mm-399 === Subject: : Re: There needs to be ANSI or ISO math standards > look > someone shoudl eb able to write the following > x + 4 + 5 + 6 = > x + 9 + 6 > x + 15 > without having a = sign behind the second statement > because solving the problem is reader/math person based > what is read or used by the math person > otherwise there needs to be a standard from maybe ANSI or ISO > Well, an ANSI or ISO standard would require an = in between at every step of > the way. When was the last time you read a standard? The last time many of us read a standard was the last we read a hard disk. And standards DO NOT require an = between every step of the way. They require an = sign at EACH AND EVERY interlude along the way. And the Standards also say that if mathema-tarts do not understand what an interlude is, they are required BY LAW, to LICENSE THE STANDARD. They CANNOT COPY OR DISTIBUTE the STANDARD, without the written permission of the STANDARDS COMMITTEE, and not their potentially moronic math teacher. > I was going to just link you to the ISO 9001 standard, but it turns out I > can't. It costs $125 (or 82 pounds) to buy it from this link: > http://www.iso-standards-international.com/iso-9001.htm. But, since you're > so excited about standards, you can shell out the bucks for the standard and > read it. It's exciting. Ranks right up there with FASB 106. No, it > doesn't. The accounting standard is more interesting. Anyway, I digress. > The ISO standard covers: Quality management systems. Requirements. QMS > requirements where an organization needs to demonstrate ability to provide > product that meets customer/regulatory requirements and aims to enhance > customer satisfaction. (This is what a liberal arts degree qualifies you to > do, in case you were wondering.) > *supporting freedom to learn and do math to get an answer, > I hate to break it to you, but the purpose of math is *not* to get an > answer. It's to understand. We demonstrate understanding by communicating > clearly and effectively. The best way to do that is to discover new ideas > and communicate your ideas in clear and simple English, but that is actually > fairly difficult with the level of maths that the average schoolteacher is > able to present to you. So you have to go out and learn things on your own, > which requires initiative and drive. And no whinging. In the meantime, > take your tests and put up with the idiots. The rest of the world does it > and seems to survive somehow. > and not supporting teachers who like to make students hate learning > and math. === === Subject: : Limit of Sequence : s(n)=(.9)(.99)(.999)...[1-(10^(-n)] The sequence below clearly converges. But can we specify the limit? .9, (.9)(.99), (.9)(.99)(.999), ... L === === Subject: : Re: Limit of Sequence : s(n)=(.9)(.99)(.999)...[1-(10^(-n)] > The sequence below clearly converges. But can we specify the > limit? > .9, (.9)(.99), (.9)(.99)(.999), ... I would be surprised if there is a closed form for the limit of that series, but, by an odd coincidence, I stumbled across a way to turn your infinite product into an infinite sum, which converges tolerably well. === === Subject: : Randi-type Challenge question: Inf. product into inf. sum For the limit of your series, I get 1 - (.1/.9) + (.1/.9)(.01/.99) - (.1/.9)(.01/.99)(.001/.999) + ... The general formula (re-worked slightly): product{n=1..inf}[ 1 - p*a^n ] = sum{k=0..inf}[ (-1)^(k-1)*p^k * product{j=1..k}[ a^k/(1=a^k) ] ] = 1 - p(a/(1-a)) + p^2(a/(1-a))(a^2/(1-a^2)) - p^3(a/(1-a))(a^2/(1-a^2))(a^3/(1-a^3)) + ... I derive it in the message noted above. === === Subject: : Re: Limit of Sequence : s(n)=(.9)(.99)(.999)...[1-(10^(-n)] Typo: > The general formula (re-worked slightly): > product{n=1..inf}[ 1 - p*a^n ] > = sum{k=0..inf}[ (-1)^k*p^k * ^^^ > product{j=1..k}[ a^k/(1=a^k) ] ] > = 1 - p(a/(1-a)) + p^2(a/(1-a))(a^2/(1-a^2)) - > p^3(a/(1-a))(a^2/(1-a^2))(a^3/(1-a^3)) + ... === === Subject: : Re: Analysis question I couldn't think that. Now, after reading your post, I try to solve following similar problem by e-d way. If g:R->R satisfies g(x+y)=g(x)*g(y) for all x,y in R and g is continuous at x=0 Then, show that g is continuous on every point x in R. But, That is not easy to me. If possible, will you please let me know how to show above problem by e-d way. === > === Subject: : Re: Analysis question > If f:R->R satisfies f(x+y)=f(x)+f(y) for all x,y in R and > lim {x->0} f(x) = L , Then prove L = 0. > lim(x->0) f(x) = lim(x->0) f(2x) > = lim(x->0) 2f(x) = 2 lim(x->0) f(x) > If f:R->R satisfies f(x+y)=f(x)+f(y) for all x,y in R and > continuous at x=c in R, then How can I prove f(x) is continous > on every point x in R? >Perhaps this will help: >consider R as a vector space over Q. Then f is an endomorphism. >The following is well-known for vector spaces over R: >- a linear map is continuous in 0 iff it is continuous everywhere >iff f maps bounded sets to bounded sets. > More to the point is similar theorem for topological groups, that > a linear function continuous somewhere is continuous everywhere. > To e-d it out if f continous at y, for all e, some d with for all x > |x - y| < d ==> |f(x) - f(y)| < e > Assume |x - z| < d, then > |x-z+y - y| < d > |f(x-z+y) - f(y)| < e > |f(x) - f(z)| < e > ---- === === Subject: : Re: Analysis question Now, after reading your post, I try to solve following similar problem by > e-d way. > If g:R->R satisfies g(x+y)=g(x)*g(y) for all x,y in R and g is continuous at > x=0 > Then, show that g is continuous on every point x in R. if some x with g(x) = 0, then g = 0 function which is continuous when g /= 0 function, then g(0) = 1 and lim(y->x) g(y) = lim(y->x) g(y-x + x) = lim(y->x) g(y-x) g(x) = g(x) lim(z->0) g(z) = g(x) g(1) = g(x) or use g(x+y) - g(x) = g(x)(g(y) - 1) and e-d stuff from lim(y->0) g(y) = g(0) = 1 to show e-d stuff for lim(y->0) g(x+y) = g(x) noting lim(y->x) g(y) = lim(y->0) g(x+y) When g is always positive, log g is defined and log g(x+y) = log g(x) + log g(y) Thus log g is linear and is continuous at 0 so use the results below In the case g is ever negative, consider extending the same log stuff and linear theorem to complex numbers. > If f:R->R satisfies f(x+y)=f(x)+f(y) for all x,y in R and > lim {x->0} f(x) = L , Then prove L = 0. > lim(x->0) f(x) = lim(x->0) f(2x) > = lim(x->0) 2f(x) = 2 lim(x->0) f(x) > If f:R->R satisfies f(x+y)=f(x)+f(y) for all x,y in R and > continuous at x=c in R, then How can I prove f(x) is continous > on every point x in R? > More to the point is similar theorem for topological groups, that > a linear function continuous somewhere is continuous everywhere. > To e-d it out if f continous at y, for all e, some d with for all x > |x - y| < d ==> |f(x) - f(y)| < e > Assume |x - z| < d, then > |x-z+y - y| < d > |f(x-z+y) - f(y)| < e > |f(x) - f(z)| < e === === Subject: : Existence of solution Would you please help me to solve following problem. Show : Every odd-degree polynomial with real coefficient has at least one sollution. let f(x)=a_(2k+1) *x^(2k+1) + a_2k *x^2k + ... + a_1 *x + a_0 , where a_(2k+1)=/=0, each a_n in R. If I can find m,n in R such that m Would you please help me to solve following problem. > Show : Every odd-degree polynomial with real coefficient has at least > one sollution. > let f(x)=a_(2k+1) *x^(2k+1) + a_2k *x^2k + ... + a_1 *x + a_0 , where > a_(2k+1)=/=0, each a_n in R. > If I can find m,n in R such that m Then, I will be able to insist such existence of solution by IVT. > But, I don't know how can I find such m,n. > If somebody can help me, please post reply. Look at the behavior of f(x) as x-> - infinity and as x-> + infinity. Hint: f(x)=x^(2k+1)*(a_(2k+1)+a_2k/x+...+a_0/x^(2k+1)) and so what remains as x-> +- infinity? === === Subject: : Re: Existence of solution > Would you please help me to solve following problem. > Show : Every odd-degree polynomial with real coefficient has at least > one sollution. > let f(x)=a_(2k+1) *x^(2k+1) + a_2k *x^2k + ... + a_1 *x + a_0 , where > a_(2k+1)=/=0, each a_n in R. > If I can find m,n in R such that m Then, I will be able to insist such existence of solution by IVT. > But, I don't know how can I find such m,n. Show that if x is large enough then you can take m = x and n = -x. === === Subject: : Bailey-Borwein-Ploufe Digit Extraction Algorithm for Pi I'm trying to understand how the BBP algorithm works, but I get stuck pretty early. (http://www.lacim.uqam.ca/~plouffe/Simon/ottawaPi.pdf) On page 6, he says (sorry, I don't know the standard notation for congruence, so I'll use ==): the k'th decimal of 1/n is given by the solution of (we are in base 10 for clarity) 10^k == r mod n. The digits are with the computation of r/n from the rank (k+1). More precisely the k'th digit is [10*r/n]. I'm having trouble seeing where he's getting the r from and what it is. Also, rank here just refers to position of the digit in relation to the decimal, right? -Don === === Subject: : Re: Bailey-Borwein-Ploufe Digit Extraction Algorithm for Pi > I'm trying to understand how the BBP algorithm works, but I get stuck > pretty early. > (http://www.lacim.uqam.ca/~plouffe/Simon/ottawaPi.pdf) > On page 6, he says (sorry, I don't know the standard notation for > congruence, so I'll use ==): the k'th decimal of 1/n is given by the solution of (we are in base > 10 for clarity) 10^k == r mod n. The digits are with the computation > of r/n from the rank (k+1). More precisely the k'th digit is > [10*r/n]. > I'm having trouble seeing where he's getting the r from and what it > is. Also, rank here just refers to position of the digit in > relation to the decimal, right? I haven't looked at the original, but from what you quote it looks like he's off by 1. E.g., take k = 3 and n = 7; then 10^k = r (mod n) gives you r = 6, and then [10r/n] gives 8, which is the 4th digit of 1/7, not the 3rd. Try some others, I think you'll get the (k + 1) digit, not the k. === === Subject: : Re: Bailey-Borwein-Ploufe Digit Extraction Algorithm for Pi >I'm trying to understand how the BBP algorithm works, but I get stuck >pretty early. >(http://www.lacim.uqam.ca/~plouffe/Simon/ottawaPi.pdf) >On page 6, he says (sorry, I don't know the standard notation for >congruence, so I'll use ==): the k'th decimal of 1/n is given by the solution of (we are in base >10 for clarity) 10^k == r mod n. The digits are with the computation >of r/n from the rank (k+1). More precisely the k'th digit is >[10*r/n]. >I'm having trouble seeing where he's getting the r from and what it >is. Also, rank here just refers to position of the digit in >relation to the decimal, right? >I haven't looked at the original, but from what you quote >it looks like he's off by 1. E.g., take k = 3 and n = 7; >then 10^k = r (mod n) gives you r = 6, and then [10r/n] >gives 8, which is the 4th digit of 1/7, not the 3rd. Try >some others, I think you'll get the (k + 1) digit, not the k. Ah, I get it now. Thank you, Gerry. === === Subject: : Re: Copyrights, fair use, and Internet realities [OT] >Oh, it's great fun, especially if your name is not too unusual. >You seem to be one of the most prominent David Ullrich's on the web, > Yeah, I noticed that. People posting things that I posted - of all the > nerve. Darn. Everything I looked at that cropped up was indeed by me, or a reference to me. Both in www.google.com and in groups.google.com. And with all combinations of dik winter, dik t winter, dik@mcvax.uucp, dik@zuring.uucp and dik@turing.uucp. I cannot hide myself. (Gives a nice history of my signature...) dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === === Subject: : Re: What is the number derivative ??? > Among the top ten Intel Science Talent Search winners just announced is one Linda Westrick, whose project concerns the number derivative -- which is described in the PR squib as a recently-defined operation> Can someone be so kind as to post the correct definition of number derivative for me? Not an adequate def'n, but in this doc http://www.sciserv.org/sts/63sts/stsfinalistbook.pdf we find Linda Brown Westrick, 18, of Mechanicsville [Virginia], submitted an Intel Science Talent Search project in mathematics concerning the number derivative, a concept introduced in a recent mathematics contest, which is analogous to the familiar calculus derivative. Linda developed fundamental properties of this new derivative and its relationship to factorization of integers. She further extended the notion to the rational numbers and analyzed solutions to the differential equations x'=a and x'=ax, where x' is the number derivative of === === Subject: : Re: What is the number derivative ??? > Among the top ten Intel Science Talent Search winners just announced is > one Linda Westrick, whose project concerns the number derivative -- which > is described in the PR squib as a recently-defined operation > Can someone be so kind as to post the correct definition of number > derivative for me? > Not an adequate def'n, but in this doc > http://www.sciserv.org/sts/63sts/stsfinalistbook.pdf > we find Linda Brown Westrick, 18, of Mechanicsville [Virginia], submitted an Intel > Science Talent > Search project in mathematics concerning the number derivative, a concept > introduced in > a recent mathematics contest, which is analogous to the familiar calculus > derivative. Linda > developed fundamental properties of this new derivative and its relationship > to factorization of > integers. She further extended the notion to the rational numbers and > analyzed solutions to the > differential equations x'=a and x'=ax, where x' is the number derivative of > x. > LH Linda's work, which I saw at the Intel STS open house last weekend, takes the same approach to the number derivative as the Waterloo paper--the derivative of a prime is 1 and the product rule for derivative holds, which implies that the derivative of any compiste integer n is the sum of n/d_i where the d_i's are the divisors of n (>1). Unfortunately, there doesn't seem to be a copy of her paper available on the Web. === === Subject: : Re: What is the number derivative ??? X-ID: XHV2EYZ1oeOh2XYc-CNTTb5Q7nAgJQX2pQ4EON8SrzkQ3IOnOUgS68 Am 18.03.04 17:37 schrieb Steve Wildstrom: >>Among the top ten Intel Science Talent Search winners just announced is >one Linda Westrick, whose project concerns the number derivative -- which >is described in the PR squib as a recently-defined operation>>Can someone be so kind as to post the correct definition of number >derivative for me? >Not an adequate def'n, but in this doc >http://www.sciserv.org/sts/63sts/stsfinalistbook.pdf >we find Linda Brown Westrick, 18, of Mechanicsville [Virginia], submitted an Intel >Science Talent >Search project in mathematics concerning the number derivative, a concept >introduced in >a recent mathematics contest, which is analogous to the familiar calculus >derivative. Linda >developed fundamental properties of this new derivative and its relationship >to factorization of >integers. She further extended the notion to the rational numbers and >analyzed solutions to the >differential equations x'=a and x'=ax, where x' is the number derivative of >x. >LH > Linda's work, which I saw at the Intel STS open house last weekend, > takes the same approach to the number derivative as the Waterloo > paper--the derivative of a prime is 1 and the product rule for > derivative holds, which implies that the derivative of any compiste > integer n is the sum of n/d_i where the d_i's are the divisors of n > (>1). Unfortunately, there doesn't seem to be a copy of her paper > available on the Web. After reading the ufranovski-paper at uwaterloo I wonder, why either Linda was not mentioned there. Ufranovski notes, that the idea was first dealt in some extension in context of a 1980-Putnam-contest. ???? Gottfried Helms ---------------------------------------------- http://www.math.uwaterloo.ca/JIS/VOL6/Ufnarovski/ufnarovski.pdf (...) 14. Concluding remarks because we found the subject to be very attractive and wanted to share our joy with others. the description of the numbers with derivatives that are divisible by 4 and his conjecture that for every n there exists a prime p such that all derivatives n(k) are divisible by p for suciently large k: In fact according to Theorems 4, 5 it is equivalent to be divisible by pp for suciently large k: Thus this conjecture is a bit stronger then Conjecture 2. The denition of the arithmetic derivative itself and its elementary properties was already in the Putnam Prize competition (it was Problem 5 of the morning session in March 25, 1950, [4] ) and probably was known in folklore even earlier. What we have done is mainly to generalize this denition in dierent directions, to solve some dierential equations, to calculate the generating function and to invite the reader to continue work in this area. We are grateful to our colleagues for useful discussion, especially to G. Almkvist, A. Chapovalov, S. Dunbar, G. Galperin, S. Shimorin and the referee, who helped to improve the text. We are especially grateful to J. Backelin, who helped us to reduce the number of conjectures by suggesting ideas that translated them into theorems. References [1] E. J. Barbeau, Remark on an arithmetic derivative, Canad. Math. Bull. 4 (1961), 117{ 122. [2] A. Buium, Arithmetic analogues of derivations, J. Algebra 198 (1997), 290{299. [3] J. H. Conway and R. K. Guy, The Book of Numbers, Springer, 1996. [4] A. M. Gleason, R. E. Greenwood, and L. M. Kelly, The William Lowell Putnam Mathe- matical Competition: Problems and Solutions 1938{1964, Mathematical Association of America, 1980. [5] J. Renze, S. Wagon, and B. Wick, The Gaussian zoo, Experiment. Math. 10:2 (2001), 161{173. 2000 Mathematics === Subject: Classication: Primary 11A25; Secondary 11A41, 11N05, 11N56, 11Y55. Keywords: Arithmetic derivative, Goldbach's Conjecture, the Twin Prime Conjecture, prime numbers, Leibnitz rule, integer sequence, generating function. -------------------------------------------------------------- -------------- - === === Subject: : Re: What is the number derivative ??? > After reading the ufranovski-paper at uwaterloo I wonder, why either > Linda was not mentioned there. Perhaps because the Ufranovski was written first? > Ufranovski notes, that the idea was > first dealt in some extension in context of a 1980-Putnam-contest. The Ufranovski has lots of open problems in it. Maybe Linda Westrick solved some of them. We cannot tell without more information. The description from STS says, analyzed solutions to the differential equations x'=a and x'=ax ...; but Ufranovski merely conjectured solutions to these. === === Subject: : Re: What is the number derivative ??? X-ID: bv3tQEZG8eT6VXBjXYnhFsdEU+WWmPe93lFha+CysnxiZ6bz7geWEZ Am 18.03.04 20:09 schrieb A N Niel: >After reading the ufranovski-paper at uwaterloo I wonder, why either >Linda was not mentioned there. > Perhaps because the Ufranovski was written first? >Ufranovski notes, that the idea was >first dealt in some extension in context of a 1980-Putnam-contest. > The Ufranovski has lots of open problems in it. Maybe Linda Westrick > solved some of them. We cannot tell without more information. > The description from STS says, analyzed solutions to the differential > equations x'=a and x'=ax ...; but Ufranovski merely conjectured > solutions to these. Yep, may be I made my words unwanted sharp. I just wondered, that the abstract/announcements of the two resources sounded very similar about Gottfried Helms === === Subject: : chinese remainder theorem Dear group Have a look at my method for the chinese remainder theorem. It's at http://www.geocities.com/dirkie6/chinese.html I would like some comments on the originality of this method. Carel === === Subject: : between easy and hard Suppose a problem P whose input size can be given as n bits can be shown not solvable in polynomial time. This means there is no algorithm that can solve P in O(n^k) operations exists, for any k. Can I say P is NP-hard? In other words, is there any class of problems between the polynomial-time solvable problems and the NP-hard problems? --Jon === === Subject: : Re: between easy and hard >Suppose a problem P whose input size can be given as >n bits can be shown not solvable in polynomial time. There are such problems, but they're not in NP. >This means there is no algorithm that can solve P in O(n^k) >operations exists, for any k. >Can I say P is NP-hard? You can say it, but it probably isn't true. Not unless every NP problem can be polynomial-time-reduced to your problem. >In other words, is there any class of problems between the >polynomial-time solvable problems and the NP-hard problems? That's not the same thing at all. Your problem could be much harder than anything in NP. Department of Mathematics http://www.math.ubc.ca/~israel === === Subject: : Re: Beall Conjecture This shows how obvious the generalization of FLT --> unequal exponents > is. It is sufficiently obvious that it was asked as an off-the-cuff > question at the end of Tate's talk. > For any single person to try to claim 'credit' for this conjecture > seems absurd. It's just too obvious a question. > Oh well, ya' gotta name the conjecture something and this > way we get some cash out of it. It's the way things go in > academia anyway. Uneducated doofuses who sank their life > savings into an oil well in the 30's or a lottery ticket > and now are wealthy get their names on the side of the new > math building even though they can't come up with either > of the square roots of 36. But that's how we get the new > building. > And maybe this could be a new source for us. Imagine, instead > of the Silverman Conjecture the: Pepsi-Silverman $2,000,000 Challenge > Now _that's_ sex appeal. > I think I'll start making conjectures and see whether Phillip- > Morris will bid on naming rights. (You must be 21 or older to > work on this conjecture.) > Bart > But he's not just trying to put his name on something mathematical, > and give money to further mathematics. He's attempting to make it > look like he's an important mathematical researcher. It's as if JSH, > after being told that his prime counting function, though original and > correct, followed obviously from known facts, and JSH responded by > offering $100,000 to anyone who can prove something related to it. > All of the sudden, the media would treat him as if he had discovered > something *major*. That's precisely what's happened here; Beal wants > to like like he made an important discovery and is a great > mathematician. Go look at his web page; not only does it belittle > modern day mathematics and make ridiculous claims about number theory > but it spends several paragraphs attempting to hammer in how remarkable and original and important the discovery of the > conjecture was. Um, right. > http://www.bealconjecture.com/ > Nathan It isn't anything new. L'Hopital supposedly paid Gauss to get some theorems to publish, like L'Hopital's rule. However, if someone proves Beal's conjecture, wouldn't it likely become whoever's theorem. If it remains unproven, and the prize money remains, I suppose Beal will continue to get his name attached to it. === === Subject: : Re: Number theory exercise Robin Chapman Larry Hammick > Let p be a prime congruent to 3 mod 8, say p = 8n+3, > and define a function f by > f(x) = x(x+1)/2 1 <= x <= 4n+1. > Show that the congruence > f(x) + f(y) = 0 mod p > has exactly n solutions (x,y) having x < y. > (Also, no solution has y = 4n+1.) > Same as (2x+1)^2 + (2y+1)^2 = 2 (mod p) > or u^2 + v^2 = 2 (mod p) > so counting points on a conic over F_p. True, but this is a hand-picked conic :) Any other way? === === Subject: : Re: Number theory exercise Robin Chapman Larry Hammick >Let p be a prime congruent to 3 mod 8, say p = 8n+3, >and define a function f by >f(x) = x(x+1)/2 1 <= x <= 4n+1. >Show that the congruence >f(x) + f(y) = 0 mod p >has exactly n solutions (x,y) having x < y. >(Also, no solution has y = 4n+1.) >Same as (2x+1)^2 + (2y+1)^2 = 2 (mod p) >or u^2 + v^2 = 2 (mod p) >so counting points on a conic over F_p. > True, but this is a hand-picked conic :) Any other way? You can't get much easier than counting points on a conic--- take any old point on it, then almost all the lines through it pass through exactly one more point. Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === === Subject: : Re: Question about associativity of cartesian product > question: is there such a thing as a set bijection which is a > homomorphism whose inverse is NOT also a homomorphism? I'm unable to > think of one. This raises a further side question: is there such a > thing as a homomorphism whose inverse is also a homomorphism, but is > NOT a bijection? I'm unable to think of one of these either. > Tell me first what algebraic structure you are talking about. Any algebraic structure(s) which would make the answer to either of my questions be yesAlso, I'm now even more confused, because your response implies that the answer to my second question above depends on the algebraic structures I'm talking about, whereas Marc Olschok responded that the answer to my second question is unconditionally no. (He answered my first question above yes, and gave an example of algebraic structures for which such homomorphisms could be defined, though I completely failed to understand.) > 3. Please explain what you mean by canonical, uniquely defined > isomorphism here. I know what the words mean, but do you mean that > there's only one isomorphism defined, or that the isomorphism which is > defined is the only possible one? > The second one, of course, since no isomorphism was defined here. Again I'm thoroughly confused. Substituting my second speculation in place of the phrase the second one, of course in your response, I get the isomorphism which is defined is the only possible one, since no isomorphism was defined here. What does _that_ mean? Is there an isomorphism, or isn't there? === === Subject: : Re: Question about associativity of cartesian product >question: is there such a thing as a set bijection which is a >homomorphism whose inverse is NOT also a homomorphism? I'm unable to >think of one. This raises a further side question: is there such a >thing as a homomorphism whose inverse is also a homomorphism, but is >NOT a bijection? I'm unable to think of one of these either. >Tell me first what algebraic structure you are talking about. > Any algebraic structure(s) which would make the answer to either of my > questions be yes> Also, I'm now even more confused, because your response implies that > the answer to my second question above depends on the algebraic > structures I'm talking about, whereas Marc Olschok responded that the > answer to my second question is unconditionally no. (He answered my > first question above yes, and gave an example of algebraic > structures for which such homomorphisms could be defined, though I > completely failed to understand.) Marc Olschok said that *if* you're thinking about the category of sets and maps between sets, *then* the answer is yes. What I said was that *if* you're thinking about a category of sets with an additional structure *then* the answer depends upon that structure. Consider, for instance, this very simple structure: the structure associated with each set is simply a subset set A' of A. Now, the homomorphisms from a set A (with a subset A') into a subset B (with a aubset B') are those maps f from A into B such that f(A') is contained in B'. For this category, the inverse of an bijective homomorphism is not, in general, a homomorphism. Take, for instance A = {1,2,3}, A' = {1}, B = {1,2,3} and B' = {1,2}. Then, the identity from A into B is a bijective homomorphism whise inverse is not a homomoprhism. >3. Please explain what you mean by canonical, uniquely defined >isomorphism here. I know what the words mean, but do you mean that >there's only one isomorphism defined, or that the isomorphism which is >defined is the only possible one? >The second one, of course, since no isomorphism was defined here. > Again I'm thoroughly confused. Substituting my second speculation in > place of the phrase the second one, of course in your response, I > get the isomorphism which is defined is the only possible one, since > no isomorphism was defined here. What does _that_ mean? Is there an > isomorphism, or isn't there? What I meant was that your first possibility, there's only one isomorphism defined, was not really a possibility, since nobody actually defined an isomorphism here. Of course there is an isomorphism; I am just saying that it has not been defined. Jose Carlos Santos === === Subject: : Re: Question about associativity of cartesian product > While the examples at wikipedia are o.k., their wording of the > definition is a bit misleading. I have yet to find any definitions of anything anywhere on the web related to mathematics that's _not_ misleading or at least so incomplete as to be essentially useless to anybody except mathematicians who already know the definitions. Kind of like unix man pages from a couple decades ago (and often times even today, but that's a rant for some other newsgroup). I tried http://mathworld.wolfram.com/Homomorphism.html and got A term used in category theory to mean a general morphism. followed by If G and H are groups, then a group homomorphism of G into H is... followed by the appropriate definition of a group homomorphism, but nothing further about homomorphisms except _group_ homomorphisms specifically, which of course I care nothing about until I first get a good grasp of _homomorphism_. Since the only definition of the word homomorphism given on the page is general morphism, let's look that up: I find http://mathworld.wolfram.com/Morphism.html which tells me A general morphism is called a homomorphismOh yes, that's so very helpful. It further tells me that A morphism is a map between two objects in an abstract category and A bijective morphism is called an isomorphism. No mention about preserving structure (or even any mention about structure). http://mathworld.wolfram.com/Isomorphism.html tells me Formally, an isomorphism is bijective morphism (sic) with no mention of structure either. It then suggests that isomorphisms of _groups_ preserve the structure of those groups, but it doesn't say anything about isomorphisms in general. Similar problems apply to mathworld's definition of category, with no formal explanation being given for the difference between an abstract category and a concrete category, etc. No need to carry on with my diatribe; suffice it to say that the formal definitions of basically everything mathematical are circular, wrong, or ridiculously incomplete, and the informal definitions are so vague as to be useless to a non-expert. Lest anybody say that I'm wanting to foundationalize mathematics, note that I'm not; I'm just asking for some _formal_ definitions of everything with an explicit list of the basic symbols/terminology for which no definition will be given (I'm perfectly happy being told that no formal definition will be given for the element-of operator in set theory, for example). I've spent many, many hours so far trying to figure out what a morphism, a homomorphism, an isomorphism, an algebra, and an algebraic structure are, and I have yet to actually find any formal definitions. All of the formal definitions I've found for any of these concepts are complete jokes, with bits and pieces written in math, and all the bits and pieces joined together by vague English. Another example is the definitions of ZFC at http://en.wikipedia.org/wiki/Zermelo-Fraenkel_set_theory though at least for ZFC I did manage to finally find formal definitions for some (but not all) of the axioms at http://www.trinity.edu/cbrown/topics_in_logic/sets/node4.html > You usually start with some notion of sets with structure and a > corresponding notion of structure preserving maps which are called homomorphisms. The latter notion should be defined in a such a way > that all identity maps are homomorphisms and the composite of two > homomorphisms is again a homomorphism. Can you state this formally? > (this situation is in fact a special case of a more > general situation idea, see e.g. > ) That page defines a category in terms of classes. Does this mean that category can't be formalized in terms of ZFC, but only in terms of NBG? > Now one defines: a homomorphism f: X --> Y is an isomorphism for which Surely you meant an isomorphism f: X --> Y is a homomorphism...? > a two sided inverse (w.r.t. composition) exists i.e., a homomorphism > g : Y ---> X such that f o g = id_Y and g o f = id_X. > If such a g exists, it follows that f is bijective and g is indeed > the inverse map of f. The important point is, that on requires that > g is indeed a homomorphism, not just any odd map. I don't understand entirely. Can you remove all the English words and put in standard mathematical symbols? > this also raises a side > question: is there such a thing as a set bijection which is a > homomorphism whose inverse is NOT also a homomorphism? > You make take topological spaces and continous maps: > Let X={0,1} with the discrete topology (all sets open) > and Y={0,1} with the indiscrete topology (only Y and the empty set are open). > Look at the continous map f: X ---> Y with f(0)=0 and f(1)=1. I know nothing about topology, and when I tried to read about it, I got bit (no, more like eaten) by a combination of vague definitions about my own stupidity, so I'll try your second example (below). > Another example is provided by ordered sets and order preserving maps > Let X=P({1,2})={ {}, {1}, {2}, {1,2} } ordered by inclusion > and Y={0,1,2,3} ordered via 0<1<2<3. > Look at the order preserving map f: X ---> Y defined via > f({})=0, f({x})=x (for x=1 or 2), and f({1,2})=3 You're saying that f is a homomorphism and a bijection, and that the inverse of f is not a homomorphism? > This raises a further side question: is there such a > thing as a homomorphism whose inverse is also a homomorphism, but is > NOT a bijection? I'm unable to think of one of these either. > Your intuition is correct this time. > As mentioned above, any isomorphism has to be bijective. This means that [(f is a homomorphism) and (f^(-1) is a homomorphism)] -> (f is an isomorphism) correct? > 3. Please explain what you mean by canonical, uniquely defined > isomorphism here. I know what the words mean, but do you mean that > there's only one isomorphism defined, or that the isomorphism which is > defined is the only possible one? [response snipped] I'm still trying to digest your response, but I anticipate complete failure. I just don't have enough brain cells. Interestingly, Jos.8e Carlos Santos responded simply, The second one, of course, since no isomorphism was defined here. And of course I have no clue how that jives with your response, and I don't think that I would have a clue even if I understood your response. === === Subject: : Re: Question about associativity of cartesian product mathematical expositions that s/he could understand. Hmm, perhaps you should try a visit to the library instead. Stephen J. Herschkorn herschko@rutcor.rutgers.edu === === Subject: : Re: Question about associativity of cartesian product >While the examples at wikipedia are o.k., their wording of the >definition is a bit misleading. > I have yet to find any definitions of anything anywhere on the web > related to mathematics that's _not_ misleading or at least so > incomplete as to be essentially useless to anybody except > mathematicians who already know the definitions. Kind of like unix man > pages from a couple decades ago (and often times even today, but > that's a rant for some other newsgroup). Actually the analogy is a good one. The old fashioned unix man pages are great if I want to lookup quickly the syntax of a particular command. But I would never expect to _learn_ unix by looking at them. Likewise, a lot of the material on the web is nice if you just need to look up a particular item and already know enough to place it into a proper context. When it comes to learning and understanding, I would certainly recommend traditional books or lecture notes; of course this include online _texts_. [ example of Mathworld-nonsense skipped ] > No need to carry on with my diatribe; suffice it to say that the > formal definitions of basically everything mathematical are circular, > wrong, or ridiculously incomplete, and the informal definitions are so > vague as to be useless to a non-expert. Lest anybody say that I'm > wanting to foundationalize mathematics, note that I'm not; I'm just > asking for some _formal_ definitions of everything with an explicit > list of the basic symbols/terminology for which no definition will be > given (I'm perfectly happy being told that no formal definition will > be given for the element-of operator in set theory, for example). > I've spent many, many hours so far trying to figure out what a > morphism, a homomorphism, an isomorphism, an algebra, and an algebraic > structure are, and I have yet to actually find any formal definitions. > All of the formal definitions I've found for any of these concepts > are complete jokes, with bits and pieces written in math, and all the > bits and pieces joined together by vague English. As mentioned above, you will probably find more useful information in the library of a nearby university. Any standard text on category theory should provide clear definitions, e.g. [1] Adamek, Herrlich, Strecker: Abstract and concrete categories [2] Freyd, Scedrov: Categories, Allegories [3] Mac Lane: Categories for the working mathematician Of course, none of them avoid English at all costs. >[...] >You usually start with some notion of sets with structure and a >corresponding notion of structure preserving maps which are called homomorphisms. The latter notion should be defined in a such a way >that all identity maps are homomorphisms and the composite of two >homomorphisms is again a homomorphism. > Can you state this formally? If pressed for formality, I would see it as equivalent to a concrete category over Set (the category of sets and mappings) (cf. [1]): A concrete category (A,V) over Set is a faithful functor V: A ---> Set. ( here faithful means that for two morphisms f,g in A with the same domain and codomain one has V(f)=V(g) ==> f=g ) Now, a set with structure would be a set of the form V(a) for some object a of A, and a structure preserving map from V(a) to V(b) would be a map of the form V(f) for some Morphism f: a ---> b. >(this situation is in fact a special case of a more >general situation idea, see e.g. >) > That page defines a category in terms of classes. Does this mean that > category can't be formalized in terms of ZFC, but only in terms of > NBG? You may formalize it without reference to any particular set theory, by introducing a suitable language and a (logical) theory (e.g. [2]). Classes are used in order to use categories, that do not have sets of objects or arrow, e.g. all sets or all groups etc. >Now one defines: a homomorphism f: X --> Y is an isomorphism for which > Surely you meant an isomorphism f: X --> Y is a homomorphism...? Sorry, I mangled the sentence; of course your are right. >a two sided inverse (w.r.t. composition) exists i.e., a homomorphism >g : Y ---> X such that f o g = id_Y and g o f = id_X. >If such a g exists, it follows that f is bijective and g is indeed >the inverse map of f. The important point is, that on requires that >g is indeed a homomorphism, not just any odd map. > I don't understand entirely. Can you remove all the English words and > put in standard mathematical symbols? What standard? In the above, o denotes composition and id_X and id_Y the identity morphisms of X and Y. I will try to rephrase this for concrete categories: first, observe that the definition of isomorphism uses only composition and identities, hence it works for any category: (1) Let A be a category, f:a--->b be a morphism in A. f is an isomorphism iff there exists a g:b--->a with fog=id_b and gof=id_a such a g is uniquely determined (if it exists) and is often called the inverse of f. (2) Let F: A ---> B be a functor between categories and f:a--->b any morphism in A. Then we have f is an isomorphism ==> F(f) is an isomorphism [ If g is the inverse of f, then then F(f) o F(g) = F(f o g) = F(id_b) = id_F(b) and F(g) o F(f) = F(g o f) = F(id_a) = id_F(a) Hence F(g) is an inverse for F(f). ] The important point is, that the converse does _not_ hold. It may happen that F(f) is an isomorphism without f being an isomorphism. (3) As a special case of (2), suppose (A,V) is concrete over Set. If f:a--->b is an isomorphism with inverse g, then V(f) is also in isomorphism with inverse V(g). In Set we have isomorphism = bijectiveBut of course V(f) can be an isomorphism without f being an isomorphism. So for structured sets, every isomorphism is bijective and its inverse set map is the only possible candidate for the inverse homomorphism. > this also raises a side >question: is there such a thing as a set bijection which is a >homomorphism whose inverse is NOT also a homomorphism? >You make take topological spaces and continous maps: >Let X={0,1} with the discrete topology (all sets open) >and Y={0,1} with the indiscrete topology (only Y and the empty set are open). >Look at the continous map f: X ---> Y with f(0)=0 and f(1)=1. > I know nothing about topology, and when I tried to read about it, I > got bit (no, more like eaten) by a combination of vague definitions > about my own stupidity, so I'll try your second example (below). >Another example is provided by ordered sets and order preserving maps >Let X=P({1,2})={ {}, {1}, {2}, {1,2} } ordered by inclusion >and Y={0,1,2,3} ordered via 0<1<2<3. >Look at the order preserving map f: X ---> Y defined via >f({})=0, f({x})=x (for x=1 or 2), and f({1,2})=3 > You're saying that f is a homomorphism and a bijection, and that the > inverse of f is not a homomorphism? Exactly. > This raises a further side question: is there such a >thing as a homomorphism whose inverse is also a homomorphism, but is >NOT a bijection? I'm unable to think of one of these either. >Your intuition is correct this time. >As mentioned above, any isomorphism has to be bijective. > This means that > [(f is a homomorphism) and (f^(-1) is a homomorphism)] -> (f is an > isomorphism) > correct? Yes, under the assumption that f is already bijective (in order to have f^(-1) available at all). >3. Please explain what you mean by canonical, uniquely defined >isomorphism here. I know what the words mean, but do you mean that >there's only one isomorphism defined, or that the isomorphism which is >defined is the only possible one? > [response snipped] > I'm still trying to digest your response, but I anticipate complete > failure. I just don't have enough brain cells. > Interestingly, Jos.8e Carlos Santos responded simply, The second one, > of course, since no isomorphism was defined here. And of course I > have no clue how that jives with your response, and I don't think that > I would have a clue even if I understood your response. Well, one could cut the story short by just defining f : (A*B)*C ---> A*(B*C) via f( ((a,b),c) ) := (a,(b,c)) g : A*(B*C) ---> (A*B)*C via g( (a,(b,c)) ) := ((a,b),c) and verify the equations from (1), but of course, this does not really explain why these mapes are chosen. === === Subject: : Re: Question about associativity of cartesian product And of course I > have no clue how that jives with your response, funky! Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === === Subject: : terminology questions: partition vs. quotient set, and graph vs. extension 1. What's the difference between a partition and a quotient set? 2. What's the difference between the graph of a relation and the extension of a relation? === === Subject: : Re: terminology questions: partition vs. quotient set, and graph vs. extension >1. What's the difference between a partition and a quotient set? If a quotient set of S is S/R, where R is an equivalence relation on S, then a quotient set of S is exactly the same thing as a partition of S - both are a pairwise-disjoint collection of subsets of S with union S. >2. What's the difference between the graph of a relation and the >extension of a relation? What is the extension of a relation? === === Subject: : Re: terminology questions: partition vs. quotient set, and graph vs. extension >1. What's the difference between a partition and a quotient set? > If a quotient set of S is S/R, where R is an equivalence relation > on S, then a quotient set of S is exactly the same thing as a > partition of S - both are a pairwise-disjoint collection of subsets > of S with union S. >2. What's the difference between the graph of a relation and the >extension of a relation? > What is the extension of a relation? The set of objects which satisfy the relation. 'cid 'ooh === === Subject: : Re: terminology questions: partition vs. quotient set, and graph vs. extension >1. What's the difference between a partition and a quotient set? >If a quotient set of S is S/R, where R is an equivalence relation >on S, then a quotient set of S is exactly the same thing as a >partition of S - both are a pairwise-disjoint collection of subsets >of S with union S. >2. What's the difference between the graph of a relation and the >extension of a relation? >What is the extension of a relation? >The set of objects which satisfy the relation. Presumably meaning the set of ordered pairs (x,y) which satisfy the relation? If so, that's exactly what the relation is (and also exactly what the graph of the relation is). >'cid 'ooh === === Subject: : Re: terminology questions: partition vs. quotient set, and graph vs. extension > 1. What's the difference between a partition and a quotient set? A quotient set is generated by an equivalence relation. A partition is any splitting a set into pairwise disjoint nonnul sets. > 2. What's the difference between the graph of a relation and the > extension of a relation? What's the difference between the graph of a function and an extension of a function? === === Subject: : confused with E[e^(a*X^2)] where X~N(0,1) Let random variable X be drawn from N(0,1), then for any a<1/2, what is the expectation of e^(a*X^2) where e is 2.71828...? I got two solutions: 1. E[e^(a*X^2)] = 1/sqrt(1-2*a) (Lemma 1 in [1]) 2. Since E[X^2]=1, I get E[e^(a*X^2)]= e^E[a*X^2]=e^(a*E[X^2])=e^a The above two solutions seem totally different, am I missing something there? [1] Rosa I. Arriaga and Santosh Vempala. An Algorithmic Theory of Learning: Robust Concepts and Random Projection === === Subject: : Re: confused with E[e^(a*X^2)] where X~N(0,1) >Let random variable X be drawn from N(0,1), then for any a<1/2, what >is the expectation of e^(a*X^2) where e is 2.71828...? >I got two solutions: >1. E[e^(a*X^2)] = 1/sqrt(1-2*a) (Lemma 1 in [1]) >2. Since E[X^2]=1, I get E[e^(a*X^2)]= e^E[a*X^2]=e^(a*E[X^2])=e^a >The above two solutions seem totally different, am I missing something >there? It is rarely the case that for any random variable Z and a function f that E(f(Z)) = f(E(Z)). If f is strictly convex (f(x)=x^2, e^x, etc.), it is never the case unless the random variable is a constant with probability one. This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === === Subject: : Re: confused with E[e^(a*X^2)] where X~N(0,1) >Let random variable X be drawn from N(0,1), then for any a<1/2, what >is the expectation of e^(a*X^2) where e is 2.71828...? >I got two solutions: >1. E[e^(a*X^2)] = 1/sqrt(1-2*a) (Lemma 1 in [1]) >2. Since E[X^2]=1, I get E[e^(a*X^2)]= e^E[a*X^2]=e^(a*E[X^2])=e^a >The above two solutions seem totally different, am I missing something >there? > It is rarely the case that for any random variable Z > and a function f that E(f(Z)) = f(E(Z)). If f is > strictly convex (f(x)=x^2, e^x, etc.), it is never > the case unless the random variable is a constant > with probability one. Thank you all very much!!! === === Subject: : Re: confused with E[e^(a*X^2)] where X~N(0,1) >Let random variable X be drawn from N(0,1), then for any a<1/2, what >is the expectation of e^(a*X^2) where e is 2.71828...? I got two solutions: >1. E[e^(a*X^2)] = 1/sqrt(1-2*a) (Lemma 1 in [1]) >2. Since E[X^2]=1, I get E[e^(a*X^2)]= e^E[a*X^2]=e^(a*E[X^2])=e^a The above two solutions seem totally different, am I missing something >there? It is rarely the case that for any random variable Z >and a function f that E(f(Z)) = f(E(Z)). If f is >strictly convex (f(x)=x^2, e^x, etc.), it is never >the case unless the random variable is a constant >with probability one. The big exception being when f is affine (= a constant + linear), in which case Ef(X) = f(EX) for all X. Stephen J. Herschkorn herschko@rutcor.rutgers.edu === === Subject: : Re: confused with E[e^(a*X^2)] where X~N(0,1) > Let random variable X be drawn from N(0,1), then for any a<1/2, what > is the expectation of e^(a*X^2) where e is 2.71828...? > 2. Since E[X^2]=1, I get E[e^(a*X^2)]= e^E[a*X^2]=e^(a*E[X^2])=e^a This is certainly incorrect. In general E[f(X)] != f(E[X]). === === Subject: : Re: Is This Right? >For a while, I have been playing with the automorphism group of the >symmetric group S_6. I believe I have now derived the sizes of its >conjugacy classes. Does anyone care to check (here I include S_6 in >Aut(S_6) by identifying g with conjugation through g)? >1440= >1=the identity >45=the nontrivial central elements in the Sylow2-subgroups, which are >the even elements of order 2 in S_6 (S_6 conjugacy class 2+2+1+1) >80=the elements of order 3 in S_6 (both 3-cycles and permutations of >S_6-conjugacy class 3+3) >90=the even elements of order 4 in S_6 (S_6-conjugacy class 4+2) >144=the 5-cycles in S_6 >30=the odd elements of order 2 in S_6 (both transpositions and >S_6-conjugacy class 2+2+2) >90=the 4-cycles in S_6 >240=the elements of order 6 in S_6 (both 6-cycles and S_6-conjugacy >class 3+2+1) >144=the elements of order 10 in Aut(S_6) >36=the elements of order 2 which arise as 5th powers of elements of >order 10 >180=the elements of order 8 in Aut(S_6) which are 8-cycles in >Aut(S_6)'s action on 10 points >180=the outer automorphisms of order 4 >180=the elements of order 8 in Aut(S_6) which are of S_10-conjugacy >class 8+2 in Aut(S_6)'s action on 10 points >The addition works out. Hopefully the group theory does also. I have grown lazy in my old age, and I do calculations like that on a computer. I can confirm for you that the above breakdown agrees with the computation done in Magma. Derek Holt. === === Subject: : re:Cantor Paradox :-) If I find the time I will post a full exposition. But meanwhile you can work it out yourself, it's not difficult. === === Subject: : Re: Cantor Paradox :-) > If I find the time I will post a full exposition. > But meanwhile you can work it out yourself, it's not difficult. It is a bit optimistic to expect a reader of news to puzzle together your various postings and attempt to extract a proof from them. Actual simple proofs of the second incompleteness theorem for ZF are certainly of interest, hypothetical proofs are not. === === Subject: : Re: Cantor Paradox :-) > Of course, at this point the proof is not as simple as I originally > planned, and further refinement might be required. What proof is that? It would be a good idea if you set out your argument in full instead of referring vaguely to modifications of an implicit argument. That is also a good way of discovering mistakes for yourself.[/quote:1b1c10b4e6] I think I have been mostly discovering and correcting mistakes on my own and at this stage I see no further mistakes. However if you're interested in the proof of unprovability of consistency, I would suggest you just go over it and see how it works. The idea is pretty simple. === === Subject: : Re: Cantor Paradox :-) > However if you're interested in the proof of unprovability of > consistency, I would suggest you just go over it and see how it > works. Sure, once you've posted what you regard as a full and correct exposition of it. === === Subject: : Re: Cantor Paradox :-) > Of course, at this point the proof is not as simple as I originally > planned, and further refinement might be required. What proof is that? It would be a good idea if you set out your argument in full instead of referring vaguely to modifications of an implicit argument. That is also a good way of discovering mistakes for yourself. === === Subject: : Measure theory proof I'm stuck with the following problem, any help would be really appreciated. Let m be a charge in (X,A) (Where (X,A) is a measurable space). And define: m^(+)(L) = sup{ m(E) | E is in A, E subset of L } Prove m^(+) is a measure in X. So the first property follows: I'll denote the empty set as {}: m^(+) ({}) = sup { m (E) | E is in A, E subset of {} } But the only subset of the empty set is the empty set itself hence: = sup { m ({}) } but since m is a charge then m{}=0 thus m^(+) ( { } ) =0. I'm having problems showing the other two properties i.e nonegative and countably additivity. === === Subject: : Re: Measure theory proof > Let m be a charge What's a charge ? === === Subject: : Re: Complex Numbers, Equilateral Triangles Mautsch: > schrieb Michael Jrgensen : > This is not a HW problem. I'm not asking for help on a HW problem :) > Let z1, z2 and z3 be complex numbers and > z1^2 + z2^2 + z3^3 = z1z2 + z2z3 + z3z1 > Prove that z1, z2 and z3 form an equilateral triangle. > Here's a hint: (z1-z2)^2 + (z1-z3)^2 + (z2-z3)^2 = 0, given the above >> hypothesis. > By the way: I don't see what use the equation > (z1-z2)^2 + (z1-z3)^2 + (z2-z3)^2 = 0 > would be for *complex* numbers z1,z2,z3. > BTW, I still don't see what is to be done next > with this equation of the form > u^2 + v^2 + z^2 = 0. You have the additional relation (z1-z2) + (z2-z3) + (z3-z1) = 0, which translates to z = u+v. Setting a = u/v we then get a^2 + a + 1 = 0 => a^3 = 1. === === Subject: : Re: Complex Numbers, Equilateral Triangles > Mautsch: >>schrieb Michael Jrgensen : >>This is not a HW problem. I'm not asking for help on a HW problem :) >>Let z1, z2 and z3 be complex numbers and >z1^2 + z2^2 + z3^3 = z1z2 + z2z3 + z3z1 >Prove that z1, z2 and z3 form an equilateral triangle. >Here's a hint: (z1-z2)^2 + (z1-z3)^2 + (z2-z3)^2 = 0, given the above >hypothesis. >>By the way: I don't see what use the equation >>(z1-z2)^2 + (z1-z3)^2 + (z2-z3)^2 = 0 >>would be for *complex* numbers z1,z2,z3. >BTW, I still don't see what is to be done next >with this equation of the form >u^2 + v^2 + z^2 = 0. > You have the additional relation (z1-z2) + (z2-z3) + (z3-z1) = 0, which > translates to z = u+v. Setting a = u/v we then get a^2 + a + 1 = 0 => a^3 = > 1. > -Michael. That's an awesome solution - really short and really neat. I was working along the same line at the begining but never had the sense to use a=u/v. Thank you very much. === === Subject: : Re: Hints for self taught topology? > Here's a good free online introduction to algebraic topology: > http://www.math.cornell.edu/~hatcher/AT/ATpage.html That's a great book, thank you very much! I'm on warm milk and laxatives Cherry-flavored antacids reverse my forename for mail! - saibot === === Subject: : Going back home charset=Windows-1252 This is basically of theoretical interest, but here goes: Assume one is dropped somewhere on the vicinity of the Milky way. Could a consistent and practical Mathematical model for navigating around be found assuming one has unlimited time to travel? Making the question more specific: Does there exist a sufficiently accurate (to be practical) Mathematical model that would allow one to calculate one's way back to Earth, once one was dropped, say, near a star which sits 6,000 light years away from Earth? It seems to me that an appropriate (linear(?) because the distances are big) transformation T: R^3->R^3 could be used to calculate the new star positions at any location, but how could one utilize such a device to find one's way back to Earth? http://users.forthnet.gr/ath/jgal/ ------------------------------------------ Eventually, _everything_ is understandable === === Subject: : Re: Going back home In sci.math, Ioannis : > This is basically of theoretical interest, but here goes: > Assume one is dropped somewhere on the vicinity of the Milky way. Could a > consistent and practical Mathematical model for navigating around be found > assuming one has unlimited time to travel? > Making the question more specific: Does there exist a sufficiently accurate > (to be practical) Mathematical model that would allow one to calculate one's > way back to Earth, once one was dropped, say, near a star which sits 6,000 > light years away from Earth? > It seems to me that an appropriate (linear(?) because the distances are big) > transformation T: R^3->R^3 could be used to calculate the new star positions > at any location, but how could one utilize such a device to find one's way > back to Earth? What information is available to the traveler, and what equipment? Isaac Asimov, in one of his Foundation series (I think it was _Second Foundation_) hypothesized a Lens, in which one can superposition the actual starfield with various theoretical