an injection. But since f is not necessarily> surjective, that does not get you all the way to g being a surjection.>Now suppose f IS NOT an injection, then there are x != x such that f(x) =>f(x'), g(f(x)) = g(f(x')), gf(x) = gf(x') --> gf is not an injection,which>is a contradiction.> This is really all you have to do. Note that you are not using that g> is an injection.> If you want a direct proof:> Let x and x' be elements of X such that f(x)=f(x'). We want to show> that x = x'. Since f(x)=f(x'), we must have gf(x) = g(f(x))=g(f(x')) => gf(x'). Since gf is injective, this implies that x=x', as desired.> Or, using the other definition of injective:> Let x and x' be different elements of X. Then gf(x) is different from> gf(x'), this because gf is injective. Thus, g(f(x)) is different from> g(f(x')), and therefore, f(x) must be different from f(x'). This> proves f is injective.>(3) Show that if gf and g are both bijections then f is also a bijection>If gf och g are both .8ar bijections we have:>gf has an inverse (gf)^-1>g has an en inverse (g)^-1>Suppose f is a bijection,> Hey! You are supposed to PROVE that. Do not suppose it!> then it has an inverse f^-1 such that (gf)^-1 =>f^-1g^-1 since gf(f^-1g^-1) = gg^-1 = i.> True, but not really relevant. You assumed what you wanted to prove.> Now suppose f is NOT a bijection>then f has no inverse -> gf has no inverse> This is false. In the example I gave above, f is not a bijection,> but gf has an inverse, since it IS a bijection.> -> gf is not a bijection, which>is a contradiction.> Try using what you proved above: since gf is a bijection, you already> know that it is an injection, and therefore that f must be> injective. So all you need to do is prove that it is surjective. Let y> in Y. Then there exists an x in X such that gf(x) = g(y), because gf> is surjective; since g is injective, this implies that f(x)=y, so f is> surjective. Since it is both injective and surjective, it is> bijective.> If you don't have this equivalence for set functions, then you were> almost on the right track: you want to show that f has an inverse. You> know that there is a function h such that h(gf) = identity_X, and> (gf)h = identity_Z. And you know there is a function k such that kg => identity_Y, and gk = identity_Z. You want to find an inverse to f.> If f had an inverse, then you know that you would have h = (gf)^-1 => f^{-1}g^{-1}, so hg = f^{-1}g^{-1}g = f^{-1}. This tells you what you> might want to try as an inverse to f: namely, hg.> First, note that hg is indeed a function from Y to X, as it should be> if it is to be the inverse of f. Next, we need to show that (hg)f => id_X, and that f(hg) = id_Y.> Let x in X. Then (hg)f(x) = h(gf(x)) = x, because h(gf)=id_X. So this> composition is the identity.> And let y in Y. Then we want to show that f(hg(y)) = y. This is the> one which is a bit more difficult; clearly we need to use k at some> point. We do so here. We can compose with the identity of Y without> changing the result, so we have:> f(hg(y)) = id_Y(f(hg(y))) = (kg)(f(hg(y)))> = k(gf(h(g(y))))> But gf(h) is the identity on Z, and g(y) is in Z, so this is equal to:> = k(g(y)) = kg(y) = y> because kg is the identity on Y. This proves that f has a two sided> inverse, namely hg, so f is bijective.> -- === ==> Arturo Magidin> magidin@math.berkeley.edu === Subject: Re: Request for comments on attempts of proofs Adjunct Assistant Professor at the University of Montana.>First let me extend a sincere thanks for taking the time to comment on my>humble proof attempts. One thing which puzzles me when trying to prove>something in general, be it graph theory or something else, is WHERE to>start. Is there some general thought procedure on how to start out a proof ?The thought procedure of course varies from person to person. But Ithink one of the most common things I see in students who seize upand say the can't come up with an answer is that they cannot see intheir heads all the steps from beginning to end, and therefore thinkthey cannot get started...Here are some suggestions that I found useful when I was student, andwhich to a large extent I still follow (as you can see in some of mypublished proofs):1. First, write down explicitly what you are given. Then write down explicitly what you want to prove. 2. If either (or both) of these conditions are given in words pursuant to a relevant definition, then rewrite them again using the definition. [If you have a number of equivalent definitions for the notion, then this is not necessarily a good idea; for example, if you know that a set function is bijective if and only if it has an inverse, but that it is also true that a set function is bijective if and only if it is both injective and surjective, then writing down either down will probably Ôlock you' onto one of the possibilities, while the other may be more useful. In that case, it is usually more constructive to leave the wording as is]3. If the conclusion you want to prove is an implication, you may assume the ->antecedent<-. E.g., if you want to prove that a certain function is surjective, surjectivity means: If y in Y, then there exists x in X such that f(x)=y So you may start your proof by assuming that you are given a y in Y. Then write down what you want to prove: that there exists an x in X such that f(x) = y.4. Try a number of things: if your assumptions have some obvious implications, try writing them down. If your conclusion would follow from something else that sounds simpler, write that down and aim for ->that<- instead of the conclusion. Etc.5. In Spanish, there is an expression for cooking up a process when you know what the answer should be, by assuming that you have an answer and figuring out what that answer should Ôlook like.' We call it spooning the answer. I have not encountered an expression like that in English, but you saw me using this in the proof that if gf is bijective and g is bijective, then f is bijective: to