ones (computed by an on-ship computer unit). A later story hypothesized a supercomputer which could automate the Jump sequence. While the Jump [*] has been discredited (a pity since it sounds like a neat way to travel :-) ), a variant of the Lens could be used for navigation purposes, given the conditions you propose. Failing that, if one has a list of stars dotting the Galaxy, one might be able to search through thems using a spectroscope. The main problem with that of course is finding them, although one might automate the spectroscope. Also, there are some considerations regarding dimensions. The Milky Way has an estimated radius of 60,000 light years, taking the black hole as the center. This is 5.67 * 10^20 m. log2(5.67 * 10^20) = 68.94, which means standard IEEE-754 precision equipment won't quite cut it, as it can only handle 52 bits in the mantissa. Of course one can shift into another coordinate system when one gets close enough to one's destination; the Earth orbits with a radius of 1AU = 1.5 * 10^11 meter, which translates into a galactic radius of about 3.78 * 10^9 AU, with log2(3.78 * 10^9) = 31.8. This is still beyond standard floating-point metrics, as one has to compute a square root of sum of squares. Pluto is about 36 AU away on average (with an eccentric orbit); I'm not sure if 100 AU is close enough, or not. Isaac Asimov also write _Star Light_, a very short story which illustrates might happen were one to automate the process: a nova definitely bollixes up things. Even without this problem, stars move. If one knows the location of a neighboring galaxy when dropped into the scenario (ideally, the galaxy would be on the extension of the Milky Way disc), one can set up a coordinate system using a line from the center of the Milky Way to this galaxy, the normal to the galactic disc, and the cross-product of the two -- a fairly standard quasi-Cartesian coordinate system. (Quasi, because space is warped by the stars and/or the galactic black hole.) You might want to clarify your question, but it is an interesting problem. :-) [.sigsnip] [*] The notion was to somehow jump into hyperspace, die for a brief fraction of a second, then emerge back into real space (and yes, in at least one story in the _I Robot_ anthology, everyone survives :-); unfortunately I can't find my copy). Variants of this idea are all over sci-fi, from the relatively prosaic (unless an unbced engine pack generates a wormhole!) high-speed travel in Star Trek to the power-assisted jumpgates used in Babylon 5 to the device in Stargate. #191, ewill3@earthlink.net It's still legal to go .sigless. === === Subject: : Re: Going back home > In sci.math, Ioannis > <1079612305.639908@athnrd02.forthnet.gr>: > This is basically of theoretical interest, but here goes: > Assume one is dropped somewhere on the vicinity of the Milky way. Could a > consistent and practical Mathematical model for navigating around be found > assuming one has unlimited time to travel? > Making the question more specific: Does there exist a sufficiently accurate > (to be practical) Mathematical model that would allow one to calculate one's > way back to Earth, once one was dropped, say, near a star which sits 6,000 > light years away from Earth? > It seems to me that an appropriate (linear(?) because the distances are big) > transformation T: R^3->R^3 could be used to calculate the new star positions > at any location, but how could one utilize such a device to find one's way > back to Earth? > What information is available to the traveler, and what equipment? > Isaac Asimov, in one of his Foundation series (I think it > was _Second Foundation_) hypothesized a Lens, in which > one can superposition the actual starfield with various > theoretical ones (computed by an on-ship computer unit). > A later story hypothesized a supercomputer which could > automate the Jump sequence. While the Jump [*] has been > discredited (a pity since it sounds like a neat way to > travel :-) ), a variant of the Lens could be used for > navigation purposes, given the conditions you propose. > Failing that, if one has a list of stars dotting the > Galaxy, one might be able to search through thems using > a spectroscope. The main problem with that of course is > finding them, although one might automate the spectroscope. Didn't the Pioneer satellite that left the solar system include some kind of map showing the relative position of the sun to nearby stars based on spectroscopy? > Also, there are some considerations regarding dimensions. > The Milky Way has an estimated radius of 60,000 light > years, taking the black hole as the center. This is > 5.67 * 10^20 m. log2(5.67 * 10^20) = 68.94, which means > standard IEEE-754 precision equipment won't quite cut it, > as it can only handle 52 bits in the mantissa. Of course > one can shift into another coordinate system when one gets > close enough to one's destination; the Earth orbits with a > radius of 1AU = 1.5 * 10^11 meter, which translates into a > galactic radius of about 3.78 * 10^9 AU, with log2(3.78 * > 10^9) = 31.8. This is still beyond standard floating-point > metrics, as one has to compute a square root of sum of > squares. Pluto is about 36 AU away on average (with an > eccentric orbit); I'm not sure if 100 AU is close enough, > or not. > Isaac Asimov also write _Star Light_, a very short story which > illustrates might happen were one to automate the process: a > nova definitely bollixes up things. Even without this problem, > stars move. > If one knows the location of a neighboring galaxy when dropped > into the scenario (ideally, the galaxy would be on the > extension of the Milky Way disc), one can set up a coordinate > system using a line from the center of the Milky Way to this > galaxy, the normal to the galactic disc, and the cross-product > of the two -- a fairly standard quasi-Cartesian coordinate system. > (Quasi, because space is warped by the stars and/or the galactic > black hole.) > You might want to clarify your question, but it is an interesting > problem. :-) > [.sigsnip] > [*] The notion was to somehow jump into hyperspace, die for > a brief fraction of a second, then emerge back into real > space (and yes, in at least one story in the _I Robot_ anthology, > everyone survives :-); unfortunately I can't find my copy). > Variants of this idea are all over sci-fi, from the relatively > prosaic (unless an unbced engine pack generates a wormhole!) > high-speed travel in Star Trek to the power-assisted jumpgates > used in Babylon 5 to the device in Stargate. === === Subject: : Re: Going back home >This is basically of theoretical interest, but here goes: >Assume one is dropped somewhere on the vicinity of the Milky way. Could a >consistent and practical Mathematical model for navigating around be found >assuming one has unlimited time to travel? [...] >It seems to me that an appropriate (linear(?) because the distances are big) >transformation T: R^3->R^3 could be used to calculate the new star positions >at any location, but how could one utilize such a device to find one's way >back to Earth? What is a mathematical model for navigating around? I propose either: visit all planets in order, stop when reach Earth (assumes number of planets is at most countably infinite) or apply linear transform to self, thus transforming self to new coordinates on Earth I'm not interested in mathematics that might have anything to do with reality. -- Easterly, in sci.math === === Subject: : Re: Going back home > Assume one is dropped somewhere on the vicinity of the Milky way. Could a > consistent and practical Mathematical model for navigating around be found > assuming one has unlimited time to travel? http://astro.isi.edu/games/dimension.html which actually started from a discussion on SAA. The introduction goes a little into that (although it doesn't mention SAA specifically). It might answer your question a bit. Brian Tung It seems to me that an appropriate (linear(?) because the distances > are big) transformation T: R^3->R^3 could be used to calculate the new > star positions at any location, but how could one utilize such a > device to find one's way back to Earth? Even if you could, you'd run into the Borg first and then you'd pretty much be screwed. (And besides, since you're giving yourself infinite time and the stars move, it's not going to be a linear transformation.) === === Subject: : Re: Going back home charset=iso-8859-7 Bart Goddard are big) transformation T: R^3->R^3 could be used to calculate the new > star positions at any location, but how could one utilize such a > device to find one's way back to Earth? > Even if you could, you'd run into the Borg first and then > you'd pretty much be screwed. (And besides, since you're giving > yourself infinite time and the stars move, it's not going to be > a linear transformation.) Um, what about if we want to caclulate the path and assume instantaneous travel instead? As far as the Borg are concerned, they are us in the distant future. So the solution to this problem is to find a way to communicate to them effectively this very fact. If they understand it, they will have no reason to assimilate us, cause they will understand that eventually we will turn into them. No need to hurry :*) http://users.forthnet.gr/ath/jgal/ ------------------------------------------ Eventually, _everything_ is understandable === === Subject: : Re: Going back home > Um, what about if we want to caclulate the path and assume > instantaneous travel instead? > As far as the Borg are concerned, they are us in the distant future. > So the solution to this problem is to find a way to communicate to > them effectively this very fact. If they understand it, they will have > no reason to assimilate us, cause they will understand that eventually > we will turn into them. No need to hurry :*) === Subject: has some useful inquisitiveness, we'd better assimilate him. Let's hurry so as to cut down on the noise. So I suppose that I know exactly what the Big Dipper looks like from Earth, and now that I'm at a random spot in the galaxy and I have a 3-D map of the stars in my computer from my new location, so I should be able to do something better than have the computer search all possible locations and test all possible points of view from those locations to see if it can see the Big Dipper from there. The transform might be represented by a 3x3 matrix, but there's still some assumptions about what the observers knows. If he's Arthur Dent, and has only a 2-D spherical surface picture of the stars from earth, (which is what I outlined in the above paragraph) then it's not a 3x3 matrix we're dealing with. The original positions of the stars are not known, but we could in theory, find our way back home since we have lots of extra information. (More than three stars. And hopefully there's no other Big Dippers in the Galaxy.) Even if you decide on the type of map you want, I'm not sure how you pin it down at even one point, since before you leave earth, you have no idea where you're going, so you don't know which of the zillions of maps it's going to be. And once you get there, you're going to need its inverse, and since you didn't know which map it was, and you don't know where you are, you sure can't come up with even a single inverse image point. (And I'm thinking you need at least 3 no matter what rules you play by.) Best case scenerio: You're stuck on an Edenic planet with Seven-of-Nine for a couple of decades. Worst case scenerio: She spends the entire time in the astrometrics lab obsessing about this very problem. === === Subject: : Re: Going back home <1079618648.943201@athnrd02.forthnet.gr Bart Goddard It seems to me that an appropriate (linear(?) because the distances > are big) transformation T: R^3->R^3 could be used to calculate the new > star positions at any location, but how could one utilize such a > device to find one's way back to Earth? > (And besides, since you're giving yourself infinite time and the stars > move, it's not going to be a linear transformation.) > Um, what about if we want to calculate the path and assume instantaneous > travel instead? Then by relativistic effects you'd arrive infinitely in the future long after good old Sol went supernova and long after the big bang universe came to it's big crunch, heat death or big rip end, looking for a space traveling society, which if they didn't self destruct within decades of your departure, would have much improbability surviving the universal end. The suggestion to the use the Hitchhikers' Guide to the Universe may be helpful, especially the improbability drive which might let you break the speed limit. As I recall somebody built a hotel at the end of time for time traveling tourists to view a very unique event. No matter how you travel from distance random location, you'll need time travel to adjust your travels back to the home you know. === === Subject: : Re: Going back home Bart Goddard It seems to me that an appropriate (linear(?) because the distances > are big) transformation T: R^3->R^3 could be used to calculate the new > star positions at any location, but how could one utilize such a > device to find one's way back to Earth? > (And besides, since you're giving yourself infinite time and the stars > move, it's not going to be a linear transformation.) > Um, what about if we want to calculate the path and assume instantaneous > travel instead? > Then by relativistic effects you'd arrive infinitely in the future long > after good old Sol went supernova and long after the big bang universe > came to it's big crunch, heat death or big rip end, looking for a space > traveling society, which if they didn't self destruct within decades of > your departure, would have much improbability surviving the universal end. > The suggestion to the use the Hitchhikers' Guide to the Universe may be > helpful, especially the improbability drive which might let you break the > speed limit. As I recall somebody built a hotel at the end of time for > time traveling tourists to view a very unique event. Hilbert? > No matter how you travel from distance random location, you'll > need time travel to adjust your travels back to the home you know. === === Subject: : Re: Going back home I'd just grab my towel and electronic thumb, and hitch a ride. Oh yeah, and maybe use the Guide too. === === Subject: : pi, 2 pi and frivolity To me it seems rather curious that the human race has chosen pi rather than 2 pi as the important number that deserves its own special symbol. It is 2 pi and not pi that appears in the most fundamental equations. Primary school children are taught that C = pi D, but the diameter of the circle is not as fundamental as the radius and isn't used in any other commonly quoted equation. We use radians, not diametans. The equation which describes a circle uses the radius not the diameter. If you can remember back to when you first learnt about radians, didn't it seem rather odd and hard to remember that the angle you were used to thinking of as 60 degrees was pi/3 radians, that 90 degrees was pi/2 radians or that 30 degrees was pi/6 radians. I know I certainly did. Supposing that we refer to 2 pi as @, for want of a better symbol, then doesn't it seem more intuitive that 90 degrees which is also known as a quadrant - note the four there in that word - is @/4 radians, than pi/2 radians? Likewise that 30 degrees which is the angle at then centre between two hours on a clock face is @/12 radians? Surely a revolution is more important than a half revolution? What's the most salient thing about the periodic functions sin and cos? The fact that they are periodic, and the period of course is 2 pi. Consider the identity cos x = sin(x - pi/2). Isn't it more obvious and easier to remember that cos x = sin(x - @/4)? The identity shifts the origin a quarter of a period. In physics the ratio between frequency and angular frequency is 2 pi. We have centripetal acceleration equal to 4 pi^2 r / T^2. I realise that I'm saying the same thing in about 6 different ways, but surely the fact that it can be said in 6 different ways is some measure of its truth. Surely exp(i 2 pi) = 1 is just as mysterious and beautiful as exp(i pi) = -1 Of course the formula for the area of a circle uses pi and not 2 pi, but that's in two dimensions not one like C = 2 pi r, and its not fundamental. When I think back to the primary school demonstration of that equation though, having a divide by 2 in the formula actually seems more natural. The demonstration I mean is the one where the circle is divided up into a large number of sectors, and the sectors are rearranged in alternating fashion to end up with a rectangle that has dimensions r by C/2. To me the choice of pi and not 2 pi seems like a historical accident, and I wonder whether it was because people didn't immediately realise, centuries ago, that the radius and not the diameter of a circle was its more fundamental measure. The diameter gives a more primitive sense of the circle's size, like the side of a square or an equalateral triangle. So my question to the group is, is there general agreement with my points? If not why not and if so how many millions of times has this been pointed out before? Perhaps it's not an important point, but nevetheless one thing appalls me. All these people who have learnt pi to thousands of digits and all the college students who march up and down the halls chanting pi in unison - if I'm right they've been learning and chanting the wrong number! :-) === === Subject: : Re: pi, 2 pi and frivolity > To me it seems rather curious that the human race has chosen pi rather than > 2 pi as the important number that deserves its own special symbol. > It is 2 pi and not pi that appears in the most fundamental equations. > Primary school children are taught that C = pi D, but the diameter of the > circle is not as fundamental as the radius and isn't used in any other > commonly quoted equation. We use radians, not diametans. The equation which > describes a circle uses the radius not the diameter. > If you can remember back to when you first learnt about radians, didn't it > seem rather odd and hard to remember that the angle you were used to > thinking of as 60 degrees was pi/3 radians, that 90 degrees was pi/2 > radians or that 30 degrees was pi/6 radians. I know I certainly did. > Supposing that we refer to 2 pi as @, for want of a better symbol, then > doesn't it seem more intuitive that 90 degrees which is also known as a > quadrant - note the four there in that word - is @/4 radians, than pi/2 > radians? Likewise that 30 degrees which is the angle at then centre between > two hours on a clock face is @/12 radians? Surely a revolution is more > important than a half revolution? > What's the most salient thing about the periodic functions sin and cos? The > fact that they are periodic, and the period of course is 2 pi. Consider the > identity cos x = sin(x - pi/2). Isn't it more obvious and easier to remember > that cos x = sin(x - @/4)? The identity shifts the origin a quarter of a > period. > In physics the ratio between frequency and angular frequency is 2 pi. We > have centripetal acceleration equal to 4 pi^2 r / T^2. > I realise that I'm saying the same thing in about 6 different ways, but > surely the fact that it can be said in 6 different ways is some measure of > its truth. > Surely exp(i 2 pi) = 1 is just as mysterious and beautiful as exp(i pi) = -1 > Of course the formula for the area of a circle uses pi and not 2 pi, but > that's in two dimensions not one like C = 2 pi r, No wonder you're confused, you got the formula wrong. It's not 2 pi r it's pi 2 r because (2 r) = d. The circumference and area both use pi, neither uses 2 pi. > and its not fundamental. When done correctly, it is fundamental. > When I think back to the primary school demonstration of that equation > though, having a divide by 2 in the formula actually seems more natural. The > demonstration I mean is the one where the circle is divided up into a large > number of sectors, and the sectors are rearranged in alternating fashion to > end up with a rectangle that has dimensions r by C/2. > To me the choice of pi and not 2 pi seems like a historical accident, and I > wonder whether it was because people didn't immediately realise, centuries > ago, that the radius and not the diameter of a circle was its more > fundamental measure. The diameter gives a more primitive sense of the > circle's size, like the side of a square or an equalateral triangle. > So my question to the group is, is there general agreement with my points? > If not why not and if so how many millions of times has this been pointed > out before? > Perhaps it's not an important point, but nevetheless one thing appalls me. > All these people who have learnt pi to thousands of digits and all the > college students who march up and down the halls chanting pi in unison - if > I'm right they've been learning and chanting the wrong number! :-) === === Subject: : Re: pi, 2 pi and frivolity if D is the diameter, the area of the sphere is piDD, and the circumference is piD. the radius is not really as fundamental, because all spherics spin (saith Bucky .-) that is to say, you don't know by inspection what is at the center of the sphre, although you can easily see the two poles of its rotation (and calculate the diameter by its angular diameter & distance). the fact that we don't use diadians is only because of the construction of circles with compasses -- blame the Greeks; eh? yes, a revolution is more fundamental than a half of one, but I'm not going to start using military time ... or it's 0.5 rev; time for lunch! http://www.rwgrayprojects.com/synergetics/plates/figs/plate02. html > Primary school children are taught that C = pi D, but the diameter of the > circle is not as fundamental as the radius and isn't used in any other > commonly quoted equation. We use radians, not diametans. The equation which > two hours on a clock face is @/12 radians? Surely a revolution is more > important than a half revolution? --Give Earth a Trickier Dick Cheeny -- out of office, after GIGA years. http://www.benfranklinbooks.com/ http://www.rand.org/publications/randreview/issues/rr.12.00/ http://members.tripod.com/~american_almanac === === Subject: : Re: pi, 2 pi and frivolity > .... > To me the choice of pi and not 2 pi seems like a historical accident, and I > wonder whether it was because people didn't immediately realise, centuries > ago, that the radius and not the diameter of a circle was its more > fundamental measure.... Yes, exactly. Euclid normally used diameters, and didn't even define a specific term for radius. Archimedes calculated bounds for the ratio of the circumference of any circle to its diameter Authors in the early modern period used many notations for that ratio. In 1652 Oughtred used pi/delta (evidently standing for periphery/diameter). Various later authors including Euler used pi (for periphery), although other notations persisted throughout the 18th century. I agree that in some ways a single letter for six-and-a bit might be more convenient than our letter pi for three-and-a-bit. However, mathematics grows organically, and we've inherited various words and symbols which may not be what we would have chosen if we were setting up mathematical notation afresh. === === Subject: : Re: pi, 2 pi and frivolity > To me it seems rather curious that the human race has chosen pi rather than > 2 pi as the important number that deserves its own special symbol. So much so that you're not the first one to have thought about it. Read 23 no. 3, pp. 7-8 (2001)). Jose Carlos Santos === === Subject: : Re: pi, 2 pi and frivolity > To me it seems rather curious that the human race has chosen pi rather than > 2 pi as the important number that deserves its own special symbol. > So much so that you're not the first one to have thought about it. Read > 23 no. 3, pp. 7-8 (2001)). I remember Serge Lang saying much the same thing in a talk he gave 20 or 30 years ago. === === Subject: : Re: pi, 2 pi and frivolity >I realise that I'm saying the same thing in about 6 different ways, but >surely the fact that it can be said in 6 different ways is some measure of >its truth. James Harris is an outstanding mathematician. James Harris' contributions to mathematics are invaluable. James Harris ranks amongst the world's top mathematicians. James Harris is second to none is mathematical ability in sci.math. James Harris has single-handedly revolutionalized mathematics. James Harris is a genius among mathematicians. Nope. I'm not interested in mathematics that might have anything to do with reality. -- Easterly, in sci.math === === Subject: : Re: pi, 2 pi and frivolity >To me it seems rather curious that the human race has chosen pi rather than >2 pi as the important number that deserves its own special symbol. ... >To me the choice of pi and not 2 pi seems like a historical accident, and I >wonder whether it was because people didn't immediately realise, centuries >ago, that the radius and not the diameter of a circle was its more >fundamental measure. The diameter gives a more primitive sense of the >circle's size, like the side of a square or an equalateral triangle. >So my question to the group is, is there general agreement with my points? >If not why not and if so how many millions of times has this been pointed >out before? Well, it was pointed out by Albert Eagle in his book _The Elliptic Functions as They Should Be_. Unless what he pointed out was the pi/2 was the important number that deserves its own special symbolCome to think of it, I guesss that *was* what he pointed out: because I remember that the special symbol in question was tau, which is of course pi chopped in two. Lee Rudolph === === Subject: : Re: pi, 2 pi and frivolity Discussion, linux) >So my question to the group is, is there general agreement with my points? >If not why not and if so how many millions of times has this been pointed >out before? > Well, it was pointed out by Albert Eagle in his book _The Elliptic > Functions as They Should Be_. Unless what he pointed out was the > pi/2 was the important number that deserves its own special symbol> Come to think of it, I guesss that *was* what he pointed out: because > I remember that the special symbol in question was tau, which is > of course pi chopped in two. Albert Eagle! Source of one of my favorite math quotes. Such behaviour is exclusively confined to functions invented by mathematicians for the sake of causing trouble. -Albert Eagle's _A Practical Treatise on Fourier's Theorem_ I know nothing about the man. I got the quote second hand. I want to say it was mentioned in Dudley's book on cranks, but I might be just wrong. All intelligent men are cowards. The Chinese are the world's worst fighters because they are an intelligent race[...] An average Chinese child knows what the European gray-haired statesmen do not know, that by fighting one gets killed or maimed. -- Lin Yutang === === Subject: : Re: pi, 2 pi and frivolity >>So my question to the group is, is there general agreement with my points? >>If not why not and if so how many millions of times has this been pointed >>out before? >Well, it was pointed out by Albert Eagle in his book _The Elliptic >Functions as They Should Be_. Unless what he pointed out was the >pi/2 was the important number that deserves its own special symbol>Come to think of it, I guesss that *was* what he pointed out: because >I remember that the special symbol in question was tau, which is >of course pi chopped in two. >Albert Eagle! Source of one of my favorite math quotes. Such behaviour is exclusively confined to functions invented by >mathematicians for the sake of causing trouble. > -Albert Eagle's _A Practical Treatise on Fourier's Theorem_ >I know nothing about the man. I got the quote second hand. I want to >say it was mentioned in Dudley's book on cranks, but I might be just >wrong. I don't know anything about Albert Eagle or his book, and I don't know what's mentioned in Dudley. But I do know something about Fourier series and transforms, and I don't see anything particularly crankish about the quote. The point: Fourier's theorem says that any periodic function is the sum of its Fourier series. He was never able to prove this, because it's not true - it's not even true for continuous periodic functions. But although it's not literally true there's a lot of truth in it - it _is_ true for most of the functions that come up in practice. People like me spend a lot of time proving theorems about exactly when it and similar things _are_ true, but that's just because we have too much time on our hands - if an engineer thinks it's simply true that's not likely to cause too much trouble. That is, showing that Fourier's theorem is _not_ true _did_ require that people invent functions to cause trouble. (Of course the problem of showing it was true, and then clarifying under what conditions it was true when it became clear that some hypotheses were needed, is where a lot of the basic ideas in analysis came from.) === === Subject: : Re: pi, 2 pi and frivolity Adjunct Assistant Professor at the University of Montana. >Albert Eagle! Source of one of my favorite math quotes. [...] >I know nothing about the man. I got the quote second hand. I want to >say it was mentioned in Dudley's book on cranks, but I might be just >wrong. Well, he's not in the index of that book (Mathematical Cranks). Didn't check the others. ======== === === Subject: : Re: pi, 2 pi and frivolity What's the point? The important thing is that everyone agrees on the same symbols. Using your arguments, why are there sixty minutes in an hour? Surely, 100 minutes/hour would be easier to deal with. I can't see why 2pi is more fundamental than pi. Sometimes 2pi appears in a formula, sometimes just pi. How do you judge the fundamentality of a number? Do you base it on subjective notions as to which formulae are easier to remember? === === Subject: : Re: pi, 2 pi and frivolity Discussion, linux) > What's the point? The important thing is that everyone agrees on the same > symbols. Using your arguments, why are there sixty minutes in an hour? > Surely, 100 minutes/hour would be easier to deal with. Imagine how much more we'd be able to accomplish, too. Leaving things always seems to fix me, Running seems to ease my worried mind. -- Bad Livers, Honey, I've Found a Brand New Way === === Subject: : Re: pi, 2 pi and frivolity > What's the point? The important thing is that everyone agrees on the same > symbols. Using your arguments, why are there sixty minutes in an hour? > Surely, 100 minutes/hour would be easier to deal with. If you had 100 seconds in a minute, 100 minutes in an hour, and 10 hours in a day, the second wouldn't have to change by all that much. === === Subject: : uncorrelated and zero mean IFF orthogonal ? By orthogonal, I mean this: E[xy*]=E[yx*]*=0 Uncorrelated is this: E[xy*] = E[x]E[y*] I know the first implies the second. Can anybody show the other way? === === Subject: : Uneven Cantor sets/ Mandelbrot-Cantor Noise I.F.S. This algorithm was suggested by the figure on page 25 of Multifractals and 1/f Noise by Benoit B. Mandelbrot I call it for that reason Mandelbrot-Cantor Noise from uneven Cantor sets. The idea is that there are six ways you can rearrange a Mandelbrot cartoon of an uneven Cantor set. This program takes each one of those states at random. I also have an IFS of this in 2d, but it is a very uninteresting dust. An alternative formulation would be on segments at (0,1/3,2/3) instead of (1/3,2/3,1). A zero symmetrical formulation would be on (-1/3,0,1/3). All these would give uneven Cantor sets related to the Mandelbrot kind of noise generator. True Basic program: SET MODE color SET WINDOW 0,1920,0,1024 SET BACKGROUND COLOR white LET x=.2 LET y=.3 LET a=0 LET b =0 LET s1=310*2 LET s2 =s1*1024/1920 RANDOMIZE PRINTUneven Cantor sets/ Mandelbrot-Cantor Noise I.F.S. PRINT three uneven line segments of lengths {1/3,2/3,1} PRINT with order taken at random: 6 permutations possible REM Mandelbrot cartoon cell types are three: REM Cantor type ( center down) REM anti-cantor type ( center up) REM staircase type ( up and down) Rem Moran dimension 6*(1/3)^s=1-- s=log(3)/Log(6)=0.613147 LET r=2 FOR n= 1 TO 1920 step 3 REM two random variables REM and two Frame type exception markers LET a=RND LET b=RND LET d=0 LET e=0 IF a <= 1/2 and d=0 and e=0 THEN LET x1=x/3+1/3 LET d=1 END IF IF a<= 2/3 AND a>1/3 and d=0 and e=0 THEN LET x1=x/3+2/3 LET d=2 END IF IF a<= 1 AND a>2/3 and d=0 and e=0 THEN LET x1=x/3+1 LET d=3 END IF SET COLOR red LET x=x1 LET y=y1 IF n>10 THEN PLOT n,1024/3+s2*x; REM end first segment: marker d REM only two possiblities left in the cycle IF b<=1/2 and d=1 and e=0 then LET x1=x/3+2/3 LET e=2 ELSE LET x1=x/3+1 LET e=3 END IF IF b<=1/2 and d=2 and e=0 then LET x1=x/3+1/3 LET e=1 ELSE LET x1=x/3+1 LET e=3 END IF IF b<=1/2 and d=3 and e=0 then LET x1=x/3+1/3 LET e=1 ELSE LET x1=x/3+2/3 LET e=2 END IF SET COLOR blue LET x=x1 LET y=y1 IF n>10 THEN PLOT n+1,1024/3+s2*x; REM end second segment: marker e REM only one possiblity left in the cycle IF (d=1 and e=2) or (d=2 and e=1) then LET x1=x/3+1 END IF IF (d=2 and e=3) or (d=3 and e=2) then LET x1=x/3+1/3 END IF IF (d=1 and e=3) or (d=3 and e=1) then LET x1=x/3+2/3 END IF SET COLOR Green LET x=x1 LET y=y1 IF n>10 THEN PLOT n+3,1024/3+s2*x; REM end third segment NEXT n END Respectfully, Roger L. Bagula tftn@earthlink.net, 11759Waterhill Road, Lakeside,Ca 92040-2905,tel: 619-5610814 : URL : http://home.earthlink.net/~tftn URL : http://victorian.fortunecity.com/carmelita/435/ === === Subject: : Re: Ability In Mathematics > TO THE SCI.MATH NEWSGROUP > I have an enquiry which people in this Mathematics newsgroup > might want to address if they so desire: > Firstly,I have to say that,personally,I have always found Mathematics > difficult.It has hindered my understandings of Astronomy and I failed > the British 11+ examination BECAUSE of it and so,consequently, > went to a British Secondary Modern School as opposed to the more > academic British Grammar School.Even though I had understanding > and competence in every other subject I studied both at school and , > later, college.So,in at least two crucial ways,Mathematics has been a > very significant subject in my life. > So,given all this, have you any ideas as to why for some people, > Mathematics ability comes so naturally whilst other people find it so > difficult?Clearly some people see patterns in Mathematics whilst > others,NO MATTER HOW MUCH THEY TRY,cannot. > Why is this so? > Thank you. > Brian > Devonald Brian, I think I can answer your question. When I was in high school I took algebra. I got an A in spite of the fact that I didn't get certain aspects of solving word problems. In particuluar I couldn't get how to do what I call mixed rate probelms. Years later I was tutoring math professionally and I was using tricks to solve word problem that my students did not know about. So, I made a list of the tricks that I used. I learned that a more appropriate name for these tricks is heuristics: To solve almost all high school word problems one needs to know how to do basic manipulation of the real numbers, of course, but one also needs to know how to use conversion factors. That's the preliminary stuff. Now for my list of heuritics: 1) Does the problem contain a sum of consecutive integers? (I include this because it is often the basis of the first word problems that students encounter and heuristics should begin as soon as possible for the student) 2) Is there a total or a part in the problem; if so, every total is equal to the sum of its parts. (This is the most important of all the rules for two reasons: First, because most word problems will have at least one equation of this form, and knowing that can help one to find it buried in the words. Second, because even beginners can quickly learn how to spot totals and parts!) 3) Is there an invariant quantity Q in a before-and-after process? If so, Q_before = Q_after. (Which is an equation, and equations are the beginning of solving systems for whatever it is you need to know.) 4) Is there a proportion? (A proportion is the claimed equality of two ratios.) 5) Any formulas called for, such the circumference of a circle or the area of a triangle? 6) Can the problem or a portion of it be solved by exhaustive search? Let's see this in action! Problem: Steve can mow a lawn in 3hrs; John can mow the same lawn in 2hr. If they both work together to mow the lawn, starting and stopping at the same time, how long will it take them to mow the lawn? I call this a mixed rate problem. First, we have to make an assumption which is usually made implicitly: The fact that Steve and John work together does not change the rates at which they work. OK, let's see how beginners are often tempted to try to solve this problem if they don't know the heuristics. They say something like: Gee, they ask for a time. How do I find an equation with time in it that gives the answer? This is called formula chasing. It is almost always counterproductive! The first thing to do to solve this problem is to NOT go on a formula chasing adventure into the jungle of formulas, but rather, to forget what you're trying to solve for in particular and just translate the word problem into a system of equations, which are then solved for the variable you need. I was never taught this; I learned it on my own by trial and error long after I got my degree in math! Now, let's use the list of heuristics above. Is there a sum of consecutive integers in the problem? No. Is there a total or a part in the problem? Yes! There is the part of the job done by Steve! Well then, we know that for every part there is a total equal to the sum of its parts. So, there is a total of one job being done: 1 job = (part of job done by Steve) + (part of job done by John) (Notice that I am in no hurry to eliminate words from the equations!!) Now, the part of job done by Steve is given by the time T Steve worked times a conversion factor (a rate) that converts time to job (just make the units come out right and you're OK -- another heuristic): part of job done by Steve = T (1job / 3hr) and similarly for John. So, 1 job = T (1job / 3hr) + T (1job / 2hr) from which we get that 1 hr = T (1/3 + 1/2) So, T = 1 /(1/3 + 1/2) hr = 6/5 hr = 1hr 12 minutes. One book I read chose to refer to this kind of problem as the fractions type, simply because of the form of (1/a + 1/b) that they contain. But this is clearly worse than pointless naming criteria. If classes of word problems are to be given names, they should be given names based on heuristic concerns, not on mere superficial arithmetic appearances! Calling the problem a mixed rate type clues in the problem solver that rates are involved! Generally speaking, the mixed rate problem has two or more machines working together at different rates, and possibly for different time periods. For more on this see (my presentation of this has evolved over the years) http://ajnpx.com/html/Math/IntroAlgebra/IntroAlgebra3.html http://ajnpx.com/pdf/AJNP/oct92c.pdf http://ajnpx.com/pdf/math/SpecialTopics/Mod_Wrd_Prb2.pdf Don't even let me get started on the best heuristics for group theory. See also http://ajnpx.com/html/MathFlowCharts.html Heuristics can make easy what would otherwise be hard. When I first took group theory I thought that Cauchy's theorem was completely out of the blue on each step. But now that I know the heuristics of it, I see the steps as almost obvious! It all has to do do with learning how to solve for what goes into the Main Logic Splitter! (My novel approach to theorem proving.) If anyone is intersted in my explanation, I will give it. So, those who do well in math have somehow learned the heuristics of it. Patrick === === Subject: : Re: Ability In Mathematics Personally, I've always found mathematics extremely easy (of course, I'm only through differential equations, still a ways to go ;)). I learned all of my elementary school mathematics in 1st and 5th grades. I easily visualize equations and word problems, and find calculations rather easy to perform internally (didn't hurt things that I didn't have a calculator until sophomore in high school - I think THIS is a large contributor to lack of mathematics ability - overuse of calculators). However, I've always been exceedingly lacking in the proofs department, and I'm mediocre at non-trigonometric geometry. Dunno, I've always just said some people have their talents, and some people have other talents. (and I'm average at chess. Started at 15, approximately 1600 USCF by 17, haven't played seriously since (20 now)). === === Subject: : Re: Ability In Mathematics > and find calculations > rather easy to perform internally (didn't hurt things that I didn't > have a calculator until sophomore in high school - I think THIS is a > large contributor to lack of mathematics ability - overuse of > calculators). I agree with you here. There are many students that are so worried about finding their calculator that they forget what they are trying to do. If you can't do basic arithmatic, you will not be able to easily focus on more advanced concepts, or truly understand them. Will Twentyman email: wtwentyman at copper dot net === === Subject: : Re: Ability In Mathematics Addendum: I've recently gotten interested in Vedic Mathematics. Quite interesting. === === Subject: : Re: Ability In Mathematics >.. Even though I had understanding and competence in every other subject I studied both at school and later, college.. Fears/inferiority associated with early failures,long held belief about self incapacity ...turning into a phobia of equations or symbols and a refusal to recognize patterns in mathematics which faculty comes so easily in other subjects or life situations. It is never too late to reverse it. It needs some relaxed fortitude, perseverance and perhaps an occasional word of encouragement from a guru/mentor to help locate and remove old stumbling blocks hidden from one's conscious experience. A joy of victory against self-doubt and pride in success gained later are not impossible. There are many who were branded poor in the early years but picked up miraculously later on, may be in different environments, but not without a strong determination. Good luck ! === === Subject: : Re: Ability In Mathematics >.. Even though I had understanding and competence in every other > subject I studied both at school and later, college.. > Fears/inferiority associated with early failures,long held belief > about self incapacity ...turning into a phobia of equations or symbols > and a refusal to recognize patterns in mathematics which faculty comes > so easily in other subjects or life situations. > It is never too late to reverse it. It needs some relaxed fortitude, > perseverance and perhaps an occasional word of encouragement from a > guru/mentor to help locate and remove old stumbling blocks hidden from > one's conscious experience. > A joy of victory against self-doubt and pride in success gained later > are not impossible. There are many who were branded poor in the early > years but picked up miraculously later on, may be in different > environments, but not without a strong determination. Good luck ! You could have been describing my wife with this statement. She was told in high school by a teacher that she would NEVER be able to succeed in the sciences due to her ability in math. It turns out that she had teachers who simply didn't care enough to explain it to her. Later, she met a teacher who explained the math to her in clear terms. She now has an associate's degree in electronics and will begin her bachelor's degree in MATH this summer. Perception of ability combined with poor teachers have ruined many students. Will Twentyman email: wtwentyman at copper dot net === === Subject: : Re: Ability In Mathematics > She now has an associate's degree in electronics and will begin her > bachelor's degree in MATH this summer. Perception of ability combined > with poor teachers have ruined many students. Did I just read that they are relaxing the proposed requirement that math teachers in high school must have degrees in math? (and science, too) It seems there is a shortage of people with math/science degrees willing to work at teachers' wages/conditions. === === Subject: : Re: Ability In Mathematics >Did I just read that they are relaxing the proposed requirement that >math teachers in high school must have degrees in math? (and science, >too) It seems there is a shortage of people with math/science >degrees willing to work at teachers' wages/conditions. I don't know who they are, or which place you're talking about. Around here AFAIK it has never been a requirement that people teaching high school math have degrees in math. If a school needs somebody to teach a math class and the gym teacher is available, guess what happens? Department of Mathematics http://www.math.ubc.ca/~israel === === Subject: : Re: Ability In Mathematics >An illustration: If you do Math and also play chess, your chess ability is >very close to your Math ability and vice versa. How depressing. Must get my chess board out and start practicing again. I'm not interested in mathematics that might have anything to do with reality. -- Easterly, in sci.math === === Subject: : Re: Ability In Mathematics charset=utf-8 Toni Lassila [CapitalEth][EDouble Dot][Micro] .b3 .b9[EDo ubleDot].b9 >An illustration: If you do Math and also play chess, your chess ability is >very close to your Math ability and vice versa. > How depressing. Must get my chess board out and start practicing > again. Um, a small clarification: The Theorem is true, assuming similar depth and time involvement on both endeavors. It's not true for example if the time you've devoted to chess, is 1% of the time you've devoted to math. It could be restated differently as follows: Suppose a person has been involved with endeavor y and is q% good at it. The Theorem states that IF the person starts involving themselves with logical endeavor z, after having devoted to it an equivalent amount of time as the time for y, in the limit, he cannot be more than q% good at z. > -- I'm not interested in mathematics that might have anything > to do with reality. -- Easterly, in sci.math http://users.forthnet.gr/ath/jgal/ ------------------------------------------ Eventually, _everything_ is understandable === === Subject: : Re: Ability In Mathematics > Suppose a person has been involved with endeavor y and is q% good at it. The > Theorem states that IF the person starts involving themselves with logical > endeavor z, after having devoted to it an equivalent amount of time as the > time for y, in the limit, he cannot be more than q% good at z. This is quite interesting. What kind of theorem is it? It doesn't sound like a mathematical one. I mean, can you prove this?!? === === Subject: : Number Theory Question Dears Do there exist two distinct prime numbers $p$ and $q$ and integers $n>1$ and $m>1$ satisfying (simultaneusly) the following equations? p^(n(n+1)/2)-p^((n^2-n+2)/2)-q^(m(m+1)/2)+q^((m^2-m+2)/2)=0 p^(n(n+1)/2)-p^((n^2-n)/2)-q^(m(m+1)/2)+q^((m^2-m)/2)=0 This is related to a Group theory problem, the positive answer will be very interesting for me!!!! All the best Alireza Abdollahi === === Subject: : Re: Packing squares into a 1 x Zeta[2] rectangle > Since the sum of the reciprocals of the squares of the positive integers > is pi^2/6, the question arises as to whether squares with sides 1, 1/2, > 1/3, etc can be packed into a rectangle of size 1 by pi^2/6. > A picture of such a packing appears at > http://www.pisquaredoversix.force9.co.uk/Tiling.htm > I know of no proof that such a packing is possible or impossible. It would be good if people interested in this thread were aware of some of the history of this problem. To that end: _Unsolved Problems in Geometry_ by Croft, Falconer and Guy (Springer-Verlag, 1991) D5. Packing unequal rectangles and squares in a square. (pp. 112-113) mentions a question similar to the one considered in this thread. D5 caused me, in 1996, to consider several packing problems of this sort. I found simple algorithms, which I implemented in Mathematica, which seemed to pack rectangles of decreasing size perfectly. The algorithms worked for classic problems, such as packing the rectangles 1/n by 1/(n+1) into the unit square, as well as for interesting new packing problems. I did not publish my results then, hoping first to prove that the algorithms were adequate for packing all of the rectangles. In November 1999, spurred by Clive Tooth, I posted to sci.math a brief description of what I had done. See both parts of Packing Rectangles of Decreasing Size at . (Readers may also be interested in looking at the packing algorithm which Clive had used back then.) Unbeknownst to me in 1999, an algorithm for such packings had appeared previously in the literature: An Algorithm for Packing Squares, Marc M. Paulhus, _J. Combin. Theory Ser. A_ 82 (1997) 147-157. His algorithm is somewhat different from any of the algorithms which I had used (but I wonder if perhaps it is the same as Clive's current algorithm). For packing squares of side lengths 1/2, 1/3, 1/4,... into a rectangle 1/2 by 2(pi^2/6 - 1), Paulhus gives two rules: Rule 1. Always place the next square in the corner of the smallest width rectangle into which it will fit. Rule 2. After placing a square into the corner of a rectangle, always cut the remaining area into two rectangular pieces by cutting from the free corner of the square tot the longer side of the original rectangle. known to me concerning such packings -- but again, I may not be up-to-date with the literature -- are the following: Perfect Square Packings, Adam Chalcraft, _J. Combin. Theory Ser. A_ 92 (2000) 158-172. Compactness Theorems for Geometric Packings, Greg Martin, . Perfect Packings of Squares Using the Stack-Pack Strategy, Johan Wastlund, . Thoughtful comments are welcome. David W. Cantrell === === Subject: : Re: Packing squares into a 1 x Zeta[2] rectangle > a few questions... >> I work with absolute precision numbers, > I assume that means you keep the sides as full precision rationals. Do you > keep them in their lowest terms? > Yes, I store each rectangle as three BigIntegers w,h,n, > where the sides are w/n and h/n , > and check whenever a rectangle is created that gcd(w,h,n) = 1. > (I don't check eg that gcd(w,n) = 1, > but I don't see that that should matter.) > I also check the orientation of each rectangle as it is created. >> I discard rectangles with h < 0.999/maxSquare > Why not with h < 1/maxSquare? > I actually keep floats x = w/n and y = h/n with each rectangle, > and use them for comparison for keeping the rectangles in order. > (I did this because I thought this might be the cause of the slowness > of my program, but I found the actual reason was that I made an error > in discarding rectangles with w < 1/n^2 rather than w < 1/n . > When I corrected this the program ran quite fast, > and could easily have reached 10^6, if it hadn't bombed out!) > I assume that you do not keep the vacant space to full precision. > No, I just keep the area of each rectangle as a float, for interest.. > Here's the main routine of my program, to see my approach. > I don't set myself up as a great programmer, > and it is all too likely that I've made a mistake. > ======================================================= > public void removeSquare(int i) { > int index = findIndex(1.0/(double)i) - 1; > if (index < 0) { > System.err.println(No room for square side 1/ + i); > System.out.println(No room for square of side 1/ + i > + = +1/(float)i); > printList(true); > System.exit(1); > } > Rectangle r = (Rectangle)rectangles.get(index); > // System.out.println(Removing + r); > rectangles.remove(index); > currentIndex = index; > BigInteger m = BigInteger.valueOf((long)i); > BigInteger w = r.w.multiply(m); > BigInteger h = r.h.multiply(m); > BigInteger n = r.n.multiply(m); > // First we construct the rectangle s = r.x x (r.y - 1/i) > // ie [w,h,n] = [r.w*i,r.h*i-r.n,r.n*i] > Rectangle s = new Rectangle(w,h.subtract(r.n),n); > s.count = r.count + 1; > // System.out.println(New + s); > s.check(); > // Next we construct the rectangle t = (r.x - 1/i) x 1/i > // ie [w,h,n] = [r.w*i-r.n,r.n,r.n*i] > Rectangle t = new Rectangle(w.subtract(r.n),r.n,n); > t.count = r.count + 1; > // System.out.println(New + t); > t.check(); > // checkOrder(); > // We determine which of the new rectangles is larger > if (s.x >= t.x) { > addRectangle(s); > addRectangle(t); > } else { > addRectangle(t); > addRectangle(s); > } > } > ======================================================= > I think you mentioned that you had 2 rectangles in your list > when n = 13045 if you discarded rectangles with w < 1/13045. > Of course I would have 0 rectangles if I did that, > since none of the rectangles in my list was large enough > to contain a square of side 1/13045 . I cannot see anything wrong with the code that you have posted, but I am making assumptions about findIndex, check, new Rectangle, addRectangle, remove etc. I am particularly curious about why it matters in which order the two calls of addRectangle are made, since you are keeping the rectangles in order. Finding the best fit by comparing floating point numbers is not guaranteed to find the theoretical best fit. In particular, it may cause you to miss some exact fits which might be vital to the packing process. It would probably be helpful if you would post the entire source code. === === Subject: : Re: Packing squares into a 1 x Zeta[2] rectangle > // We determine which of the new rectangles is larger > if (s.x >= t.x) { > addRectangle(s); > addRectangle(t); > } else { > addRectangle(t); > addRectangle(s); > } > I am particularly curious about why it matters in which order the two > calls of addRectangle are made, since you are keeping the rectangles in > order. It is unimportant. It was meant to speed up the search for the correct position to insert the new rectangle. But now I come to think of it, it doesn't even do that. I keep a record of the last position from which a rectangle was removed (currentIndex) and start the search for the position to put a new rectangle at currentIndex - 1,moving right from there. It's really quite likely I've made a logical error here. (I would imagine the probability of my having made an error, as opposed to you, is 0.9.) > Finding the best fit by comparing floating point numbers is not guaranteed > to find the theoretical best fit. In particular, it may cause you to miss > some exact fits which might be vital to the packing process. True, though improbable I think with squares of order 10000. I did look through a couple of lists, and there did not appear to be any near repetitions of floats. But I'll try using exact comparison instead > It would probably be helpful if you would post the entire source code. I'll do that. But first I'll make the change above. Also I have a debug module to check that the list is in correct order which I'll run after inserting a rectangle. On the other hand, I was thinking that you were sailing pretty near the wind to have only two rectangles in your list, admittedly with maximum cut-off, and even though one of them was very large. This made me think that it could conceivably be possible to prove the result by showing that the largest recangle left was always more than a given multiple of the next square to be removed. (In my earlier experiments I thought this ratio was fairly steadily increasing.) I was just wondering too -- suppose one could prove that it was always possible to insert n rectangles, would that show that one could insert them all at once? Timothy Murphy e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ie tel: +353-86-2336090, +353-1-2842366 s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland === === Subject: : Re: Packing squares into a 1 x Zeta[2] rectangle >It would probably be helpful if you would post the entire source code. Here's my Java program. I changed the program to use exact arithmetic to compare two rectangles. I should say again that I do not claim to be a programmer, certainly not a Java programmer. Anyone interested can find the code at http://www.maths.tcd.ie/pub/Maths/PackingSquares/SquarePack.w However, since the program presumably doesn't work properly this is probably not a good idea! I'm reasonably sure I've made a simple logical error somewhere. I'd be very grateful if someone could point it out. In principle the program should be compiled with something like javac SquarePack.java Then it might be run with java SquarePack 20000 > out =========================================================== import java.math.*; import java.io.*; import java.util.