figure out what the inverse of f should be, we assume it is bijective, in which case we would necessarily have (gf)^{-1} = f^{-1}g^{-1}, so f^{-1} = f^{-1}g^{-1}g = (gf)^{-1}g. This tells you what function you should try for the inverse of f: (gf)^{-1}g. What you need to be careful here is that this sort of thing is what you use to come up with a proof, but is not usually part of the proof (it is part of the intuition behind the proof; in a properly organized proof, you would make this observation as an aside, not part of the proof, to explain what you are doing, but your proof would NOT contain any mention of f^{-1} until you have proven that such a thing exists). The point being that Ôspooning the answer' can be useful when you don't really know how to get to the answer.It is important not to be discouraged if you cannot see the wholeproof from beginning to end from the start. Just start taking one stepat a time from the premises in the (general) direction of theconclusion you want. With some luck and experience, somewhere alongthe line you will finally see the full connecting trail. When you aredone, go back through your proof and see if you took some unnecessarydetour, or some extra assumption.(Another important thing is to remember that, usually, ALL hypothesisshould come into play. It is certainly possible that you are assumingmore than you need for a proof, but especially when you are stilllearning, all hypothesis turn out to be important. So if you managedto complete the proof of an exercise in the book without using all thehypothesis, go through it with particular care to make sure that youreally did not need it. More often than not, you've made a mistake.)-- === Subject: Re: Request for comments on attempts of proofs>First let me extend a sincere thanks for taking the time to comment on my>humble proof attempts. One thing which puzzles me when trying to prove>something in general, be it graph theory or something else, is WHERE to>start. Is there some general thought procedure on how to start out a proof ?The first thing would be to figure out exactly what it is you'retrying to proof. Then see what you're given and try to step from onelogical statement to the next one. Look at Arturo Magidin's proof:>> Say f:X->Y, and g:Y->Z. To show that g is a surjection, let z in Z.[We>> want to show that there exists y in Y such that g(y)=z.]Here he tells what we are trying to accomplish.>> Since gf is a surjectionThis is what we're given.>> there exists x in X such that gf(x)=z. Then f(x)=y is in>> Y, and g(y)=g(f(x))=z, so g is a surjection.And this is what steps we have to take. Usually this part involves afew pages of scribblings, a lot of coffee and starting over a coupleof times. Then you write down the proof and proceed to spend an hourconvincing yourself it can't possibly be right.-- I'm not interested in mathematics that might have anythingto do with reality. -- Easterly, in sci.math === Subject: Re: Request for comments on attempts of proofs>In the ever lasting quest to learn, i hereby request those who arewilling ,>to comment on the following proofs i constructed. If i errored, im happyto>receive corrected versions :)>(1) if gf is surjection, show that g is a surjection>If gf is surjection then we have gf(x) = z. Let f(x) = y. We then have>gf(x) = g(y) = z and thus g is a surjection> The idea is right, but if I were grading this I would deduct points> for being unclear. Here is how I would write it in solutions handed> out to students (which means it is much more verbose than it really> needs to be; I will enclose in brackets things that probably don't> have to be there, but make it for easier reading if they are, IMHO):> Say f:X->Y, and g:Y->Z. To show that g is a surjection, let z in Z. [We> want to show that there exists y in Y such that g(y)=z.] Since gf is a> surjection, there exists x in X such that gf(x)=z. Then f(x)=y is in> Y, and g(y)=g(f(x))=z, so g is a surjection.>(2) if gf is injection, show that f is a injection>Om gf is injection we have>gf(x) = gf(x') --> f(x) = f(x')>Let f(x) = y, Let f(x') = y' and we get>g(y) = g(y') --> y = y' thus g is an injection> This is false. Consider X={1}, Y = {2,3}, Z = {4}, f sends 1 to 2; and> g sends 2 and 3 to 4. Then gf is an injection, but g is not an> injection.> Your error is that the only thing you can prove is that g ->RESTRICTED> TO THE IMAGE OF f<- is an injection. But since f is not necessarily> surjective, that does not get you all the way to g being a surjection.>Now suppose f IS NOT an injection, then there are x != x such that f(x) =>f(x'), g(f(x)) = g(f(x')), gf(x) = gf(x') --> gf is not an injection,which>is a contradiction.> This is really all you have to do. Note that you are not using that g> is an injection.> If you want a direct proof:> Let x and x' be elements of X such that f(x)=f(x'). We want to show> that x = x'. Since f(x)=f(x'), we must have gf(x) = g(f(x))=g(f(x')) => gf(x'). Since gf is injective, this implies that x=x', as desired.> Or, using the other definition of injective:> Let x and x' be different elements of X. Then gf(x) is different from> gf(x'), this because gf is injective. Thus, g(f(x)) is different from> g(f(x')), and therefore, f(x) must be different from f(x'). This> proves f is injective.>(3) Show that if gf and g are both bijections then f is also a bijection>If gf och g are both .8ar bijections we have:>gf has an inverse (gf)^-1>g has an en inverse (g)^-1>Suppose f is a bijection,> Hey! You are supposed to PROVE that. Do not suppose it!> then it has an inverse f^-1 such that (gf)^-1 =>f^-1g^-1 since gf(f^-1g^-1) = gg^-1 = i.> True, but not really relevant. You assumed what you wanted to prove.> Now suppose f is NOT a bijection>then f has no inverse -> gf has no inverse> This is false. In the example I gave above, f is not a bijection,> but gf has an inverse, since it IS a bijection.> -> gf is not a bijection, which>is a contradiction.