*; public class SquarePack{ int currentIndex = 0; int lastIndex = 0; LinkedList rectangles; static int maxCount = 0; static int sumCount = 0; static float vacantArea = 0; public SquarePack() { rectangles = new LinkedList(); } public SquarePack(long w,long h,long n) { rectangles = new LinkedList(); Rectangle r = new Rectangle(w,h,n); addRectangle(r); } public SquarePack(double x,double y) { rectangles = new LinkedList(); long w = (long)(x*1000000); long h = (long)(y*1000000); long n = 1000000; Rectangle r = new Rectangle(w,h,n); addRectangle(r); } public int findIndex(Rectangle r) { Rectangle s; int index; int size = rectangles.size(); for(index = currentIndex; index < size; index++) if(((Rectangle)rectangles.get(index)).compareTo(r) < 0) break; return index; } public void checkOrder() { float lastx = (float)10.0; for(int index = 0; index lastx) System.out.println(List out of order at index + index); } public void printList(boolean verbose) { int maxCount = 0; int sumCount = 0; double vacantArea = 0; for(int index = 0; index < rectangles.size(); index++) { Rectangle r = (Rectangle)rectangles.get(index); if(verbose) System.out.println(r + + index + + r.count); sumCount += r.count; vacantArea += r.area; if(r.count > maxCount) maxCount = r.count; } if(first) { System.out.println( squaret list lengtht max split(average split)t vacant space); first = false; } System.out.println(square + t + rectangles.size() + ttt + maxCount + ( +sumCount/rectangles.size() + )ttt + (float)vacantArea); } public void addRectangle(Rectangle r) { int index = findIndex(r); if(r.x > epsilon) rectangles.add(index,r); } public void removeSquare(int i) { Rectangle square = new Rectangle(BigInteger.ONE,BigInteger.ONE, BigInteger.valueOf((long)i)); int index = findIndex(square)-1; if(index < 0) { System.err.println(No room for square side 1/ + i); System.out.println(No room for square of side 1/ + i + = + 1/(float)i); printList(true); System.exit(1); } Rectangle r = (Rectangle)rectangles.get(index); rectangles.remove(index); currentIndex = index; BigInteger m = BigInteger.valueOf((long)i); BigInteger w = r.w.multiply(m); BigInteger h = r.h.multiply(m); BigInteger n = r.n.multiply(m); Rectangle s = new Rectangle(w,h.subtract(r.n),n); s.count = r.count + 1; s.check(); addRectangle(s); Rectangle t = new Rectangle(w.subtract(r.n),r.n,n); t.count = r.count + 1; addRectangle(t); t.check(); checkOrder(); } static int square = 0; static float epsilon; static boolean first = true; public static void main(String[]args) { int count = 1000; if(args.length> 0) { count = Integer.parseInt(args[0]); } epsilon = (float)0.999/(float)count; SquarePack squarePack = new SquarePack( 6L*33102L*33102L,103993L*103993L,6L*33102L*33102L); for(square = 1; square <= count; square++) { squarePack.removeSquare(square); if(square%16 == 0) System.err.print(.); if(square%1024 == 0) squarePack.printList(false); } squarePack.printList(true); } class Rectangle{ BigInteger w,h,n; double x,y; double area; int count = 0; Rectangle(BigInteger w,BigInteger h,BigInteger n) { this.w = w; this.h = h; this.n = n; x = w.doubleValue()/n.doubleValue(); y = h.doubleValue()/n.doubleValue(); area = x*y; } Rectangle(long w,long h,long n) { this.w = BigInteger.valueOf(w); this.h = BigInteger.valueOf(h); this.n = BigInteger.valueOf(n); x = (double)w/(double)n; y = (double)h/(double)n; area = x*y; } public void check() { BigInteger d = w.gcd(n); d = d.gcd(h); if(d.compareTo(BigInteger.ONE)> 0) { w = w.divide(d); h = h.divide(d); n = n.divide(d); } if(w.compareTo(h)> 0) { w = w.divide(d); h = h.divide(d); n = n.divide(d); } if(w.compareTo(h)> 0) { BigInteger t = w; w = h; h = t; double z = x; x = y; y = z; } } public String toString() { return(( + (float)x + , + (float)y + )); } public int compareTo(Rectangle s) { return this.w.multiply(s.n).compareTo(s.w.multiply(this.n)); } } } =========================================================== Timothy Murphy e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ie tel: +353-86-2336090, +353-1-2842366 s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland === === Subject: : Re: Packing squares into a 1 x Zeta[2] rectangle > Here's my Java program. > I changed the program to use exact arithmetic to compare two rectangles. Apologies .. my news program screwed up slightly. Lines 177-181 in the routine check() should be omitted, ie the lines between ========= below. > public void check() { > BigInteger d = w.gcd(n); > d = d.gcd(h); > if(d.compareTo(BigInteger.ONE)> 0) { > w = w.divide(d); > h = h.divide(d); > n = n.divide(d); > } ======================================== > if(w.compareTo(h)> 0) { > w = w.divide(d); > h = h.divide(d); > n = n.divide(d); > } ======================================== > if(w.compareTo(h)> 0) { > BigInteger t = w; > w = h; > h = t; > double z = x; > x = y; > y = z; > } > } Timothy Murphy e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ie tel: +353-86-2336090, +353-1-2842366 s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland === === Subject: : Re: Packing squares into a 1 x Zeta[2] rectangle > Am 14.03.04 14:41 schrieb The Last ish Pastry: > Since the sum of the reciprocals of the squares of the positive integers is > pi^2/6, the question arises as to whether squares with sides 1, 1/2, 1/3, > etc can be packed into a rectangle of size 1 by pi^2/6. > A picture of such a packing appears at > http://www.pisquaredoversix.force9.co.uk/Tiling.htm > I know of no proof that such a packing is possible or impossible. > Nice, indeed. > Is it true, that the squares with contents 1,6,24,... (factorials) > are placed sequentially from the left at the bottom? > Gottfried Helms http://tinylink.com/?0g6QpfQTMZ It does not use the same method as I am now using. I dug out the old program but it now does not produce the picture at http://www.pisquaredoversix.force9.co.uk/Tiling.htm the 24 and 25 are interchanged, and possibly other changes. So, I am afraid I cannot answer your question at the moment. ~~~~ See the thread at http://tinylink.com/?N4zBhRwvDE for two interesting posts by David W. Cantrell. === === Subject: : Re: Packing squares into a 1 x Zeta[2] rectangle > http://tinylink.com/?0g6QpfQTMZ > It does not use the same method as I am now using. I dug out the old > program but it now does not produce the picture at > http://www.pisquaredoversix.force9.co.uk/Tiling.htm > the 24 and 25 are interchanged, and possibly other changes. So you've been doing this for years ... I said in my last posting that the probability of my being right, and you wrong, was 0.1. Make that 0.01! Timothy Murphy e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ie tel: +353-86-2336090, +353-1-2842366 s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland === === Subject: : Re: Packing squares into a 1 x Zeta[2] rectangle > [exaple explaining how exact calculations are carried on] >It will hold a rational quantity as an exact fraction (reduced to its lowest >terms). It will hold a value such as 840/1010+pi as the rational number >84/101 plus the known constant pi. If it needs to compare that value with, >say, 10429/8310+e it will subtract one from the other to give -355289/839310-e+pi and then compute that quantity with sufficient >precision to determine if it is positive or negative. >> Incidentally I took pi=355/113, which I think is just good enough >> for the range I was considering. >It is difficult to say what approximation to pi is good enough for packing a >given set of squares. Since 355/133 is a little greater than pi it is >possible that a square might just get squeezed in somewhere when it really >should not have fitted. >Taking one more term in the continued fraction for pi gives the rational >approximation 103993/33102 which is about 6*10^-10 less than pi. Using my >Mathematica program I can pack 100,000 squares into a 1 by >(103993/33102)^2/6 rectangle in about 3 minutes. [As opposed to about 12 >minutes when using the exact value for pi.] > Not to contradict your claim, by any means, since it sounds percfectly > reasonable... but couldn't the differences in computation times as > reported both here and in the next post be due to the differences in > how each single calculation is done? (Taking into account that in the > approximed situation one has to deal with rational numbers only) Certainly. I expect it is faster for Mathematica to compare two rationals than a rational and an irrational. === === Subject: : Re: Packing squares into a 1 x Zeta[2] rectangle > Certainly. I expect it is faster for Mathematica to compare two rationals > than a rational and an irrational. As a matter of interest, is there any way of determining if a rational function f(theta)/g(theta) is >0 or <0 knowing the continued fraction for theta? Or does mathematica in this case know the continued fraction for pi^2? (Is there a simple continued fraction for pi^2?) I know there is an exact generalised continued fraction for pi, with numerators != 1. I wonder if mathematica uses this, or if it remembers the first n integers in the standard continued fraction? I don't use mathematica -- does it give you information like that, or is it secret? Timothy Murphy e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ie tel: +353-86-2336090, +353-1-2842366 s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland === === Subject: : Re: Packing squares into a 1 x Zeta[2] rectangle > As a matter of interest, is there any way of determining > if a rational function f(theta)/g(theta) is >0 or <0 > knowing the continued fraction for theta? I believe this is undecidable. If you restrict the form of the continued fraction, theta, you could probably make that weaker version decidable. Let me try to demonstrate the undecidability of the general problem: if we have a Turing machine theta which never loops and always outputs the nth continued fraction digit of our number, can we always decide the sign of f(theta), where f is some polynomial? I'll reduce the halting problem to this. Suppose we have a machine m, and we want to see if it halts on the empty input or not. Create machine theta=theta(m) which works in the following way: it internally simulates m running on the empty input. When theta is supposed to output the nth digit of its continued fraction, it simulates m for n steps. If m halts within that time, it ouputs a 2. Otherwise, it outputs a 1. The important thing to notice is that: machine theta outputs all 1's iff m never halts. Let f(theta) = theta - phi, where phi is the golden ratio (whose continued fraction expansion is all 1's). If we could decide the sign of f(theta), then we could decide the halting problem. Last time I checked, this was a contradiction. -Tyler === === Subject: : Re: Packing squares into a 1 x Zeta[2] rectangle > Certainly. I expect it is faster for Mathematica to compare two rationals > than a rational and an irrational. > As a matter of interest, is there any way of determining > if a rational function f(theta)/g(theta) is >0 or <0 > knowing the continued fraction for theta? I am sorry, I do not know. > Or does mathematica in this case know the continued fraction for pi^2? I doubt that it does. In my program I specify the dimensions of the starting rectangle as 1 and Zeta[2]. I imagine that Mathematica knows the formula for Zeta[n], where n is even, in terms of the n'th Bernoulli number. It can tell me the value of Zeta[1000] (as an exact rational multiple of pi^1000) in 10.0 seconds. But it can then tell me what Zeta[1002] is in 0.07 seconds. I imagine that it had computed, and remembered, the values of all the Bernoulli numbers up to B_1000. > (Is there a simple continued fraction for pi^2?) I do not know of one. > I know there is an exact generalised continued fraction for pi, > with numerators != 1. ... you may be referring to one discovered by William Brouncker ... > I wonder if mathematica uses this, or if it remembers the first n integers > in the standard continued fraction? It computes the numerical value of pi using equation (63) at http://mathworld.wolfram.com/PiFormulas.html I imagine that it then computes the continued fraction from the numerical value. > I don't use mathematica -- does it give you information like that, > or is it secret? There is an Implementation section in the Help. It gives, in general terms, the methods used. For example, under Number-theoretical functions we read To find a requested number of terms ContinuedFraction uses a modification of Lehmer's indirect method, with a self-restarting divide-and-conquer algorithm to reduce the numerical precision required at each step. Here is another quotation: Pi uses the Chudnovsky formula for computations up to ten million digits. === === Subject: : Re: Cantor's Diagonal Argument In sci.logic, Virgil In sci.logic, |-|erc ><4056d6db$0$3957$afc38c87@news.optusnet.com.au>: > oo > ____|mn > / /_/ / _ > / K-9/ /_/ - www.YeOldeCoffeeShoppe.com - >/____/_____ >-------------- > I have : > for all n, for all i, exists j > b(n)_i = c(j)_i >> & b(n)_1 = c(j)_1 >> & b(n)_2 = c(j)_2 >> up to _i >> No, for *all* i, not just i <= n. >non comprendez, n is irrelevant its just the particular b. >if you said iFor the standard construction one can compute an infinite number >of b's. The main requirement is that b(n)_i is not equal >to c(i)_i or 9 -- for decimal expansions, anyway. > One should avoid b(n)_i = 0 on all bases,as well as b(n)_i = (base-1) in > all bases, since either may be part of one of the dual representations > of some real. This restrict the useful range of values of b(n)_i to > 1 <= b(n)_i <= (base - 2). > This effectively bars use of base two and base three from using > precisely this method (though there are legitimate variations on this > method that will work even for those bases). Good advice, and I've already posted how to get around the problem in base 2 (by effectively converting the problem to base 4); the same technique works for base 3 (converting it to base 9). As it is, I think Cantor's first proof: http://en.wikipedia.org/wiki/Cantor%27s_first_uncountability_ proof which does not depend on infinite decimal expansions (though it does depend on infinite sequences) has barely been touched by all this. :-) > You want : > for all n, exists j, for all i > b(n)_i = c(j)_i > where > b = DIAG > c = COMPUTABLES > That's right, and they're different. >> agreed, but is it significant? >> Quite significant. >> Suppose, just for whimsy's sake, I generate a series of binary >> numbers d(j). These numbers are as follows: >> j <-> d(j) >> 0 <-> 0.000000... >> 1 <-> 0.100000... >> 2 <-> 0.010000... >> 3 <-> 0.110000... >> 4 <-> 0.001000... >> 5 <-> 0.101000... >> 6 <-> 0.011000... >> 7 <-> 0.111000... >> 8 <-> 0.000100... >> As you can see, I'm writing the number backwards on the right. >> I can formalize this with a little work but the idea should be >> clear enough. >> Now take b = 1/3, and express it in binary: >> b = .010101010101... >> It's clear that for all i, there exists a j such that >> c(j)_i = b_i >> (one can take j = 0 if i is odd, and j = 2^(i-1) if i is even) >> but there is no j such that, for all i, >> c(j)_i = b_i >> as that would require 1/3 = 2^(-k) * m for some k and m, which >> is not possible. >granted, but examine this list >> 0 <-> 0.000000... >> 1 <-> 0.1... >> 2 <-> 0.01... >> 3 <-> 0.11... >> 4 <-> 0.001... >> 5 <-> 0.101... >> 6 <-> 0.011... >> 7 <-> 0.111... >> 8 <-> 0.0001... >where the initial sequence is repeating, the second number is 0.010101010.. >Erm...is 0.1... a specification of the open interval >(0.5, 0.75), the half-open interval [0.5, 1.0), some >arbitrary number within that interval, or what? >> they both mean the number is on the list to any degree required. >> the 1st digit is on the list, >> and the 2nd digit is on the list too, >> and the 3rd digit is on the list too >> .... >> all digits are on the list >> therefore the number is on the list. > The computable numbers are countable. The real numbers are not. >> fantacy, non computable number is a misnomer. >> I can show non-computable numbers without difficulty, >> by noting that a UTM must accept at most a finite machine >> specifier which is therefore mappable to an integer. >> Since the reals are not mappable to the set of integers >> (pick one of Cantor's proofs), uncomputable numbers exist. >> A nasty situation but what can I do? :-) >You do OK, but that is purely arguing backwards from the task at hand. >What I need is a pure functional program to demonstrate my point. It >uses the First function that takes input from a stream even if infinite. >first(3, <5,4,3,2,2>) = <5,4,3 >first(2, N) = <1,2 >This is normally used on the keyboard as input to allow for interaction >even >with pure functional constructs. >It relies on the language parser being lazy. Lazy means it wont try to >store >N in memory, complete that task(!), then copy the 1st 2 numbers. All the >parameters >are utilised as needed. >Would a pure functional program find a 1 to 1 correspondence between >N and R? I'm sure it would. Given the input stream b_n, as the digits >are processed it would just calculate what computable number it maps to. >As more digits are processed the number is found further down the list. >At any point in time the progress can be tested and any number input >will be found. >1/3 forms a pattern in binary, we can systematically determine its not >on that binary list. But how can you determine a number is not on the >list of computables? Every digit sequence is covered WITHOUT inherent >patterns partitioning classes. >Also, different UTMs provide different orderings to the same list >of computable numbers. How can you determine that a number fails >on all of them? >You don't. The number of computable numbers is countable. >That's not the problem. >Herc #191, ewill3@earthlink.net It's still legal to go .sigless. === === Subject: : Re: Cantor's Diagonal Argument In sci.logic, |-|erc : >In sci.logic, |-|erc ><4056d6db$0$3957$afc38c87@news.optusnet.com.au>: > oo > ____|mn > / /_/ / _ > / K-9/ /_/ - www.YeOldeCoffeeShoppe.com - >/____/_____ >-------------- > I have : > for all n, for all i, exists j > b(n)_i = c(j)_i >> & b(n)_1 = c(j)_1 >> & b(n)_2 = c(j)_2 >> up to _i >> No, for *all* i, not just i <= n. >non comprendez, n is irrelevant its just the particular b. >if you said iFor the standard construction one can compute an infinite number >of b's. The main requirement is that b(n)_i is not equal >to c(i)_i or 9 -- for decimal expansions, anyway. > n hasn't been defined to have any constraints, its just ranges over N. > b(1) = 0.937365764837... > b(2) = 0.3837364748493.. > there are infinite modifications to the diagonal, why does the particular > one you select have bearing on the number of digits that can be computed (i)? > You want : > for all n, exists j, for all i > b(n)_i = c(j)_i > where > b = DIAG > c = COMPUTABLES > That's right, and they're different. >> agreed, but is it significant? >> Quite significant. >> Suppose, just for whimsy's sake, I generate a series of binary >> numbers d(j). These numbers are as follows: >> j <-> d(j) >> 0 <-> 0.000000... >> 1 <-> 0.100000... >> 2 <-> 0.010000... >> 3 <-> 0.110000... >> 4 <-> 0.001000... >> 5 <-> 0.101000... >> 6 <-> 0.011000... >> 7 <-> 0.111000... >> 8 <-> 0.000100... >> As you can see, I'm writing the number backwards on the right. >> I can formalize this with a little work but the idea should be >> clear enough. >> Now take b = 1/3, and express it in binary: >> b = .010101010101... >> It's clear that for all i, there exists a j such that >> c(j)_i = b_i >> (one can take j = 0 if i is odd, and j = 2^(i-1) if i is even) >> but there is no j such that, for all i, >> c(j)_i = b_i >> as that would require 1/3 = 2^(-k) * m for some k and m, which >> is not possible. >granted, but examine this list >> 0 <-> 0.000000... >> 1 <-> 0.1... >> 2 <-> 0.01... >> 3 <-> 0.11... >> 4 <-> 0.001... >> 5 <-> 0.101... >> 6 <-> 0.011... >> 7 <-> 0.111... >> 8 <-> 0.0001... >where the initial sequence is repeating, the second number is 0.010101010.. >Erm...is 0.1... a specification of the open interval >(0.5, 0.75), the half-open interval [0.5, 1.0), some >arbitrary number within that interval, or what? > its interpreted exactly the same as you originally suggested, except the > initial reverse binary sequence repeats. > Instead of > 2 <-> 0.0100000000000000000000... > 2 <-> 0.0101010101010101010101... > by 0.01.. I indicated I mean > __ > 0.01.. In that case, I can replace b with the sequence b = 0.0110111001011101111000100110101011... where I'm simply concatenating consecutive integers into the expansion: b = 0.0 1 10 11 100 101 110 111 1000 1001 1010 1011... Or I can use the far simpler b = 0.01011010110101101011... = 11/31. >> they both mean the number is on the list to any degree required. >> the 1st digit is on the list, >> and the 2nd digit is on the list too, >> and the 3rd digit is on the list too >> .... >> all digits are on the list >> therefore the number is on the list. > The computable numbers are countable. The real numbers are not. >> fantacy, non computable number is a misnomer. >> I can show non-computable numbers without difficulty, >> by noting that a UTM must accept at most a finite machine >> specifier which is therefore mappable to an integer. >> Since the reals are not mappable to the set of integers >> (pick one of Cantor's proofs), uncomputable numbers exist. >> A nasty situation but what can I do? :-) >You do OK, but that is purely arguing backwards from the task at hand. >What I need is a pure functional program to demonstrate my point. It >uses the First function that takes input from a stream even if infinite. >first(3, <5,4,3,2,2>) = <5,4,3 >first(2, N) = <1,2 >This is normally used on the keyboard as input to allow for interaction even >with pure functional constructs. >It relies on the language parser being lazy. Lazy means it wont try to store >N in memory, complete that task(!), then copy the 1st 2 numbers. All the parameters >are utilised as needed. >Would a pure functional program find a 1 to 1 correspondence between >N and R? I'm sure it would. Given the input stream b_n, as the digits >are processed it would just calculate what computable number it maps to. >As more digits are processed the number is found further down the list. >At any point in time the progress can be tested and any number input >will be found. >1/3 forms a pattern in binary, we can systematically determine its not >on that binary list. But how can you determine a number is not on the >list of computables? Every digit sequence is covered WITHOUT inherent >patterns partitioning classes. >Also, different UTMs provide different orderings to the same list >of computable numbers. How can you determine that a number fails >on all of them? >You don't. The number of computable numbers is countable. >That's not the problem. > Well if a diag number doesn't fail to belong to {UTM'(N)} then diag is > countable in the classes of UTM. > Anyway I've demonstrated that given a diagonal construct real number, > I can always provide a natural number that maps to it. Perhaps not a > bijection but countable in a sense. We can always go from any real > to N. > Herc #191, ewill3@earthlink.net It's still legal to go .sigless. === === Subject: : Re: Cantor's Diagonal Argument > Instead of > 2 <-> 0.0100000000000000000000... > 2 <-> 0.0101010101010101010101... > by 0.01.. I indicated I mean > __ > 0.01.. > In that case, I can replace b with the sequence > b = 0.0110111001011101111000100110101011... > where I'm simply concatenating consecutive integers > into the expansion: > b = 0.0 1 10 11 100 101 110 111 1000 1001 1010 1011... That's back to the original problem with irrationals, 1st you have to define_the_irrational, my point is definitions are countable, since definition = program, numbers are countable. All irrarationals are computable yes? To negate this you are insisting that there is a massive set of NON COMPUTABLE IRRATIONALS. What does that mean exactly? Can I see one? > Or I can use the far simpler > b = 0.01011010110101101011... = 11/31. _____ b = 0.01011.. number 26 on the list! ZZZZ Herc >> they both mean the number is on the list to any degree required. > the 1st digit is on the list, > and the 2nd digit is on the list too, > and the 3rd digit is on the list too > .... > all digits are on the list > therefore the number is on the list. >> The computable numbers are countable. The real numbers are not. > fantacy, non computable number is a misnomer. > I can show non-computable numbers without difficulty, > by noting that a UTM must accept at most a finite machine > specifier which is therefore mappable to an integer. > Since the reals are not mappable to the set of integers > (pick one of Cantor's proofs), uncomputable numbers exist. > A nasty situation but what can I do? :-) > You do OK, but that is purely arguing backwards from the task at hand. >> What I need is a pure functional program to demonstrate my point. It >> uses the First function that takes input from a stream even if infinite. > first(3, <5,4,3,2,2>) = <5,4,3 >> first(2, N) = <1,2 > This is normally used on the keyboard as input to allow for interaction even >> with pure functional constructs. >> It relies on the language parser being lazy. Lazy means it wont try to store >> N in memory, complete that task(!), then copy the 1st 2 numbers. All the parameters >> are utilised as needed. > Would a pure functional program find a 1 to 1 correspondence between >> N and R? I'm sure it would. Given the input stream b_n, as the digits >> are processed it would just calculate what computable number it maps to. >> As more digits are processed the number is found further down the list. >> At any point in time the progress can be tested and any number input >> will be found. > 1/3 forms a pattern in binary, we can systematically determine its not >> on that binary list. But how can you determine a number is not on the >> list of computables? Every digit sequence is covered WITHOUT inherent >> patterns partitioning classes. > Also, different UTMs provide different orderings to the same list >> of computable numbers. How can you determine that a number fails >> on all of them? > You don't. The number of computable numbers is countable. >> That's not the problem. Well if a diag number doesn't fail to belong to {UTM'(N)} then diag is > countable in the classes of UTM. > Anyway I've demonstrated that given a diagonal construct real number, > I can always provide a natural number that maps to it. Perhaps not a > bijection but countable in a sense. We can always go from any real > to N. > Herc > #191, ewill3@earthlink.net > It's still legal to go .sigless. === === Subject: : Re: Cantor's Diagonal Argument >That's back to the original problem with irrationals, 1st you >have to define_the_irrational, my point is definitions are countable, >since definition = program, numbers are countable. All irrarationals >are computable yes? To negate this you are insisting that there >is a massive set of NON COMPUTABLE IRRATIONALS. >What does that mean exactly? Can I see one? No you can't, that's the point. Most real numbers are indescribable irrational numbers. They have infinitely many digits, and cannot be computed by a finite-sized program or described in a finite-sized Usenet message. --Keith Lewis klewis {at} mitre.org The above may not (yet) represent the opinions of my employer. === === Subject: : Re: Cantor's Diagonal Argument : That's back to the original problem with irrationals, 1st you : have to define_the_irrational I'm sorry, you just CAN'T SAY that. Seriously, in that sentence, define is hopelessly ambiguous. You have to invent 2 new different words for the two meanings BETWIXT which define is ambiguous, and then REPLACE define with whichever one of them is appropriate, wherever it occurs, in order to talk coherently about this. I mean, I can DEFINE the irrational, pi as the ratio of the circumference of a circle to its diameter, but that just plain ISN'T HELPING anybody. I can define it as the output of a certain Turing Machine but even that is a little dodgy since that machine would have to run forever in order for its output to be the actual value of pi as opposed to some finite approximation thereof. In order to clarify the difference between the two meanings of define, let's eliminate confusions caused by infinity from the scenario. Let's just pick some random expression that refers to a number that YOU DON'T KNOW, such as the number of Jefferson's descents. That is just as good a definition of 227 as 227 is, if you KNOW that Jefferson has 227 descents. But if you DON'T KNOW that, then this ISN'T a good definition of 227. The phrase I chose INDICATES 227: it MEANS 227; it REFERS to 227, whether you know that or not. But it doesn't EXPLICATE 227; it doesn't REPRESENT 227; it doesn't SPECIFY 227. 