> Try using what you proved above: since gf is a bijection, you already> know that it is an injection, and therefore that f must be> injective. So all you need to do is prove that it is surjective. Let y> in Y. Then there exists an x in X such that gf(x) = g(y), because gf> is surjective; since g is injective, this implies that f(x)=y, so f is> surjective. Since it is both injective and surjective, it is> bijective.> If you don't have this equivalence for set functions, then you were> almost on the right track: you want to show that f has an inverse. You> know that there is a function h such that h(gf) = identity_X, and> (gf)h = identity_Z. And you know there is a function k such that kg => identity_Y, and gk = identity_Z. You want to find an inverse to f.> If f had an inverse, then you know that you would have h = (gf)^-1 => f^{-1}g^{-1}, so hg = f^{-1}g^{-1}g = f^{-1}. This tells you what you> might want to try as an inverse to f: namely, hg.> First, note that hg is indeed a function from Y to X, as it should be> if it is to be the inverse of f. Next, we need to show that (hg)f => id_X, and that f(hg) = id_Y.> Let x in X. Then (hg)f(x) = h(gf(x)) = x, because h(gf)=id_X. So this> composition is the identity.> And let y in Y. Then we want to show that f(hg(y)) = y. This is the> one which is a bit more difficult; clearly we need to use k at some> point. We do so here. We can compose with the identity of Y without> changing the result, so we have:> f(hg(y)) = id_Y(f(hg(y))) = (kg)(f(hg(y)))> = k(gf(h(g(y))))> But gf(h) is the identity on Z, and g(y) is in Z, so this is equal to:> = k(g(y)) = kg(y) = y> because kg is the identity on Y. This proves that f has a two sided> inverse, namely hg, so f is bijective.> -- === ==> Arturo Magidin> magidin@math.berkeley.eduIsn't it great that within the mud, diamonds like you, amongst others,freely give your time for absolutely zilch? The universe must befundamentally benevolent and no, I'm not a space cadet by artificial means;)jmc === Subject: Anyone have a user manual for Sharp EL-9200C Graphing Scientific calculator?Anyone have a spare user manual for Sharp EL-9200C Graphing Scientific calculator?I would love to buy one, but I'm having trouble locating one. Maybe you have a manual andlost your calc? ;)It's an amazing machine, and I want to maximize its potential for my application-- Myron SamilaToronto, ON CanadaSamila Racinghttp://204.101.251.229/myronx19 === Subject: Re: Anyone have a user manual for Sharp EL-9200C Graphing Scientific calculator?Nowadays manuals are often put on line by the manufacturers. So checkSharp's web page and see. === Subject: Re: Anyone have a user manual for Sharp EL-9200C Graphing Scientific calculator?> Nowadays manuals are often put on line by the manufacturers. So check> Sharp's web page and see.The manuals that Sharp provide online are for only their current line up, they do notsupport older machines at all. === Subject: Re: JSH: Factor of x > Yes. b_2(x) is a root of the polynomial: > (b-1)^2 - (x - 1)(b - 1) + 7(x^2 + x) > b^2 - 2b + 1 - (x - 1)b + x - 1 + 7(x^2 + x) > b^2 - (x + 1)b + (7x^2 + 2x) > So what? Both a's and both b's have a factor in common with x. > Nothing strange here. Oh, you are wondering how it is possible > that both a_2(x) and b_2(x) = a_2(x) + 1 can have a factor in > common with x. Nope.In that case I am wondering about the reason of your posting. > And something similar is true in your example. The factor a_1(x) has in > common with x is the same as the factor b_2(x) has in common with x, and > is in general coprime to the factor a_2(x) has in common with x. > What about at x=0?What about it? Oh, 0 is not a factor of anything, and anything is a factorof 0. So the factor a_2(0) has in common with 0 is a_2(0). The factorb_2(0) has in common with 0 is b_2(0). Moreover, a_2(0) and b_2(0) arecoprime (by your definition of b_2(x)). But indeed, my statement thatthe factor a_1(x) has in common with x is the same as the factor b_2(x)has in common with x is too sweeping. It is true when x is coprime to allother factors of the constant term of the quadratic. It may not be truewhen that is false. > BTW, also something similar occurs in the integers. Both 15 and 16 have > a factor in common with 6. In the algebraic integers the situation is > a bit more complicated, because there are so many divisors and no primes. > You're babbling. What you just said just doesn't relate to the issue at > hand.How would I know when you do not tell what the issue at hand is? > Yup, has already been shown. Given an irreducible, primitive, monic > polynomial with integer coefficients: > x^n + c_(n-1).x^(n-1) + ... + c1.x + c0 > each of the roots has a factor in common with each of the prime divisors > of c0. > What about at x=0?Is x = 0 a root? Yes, it is when c0 = 0. In that case the polynomial isnot irreducible, primitive, monic. You may take the first word from thatlist. > It might seem too esoteric with x there, so let's put in a value for > x, and let x=13. Then > a^2 - 12a + 7(13)(14) > and we already know that *one* of the a's, is coprime to 13, or wait, > do we? > No, we don't, *both* are not coprime to 13. > So you say that *both* a_1(13) and a_2(13) have some factor in common > with 13. > Interesting.Why? The proof has been given a number of times. Only you do not believeit.-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: JSH: Factor of x> Yes. b_2(x) is a root of the polynomial:> (b-1)^2 - (x - 1)(b - 1) + 7(x^2 + x)> b^2 - 2b + 1 - (x - 1)b + x - 1 + 7(x^2 + x)> b^2 - (x + 1)b + (7x^2 + 2x)> So what? Both a's and both b's have a factor in common with x.> Nothing strange here. Oh, you are wondering how it is possible> that both a_2(x) and b_2(x) = a_2(x) + 1 can have a factor in> common with x. Nope.In that case I am wondering about the reason of your posting.Ok, so you're wondering. > And something similar is true in your example. The factor a_1(x) has in> common with x is the same as the factor b_2(x) has in common with x, and> is in general coprime to the factor a_2(x) has in common with x.> What about at x=0?What about it? Oh, 0 is not a factor of anything, and anything is a factor> of 0. So the factor a_2(0) has in common with 0 is a_2(0). The factorGo the other way, as the question is about how a_1(x) and b_2(x) havefactors in common with x such that they *both* equal 0 when x=0On the other hand a_2(0) = -1, which is as far from 0 as needed DikWinter.The point is that *most* people with some mathematical knowledge wouldaccept that if you have a function f(x), where f(0) = 0, then hey,maybe that means that it has some factor in common with, like youcould havef(x) = x^{1/3} where x^{1/3} is, of course, a factor in common with x.My point here is that posters like you are challenging what mostpeople accept as rather basic, and your coy answer here, I think,indicates that you *know* what you're doing, but are doing it anyway.> b_2(0) has in common with 0 is b_2(0). Moreover, a_2(0) and b_2(0) are> coprime (by your definition of b_2(x)). But indeed, my statement that> the factor a_1(x) has in common with x is the same as the factor b_2(x)> has in common with x is too sweeping. It is true when x is coprime to all> other factors of the constant term of the quadratic. It may not be true> when that is false.> There's one point here which is why it's in the subject line: Factorof xI think MOST people with some mathematical understanding will acceptthat if you have a function of x, f(x) and that function equals 0 whenx=0, then it makes sense that f(x) has some factor of x in common withx.Now I've found myself thinking about that a lot, and hopefully here Ican make a case for why I keep at this when people like Dik Winter orArturo Magidin or all these other posters come after me and claim it'sall settled and I'm wrong.What I think is not wrong is that with f(x), if f(0)=0, then there'ssome factor of x in there!!!> BTW, also something similar occurs in the integers. Both 15 and 16 have a factor in common with 6. In the algebraic integers the situation is> a bit more complicated, because there are so many divisors and no primes.> You're babbling. What you just said just doesn't relate to the issue at> hand.How would I know when you do not tell what the issue at hand is?> The subject line *gives* the issue at hand.I'm making a point where I hope to rely on what mathematically awarereaders know themselves happens around 0.> Yup, has already been shown. Given an irreducible, primitive, monic> polynomial with integer coefficients:> x^n + c_(n-1).x^(n-1) + ... + c1.x + c0> each of the roots has a factor in common with each of the prime divisors> of c0.> What about at x=0?Is x = 0 a root? Yes, it is when c0 = 0. In that case the polynomial is> not irreducible, primitive, monic. You may take the first word from that> list.> Basically the functions a_1(x) and a_2(x) as defined simply travelover an interesting path where the polynomial that defines them goesfrom being in general irreducible over *integers* with an integer x,and your position is that the *functions* themselves are aware of thisin some way.My position is that functions are functions.Functional behavior is not that intelligent as to worry about whetheror not some polynomial has integer roots are not!I mean for readers who don't get it yet, the distinction Dik Winter ismaking here is the difference betweenx^2 + 2x + 1, which is reducible, and x^2 + 2x + 2, which, of course is not.> It might seem too esoteric with x there, so let's put in a value for> x, and let x=13. Then> a^2 - 12a + 7(13)(14)> and we already know that *one* of the a's, is coprime to 13, or wait,> do we?> No, we don't, *both* are not coprime to 13.> So you say that *both* a_1(13) and a_2(13) have some factor in common> with 13.> Interesting.Why? The proof has been given a number of times. Only you do not believe> it.Well, knowing that with *functions* a_1(x) and a_2(x) that a_1(0) = 0,and a_2(0) = -1, it is NOT a great leap to at least consider thepossibility that a_1(0) has some factor in common with x, while a_2(0)does not. === Subject: Re: JSH: Factor of xof x>I think MOST people with some mathematical understanding will accept>that if you have a function of x, f(x) and that function equals 0 when>x=0, then it makes sense that f(x) has some factor of x in common with>x.People with little mathematical understanding might think that. >Now I've found myself thinking about that a lot, and hopefully here I>can make a case for why I keep at this when people like Dik Winter or>Arturo Magidin or all these other posters come after me and claim it's>all settled and I'm wrong.>What I think is not wrong is that with f(x), if f(0)=0, then there's>some factor of x in there!!!Let f(x) be the function yielding the count of rational primes psuch that 0 <= p <= x. Then f(0) = 0, but f(3) = 2 and in anysubring of the complex numbers containing 2 and 3 their only commondivisors are the units of that ring. Since no leading expert in number theory could possibly commit theschoolboy howler of thinking that x and f(x) had any nontrivialcommon factor, this is all a joke, right?I'm really glad to see that you are recovering your sense of humourbut perhaps you should consider changing your gag writer.It's not a very funny joke.John Roberts-Jones === Subject: Re: JSH: Factor of x> Yes. b_2(x) is a root of the polynomial:> (b-1)^2 - (x - 1)(b - 1) + 7(x^2 + x)> b^2 - 2b + 1 - (x - 1)b + x - 1 + 7(x^2 + x)> b^2 - (x + 1)b + (7x^2 + 2x)> So what? Both a's and both b's have a factor in common with x.> Nothing strange here. Oh, you are wondering how it is possible> that both a_2(x) and b_2(x) = a_2(x) + 1 can have a factor in> common with x. Nope.In that case I am wondering about the reason of your posting.Ok, so you're wondering.