227, on the other hand, if you know what number-base you're in, and you know we're using the place-value paradigm for expressing natural numbers, is as full an explication of this number as you could ask. Now, on one level, you could say that Pi could be indicated (since the finite TM program for the TM that outputs it is finitely describable and understandable), but that it can't be represented or specified because its specification in numerical-terms-we-can-use is infinite, and therefore pragmatically UNuseable. In real life, however, we won't be that uncharitable; we will allow the finite specification of the TM program to COUNT as a specification of Pi. Some irrationals, though, don't have ANY TM program that will specify them, not EVEN when it runs forever. What you are asking is for us to INDICATE one of those irrationals that we *can't* EXPLICATE, NOT EVEN via some lame dodge like calling it the infinite output of a finite TM. Here is 1 such indication: The irrational in [0,1] represented by the bit-string that has a 1 in its nth place whenever n encodes a TM that doesn't halt when its input is n, and has an 0 in all other places. In other words, DON'T SAY define when you are trying to discuss this stuff. Say point at for a descriptive definition and say specify for a representative one. === === Subject: : most practical way to solve open path travelling salesman with trianlge inequality Hello everyone! What is the most practical way to solve the open path version of travelling salesman with triangle inequality? I mean fast to compute. Alex. PS. To email me, remove loeschedies from the email address given. === === Subject: : most practical way to solve open path travelling salesman with trianlge inequality and starting node Hello everyone! What is the most practical way to solve the open path version of travelling salesman with triangle inequality and starting node? I mean a fast to compute version. Alex. PS. To email me, remove loeschedies from the email address given. === === Subject: : Re: Hypocritical Fortune Tellers (What are the odds?) > you might want to consider to continue this discussion per e-mail, so that > we're not bothered by it? The mathematicians are among the most closed minded of the so-called scientists, they refuse to see spiritual influences that effect outcomes in this world, such as the harmony in the names and birthdays of family units. Many beautiful patterns exist in the mathematical makeup of the Bible, these patterns extend into the mathematical makeup of the family units, for all are gifts from God, they should be related. The mathematicians have tools to analyse sets of numbers but their tools will not for example, show us that the 29x29th chapter of the Bible contains 29 verses. The mathematicans are too closed minded to stop and look at some of the patterns, and here one of them simply doesn't want the topic discussed. And you call that science?!!! No, your version of mathematics is nothing less than Satanism. -Daryl S. Kabatoff ############################################################## ###### HERE IS HOW THIS THREAD BEGAN: > are you a mathematician or a numerologist? > is what you're posting anything more than a > glorified horoscope? I am not looking to the future but goink back to the past, to the day the person was born. God provided the individual with his or her life and also their name, back then. Horoscopes deal with astrology, can you see any discussions of planets and stars in my postings? The mainstream Christian churches embrace pagan holidays that are linked to sun whoreship and merge them in with their Christian worship, you are likely a member of one of these cults and now you hypocritically condemn me and my work as being astrology. For example, by December 25th the sun is visibly returning from the south, calling this pagan sunwhoreshipping holiday Christmas is a violation of God's Third Commandment, for it is not Christ's Mass but is instead a pagan mass, it is the use of God's name in vain. And for example, the mainstream Christians embrace the pagan Easter holiday at a time based upon the phase of the moon. And then look at all the predictions that the mainstream mathematicians attempt, you people are tossing coins and looking for the probababbility of them landing as heads or tails (while disavowing Satanic influences that can and do manipulate these coins). You mathe- maticians are looking for the probababbility of events on earth (while disavowing spiritual influences), and then have the audacity to call me a fortune teller. I am a mathematician who deals with God and little primes, while the mainstream mathematicians are Godless numerologists who salivate when they discover a prime that is large enough to fill a telephone book. -Daryl S. Kabatoff === === Subject: : Re: Hypocritical Fortune Tellers (What are the odds?) > I'm not about to get into this, but the creator of the thread > mentioned something interesting - the 29x29th chapter has 29 verses? > There are several numerical recipes for amusement within the bible, and > in fact in many other places, in fact just about everywhere if you let > your mind freewheel for a bit. You could even try looking at the night > sky for patterns, although good luck finding any...oh, hang on... > Rumour has it that Shakespeare was involved with the King James Bible, > and Psalm 48 has shake 48 words from the start and spear 48 words > from the end. I heard it was Psalm 46, not 48, since Shakespeare was 46 when the KJV came out. I also heard the same pattern appears in the previous version of the Bible (the Vulgate?), which was written well before Shakespeare was born. > [..] > What Daryl has done is to stare at a bunch of numbers until some sort of > pattern emerged, then to deduce from that that God must be saying > something. This is not a safe approach, because as the above experiment > shows your brain will make patterns from /anything/ - meaningful or > otherwise. This is the Law of Fives, as known to Discordians (which does not depend explicitly on the number five). > Another mistake Daryl made was in stating that because one of the books > was called Numbers, God must be saying something about numbers. I was at a math conference a few years ago, where Bren McKay (mathematician/computer scientist from Australia) was debunking the claims in _The Bible Code_. This specific lecture was opened to the public, and you should have seen the circus ... Anyway, a guy sat next to me who said something about the book of Numbers. The numbers which appear are supposed to be a census of a certain number of tribes; this fact is repeated over and over, that these numbers have not been rounded off. But the numbers are all evenly divisible by 50, and most are divisible by 100. This census was taken before a war, and another one a few decades after the war, and this divisibility pattern shows up there too. > There is a book called Genesis, but I doubt God is making a > statement about Phil Collins. Genesis means life, you dope. The band took its name from the Bible, just like many of the common names in the Western world come from there (including Dave, but not Christopher, interestingly enough. The Catholics de-canonized St. Christopher, saying there was no evidence he had existed.) > [...] Unfortunately you're in my killfile > Daryl, so I won't get your reply, assuming you bother.) Then you didn't write this letter with a response from him in mind? -- Christopher Heckman === === Subject: : Re: Hypocritical Fortune Tellers (What are the odds?) >Rumour has it that Shakespeare was involved with the King James Bible, >and Psalm 48 has shake 48 words from the start and spear 48 words >from the end. > I heard it was Psalm 46, not 48, since Shakespeare was 46 when the KJV > came out. I also heard the same pattern appears in the previous version > of the Bible (the Vulgate?), which was written well before Shakespeare > was born. feh...what's two psalms between friends :-) Point is - patterns are easy to find if you look for them, but they don't necessarily mean anything. > I was at a math conference a few years ago, where Bren McKay > (mathematician/computer scientist from Australia) was debunking the claims > in _The Bible Code_. This specific lecture was opened to the public, > and you should have seen the circus ... > Anyway, a guy sat next to me who said something about the book of > Numbers. The numbers which appear are supposed to be a census of a certain > number of tribes; this fact is repeated over and over, that these numbers > have not been rounded off. But the numbers are all evenly divisible by 50, > and most are divisible by 100. This census was taken before a war, and > another one a few decades after the war, and this divisibility pattern shows > up there too. I haven't looked into the Bible Code, which is why I was careful not to mention it. I figure it has to fall into one of three camps: (a) this fancy pattern shows the Bible means what it says, (b) this fancy pattern shows the Bible doesn't mean what it says, or (c) this fancy pattern states nothing about the meaning of the Bible text. Whichever way the code is irrelevant to me as a Christian, although possibly interesting as an intellectual exercise. However I have lots of other interesting exercised lined up so I really can't be bothered with it. Numbers...ok. Here's one such line: The children of Reuben, Israels firstborn, their generations, by their families, by their fathers houses, according to the number of the names, one by one, every male from twenty years old and upward, all who were able to go out to war; those who were numbered of them, of the tribe of Reuben, were forty-six thousand five hundred. (taken from www.ebible.org) Were there exactly 46500 able-bodied men? I doubt it. Warmaking isn't an interest of mine, but there are several questions that arise from this from simple logic: (1) Um, so what? if it was 46523, this is still accurate to 3 significant figures, which is accurate enough even for many 21st century calculations. (2) all who were able to go out to war. If there were 46523 able-bodied men, perhaps the 23 couldn't go out for logistical reasons. If they could only go out in groups of 50, that could explain why everything is equally divisible by 50 or 100. A cohort of one would be silly. As would a cohort of two. So what's the cutoff point? Would a cohort of 49 go out (assuming cohort means 50, which it probably doesn't)? Would a cohort of 49 be counted as a full cohort, and the total number of men not actually counted, but the number of cohorts multipled by the number of men in a cohort? (3) those who were numbered of them. Perhaps there really were 46523, but only 46500 were counted (which brings the word circular to my mind). You wouldn't be able to count 46500 if there were only 27364. I Googled for soldiers sent to Iraq, just for interest. The first The headline states 120 soldiers sent to Iraq. The body states about 120 soldiers. So was it 120 or wasn't it? If news reports from only last year are not accurate to the last digit, why should Numbers need to be? Dave. === === Subject: : Re: Hypocritical Fortune Tellers (What are the odds?) >Rumour has it that Shakespeare was involved with the King James Bible, >>and Psalm 48 has shake 48 words from the start and spear 48 words >>from the end. > I heard it was Psalm 46, not 48, since Shakespeare was 46 when the KJV > came out. I also heard the same pattern appears in the previous version > of the Bible (the Vulgate?), which was written well before Shakespeare > was born. > feh...what's two psalms between friends :-) Point is - patterns are > easy to find if you look for them, but they don't necessarily mean anything. > I was at a math conference a few years ago, where Bren McKay > (mathematician/computer scientist from Australia) was debunking the claims > in _The Bible Code_. This specific lecture was opened to the public, > and you should have seen the circus ... > Anyway, a guy sat next to me who said something about the book of > Numbers. The numbers which appear are supposed to be a census of a certain > number of tribes; this fact is repeated over and over, that these numbers > have not been rounded off. But the numbers are all evenly divisible by 50, > and most are divisible by 100. This census was taken before a war, and > another one a few decades after the war, and this divisibility pattern shows > up there too. > I haven't looked into the Bible Code, which is why I was careful not to > mention it. I figure it has to fall into one of three camps: (a) this > fancy pattern shows the Bible means what it says, (b) this fancy pattern > shows the Bible doesn't mean what it says, or (c) this fancy pattern > states nothing about the meaning of the Bible text. Whichever way the > code is irrelevant to me as a Christian, although possibly interesting > as an intellectual exercise. However I have lots of other interesting > exercised lined up so I really can't be bothered with it. > Numbers...ok. Here's one such line: The children of Reuben, Israels > firstborn, their generations, by their families, by their fathers > houses, according to the number of the names, one by one, every male > from twenty years old and upward, all who were able to go out to war; > those who were numbered of them, of the tribe of Reuben, were forty-six > thousand five hundred. (taken from www.ebible.org) > Were there exactly 46500 able-bodied men? I doubt it. > Warmaking isn't an interest of mine, but there are several questions > that arise from this from simple logic: > (1) Um, so what? if it was 46523, this is still accurate to 3 > significant figures, which is accurate enough even for many 21st century > calculations. > (2) all who were able to go out to war. If there were 46523 > able-bodied men, perhaps the 23 couldn't go out for logistical reasons. > If they could only go out in groups of 50, that could explain why > everything is equally divisible by 50 or 100. A cohort of one would be > silly. As would a cohort of two. So what's the cutoff point? Would a > cohort of 49 go out (assuming cohort means 50, which it probably > doesn't)? Would a cohort of 49 be counted as a full cohort, and the > total number of men not actually counted, but the number of cohorts > multipled by the number of men in a cohort? > (3) those who were numbered of them. Perhaps there really were 46523, > but only 46500 were counted (which brings the word circular to my > mind). You wouldn't be able to count 46500 if there were only 27364. > I Googled for soldiers sent to Iraq, just for interest. The first > The headline states 120 soldiers sent to Iraq. The body states about > 120 soldiers. So was it 120 or wasn't it? If news reports from only > last year are not accurate to the last digit, why should Numbers need to be? > Dave. A) The work I do with math has nothing to do with the Bible Code B) The Jews were very diciplined, their tents were perfectly set into rows, they wore clean garments, they numbered their soldier accurately. C) There are many different ways to give testimony of God's power, why don't you continue with the many excercizes you have lined up without trying to denigrate my work? -Daryl S. Kabatoff === === Subject: : Re: q: linear algebra / matrices >>Derek, your previous answers look promising however I didn't not >>understand them >>because I know basic linear algebra, perhaps you can give me a more >>practical solution/explanation? >Calculate the adjoint of the 3x3 matrix -I (diagonal matrix with all >diagonal entries -1). You will find that you get the identity matrix, >which is also the adjoint of I. So both I and -I have the same >adjoint. >So, if you are told that adj(A) = I, then you cannot find A^-1, >because >you don't known whether A = I or A = -I. >The same is true for nxn matrices for any odd n. >But for n even, adj(-I) = -I. If you are dealing only with real >matrices, >then for nxn matrices with n even, it is possible to compute A^-1 >from >adj(A). For complex matrices it is not. >How to calculate A^-1 from adj(A) when we have promising conditions >(or as you described above)? (and knowing that A is not related to I) You could use the equation A adj(A) = det(A) I, which implies det(A) det(adj(A)) = det(A)^n for an n x n matrix. Derek Holt. === === Subject: : Re: sequence of polynomials on [a,b] >Let {P_n} be a sequence of polynomials that converges uniformly on the >interval [a, b] to a function f. Show that if f is not a polynomial, >then G_n => infinity, where {G_n} is the sequence formed by the >degrees of the polynomials, that is, G_n = degree(P_n). I'm a little confused here - is this problem really that hard? Is it not clear that {|p - f| : deg p <= n} is compact and bounded away from 0? |.| indicates the sup norm. I'm not sure about the compact part as stated; if not, use {|p - f|: deg p <= n, |p - f| <= |f|}. What am I missing? Stephen J. Herschkorn herschko@rutcor.rutgers.edu === === Subject: : Re: sequence of polynomials on [a,b] >Let {P_n} be a sequence of polynomials that converges uniformly on the >interval [a, b] to a function f. Show that if f is not a polynomial, >then G_n => infinity, where {G_n} is the sequence formed by the >degrees of the polynomials, that is, G_n = degree(P_n). > I'm a little confused here - is this problem really that hard? It may be hard, it may be easy. Depends on what you know. It's a nice problem at the baby Rudin level. > Is it > not clear that {|p - f| : deg p <= n} is compact and bounded away from > 0? |.| indicates the sup norm. I'm not sure about the compact part as > stated; It's obviously not compact as stated. > if not, use {|p - f|: deg p <= n, |p - f| <= |f|}. What am > I missing? The proof. === === Subject: : Re: sequence of polynomials on [a,b] > I would like some help with this problem. So far,I couldn't start out. > Let {P_n} be a sequence of polynomials that converges uniformly on the > interval [a, b] to a function f. Show that if f is not a polynomial, > then G_n => infinity, where {G_n} is the sequence formed by the > degrees of the polynomials, that is, G_n = degree(P_n). Before some eager beaver gives you the full detailed solution, is this homework? === === Subject: : Re: sequence of polynomials on [a,b] > I would like some help with this problem. So far,I couldn't start out. > Let {P_n} be a sequence of polynomials that converges uniformly on the > interval [a, b] to a function f. Show that if f is not a polynomial, > then G_n => infinity, where {G_n} is the sequence formed by the > degrees of the polynomials, that is, G_n = degree(P_n). > Before some eager beaver gives you the full detailed solution, is this > homework? No, it's not. Amanda === === Subject: : Re: sequence of polynomials on [a,b] > I would like some help with this problem. So far,I couldn't start out. > Let {P_n} be a sequence of polynomials that converges uniformly on the > interval [a, b] to a function f. Show that if f is not a polynomial, > then G_n => infinity, where {G_n} is the sequence formed by the > degrees of the polynomials, that is, G_n = degree(P_n). Maybe start by noting that if not then there is a subsequence of polynomials of bounded degree - of constant degree, even - and that sequence also converges to f. === === Subject: : Re: sequence of polynomials on [a,b] > I would like some help with this problem. So far,I couldn't start out. > Let {P_n} be a sequence of polynomials that converges uniformly on the > interval [a, b] to a function f. Show that if f is not a polynomial, > then G_n => infinity, where {G_n} is the sequence formed by the > degrees of the polynomials, that is, G_n = degree(P_n). > Maybe start by noting that if not then there is a subsequence > of polynomials of bounded degree - of constant degree, even - > and that sequence also converges to f. Yes, since the convergence is uniform it follows f is continuous on [a, b] and Bernstein's approximation theorem shows there's a sequence of constant polynomials that also converges uniformly to f. But this doesn't imply the cesired conclusion, right? Amanda === === Subject: : Re: sequence of polynomials on [a,b] >Let {P_n} be a sequence of polynomials that converges uniformly on the >interval [a, b] to a function f. Show that if f is not a polynomial, >then G_n => infinity, where {G_n} is the sequence formed by the >degrees of the polynomials, that is, G_n = degree(P_n). >Maybe start by noting that if not then there is a subsequence >of polynomials of bounded degree - of constant degree, even - >and that sequence also converges to f. > Yes, since the convergence is uniform it follows f is continuous on > [a, b] and Bernstein's approximation theorem shows there's a sequence > of constant polynomials that also converges uniformly to f. But this > doesn't imply the cesired conclusion, right? Robert Israel has already answered to that hours ago. Here's his reply again: Umm, a convergent sequence of constant polynomials converges to a constant. I don't think you have Bernstein's approximation theorem right. You might note that the polynomials of degree <= n form a finite-dimensional linear space, so a sequence of such polynomials that is uniformly Cauchy must converge to a polynomial of degree <= n. Jose Carlos Santos === === Subject: : Re: sequence of polynomials on [a,b] > I would like some help with this problem. So far,I couldn't start out. > Let {P_n} be a sequence of polynomials that converges uniformly on the > interval [a, b] to a function f. Show that if f is not a polynomial, > then G_n => infinity, where {G_n} is the sequence formed by the > degrees of the polynomials, that is, G_n = degree(P_n). > Maybe start by noting that if not then there is a subsequence > of polynomials of bounded degree - of constant degree, even - > and that sequence also converges to f. Yes, since the convergence is uniform it follows f is continuous on [a, b] and Bernstein's approximation theorem shows there's a sequence of constant polynomials that also converges uniformly to f. But this doesn't imply the cesired conclusion, right? Amanda === === Subject: : Re: sequence of polynomials on [a,b] >Let {P_n} be a sequence of polynomials that converges uniformly on the >interval [a, b] to a function f. Show that if f is not a polynomial, >then G_n => infinity, where {G_n} is the sequence formed by the >degrees of the polynomials, that is, G_n = degree(P_n). >Maybe start by noting that if not then there is a subsequence >of polynomials of bounded degree - of constant degree, even - >and that sequence also converges to f. >Yes, since the convergence is uniform it follows f is continuous on >[a, b] and Bernstein's approximation theorem shows there's a sequence >of constant polynomials that also converges uniformly to f. But this >doesn't imply the cesired conclusion, right? Umm, a convergent sequence of constant polynomials converges to a constant. I don't think you have Bernstein's approximation theorem right. You might note that the polynomials of degree <= n form a finite-dimensional linear space, so a sequence of such polynomials that is uniformly Cauchy must converge to a polynomial of degree <= n. Department of Mathematics http://www.math.ubc.ca/~israel === === Subject: : Re: sequence of polynomials on [a,b] > You might note that the polynomials of degree <= n form a > finite-dimensional linear space, so a sequence of such polynomials that > is uniformly Cauchy must converge to a polynomial of degree <= n. Do you have something other than the equivalence of norms on a finite dimensional vector space in mind here? That may be something the OP doesn't know. Anyway, here's an elementary approach: Suppose deg(P_n) <= D for all n, where D is an integer, and P_n converges pointwise at D + 1 distinct points. Then there exists a polynomial P, deg(P) <= D, that is the uniform limit of the P_n's on every bounded interval. Proof: This is obvious if D = 0. Now use induction, noting that if P_n(x_0) converges, then P_n(x) - P_n(x_0) = (x-x_0)Q_n(x), and Q_n will converge at the remaining points. === === Subject: : Re: sequence of polynomials on [a,b] >You might note that the polynomials of degree <= n form a >finite-dimensional linear space, so a sequence of such polynomials that >is uniformly Cauchy must converge to a polynomial of degree <= n. >Do you have something other than the equivalence of norms on a finite >dimensional vector space in mind here? That may be something the OP doesn't >know. >Anyway, here's an elementary approach: Suppose deg(P_n) <= D for all n, >where D is an integer, and P_n converges pointwise at D + 1 distinct >points. Then there exists a polynomial P, deg(P) <= D, that is the uniform >limit of the P_n's on every bounded interval. Proof: This is obvious if D = >0. Now use induction, noting that if P_n(x_0) converges, then P_n(x) - >P_n(x_0) = (x-x_0)Q_n(x), and Q_n will converge at the remaining points. Or you could use the fact that there's an explicit formula (Lagrange interpolation) for P_n(x) in terms of the values of P_n at the D+1 points, say P_n(x) = L(x, P_n(p_0),...,P_n(p_D)) where p_0,...,p_D are your D+1 points. For each fixed x, L is a continuous function of its last D+1 arguments. So if P_n(p_j) converges to y_j as n -> infinity for each j, P_n(x) converges to L(x, y_0,...,y_D) which is a polynomial of degree <= D. Department of Mathematics http://www.math.ubc.ca/~israel === === Subject: : Re: sequence of polynomials on [a,b] >> Let {P_n} be a sequence of polynomials that converges uniformly on the >> interval [a, b] to a function f. Show that if f is not a polynomial, >> then G_n => infinity, where {G_n} is the sequence formed by the >> degrees of the polynomials, that is, G_n = degree(P_n). >> Maybe start by noting that if not then there is a subsequence >> of polynomials of bounded degree - of constant degree, even - >> and that sequence also converges to f. >Yes, since the convergence is uniform it follows f is continuous on >[a, b] and Bernstein's approximation theorem shows there's a sequence >of constant polynomials that also converges uniformly to f. But this >doesn't imply the cesired conclusion, right? > Umm, a convergent sequence of constant polynomials converges to a > constant. I don't think you have Bernstein's approximation theorem right. Oh, what I meant was, there's a sequence of polynomials with constant degrees that converges to f. > You might note that the polynomials of degree <= n form a > finite-dimensional linear space, so a sequence of such polynomials that > is uniformly Cauchy must converge to a polynomial of degree <= n. === === Subject: : Re: sequence of polynomials on [a,b] >> Let {P_n} be a sequence of polynomials that converges uniformly on the >> interval [a, b] to a function f. Show that if f is not a polynomial, >> then G_n => infinity, where {G_n} is the sequence formed by the >> degrees of the polynomials, that is, G_n = degree(P_n). >> Maybe start by noting that if not then there is a subsequence >> of polynomials of bounded degree - of constant degree, even - >> and that sequence also converges to f. >Yes, since the convergence is uniform it follows f is continuous on >[a, b] and Bernstein's approximation theorem shows there's a sequence >of constant polynomials that also converges uniformly to f. But this >doesn't imply the cesired conclusion, right? >Umm, a convergent sequence of constant polynomials converges to a >constant. I don't think you have Bernstein's approximation theorem right. >Oh, what I meant was, there's a sequence of polynomials with constant >degrees that converges to f. That's what _I_ assumed you meant. Like he said, I don't think you have Bernstein's theorem right. Did you read the following paragraph? >You might note that the polynomials of degree <= n form a >finite-dimensional linear space, so a sequence of such polynomials that >is uniformly Cauchy must converge to a polynomial of degree <= n. Department of Mathematics http://www.math.ubc.ca/~israel >University of British Columbia >Vancouver, BC, Canada V6T 1Z2 X-mailer: xrn 9.02 === === Subject: : Proving associativity Mail-To-News-Contact: postmaster@nym.alias.net Back in 2002, I posted a question about how to prove that all groups of size <= 5 were Abelian. With the hints that I received then (right before losing my Usenet access), I made some progress. I had thought that I had proved the cases for groups of size 1, 2, and 3. However, I just realized that there was a biiig hole in my proofs. I hadn't shown that the objects with which I was working had the associative property, and therefore had no right to claim that they had anything to do with groups! For instance, take the following multiplication table: * | e a --------- e | e a a | a e In order for me to show that this is a group, I'd need to perform 2*(2^3) multiplications before I'd really confirmed that it has the associative property. That's not too bad. But, what about: * | e a b ----------- e | e a b a | a e b b | b a e As far as I can tell, I'd need to perform 2*(3^3) multiplications before I could claim that this associated. Now I know that in most cases, an operation isn't defined in terms of a table, but in terms of elementary arithmetic operations. In such cases, you can prove the associative nature of the operation. But, what do you do when all that you have is a table? Is there anything other than exhaustive brute force available? The book that I'm using as a text (Landin) poses this as a question; one which I've never been able to answer. Commutativity is easy to see from a table -- just look for symmetry about the major diagonal. Is there a simple way to check for associativity? Michael F. Stemper #include Back in 2002, I posted a question about how to prove that all groups > of size <= 5 were Abelian. With the hints that I received then (right > before losing my Usenet access), I made some progress. I had thought > that I had proved the cases for groups of size 1, 2, and 3. However, > I just realized that there was a biiig hole in my proofs. I hadn't > shown that the objects with which I was working had the associative > property, and therefore had no right to claim that they had anything > to do with groups! Your original problem is a fairly well-known exercise which can be solved (rather ponderously) in very elementary terms by being a bit careful with logic. Normally I'd just give a hint, but since you've already put in so much effort, here are the main ideas of a proof. To prove that (the order of group G <= 5) implies (G is Abelian), it's good enough to prove the contrapositive (G is not Abelian) implies (the order of G > 5). So suppose that G is not Abelian. That means: not (for every a, b in G) (ab = ba) i.e. (there exist a, b in G) (ab != ba). Hence G must contain at least the elements e, a, b, ab, ba. The tedious part of the proof now is to show that all five of them are distinct. If e = a then ab = eb = b = be = ba which is false. Therefore e != a. You need ten little proofs like that to show that no pair of e, a, b, ab, ba can be equal. I'll leave the quite easy but rather tedious details to you. We now know that the order of G is at least 5. To show it can't actually be 5 is not quite so easy, and I know two approaches. You could appeal to the less elementary theorem that all groups of prime order (such as 5) are cyclic (so Abelian). Alternatively, you could continue with the method as above by showing either aba is different from each of e, a, b, ab, ba, or bab is different from each of e, a, b, ab, ba, or aba = b and bab = a, in which case a^2 is different from each of e, a, b, ab, ba. Each case gives at least six elements in G, so the proof is complete and it's definitely time for coffee. === === Subject: : Re: Proving associativity > Back in 2002, I posted a question about how to prove that all groups > of size <= 5 were Abelian. With the hints that I received then (right > before losing my Usenet access), I made some progress. I had thought > that I had proved the cases for groups of size 1, 2, and 3. However, > I just realized that there was a biiig hole in my proofs. I hadn't > shown that the objects with which I was working had the associative > property, and therefore had no right to claim that they had anything > to do with groups! If the object is a group then the binary operation is associative by definition of a group. IOW for any group of any order, it is not necessary to test for associativity. If your proof neglected to do this it is OK as long as you started with a group. Of course it is different to take a set of elements with a binary operation and prove that that is a group. > For instance, take the following multiplication table: > * | e a > --------- > e | e a > a | a e > In order for me to show that this is a group, I'd need to perform > 2*(2^3) multiplications before I'd really confirmed that it has > the associative property. That's not too bad. Only have to test one product here (a*a)*a = a*(a*a). >...But, what about: > * | e a b > ----------- > e | e a b > a | a e b > b | b a e > As far as I can tell, I'd need to perform 2*(3^3) multiplications > before I could claim that this associated. > Now I know that in most cases, an operation isn't defined in terms > of a table, but in terms of elementary arithmetic operations. In > such cases, you can prove the associative nature of the operation. > But, what do you do when all that you have is a table? Is there > anything other than exhaustive brute force available? The book > that I'm using as a text (Landin) poses this as a question; one > which I've never been able to answer. Commutativity is easy to > see from a table -- just look for symmetry about the major diagonal. > Is there a simple way to check for associativity? > Michael F. Stemper > #include There is three erors in this sentence. For any size binary operator table if there is an identity, then the product (X*Y)*Z reduces to the product of two elements no matter which one is the identity and X*(Y*Z) reduces to the same product. So it's not necessary to check any triplets that involve the identity. There will thus be (n-1)^3 triplet pairs left to check. I don't think there is a way to get around checking them all. Each (X*Y)*Z needs 2 multiplications to evaluate and 2 for X*(Y*Z) so the number of multiplications is 4 * (n-1)^3. if n = 3 there will be 4*8 = 32 multiplications. A binary operation may be associative and have an identity element and fail to be a group: * | e a --------- e | e a a | a a Since e is the identity, just have to check if (a*a)*a = a*(a*a). Hanford === === Subject: : Re: Proving associativity > Back in 2002, I posted a question about how to prove that all groups > of size <= 5 were Abelian. With the hints that I received then (right > before losing my Usenet access), I made some progress. I had thought > that I had proved the cases for groups of size 1, 2, and 3. However, > I just realized that there was a biiig hole in my proofs. I hadn't > shown that the objects with which I was working had the associative > property, and therefore had no right to claim that they had anything > to do with groups! > For instance, take the following multiplication table: > * | e a > --------- > e | e a > a | a e > In order for me to show that this is a group, I'd need to perform > 2*(2^3) multiplications before I'd really confirmed that it has > the associative property. That's not too bad. But, what about: > * | e a b > ----------- > e | e a b > a | a e b > b | b a e > As far as I can tell, I'd need to perform 2*(3^3) multiplications > before I could claim that this associated. > Now I know that in most cases, an operation isn't defined in terms > of a table, but in terms of elementary arithmetic operations. In > such cases, you can prove the associative nature of the operation. > But, what do you do when all that you have is a table? Is there > anything other than exhaustive brute force available? The book > that I'm using as a text (Landin) poses this as a question; one > which I've never been able to answer. Commutativity is easy to > see from a table -- just look for symmetry about the major diagonal. > Is there a simple way to check for associativity? http://www.ajnpx.com/pdf/math/SpecialTopics/Morphism.pdf Patrick === === Subject: : Re: Proving associativity > Back in 2002, I posted a question about how to prove that all groups > of size <= 5 were Abelian. With the hints that I received then (right > before losing my Usenet access), I made some progress. I had thought > that I had proved the cases for groups of size 1, 2, and 3. However, > I just realized that there was a biiig hole in my proofs. I hadn't > shown that the objects with which I was working had the associative > property, and therefore had no right to claim that they had anything > to do with groups! I don't know how much detail you're looking to include in your proof, but this problem really isn't too hard. For instance, every group of prime order is cyclic, hence Abelian. So the 2, 3, 5 cases are taken care of. The 1 case is obvious. So now all you have to take care of is the order 4 case. To do this quickly, you'll probably want to look into the Sylow p-subgroup theorems. If you want to prove this using more elementary facts, you've got your work cut out for you. 'cid 'ooh === === Subject: : Re: Proving associativity > Back in 2002, I posted a question about how to prove that all groups > of size <= 5 were Abelian. With the hints that I received then (right > before losing my Usenet access), I made some progress. I had thought > that I had proved the cases for groups of size 1, 2, and 3. However, > I just realized that there was a biiig hole in my proofs. I hadn't > shown that the objects with which I was working had the associative > property, and therefore had no right to claim that they had anything > to do with groups! If you showed commutativity for all operations * on sets with 1,2, or 3 elements, then you also showed it for all groups with 1,2, or 3 elements. Alternatively, you could show that all the non-commutative operations * are NOT group operations. Not sure if this helps. Will Twentyman email: wtwentyman at copper dot net === === Subject: : Re: Proving associativity >* | e a b >----------- >e | e a b >a | a e b >b | b a e >As far as I can tell, I'd need to perform 2*(3^3) multiplications >before I could claim that this associated. >Now I know that in most cases, an operation isn't defined in terms >of a table, but in terms of elementary arithmetic operations. In >such cases, you can prove the associative nature of the operation. >But, what do you do when all that you have is a table? Is there >anything other than exhaustive brute force available? The book >that I'm using as a text (Landin) poses this as a question; one >which I've never been able to answer. Commutativity is easy to >see from a table -- just look for symmetry about the major diagonal. >Is there a simple way to check for associativity? If you can map the operation to something that is known to be associative, for example modulo 3 addition, or a congruent group you've already checked, you're home free. But checking for congruence is probably as much trouble as doing the operations... You can pare down the number of table lookups by doing the one out of every four inside the second (as opposed to third) nested loop in your algorithm, i.e. for A = e to b for B = e to b AB = lookup(A,B) for C = e to b P1 = lookup(AB,C) BC = lookup(B,C) P2 = lookup(A,BC) if (P1 != P2) fail end end end Which would mean 90 lookups instead of 108. --Keith Lewis klewis {at} mitre.org The above may not (yet) represent the opinions of my employer. Message-Id: <405a001a$0$771$ba620e4c@news.skynet.be === === Subject: : Problem understanding a proof for compactification I'm trying to understand the proof of lemma 38.1 in Topology from James R. Munkres. They start with a imbedding h from a space X to a compact hausdorf space Z and define then X0 as the space h(X) and Y0 as the closure of X0. Now for the part that I don't understand, they choose then a set A disjoint from X that is in bijective correspondence k with the set Y0-X0, can some one point me to a proof that such a bijection indeed exist? === === Subject: : Re: Problem understanding a proof for compactification Adjunct Assistant Professor at the University of Montana. > I'm trying to understand the proof of lemma 38.1 in Topology from James >R. Munkres. They start with a imbedding h from a space X to a compact >hausdorf space Z and define then X0 as the space h(X) and Y0 as the closure >of X0. Now for the part that I don't understand, they choose then a set A >disjoint from X that is in bijective correspondence k with the set Y0-X0, >can some one point me to a proof that such a bijection indeed exist? Are you asking, how do you know you can find a set which is both disjoint from X and in bijective correspondence with Y0-X0? I.e., are you asking how can one show that: Given two sets A and B, there is always a set C which is (a) disjoint from A; and (b) in bijective correspondence with B ? First: given any set A, there is a set c which is not in A. Proof: Let c = {x in A: x is not in x}. This is a set; if c were an element of A, then either c is an element of c or it is not an element of c. If c is not an element of c, then by construction it must be an element of c. If c is an element of c, then by construction it is not an element of c. This is a contradiction, so therefore c is not an element of A. QED Now let A and B be sets; we want to show there is a set C which is disjoint from A and in bijective corresponce with B. (Inspired by a post from Fred Galvin, ID:<0311111418320.8057-100000@gandalf.math.ukans.edu>): Let Z be the set of all elements of elements of A. Let c be something which is not in Z (by the above, c exists). Now let C = { {b,c}: b in B}. First, C is disjoint from A: for if {b,c} is in C and in A, then c is an element of an element of A, hence c would be an element of Z. But we chose c specifically so it would not be in Z. Second, C is in bijective correspondence to B: define f:B->C by f(b) = {b,c}. This is clearly surjective; it is also injective. If {b,c} = {b',c}, then b'= c or b'=b. If b'=b, we are done; if b'=c, then {b',c} has only one element, hence {b,c} has only one element, so b=c as well, hence b'=c'=b. In either case, b=b' so f is injective. Thus, C satisfies the requirements. ======== === === Subject: : Re: expectation of the product of two dependent random variables >Hi sorry to bother you guys here. >Let random variable X = a_1r_1+a_2r_2+...+a_nr_n, and >random variable Y = b_1r_1+b_2r_2+...+b_nr_n, where a_i and b_i >(i=1...n) are >constants, and r_i is i.i.d. random variable chosen from N(0,1). Note >that r_i in both X and Y are the same. >So how to compute the expectation of exp(p times X times Y), that >is >E[exp(p times X times Y)] = ? > X and Y are jointly normal random variables with means 0, variances > E[X^2] = sum_i (a_i)^2 and E[Y^2] = sum_i (b_i)^2, and covariance > E[XY] = sum_i a_i b_i. By diagonalizing the covariance matrix > ( E[X^2] E[XY] ) > ( E[XY] E[Y^2] ) > you can reduce to the case where n=2. Thank you very much, Robert. I am sorry but I didn't get your point. Given the covariance matrix, how do you compute the E[e^(pXY)] where p is a constant? thanks a lot Roy === === Subject: : Re: expectation of the product of two dependent random variables >Let random variable X = a_1r_1+a_2r_2+...+a_nr_n, and >random variable Y = b_1r_1+b_2r_2+...+b_nr_n, where a_i and b_i >(i=1...n) are >constants, and r_i is i.i.d. random variable chosen from N(0,1). Note >that r_i in both X and Y are the same. >So how to compute the expectation of exp(p times X times Y), that >is >E[exp(p times X times Y)] = ? >X and Y are jointly normal random variables with means 0, variances >E[X^2] = sum_i (a_i)^2 and E[Y^2] = sum_i (b_i)^2, and covariance >E[XY] = sum_i a_i b_i. By diagonalizing the covariance matrix >( E[X^2] E[XY] ) >( E[XY] E[Y^2] ) >you can reduce to the case where n=2. >Thank you very much, Robert. >I am sorry but I didn't get your point. Given the covariance matrix, >how do you compute the E[e^(pXY)] where p is a constant? If X = a_1 r_1 + a_2 r_2 and Y = b_1 r_1 + b_2 r_2, and |p| is small enough, E[ exp(pXY)] = 1/(2 pi) int int exp(-(t_1^2 + t_2^2)/2) exp(p(a_1 t_1 + a_2 t_2)(b_1 t_1 + b_2 t_2)) dt_1 dt_2 = 1/(2 pi) int int exp(- C ^T) dt_1 dt_2 = 1/(2 sqrt(det(C))) where C is the matrix [ 1/2 - p a_1 b_1 -p(a_1 b_2 + a_2 b_1)/2 ] [ -p(a_1 b_2 + a_2 b_1)/2 1/2 - p a_2 b_2 ] and det(C) = 1/4 - (a_2 b_2 + a_1 b_1) p/2 - (a_1 b_2 - a_2 b_1)^2 p^2/4 Department of Mathematics http://www.math.ubc.ca/~israel === === Subject: : anthyphairetic ratio Mathworld says: An archaic word for a continued fraction is anthyphairetic ratio. from: http://mathworld.wolfram.com/ContinuedFraction.html But I can find the term in very few other places on-line. And I cannot find the word anthyphairetic, without ratio, at all. I assume it had a traditional non-mathematical meaning once, but I have no idea what this was, if it did exist as a word before it was applied to continued fractions. Did Mathworld misspell this phrase? What is its etymology? (By the way, the term is part of the solution of my rec.puzzles coded-phrase puzzle at: Leroy Quet === === Subject: : Re: anthyphairetic ratio Leroy Quet schrieb in Nachricht ... >Mathworld says: An archaic word for a continued fraction is anthyphairetic ratio. >from: >http://mathworld.wolfram.com/ContinuedFraction.html >But I can find the term in very few other places on-line. >And I cannot find the word anthyphairetic, without ratio, at all. >I assume it had a traditional non-mathematical meaning once, but I >have no idea what this was, if it did exist as a word before it was >applied to continued fractions. >Did Mathworld misspell this phrase? No, it's correct. It is the Greek expression for the process of obtaining a sequence of fractions bei Euclid's algorithm. Here are some links: MAA review: http://www.maa.org/reviews/mpa.html two threads from the Historia Matematica mailing list: http://mathforum.org/epigone/historia_matematica/dwimandphex/ http://mathforum.org/epigone/math-history-list/stayprixtherm/ conference abstracts: http://www.iwr.uni-heidelberg.de/transmath/proceedings/ abstracts.html --> Jan Hogendijk -- http://www.math.ruu.nl/people/hogend/mahani.html >What is its etymology? See Ioannis' posting. There is the similar expression prosthaphairesis, which stood for a forerunner of the logarithms by using the addition theorems of the sine and cosine for multiplying two numbers: a*b = ? Look up u = arccos(a), v = arccos(b) in a table, Look up r = cos(u+v), s = cos(u-v) in the table, Then a*b = (r + s)/2. Proof: cos(u)*cos(v) = [ cos(u+v) + cos(u-v) ]/2 Hermann >(By the way, the term is part of the solution of my rec.puzzles >coded-phrase puzzle at: 5.466293%40socal.rr.com&prev=) >Leroy Quet === === Subject: : Re: anthyphairetic ratio charset=utf-8 Leroy Quet [CapitalEth][EDouble Dot][Micro] .b3 .b9[EDo ubleDot].b9 > Mathworld says: An archaic word for a continued fraction is anthyphairetic ratio. > from: > http://mathworld.wolfram.com/ContinuedFraction.html > But I can find the term in very few other places on-line. > And I cannot find the word anthyphairetic, without ratio, at all. > I assume it had a traditional non-mathematical meaning once, but I > have no idea what this was, if it did exist as a word before it was > applied to continued fractions. > Did Mathworld misspell this phrase? > What is its etymology? Anti + hyphairesis Anti=opposite. Antiphairesis -> Nu and Tau transform to Theta->Anthyphairesis. Hyphairesis: In legislative and legal use the term denotes extraction, fraud, stealing or otherwise illegal acquisition of money/goods which happens between: 1) relatives (by blood or otherwise). 2) Parents and their adopted children, 3) Husband and wife or engaged people, 4) brothers and their spouses, 5) spouse that survived and spouse that died, about the property of the one who died, 6) higher public official, supervisor and supervisee. In financial use the tax that's being substracted from a monetary token when it is being liquidated in advance. (I tried to translate the lemma as well as I could). A very rare Greek word indeed. That's probably the first or second time I afairesis (deduction/subtraction), that's been reserved for a specific arithmetical operation. It appears that anthyphairesis probably means something akin to opposite subtraction/deductionIt's worth mentioning that I know recall that I have heard of hyphairesis being in the set of Arithmetic Operations long time ago, but can't recall the details: {Prosthesis, Afairesis, Pollaplasiasmos, Diairesis, Hyphairesis} = {Addition, Subtraction, Multiplication, Division, Hyphairesis(?)} I have no idea why Continued Fractions would be called Anthyphairetic Ratios, though. Perhaps they approximate well some sort of funny looking subtraction-like operation in specific applications. >Leroy Quet --- http://users.forthnet.gr/ath/jgal/ ------------------------------------------ Eventually, _everything_ is understandable === === Subject: : Re: anthyphairetic ratio I don't suppose I could drum up any support for a consensus on the (mis)use of UTF-8, or whatever encoding that was, in an English text newsgroup... Pine must default to (I guess) Latin-1, which makes actual Greek encoded text look really weird. > Mathworld says: An archaic word for a continued fraction is anthyphairetic ratio. > What is its etymology? > Anti + hyphairesis > A very rare Greek word indeed. That's probably the first or second time I afairesis (deduction/subtraction), that's been reserved for a specific > arithmetical operation. > It appears that anthyphairesis probably means something akin to opposite > subtraction/deduction> It's worth mentioning that I know recall that I have heard of hyphairesis > being in the set of Arithmetic Operations long time ago, but can't recall > the details: {Prosthesis, Afairesis, Pollaplasiasmos, Diairesis, > Hyphairesis} = {Addition, Subtraction, Multiplication, Division, > Hyphairesis(?)} > I have no idea why Continued Fractions would be called Anthyphairetic > Ratios, though. Perhaps they approximate well some sort of funny looking > subtraction-like operation in specific applications. Well, anti-hyphairetic implies opposite subtraction. And that's exactly what continued fractions are all about: /subtract/ the integer part of the number, and then take the multiplicative /opposite/, or reciprocal, of the remainder. Repeat ad infinitum or until the remainder becomes zero. Anthyphairesis, from what I read, seems to have been the name for one stage of the anthyphairetic process of constructing these continued fractions. And yes, I think the guys who say the Greeks had continued fractions are probably right. See http://www.maa.org/reviews/mpa.html The ancient Greeks might not have considered continued fractions to be ratios or fractions in the same sense we do today, but they definitely seem to have used them. (All IMH and ill-informed O, of course.) -Arthur === === Subject: : Re: anthyphairetic ratio Ioannis > ? Leroy Quet ?????? ??? ?????? > Mathworld says: An archaic word for a continued fraction is anthyphairetic ratio. > from: > http://mathworld.wolfram.com/ContinuedFraction.html > But I can find the term in very few other places on-line. > And I cannot find the word anthyphairetic, without ratio, at all. > I assume it had a traditional non-mathematical meaning once, but I > have no idea what this was, if it did exist as a word before it was > applied to continued fractions. > Did Mathworld misspell this phrase? > What is its etymology? ... > Hyphairesis: ... I found this def'n in the massive New International Dictionary (Webster's): hyphaeresis Gk hyphairesis from hyphairein to take from under, subtract, from hypo- + hairein to take. : the omission of a sound, letter, or syllable from the body of a word === === Subject: : Re: anthyphairetic ratio Larry Hammick > Mathworld says: An archaic word for a continued fraction is anthyphairetic ratio. > I found this def'n in the massive New International Dictionary (Webster's): > hyphaeresis > Gk hyphairesis from hyphairein to take from under, subtract, from hypo- + > hairein to take. > : the omission of a sound, letter, or syllable from the body of a word An afterthought... doctors speak of pheresis and plamapheresis -- a process of taking out blood, filtering some components out, and putting the rest back. === === Subject: : Re: anthyphairetic ratio > Mathworld says: An archaic word for a continued fraction is anthyphairetic ratio. > from: > http://mathworld.wolfram.com/ContinuedFraction.html > But I can find the term in very few other places on-line. > And I cannot find the word anthyphairetic, without ratio, at all. > I assume it had a traditional non-mathematical meaning once, but I > have no idea what this was, if it did exist as a word before it was > applied to continued fractions. > Did Mathworld misspell this phrase? Nope. > What is its etymology? Greek, of course! Anthyphairesis (Google it!) is apparently the process of constructing a continued fraction by subtraction of fractional parts. Ask a Greek scholar; the only derivation I can find has something to do with carrying flowers. :) -Arthur === === Subject: : Re: anthyphairetic ratio >What is its etymology? Well, it seems to be the adjectival form of the noun anthyphairesis, for what that's worth. Lee Rudolph === === Subject: : Re: anthyphairetic ratio >What is its etymology? > Well, it seems to be the adjectival form of the noun anthyphairesis, for what that's worth.... Yes. Anthyphaeresis was the Greek mathematicians' name for what we (and not they) call the Euclidean algorithm. In modern terms of course you can link that with continued fractions, but the Greeks didn't use those. The Mathworld statement is misleading. There is some evidence of an early theory of irrational numbers based on infinitely continued anthyphaeresis. To prove that magnitudes a and b are in the same ratio as c and d, you can consider two Euclidean algorithms, one starting from a and b, the other from c and d. If you can prove that both algorithms produce the same (possibly infinite) sequence of quotients, then you have equal ratios. It's not hard to see the connection with modern continued fractions. Just how much that anthyphaeretic method was used is uncertain, and a matter of discussion by historians. Anyway, it was certainly displaced around 350 B.C. by Eudoxus's treatment of irrationals as in Euclid Book V, whose modern version is Dedekind cuts. === === Subject: : Number-Of-Iteration Sequences Consider the sequence where a(n) = sum{k>=1} k*c(k), if n = product{k>=1} p(k)^c(k), each c = a nonnegative integer, and p(k) is the k_th prime. {a(k)} is: http://www.research.att.com/projects/OEIS?Anum=A056239 (0,1,2,2,3,3,4,3,4,4,5,4,6,5,5,4,7,...) Now, define b(n) as the number of iterations needed to get a(a(...a(n)..)) = 0. (same as, for n>= 2, number of a's until 1.) In other words, b(1) = 0; b(n) = b(a(n)) + 1: 0,1,2,2,3,3,3,3,3,3,4,3,4,4,4,3,4,... (unless I goofed) My choice of {a(j)} = A056239 was almost arbitrary. For this b-sequence, it might be more natural to add one or subtract one from each term for whatever reason. I originally had wondered about the number-of-iterations sequence when {a(k)} was A001414. (a sequence similar to A056239: A001414: Name: Integer log of n: sum of primes dividing n (with repetition).) But this b-sequence is already in the EIS: http://www.research.att.com/projects/OEIS?Anum=A002217 Generally, I wonder about interesting sequences formed from other sequences, where the original sequences consist of positive integer terms <= their indexes, and the derived sequences are the number of iterations needed to achieve a fixed-point or 0 or 1 or whatever specified value. I am guessing that there are some interesting number-of-iteration sequences of this kind which are derived from EIS sequences. Leroy Quet === === Subject: : Integrating with trig substitutions For integrating certain quadratics, many authors suggest the following substitutions : SQRT(a^2 x^2)tx = a sin(z) SQRT(a^2 + x^2)tx = a tan(z) SQRT(x^2 a^2)tx = a sec(z) But at least one author suggests the same substitutions for the above expressions without the SQRT. So out of curiosity, I applied x = a tan(z) to the fundamental formula Integral(dx/(a^2 + x^2)) and arrived at the correct answer i.e. (1/a)arctan(x/a) + C. Now applying the same process to the fundamental formula Integral(dx/(x^2 a^2)) with the substitution x = a sec(z), I don't arrive at the table answer' which is (1/2a)ln abs((x-a)/(x+a)) + C. Instead, I arrived at (1/a)ln abs(sec(SQRT((a^2+x^2)/a)) + tan(x/a)) + C. Why does my approach not yield the correct answer? P.S. I know I can arrive at the correct answer with partial franctions. But my objective is to make it work' with the trig. substitution. === === Subject: : Re: Integrating with trig substitutions Adjunct Assistant Professor at the University of Montana. >For integrating certain quadratics, many authors suggest the following >substitutions : >SQRT(a^2 - x^2) x = a sin(z) >SQRT(a^2 + x^2) x = a tan(z) >SQRT(x^2 - a^2) x = a sec(z) >But at least one author