> He's right to wonder. It looks like an excursion which haslittle to do with your main difficulty with Rick Decker'spolynomial. The Decker quadratic is 7*(25*x^2 + 30*x + 2). Note that in the Decker quadratic, as in your previouscubic, the polynomial variable is 5. Note that if youreplace this with t, you have Q(x) = 7*(x^2*t^2 +6*x*t + 2).When 5 was there instead of t, Decker was able to change the form of the polynomial to look like: 7*((x^2 + x)*5^2 + (x - 1)*5 + 7), and to consider a factorization of the form (5*a_1(x) + 7)*(5*a_2(x) + 7).Note that this doesn't work when you substitute tfor 5: Q(x) would become 7*((x^2 + x)*t^2 + (-x*t + 6*x)*t + 2),which essentially goes nowhere. Using 5 as a polynomial variable in either yourcubic or Decker's quadratic makes essentially nosense. The derivation of the fact that a_1(x) and a_2(x)are roots of a^2 - (x - 1)*a + 7*(x^2 + x)goes back to a factorization in terms of a *polynomial*variable like t, not a *fixed constant* like 5. Perhapsyou can escape the Decker example by simply and logicallyreplacing 5 by t or some other polynomial variable. And something similar is true in your example. The factor a_1(x) has in> common with x is the same as the factor b_2(x) has in common with x, and> is in general coprime to the factor a_2(x) has in common with x.> What about at x=0?What about it? Oh, 0 is not a factor of anything, and anything is a factor> of 0. So the factor a_2(0) has in common with 0 is a_2(0). The factorGo the other way, as the question is about how a_1(x) and b_2(x) have> factors in common with x such that they *both* equal 0 when x=0On the other hand a_2(0) = -1, which is as far from 0 as needed Dik> Winter.The point is that *most* people with some mathematical knowledge would> accept that if you have a function f(x), where f(0) = 0, then hey,> maybe that means that it has some factor in common with, like you> could havef(x) = x^{1/3} where x^{1/3} is, of course, a factor in common with x. Let f(x) = sqrt(x^3 + 1) - sqrt(x^4 + 1). Note that f(0) = 0. Now what factor does f(x) have in common with x ??? > My point here is that posters like you are challenging what most> people accept as rather basic, Your thinking here is transparent, intuitive, and wrong. Youthink that if you have a function f(x) such that f(0) = 0,it must be the case that f(x) has a factor like, say, x^(1/43), or (x^2 - x)^67, or some such; that is, you can divide out something that lookslike a power of x. The example above and many others show thatthis is not true. Not to mention functions with various kindsof discontinuity and non-differentiability.>and your coy answer here, I think,> indicates that you *know* what you're doing, but are doing it anyway.b_2(0) has in common with 0 is b_2(0). Moreover, a_2(0) and b_2(0) are> coprime (by your definition of b_2(x)). But indeed, my statement that> the factor a_1(x) has in common with x is the same as the factor b_2(x)> has in common with x is too sweeping. It is true when x is coprime to all> other factors of the constant term of the quadratic. It may not be true> when that is false.> There's one point here which is why it's in the subject line: Factor> of x What is the factor of x in a function like f(x) = (x^13 + 5)^(1/5) - (x^(1/3) + 5^4)^(1/20),where of course f(0) = 0 ??? Nora B. > I think MOST people with some mathematical understanding will accept> that if you have a function of x, f(x) and that function equals 0 when> x=0, then it makes sense that f(x) has some factor of x in common with> x.Now I've found myself thinking about that a lot, and hopefully here I> can make a case for why I keep at this when people like Dik Winter or> Arturo Magidin or all these other posters come after me and claim it's> all settled and I'm wrong.> You are almost invariably wrong every time you launch a new idea.You have an amazing instinct for wrong mathematics. Do you rememberthe episode a few days ago when you were trying to factor a product of two primes ? Immediate counterexamples, and a whole series of unsuccessful second and third and fourth guesses by you, every one of them wrong. And here, with this factor of x idea, you are wrong again!> What I think is not wrong is that with f(x), if f(0)=0, then there's> some factor of x in there!!!> See above.> BTW, also something similar occurs in the integers. Both 15 and 16 have> a factor in common with 6. In the algebraic integers the situation is> a bit more complicated, because there are so many divisors and no primes.> You're babbling. What you just said just doesn't relate to the issue at> hand.How would I know when you do not tell what the issue at hand is?> The subject line *gives* the issue at hand.I'm making a point where I hope to rely on what mathematically aware> readers know themselves happens around 0. > Yup, has already been shown. Given an irreducible, primitive, monic> polynomial with integer coefficients:> x^n + c_(n-1).x^(n-1) + ... + c1.x + c0> each of the roots has a factor in common with each of the prime divisors> of c0.> What about at x=0?Is x = 0 a root? Yes, it is when c0 = 0. In that case the polynomial is> not irreducible, primitive, monic. You may take the first word from that> list.> Basically the functions a_1(x) and a_2(x) as defined simply travel> over an interesting path where the polynomial that defines them goes> from being in general irreducible over *integers* with an integer x,> and your position is that the *functions* themselves are aware of this> in some way.My position is that functions are functions.> Right. That explains a lot.> Functional behavior is not that intelligent as to worry about whether> or not some polynomial has integer roots are not!> ???> I mean for readers who don't get it yet, the distinction Dik Winter is> making here is the difference betweenx^2 + 2x + 1, which is reducible, and x^2 + 2x + 2, which, of course is not. > It might seem too esoteric with x there, so let's put in a value for> x, and let x=13. Then> a^2 - 12a + 7(13)(14)> and we already know that *one* of the a's, is coprime to 13, or wait,> do we?> No, we don't, *both* are not coprime to 13.> So you say that *both* a_1(13) and a_2(13) have some factor in common> with 13.> Interesting.Why? The proof has been given a number of times. Only you do not believe> it.Well, knowing that with *functions* a_1(x) and a_2(x) that a_1(0) = 0,> and a_2(0) = -1, it is NOT a great leap to at least consider the> possibility that a_1(0) has some factor in common with x, while a_2(0)> does not.> a_1(x) = 0 when x = 0. What you may have intended to say here was that you think it is reasonable to expect that a_1(x) has some kind of factor in common with x: I believe you think it must be somekind of power of x. That is not true. There is not even goodreason to think that a_1(x) in the examples considered hereis continuous. Nora B. === Subject: Re: JSH: Factor of xThe point is that *most* people with some mathematical knowledge would> accept that if you have a function f(x), where f(0) = 0, then hey,> maybe that means that it has some factor in common with, like you> could havef(x) = x^{1/3} where x^{1/3} is, of course, a factor in common with x.> Some example of functions f(x) with f(0)=0f(x) = sin(x)f(x) = 2 - sqrt(4+x)f(x) = 0, if x=0; 47 if x !=0None of these has a factor in common with x.There are more things on heaven and earth than polynomials inpowers of x. - William Hughes > My point here is that posters like you are challenging what most> people accept as rather basic, and your coy answer here, I think,> indicates that you *know* what you're doing, but are doing it anyway.b_2(0) has in common with 0 is b_2(0). Moreover, a_2(0) and b_2(0) are> coprime (by your definition of b_2(x)). But indeed, my statement that> the factor a_1(x) has in common with x is the same as the factor b_2(x)> has in common with x is too sweeping. It is true when x is coprime to all> other factors of the constant term of the quadratic. It may not be true> when that is false.> There's one point here which is why it's in the subject line: Factor> of xI think MOST people with some mathematical understanding will accept> that if you have a function of x, f(x) and that function equals 0 when> x=0, then it makes sense that f(x) has some factor of x in common with> x.Now I've found myself thinking about that a lot, and hopefully here I> can make a case for why I keep at this when people like Dik Winter or> Arturo Magidin or all these other posters come after me and claim it's> all settled and I'm wrong.What I think is not wrong is that with f(x), if f(0)=0, then there's> some factor of x in there!!! > BTW, also something similar occurs in the integers. Both 15 and 16 have> a factor in common with 6. In the algebraic integers the situation is> a bit more complicated, because there are so many divisors and no primes.> You're babbling. What you just said just doesn't relate to the issue at> hand.How would I know when you do not tell what the issue at hand is?> The subject line *gives* the issue at hand.I'm making a point where I hope to rely on what mathematically aware> readers know themselves happens around 0. > Yup, has already been shown. Given an irreducible, primitive, monic> polynomial with integer coefficients:> x^n + c_(n-1).x^(n-1) + ... + c1.x + c0> each of the roots has a factor in common with each of the prime divisors> of c0.> What about at x=0?Is x = 0 a root? Yes, it is when c0 = 0. In that case the polynomial is> not irreducible, primitive, monic. You may take the first word from that> list.> Basically the functions a_1(x) and a_2(x) as defined simply travel> over an interesting path where the polynomial that defines them goes> from being in general irreducible over *integers* with an integer x,> and your position is that the *functions* themselves are aware of this> in some way.My position is that functions are functions.Functional behavior is not that intelligent as to worry about whether> or not some polynomial has integer roots are not!I mean for readers who don't get it yet, the distinction Dik Winter is> making here is the difference betweenx^2 + 2x + 1, which is reducible, and x^2 + 2x + 2, which, of course is not. > It might seem too esoteric with x there, so let's put in a value for> x, and let x=13. Then> a^2 - 12a + 7(13)(14)> and we already know that *one* of the a's, is coprime to 13, or wait,> do we?> No, we don't, *both* are not coprime to 13.> So you say that *both* a_1(13) and a_2(13) have some factor in common> with 13.> Interesting.Why? The proof has been given a number of times. Only you do not believe> it.Well, knowing that with *functions* a_1(x) and a_2(x) that a_1(0) = 0,> and a_2(0) = -1, it is NOT a great leap to at least consider the> possibility that a_1(0) has some factor in common with x, while a_2(0)> does not.> === Subject: Re: JSH: Factor of x> I mean for readers who don't get it yet, the distinction Dik Winter is> making here is the difference betweenx^2 + 2x + 1, which is reducible, and x^2 + 2x + 2, which, of course is not.And, as a polynomial in x, the reducability of x^2 + 2*x + y depends on the value of y, which is part of what Dik has been trying to teach you, but which you seem unable to learn. === Subject: Re: JSH: Factor of x> What I think is not wrong is that with f(x), if f(0)=0, then there's> some factor of x in there!!!Consider f(x) = sin(x), or f(x) = e^x - 1, etc.While these are O(x) as x -> 0, and have f(0) = 0, they do not, in any reasonable sense, have any factor in common with x.Or even worse, let f(x) equal0 at x = 0, but otherwise equal 1. Now if you want f(x) to be a polynomial function,... === Subject: Re: JSH: Factor of x>>I'll admit that I'm still hoping on that math journal or some contacts>>that I have out there who are supposed to get back to me soon, Guffaw.[...]It's not too farfetched that he will end up in a journal. Perhaps someonewill need more subjects for a followup to this paper: http://www.apa.org/journals/psp/psp7761121.html:-)-- --Tim Smith === Subject: Re: JSH: Factor of x> I'll admit that I'm still hoping on that math journal or some contacts> that I have out there who are supposed to get back to me soon, but I'm> bored, so I'll see what happens here.Recently Rick Decker, a professor at Hamilton College, apparently> trying to refute my research came up with a quadratic example, which I> like because it's a quadratic, and easier to manipulate than the> cubics I've used before.If you wish to see his original post here are some headers which also> show that he posts from Hamilton College:> === > Subject: Re: Mathematical consistency, courageDecker put forward the quadratic(5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) where his a's are roots of a^2 - (x - 1)a + 7(x^2 + x).The factors (5a_1(x) + 7) and (5a_2(x) + 7) are examples of> non-polynomial factors.Notice that despite not being polynomials they are algebraic integers> if x is an algebraic integer because a_1(x) and a_2(x) are the two> roots ofa^2 - (x - 1)a + 7(x^2 + x).However, there's something odd here as if you let a=0, you have one of> the roots equals 0, but the other equals -1, so it makes sense to use> b_2(x), wherea_2(x) = b_2(x) - 1which gives(5a_1(x) + 7)(5b_2(x) + 2) = 7(25x^2 + 30x + 2)where a_1(0) = b_2(0) = 0.But that implies that a_1(x) and b_2(x) *both* have some factor in> common with x with algebraic integer x. If x = 0, that is. When x <> 0, in general b_2(x) does not havea factor in common with x. > Now given that a_1(x) a_2(x) = 7(x^2 + x) = 7x(x+1)that's especially odd, as what about the one that does versus the one> that doesn't? Or doesn't it? Do *both* somehow have factors in> common with x?> YOU GOT IT!!!> It might seem too esoteric with x there, so let's put in a value for> x, and let x=13. Then> [*] a^2 - 12a + 7(13)(14)and we already know that *one* of the a's, is coprime to 13, or wait,> do we?> No. Neither is coprime to 13. This follows from a very basicresult in Galois theory or the argument shown below.> Can any of you explain how it all works?> Yes, but you either cannot understand or will not listen. Here goes anyway. If as you say, one of the a's is coprime to 13, the other must be divisible by 13. Thereforeassume a_1(13) is divisible by 13. Lete a_1(13) = 13*b.Substituting in [*], one has 13^2 b^2 - 12*13*b + 7*13*14 = 0.Factoring out 13 gives 13*b^2 - 12*b + 7*14 = 0.where the left side is a non-monic, non-primitive,irreducible polynomial in b. Therefore b cannot bean algebraic integer. Therefore the assumption thaa_1(13) is divisible by 13 is not correct. The sameargument exactly works for a_2(13). Therefore neitheris divisible by 13 and neither is coprime to 13. So that shows how it DOESN'T work. To see how 13 can be factored out, define: r = GCD(a_1(13), 13) and s = GCD(a_2(13), 13). Since a_1(13)*a_2(13) = 7*13*14,and 7*14 is coprime to 13, we must have (a_1(13)/r) * (a_2(13)/s) = 7*14,and both a_1(13)/r and a_2(13)/s are algebraic integers. This is a side-issue anyway. The main argument has been whether7 was or was not a factor of a_1(x) and a_2(x). That however can be handled in very much the same way. For x > 0, 7 isNOT a factor of a_1(x) or a_2(x), but both have nonunitfactors in common with 7.> Note that the a's are (12 + sqrt(-4952))/2, and I didn't put indices> on them as how do you know?> Doesn't matter. See above. === ===================================================== === ==================== What is interesting here is that it is now clear that youactually understand what we have been saying all along. You areeither putting on an act to continue annoying us or you are simplyrefusing to accept what you now know to be true. I think thelatter is the case. Accepting this is tantamount to admitting that the main argument that you have been fighting for since lastsummer is wrong, and you lose everything: FLT, Advanced PolynomialFactorization and all the rest. That outcome does such violenceto your hyperinßated ego that I think you cannot stand it. Thismay go on forever: denial of the obvious, paranoia, etc.. Have fun, Nora B. === Subject: Re: JSH: Factor of x... > a^2 - (x - 1)a + 7(x^2 + x).... > a_2(x) = b_2(x) - 1... > But that implies that a_1(x) and b_2(x) *both* have some factor in > common with x with algebraic integer x. > If x = 0, that is. When x <> 0, in general b_2(x) does not have > a factor in common with x. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: JSH: Factor of x> Decker put forward the quadratic(5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) where his a's are roots of a^2 - (x - 1)a + 7(x^2 + x).The factors (5a_1(x) + 7) and (5a_2(x) + 7) are examples of> non-polynomial factors.Notice that despite not being polynomials they are algebraic integers> if x is an algebraic integer because a_1(x) and a_2(x) are the two> roots ofa^2 - (x - 1)a + 7(x^2 + x).However, there's something odd here as if you let a=0, you have one of> the roots equals 0, but the other equals -1, so it makes sense to use> b_2(x), wherea_2(x) = b_2(x) - 1which gives(5a_1(x) + 7)(5b_2(x) + 2) = 7(25x^2 + 30x + 2)where a_1(0) = b_2(0) = 0.But that implies that a_1(x) and b_2(x) *both* have some factor in> common with x with algebraic integer x.> Certainly, but you presumably want a non-unit factor of x, in which> case when x = 1, a_1(1) = sqrt(-14) and b_2(1) = 1 + sqrt(-14) and> I don't see either of those numbers having a non-unit factor> in common with x (= 1).> How can there a be a non-unit factor of x for a value of x that is a unit?That's nonsensical. === Subject: Re: JSH: Factor of x (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) > where his a's are roots of > a^2 - (x - 1)a + 7(x^2 + x).> The factors (5a_1(x) + 7) and (5a_2(x) + 7) are examples of> non-polynomial factors.> Notice that despite not being polynomials they are algebraic integers> if x is an algebraic integer because a_1(x) and a_2(x) are the two> roots of> a^2 - (x - 1)a + 7(x^2 + x).> However, there's something odd here as if you let a=0, you have one of> the roots equals 0, but the other equals -1, so it makes sense to use> b_2(x), where> a_2(x) = b_2(x) - 1> which gives> (5a_1(x) + 7)(5b_2(x) + 2) = 7(25x^2 + 30x + 2)> where a_1(0) = b_2(0) = 0.> But that implies that a_1(x) and b_2(x) *both* have some factor in> common with x with algebraic integer x.> Certainly, but you presumably want a non-unit factor of x, in which> case when x = 1, a_1(1) = sqrt(-14) and b_2(1) = 1 + sqrt(-14) and> I don't see either of those numbers having a non-unit factor> in common with x (= 1).> How can there a be a non-unit factor of x for a value of x that is a unit?That's nonsensical.Then why do you want it? === Subject: Fourier transform of log-spaced impulse train?It is well-known that the Fourier of a series of reguarly-spacedimpulses is also a series of impulses in the frequency domain. I needto find the Fourier Transform of a series log-spaced of impulsesdescribed byv(t) = sum(Dirac(t - ln(K)), K = an integer from 1 to infinitywhere Dirac(x) = 0 for x != 0, and Dirac(x) = a unit-area impulse forx = 0. In other words, v(t) is an impulse train with impulses occuringat t = 0, ln(2), ln(3), ln(4) ...Any ideas on how to approach this?Bob Adams === Subject: Re: Fourier transform of log-spaced impulse train?>It is well-known that the Fourier of a series of reguarly-spaced>impulses is also a series of impulses in the frequency domain. I need>to find the Fourier Transform of a series log-spaced of impulses>described by>v(t) = sum(Dirac(t - ln(K)), K = an integer from 1 to infinity>where Dirac(x) = 0 for x != 0, and Dirac(x) = a unit-area impulse for>x = 0. In other words, v(t) is an impulse train with impulses occuring>at t = 0, ln(2), ln(3), ln(4) ...Caution: this is not a tempered distribution. For example, if f(t) = exp(-t) for t > 0, then int f(t) v(t) dt involves the sum of 1/j from j=1 to infinity, which diverges. But you can define the complex Fourier transform F(p) = sum_{j=1}^infinity exp(-i p ln(j)) = sum_{j=1}^infinity j^(-ip)for Im(p) < -1, where it is zeta(ip) (zeta being the Riemann zeta function). F has an analytic continuation to real p, but I don'tknow if that would be useful to you.Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Muses bi-matricesOf late I have been reading quite a bit from Muses and his definition ofOctonions and higher algebra's using bi-matrices. There are two papersby Muses where he explained the rules, but they are in conßict witheach other, and they are (as far as I can see) not consistent. Has heever given consistent rules about the arithmetic with his bi-matrices?If so, where?-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: cauchy sequenceThis fact was stated in a proof I'm reading but I don't see how theyThe fact: Suppose {f_n} is a Cauchy sequence, i.e., given epsilon > 0, thenthere exists a positive integer N such that ||f_n - f_m|| < epsilon for n,m>= N. It then follows that there exists a subsequence f_n(k) with n(k)increasing such that ||f_n(k+1) - f_n(k)|| < 1/2^k. === Subject: Re: cauchy sequenceContent-transfer-encoding: 8bit> This fact was stated in a proof I'm reading but I don't see how theyThe fact: Suppose {f_n} is a Cauchy sequence, i.e., given epsilon > 0, then> there exists a positive integer N such that ||f_n - f_m|| < epsilon for n,m>= N. It then follows that there exists a subsequence f_n(k) with n(k)> increasing such that ||f_n(k+1) - f_n(k)|| < 1/2^k.Take epsilon = 1/2^k. Then there exists an N = N(k) such that(*) m, n >= N(k) ==> |f(n) - f(m)| < eps.Furthermore, N(k) can be replaced by any larger number withoutaffecting the truth of (*).We'd like to take n_k = N(k), but there's no guarantee the N(k) areincreasing. That's why the Furthermore... remark above is important.Now let n_1 = N(1), n_2 = max(n_1, N(2)) + 1, n_3 = max(n_2, N(3)) + 1,... We have n_1 < n_2 < n_3 < ...and n_k >= N(k). Thus m, n >= n_k ==> |f(m) - f(n)| < 1/2^k.Since n_k and n_{k+1} are both >= n_k we have |f(n_k) - f(n_{k+1})| < 1/2^k.--Ron Bruck === Subject: How Does the Fibonacci Sequence Start from ZeroIt has always bothered me logically that everything about theFibonacci sequence is orderly except how it ever gets started fromzero to reach the first number one:0 + 0 + 1 + 1 + 2 + 3 + 5 + 8 + 13 + 21 . . .So, how did the 0 + 0 ever add up to the first 1 in the sequence?Well, I was truly bored today, and I did some further reading in RudyRucker's book about infinity and the mind, and he said something atpage 212 (softcover edition) that tee'd me off for taking so long tosay something fundamental, but which I found interesting regardless:He says that zero in set theory is represented as the null set, {},and that the number one can be represented as the first set thatcontains zero, {{}}. Thus, logically, the symbology seems to besaying that the simple recognition of zero implies the existence ofthe number 1.Thinking further, it could be argued that the addition of {} + {}implies a recognition of zero that further implies 1, since to addanything to anything requires the mental process of recognizingquantity.Thus, symbolically, {} + {} could equal {{}}, or 1.This, of course, raises the objection that 0 + 0 = 0, not 1.My guess is that {{}} may be translated as either 0 or 1 and that theobjection is therefore semantic, not real. In other words, both 0 and1 are correct translations of a more fundamental process.If this is nuts, then how does the Fibonacci sequence otherwise getstarted? === Subject: Re: How Does the Fibonacci Sequence Start from Zero|It has always bothered me logically that everything about the|Fibonacci sequence is orderly except how it ever gets started from|zero to reach the first number one:||0 + 0 + 1 + 1 + 2 + 3 + 5 + 8 + 13 + 21 . . .||So, how did the 0 + 0 ever add up to the first 1 in the sequence?As someone else explained, it doesn't. Either you start with two termsthat you just choose, or if you extend the sequence back, you have a 1for the term in the -1 position, and so on.