suggests the same substitutions for the above >expressions without the SQRT. Yes; you would convert it into a trigonometric integral either way. > So out of curiosity, I applied x = a >tan(z) to the fundamental formula >Integral(dx/(a^2 + x^2)) and arrived at the correct answer i.e. >(1/a)arctan(x/a) + C. >Now applying the same process to the fundamental formula >Integral(dx/(x^2 - a^2)) with the substitution x = a sec(z), I don't >arrive at the table answer' which is >(1/2a)ln abs((x-a)/(x+a)) + C. Instead, I arrived at >(1/a)ln abs(sec(SQRT((a^2+x^2)/a)) + tan(x/a)) + C. >Why does my approach not yield the correct answer? Well, first, it is unclear whether it is the right answer or not: the two answers could differ by a constant... But how did you arrive at that answer? Subsituting x = a sec(z) into int(dx/(x^2-a^2)) I get the integral of [a sec(z) tan(z)]/[a^2(sec^2(z) - 1)] = sec(z)tan(z)/a tan^2(z) = (1/a) sec(z)/tan(z) = (1/a) csc(z) and the integral of (1/a) csc(z)dz is (1/a)*ln |csc z - cot z| + C = (1/2)ln |tan(z/2)| + C. Since sec(z) = x/a, I get that csc(z) = x/sqrt(x^2-a^2), cot(z) = a/sqrt(x^2-a^2), so csc(z) - cot(z) = (x-a)/sqrt(x^2-a^2) = (x-a)/sqrt(x-a)sqrt(x+a) = sqrt((x-a)/(x+a)), so we get (1/a)*ln|sqrt((x-a)/(x+a))| + C = (1/2)(1/a)ln|(x-a)/(x+a)| + C = (1/2a)ln |(x-a)/(x+a)| + C, the same answer as your table. ======== === === Subject: : Product-Over-Integers Expansion Of Real Let A be a strictly increasing sequence of positive integers where sum{k=elements of A} 1/k diverges. For x = any real > 1, we can find an infinite number of expansions (each a sub-sequence of A) such that: x = product{k=some elements of A} 1/(1-1/k) But we can find the specific greedy-algorithm subsequence, where each integer is chosen so that the partial-product remains <= x, and each new term is the lowest unpicked element of A such that the partial product is <= x. (Such an expansion is, obviously, infinite if x is irrational.) For example, either if A consists of all positive integers >= 2 or consists of just the primes, then, for x = pi, we have the expansion's terms 2, 3, 23,... (pi = 1/(1-1/2) *1/(1-1/3) *1/(1-1/23) *..., if I calculated right.) I guess we can also ask about the subsequences of A such that x = product{k=some elements of A} (1 + 1/k). I bet this idea is not new. Are any such expansions already in the Encyclopedia of Integer Sequences? Is there anything else known about such expansions? Leroy Quet === === Subject: : Re: (-2/3)^(-2/3) = (3/2)^(2/3)? > ... ((-1)^sqrt(2))^sqrt(2), as being multivalued in the complex numbers, > it is e^(i*2*Sqrt(2)*N*pi) for integer N. Among those values, the only real > value is +1 (using N=0), rather than -1. I got different result for the above expression: |x = ((-a)^sqrt(2))^sqrt(2) |a >= 0 |a belongs to R => x = ((|-a|*e^(i*arg(-a))^sqrt(2))^sqrt(2) = ((a^sqrt(2))*e^(i*pi*sqrt(2)))^sqrt(2) = (a^(sqrt(2) * sqrt(2))*e^(i*pi*sqrt(2) * sqrt(2)) = (a^2)*e^(2*pi*i) = (a^2)*1 = a^2 = a * a So for a = 1 => x = 1 * 1 = 1 and therefore this is the only one solution in the complex plane. Right? > Suppose that you ask some mathematicians for the implied domain of the > function > f(x) = Sqrt(x-3)*Sqrt(1-x) > with the stipulation that f must map reals to reals. I would guess that > some would say that the implied domain is empty, while others would say > that it is the closed interval [1,3]. Why you are not completely certain about this? You suppose that the following system: |y = sqrt(x-3) * sqrt(1-x) |1 <= x <= 3 |x and y belong to R will be equivalent to: |y = sqrt(4x - x^2 - 3) |1 <= x <= 3 |x and y belong to R only for some mathematicians? Calvin === === Subject: : Re: (-2/3)^(-2/3) = (3/2)^(2/3)? > ... ((-1)^sqrt(2))^sqrt(2), as being multivalued in the complex > numbers, it is e^(i*2*Sqrt(2)*N*pi) for integer N. Among those values, > the only real value is +1 (using N=0), rather than -1. > I got different result for the above expression: > |x = ((-a)^sqrt(2))^sqrt(2) > |a >= 0 > |a belongs to R > => x = ((|-a|*e^(i*arg(-a))^sqrt(2))^sqrt(2) = > ((a^sqrt(2))*e^(i*pi*sqrt(2)))^sqrt(2) = (a^(sqrt(2) * > sqrt(2))*e^(i*pi*sqrt(2) * sqrt(2)) = (a^2)*e^(2*pi*i) = (a^2)*1 = a^2 > = a * a At no point above did you consider that it can take on multiple values. > So for a = 1 => x = 1 * 1 = 1 and therefore this is the only one > solution in the complex plane. Right? No, there are infinitely many different values. For a = 1: Suppose you ask a computer algebra system to evaluate ((-1)^sqrt(2))^sqrt(2). It probably won't give you 1. Rather, it will probably give you e^(-i*2*Sqrt(2)*pi), approx. -0.858216 - 0.513288*i, which can be obtained from the formula I'd given earlier by taking N=-1. But that's just one of infinitely many complex values for the expression. Among them, 1 is the only real value. > Suppose that you ask some mathematicians for the implied domain of the > function > f(x) = Sqrt(x-3)*Sqrt(1-x) > with the stipulation that f must map reals to reals. I would guess that > some would say that the implied domain is empty, while others would say > that it is the closed interval [1,3]. > Why you are not completely certain about this? For one thing, I haven't taken a poll. Please, feel free to take one yourself! (Sci.math might be a good place to do it.) > You suppose that the following system: > |y = sqrt(x-3) * sqrt(1-x) > |1 <= x <= 3 > |x and y belong to R > will be equivalent to: > |y = sqrt(4x - x^2 - 3) > |1 <= x <= 3 > |x and y belong to R > only for some mathematicians? I'm not sure. But note that, in the scenario I'd described earlier, there was no statement of the form 1 <= x <= 3. Such a statement is surely prejudicial if the question is What is the implied domain? David Cantrell === === Subject: : simple integration question (...it's been a while) I have a simple integration question: S (y=n^x dx) I need to integrate this term. Unfortunately i don't remember much... Any help is appreciated. Nicolas === === Subject: : Re: simple integration question (...it's been a while) You could let u = n^x Then, ln u = x ln n So, du / u = ln n dx. Which means that [ln n) / u] du = dx Now, substituting into the integral yield Int( n^x dx) = Int( u [ln n) / u] du ) = Int( ln n du) = u ln n + C Now, substitute n^x in for u. Brian >I have a simple integration question: >S (y=n^x dx) >I need to integrate this term. Unfortunately i don't remember much... Any >help is appreciated. >Nicolas === === Subject: : Re: simple integration question (...it's been a while) > I have a simple integration question: > S (y=n^x dx) > I need to integrate this term. Unfortunately i don't remember much... Any > help is appreciated. > Nicolas The formula is n^x(ln n)+C David Moran === === Subject: : Re: simple integration question (...it's been a while) Adjunct Assistant Professor at the University of Montana. >I have a simple integration question: >S (y=n^x dx) >I need to integrate this term. Unfortunately i don't remember much... Any >help is appreciated. n^x = e^{x*ln(n)}. So Int (n^x)dx = Int(e^{x*ln(n)})dx = (1/ln(n))e^{x*ln(n)} + C = n^x/ln(n) + C. Alternatively, remember that d -- (n^x) = n^x*ln(n). dx ======== === === Subject: : Re: Akcerman's function > I want to relate ackerman's function to complexity classes (p,np), for > a homework question. > I know that Ackerman's function is not a polynom. > But how such assertion can be proven ? According to: http://mathworld.wolfram.com/AckermannFunction.html Ackermann's function, A(m,n), is a polynomial for m= 0,1,2. But for m >= 3, it simply rises faster than ANY polynomial of fixed order, sometimes MUCH faster, as n increases. Leroy Quet === === Subject: : Re: Double-Sum Over Coprime Indexes (puzzle problem) Hint below: > I bet this is one of the easiest math puzzle, but it seems worth posting anyway. > Evaluate: > oo oo > --- --- 1 > ----------- > / / k j (k+j) > --- --- > k=1 j=1 > GCD(k,j)=1 > In linear mode: > sum{k=1 to oo} sum {j=1 to oo, GCD(k,j)=1} 1/(kj(k+j)). > Leroy Quet Hint: | | V | | V | | V | | V the double-sum evaluates to a rational. Bigger hint below: | | V | | V | | V | | V | | V the double-sum is equal to an integer. (if I did my math right...) Leroy Quet === === Subject: : Re: Double-Sum Over Coprime Indexes (puzzle problem) Evaluate it? I can't even read the friggin' thing. > Hint below: > I bet this is one of the easiest math puzzle, but it seems worth posting anyway. > Evaluate: > oo oo > --- --- 1 > ----------- > / / k j (k+j) > --- --- > k=1 j=1 > GCD(k,j)=1 > In linear mode: > sum{k=1 to oo} sum {j=1 to oo, GCD(k,j)=1} 1/(kj(k+j)). > Leroy Quet > Hint: > | > | > V > | > | > V > | > | > V > | > | > V > the double-sum evaluates to a rational. > Bigger hint below: > | > | > V > | > | > V > | > | > V > | > | > V > | > | > V > the double-sum is equal to an integer. > (if I did my math right...) > Leroy Quet === === Subject: : Interpolating Function If I know my function f has this property: f(0) = a(0) f'(0) = a(1) f''(0) = a(2) .... where a(0), a(1), ... are parameters I wonder whether there is any method so that I can derive my function f. Does anyone know or have any ideas on this? Thank you! Khoa Tran === === Subject: : Re: Interpolating Function > If I know my function f has this property: > f(0) = a(0) > f'(0) = a(1) > f''(0) = a(2) > .... > where a(0), a(1), ... are parameters > I wonder whether there is any method so that I can derive my function f. > Does anyone know or have any ideas on this? Thank you! f(x) = the sum, from k = 0 to whatever, of a(k) times x-to-the-k, divided by k-factorial. === === Subject: : Re: Interpolating Function >If I know my function f has this property: > f(0) = a(0) > f'(0) = a(1) > f''(0) = a(2) > .... >where a(0), a(1), ... are parameters >I wonder whether there is any method so that I can derive my function f. >Does anyone know or have any ideas on this? Thank you! >f(x) = the sum, from k = 0 to whatever, of a(k) times x-to-the-k, > divided by k-factorial. I'm reminded of that quote in a post I just replied to: Such behaviour is exclusively confined to functions invented by >mathematicians for the sake of causing trouble. > -Albert Eagle's _A Practical Treatise on Fourier's Theorem_ It's my job to cause trouble here by pointing out that what you said doesn't necessarily reproduce f; in fact knowing that f is infinitely differentiable and knowing all the derivatives at the origin says nothing at all about f(1). On the other hand, of course what you said is (by definition) correct if f is analytic, and there's a good chance that the OP's function _is_ analytic. === === Subject: : Re: Sum, Over Some Prime Multiples, Is Always m > Let P be any subset of the primes. > (example: P contains all primes of even-index, > or those congruent to 1 (mod 6).) > Let A be the set of positive integers: including 1 and every positive > integer which is a multiple of only the primes in P, and A contains > no member which is divisible by a prime not in P. > Let B be the set of positive integers: including 1 and every positive > integer which is a multiple of only the primes *not* in P, and B > contains no member which is divisible by a prime *in* P. > (Yes, there are positive integers in neither set, as long as P does > not contain every prime.) > Let g(x) be the number of distinct elements of A which are <= x, > for x = a positive real. > So then, for m = any positive integer: > --- > / g(m/k) > --- > k=elements of B, k <= m > always = m. > In linear-mode: > sum{k=elements of B, k<=m} g(m/k) > always equals m. > (Right?) > Leroy Quet > A consequence (which probably is trivially shown by a more direct > method) of the above: > If, for a given P (and A), > limit{m->oo} g(m)/m = x > exists and is finite nonzero, then > sum{k=elements of B} 1/k > ( = product{p=primes not in P} 1/(1 -1/p) ) > converges to 1/x. > And so, from this we see that > sum{p=primes not in P} 1/p converges > if the limit of g(m)/m approaches a limit which is finite nonzero. > But...I am being VERY nonrigorous above, and could very well be wrong. > Have I posted truth anyway? > Leroy Quet I doubt I have posted truth, I now believe, regarding the x-limit result. My math is SO unrigorous as to be laughable. :/ Leroy Quet === === Subject: : KENO BOARD AS A FIXED FORMAT 2D PROJECTION I used to be a Keno writer about 25 years ago, and I had seen at least one regular player who knew how to regularly beat the game. He would play a 3-way four (a two-spot and three one-spots) for a $1 each way, and he would always eventually hit a four out of four for about a $250 payout (my local casino is a cheap tightwad by comparison). I tried to fathom the player's method, but he had no set pattern, and he would sometimes peer at the board intently, not like someone waiting for luck. I realized that he was able to see the pattern of where the balls would eventually show up on the Keno board. The best I could do, in mimicking him, would be to wait for Keno boards where the patterns were showing up in contigious block groups. I saw that any given pattern or texture will continue over many games. So, during heavy block-ish Keno boards, I would play block-ish tickets covering the blank spaces of block-filled Keno boards. I broke even! (So what! But given the odds -- fantastic!). I found that I would NOT hit right away, and that hits in the spaces previously chosen would come at unexpected times. Since those many years ago, I've come to new ideas and I've gotten better at predicting the movements of patterns on Keno boards, but I'm thinking about becoming horribly serious about it -- I'm considering studying MATH. First, let's dispense with standard probability models and statistics: a physical, closed system that shuffles its elements the same way every time is NOT random. Random means that any element can occur in combination with any other element at any time. In a PHYSICAL CLOSED SYSTEM, all the elements stand as barriers to each other, preventing pure mathematical randomness. For example, if the 1-ball is at the top of the pile of balls when the blower is turned on, it is NOT going to magically disappear and then reappear at the bottom of the pile. It will travel, but not without travelling with adjacent balls affected in similar ways. THAT'S WHY PATTERNS PERSIST IN ANY PHYSICAL CLOSED SYSTEM SUBJECTED TO ANY STANDARDIZED SHUFFLE PROCESS, NO MATTER HOW SEEMINGLY CHAOTIC THAT SHUFFLE PROCESS SEEMS. There's math, and then there's STUPID MATH. Does anyone have any intelligent ideas about what math might work for analyzing Keno boards? === === Subject: : FIFO problem - yet another .sig rot script... ', $fifo or die Can't write to `$fifo': $!n; print $fh do { # [SNIP - actual content generation] }; } sleep 1; } __END__ Michele > Comments should say _why_ something is being done. Oh? My comments always say what _really_ should have happened. :) - Tore Aursand on comp.lang.perl.misc === === Subject: : Re: FIFO problem - yet another .sig rot script... >I'm trying to write a .sig rotation script for client-independet use Sorry, wrong group!! Michele > Comments should say _why_ something is being done. Oh? My comments always say what _really_ should have happened. :) - Tore Aursand on comp.lang.perl.misc === === Subject: : Re: FIFO problem - yet another .sig rot script... >I'm trying to write a .sig rotation script for client-independet use >Sorry, wrong group!! Oh, heck, and here I was waiting for someone to tell you to pre- and postmultiply your .sig by a unit quaternion. Lee Rudolph === === Subject: : Re: FINAL SOLUTIONS To Astrophysical Mysteries Is there something wrong with your CAPSLOCK key? If so please fix it as it is very irritating to read, rather like someone fiddling with the volme control of a stereo. Robert skrev i meddelandet > (Book Review): > THE UNIVERSE OF MOTION, by Dewey B. Larson, 1984, North > Pacific Publishers, Portland, Oregon, 456 pages, indexed, > hardcover. > THE UNIVERSE OF MOTION contains FINAL SOLUTIONS to > most ALL astrophysical mysteries. > This book is Volume III of a revised and enlarged > edition of THE STRUCTURE OF THE PHYSICAL UNIVERSE, 1959. > Volume I is NOTHING BUT MOTION (1979), and Volume II is > THE BASIC PROPERTIES OF MATTER (1988). > astrophysics are bristling with integrals, partial > differentials, and other FANCY MATHEMATICS. In this book, by > contrast, mathematics is conspicuous by its absence, except > for some relatively simple formulas imbedded in the text. > Larson emphasizes CONCEPTS and declares that mathematical > agreement with a theory does NOT guarantee its conceptual > validity. > Dewey B. Larson was a retired engineer with a Bachelor > of Science Degree in Engineering Science from Oregon State > University. He developed the Theory described in his books > while trying to find a way to MATHEMATICALLY CALCULATE the > properties of chemical compounds based ONLY on the elements > they contain. > THE UNIVERSE OF MOTION describes the astrophysical > portions of Larson's CONSISTENT, INTEGRATED, COMPREHENSIVE, > GENERAL UNIFIED Theory of the Physical Universe, a kind of > grand unified field theory that orthodox physicists and > astro-physicists CLAIM to be looking for. It is built on two > postulates about the physical and mathematical nature of > space and time: > (1) The Physical Universe is composed ENTIRELY of ONE > component, MOTION, existing in THREE dimensions, in DISCRETE > units, and with two RECIPROCAL aspects, SPACE and TIME. > (2) The Physical Universe conforms to the relations of > ORDINARY COMMUTATIVE mathematics, its primary magnitudes are > ABSOLUTE, and its geometry is EUCLIDEAN. > COMPLETE theoretical universe, from photons and subatomic > concept of INWARD and OUTWARD SCALAR MOTIONS with > translational, vibrational, rotational, and rotational- > vibrational motions. At each step in the development, he was > able to match parts of his theoretical universe with > corresponding parts in the real Physical Universe, including > EVEN THINGS NOT YET DISCOVERED. For example, in his 1959 > book, he first predicted the existence of EXPLODING GALAXIES, > several years BEFORE astronomers started finding them. They > are a NECESSARY CONSEQUENCE of his comprehensive Theory. And > when quasars were discovered, he had a related explanation > ready for those also. > As a result of his theory, which he called THE > RECIPROCAL SYSTEM, Larson TOTALLY REJECTED many of the > sacred doctrines of orthodox physicists and astrophysicists, > including black holes, neutron stars, degenerate matter, > nuclear physics, general relativity, relativistic mass > increases, relativistic Doppler shifts, nuclear fusion in > stars, and the big bang, all of which he considered to be > nothing more than MATHEMATICAL FANTASIES. He was very > critical of the AD HOC assumptions, uncertainty principles, > solutions in principle, no other way declarations, etc., > used to maintain them. > THE UNIVERSE OF MOTION is divided into 31 chapters. > It begins with a description of how galaxies are built from > the gravitational attraction between globular star clusters, > which are formed from intergalactic gas and dust clouds that > accumulate from the decay products of cosmic rays coming in > from the ANTI-MATTER HALF of the physical universe. (Galaxy > formation from the MYTHICAL big bang is a big mystery to > orthodox astronomers.) He then goes on to describe the life > cycles of stars and how binary and multiple star systems and > solar systems result from Type I supernova explosions of > SINGLE stars. > Several chapters are devoted to quasars which, according > to Larson, are densely-packed clusters of stars that have > been ejected from the central bulges of exploding galaxies > and are actually traveling FASTER THAN THE SPEED OF LIGHT > (although most of that speed is AWAY FROM US IN TIME). > Astronomers and astrophysicists who run up against > observations that contradict their theories would find > Larson's explanations quite valuable if considered with an > BURSTS originated from pulsars, which exist primarily in the > plane or central bulge of our galaxy. But the new Gamma Ray > Telescope in earth orbit observed that the bursts come from > ALL DIRECTIONS UNIFORMLY and do NOT correspond with any > visible objects, (except for a few cases of directional > coincidence). Larson's explanation is that the gamma ray > bursts originate from SUPERNOVA EXPLOSIONS in the ANTI-MATTER > HALF of the Physical Universe, which Larson calls the cosmic > sector. Because the anti-matter universe exists in a > RECIPROCAL RELATION to our material universe, with the SPEED > OF LIGHT as the BOUNDARY between them, and has THREE > dimensions of TIME, and ONLY ONE dimension of space, the > bursts can pop into our material universe ANYWHERE seemingly > at random. > Larson heavily quotes or paraphrases statements from > astronomers. In this book, 351 of them are superscripted > with numbers identifying entries in the reference list at the > end of the book. For example, a quote from the book > Astronomy: The Cosmic Journey, by William K. Hartmann, > says, Our hopes of understanding all stars would brighten if > we could explain exactly how binary and multiple stars > form.... Unfortunately we cannot. Larson's book contains > LOGICAL CONSISTENT EXPLANATIONS of such mysteries that are > WORTHY OF SERIOUS CONSIDERATION by ALL physicists, > astronomers, and astrophysicists. > A WEALTH of Information about the GENERAL UNIFIED Theory > of the Physical Universe developed by the late Physicist > Dewey B. Larson can be found at the web site > http://www.rsystem.org/isus/index.htm or /dbl/index.htm . > MOST of the TEXT of the book THE UNIVERSE OF MOTION can be > found at the web site http://www.rsystem.org/um/index.htm . > Robert E. McElwaine > B.S., Physics and Astronomy, UW-EC > http://members.aol.com/rem547 PLUS > http://members.aol.com/rem460 > P.S.: PASS IT ON ! EVERYTHING you know is WRONG. The Truth IS stranger than fiction. The Truth is ALWAYS the FIRST CASUALTY OF WAR. OFFICIAL LIES are ALWAYS the BIGGEST LIES OF ALL. === === Subject: : Re: FINAL SOLUTIONS To Astrophysical Mysteries Is there something wrong with your CAPSLOCK key? If so please fix it as it s very irritating to read, rather like someone continually fiddling with the volume control of a stereo. Robert skrev i meddelandet > (Book Review): > THE UNIVERSE OF MOTION, by Dewey B. Larson, 1984, North > Pacific Publishers, Portland, Oregon, 456 pages, indexed, > hardcover. > THE UNIVERSE OF MOTION contains FINAL SOLUTIONS to > most ALL astrophysical mysteries. > This book is Volume III of a revised and enlarged > edition of THE STRUCTURE OF THE PHYSICAL UNIVERSE, 1959. > Volume I is NOTHING BUT MOTION (1979), and Volume II is > THE BASIC PROPERTIES OF MATTER (1988). > astrophysics are bristling with integrals, partial > differentials, and other FANCY MATHEMATICS. In this book, by > contrast, mathematics is conspicuous by its absence, except > for some relatively simple formulas imbedded in the text. > Larson emphasizes CONCEPTS and declares that mathematical > agreement with a theory does NOT guarantee its conceptual > validity. > Dewey B. Larson was a retired engineer with a Bachelor > of Science Degree in Engineering Science from Oregon State > University. He developed the Theory described in his books > while trying to find a way to MATHEMATICALLY CALCULATE the > properties of chemical compounds based ONLY on the elements > they contain. > THE UNIVERSE OF MOTION describes the astrophysical > portions of Larson's CONSISTENT, INTEGRATED, COMPREHENSIVE, > GENERAL UNIFIED Theory of the Physical Universe, a kind of > grand unified field theory that orthodox physicists and > astro-physicists CLAIM to be looking for. It is built on two > postulates about the physical and mathematical nature of > space and time: > (1) The Physical Universe is composed ENTIRELY of ONE > component, MOTION, existing in THREE dimensions, in DISCRETE > units, and with two RECIPROCAL aspects, SPACE and TIME. > (2) The Physical Universe conforms to the relations of > ORDINARY COMMUTATIVE mathematics, its primary magnitudes are > ABSOLUTE, and its geometry is EUCLIDEAN. > COMPLETE theoretical universe, from photons and subatomic > concept of INWARD and OUTWARD SCALAR MOTIONS with > translational, vibrational, rotational, and rotational- > vibrational motions. At each step in the development, he was > able to match parts of his theoretical universe with > corresponding parts in the real Physical Universe, including > EVEN THINGS NOT YET DISCOVERED. For example, in his 1959 > book, he first predicted the existence of EXPLODING GALAXIES, > several years BEFORE astronomers started finding them. They > are a NECESSARY CONSEQUENCE of his comprehensive Theory. And > when quasars were discovered, he had a related explanation > ready for those also. > As a result of his theory, which he called THE > RECIPROCAL SYSTEM, Larson TOTALLY REJECTED many of the > sacred doctrines of orthodox physicists and astrophysicists, > including black holes, neutron stars, degenerate matter, > nuclear physics, general relativity, relativistic mass > increases, relativistic Doppler shifts, nuclear fusion in > stars, and the big bang, all of which he considered to be > nothing more than MATHEMATICAL FANTASIES. He was very > critical of the AD HOC assumptions, uncertainty principles, > solutions in principle, no other way declarations, etc., > used to maintain them. > THE UNIVERSE OF MOTION is divided into 31 chapters. > It begins with a description of how galaxies are built from > the gravitational attraction between globular star clusters, > which are formed from intergalactic gas and dust clouds that > accumulate from the decay products of cosmic rays coming in > from the ANTI-MATTER HALF of the physical universe. (Galaxy > formation from the MYTHICAL big bang is a big mystery to > orthodox astronomers.) He then goes on to describe the life > cycles of stars and how binary and multiple star systems and > solar systems result from Type I supernova explosions of > SINGLE stars. > Several chapters are devoted to quasars which, according > to Larson, are densely-packed clusters of stars that have > been ejected from the central bulges of exploding galaxies > and are actually traveling FASTER THAN THE SPEED OF LIGHT > (although most of that speed is AWAY FROM US IN TIME). > Astronomers and astrophysicists who run up against > observations that contradict their theories would find > Larson's explanations quite valuable if considered with an > BURSTS originated from pulsars, which exist primarily in the > plane or central bulge of our galaxy. But the new Gamma Ray > Telescope in earth orbit observed that the bursts come from > ALL DIRECTIONS UNIFORMLY and do NOT correspond with any > visible objects, (except for a few cases of directional > coincidence). Larson's explanation is that the gamma ray > bursts originate from SUPERNOVA EXPLOSIONS in the ANTI-MATTER > HALF of the Physical Universe, which Larson calls the cosmic > sector. Because the anti-matter universe exists in a > RECIPROCAL RELATION to our material universe, with the SPEED > OF LIGHT as the BOUNDARY between them, and has THREE > dimensions of TIME, and ONLY ONE dimension of space, the > bursts can pop into our material universe ANYWHERE seemingly > at random. > Larson heavily quotes or paraphrases statements from > astronomers. In this book, 351 of them are superscripted > with numbers identifying entries in the reference list at the > end of the book. For example, a quote from the book > Astronomy: The Cosmic Journey, by William K. Hartmann, > says, Our hopes of understanding all stars would brighten if > we could explain exactly how binary and multiple stars > form.... Unfortunately we cannot. Larson's book contains > LOGICAL CONSISTENT EXPLANATIONS of such mysteries that are > WORTHY OF SERIOUS CONSIDERATION by ALL physicists, > astronomers, and astrophysicists. > A WEALTH of Information about the GENERAL UNIFIED Theory > of the Physical Universe developed by the late Physicist > Dewey B. Larson can be found at the web site > http://www.rsystem.org/isus/index.htm or /dbl/index.htm . > MOST of the TEXT of the book THE UNIVERSE OF MOTION can be > found at the web site http://www.rsystem.org/um/index.htm . > Robert E. McElwaine > B.S., Physics and Astronomy, UW-EC > http://members.aol.com/rem547 PLUS > http://members.aol.com/rem460 > P.S.: PASS IT ON ! EVERYTHING you know is WRONG. The Truth IS stranger than fiction. The Truth is ALWAYS the FIRST CASUALTY OF WAR. OFFICIAL LIES are ALWAYS the BIGGEST LIES OF ALL. === === Subject: : Re: This Week's Finds in Mathematical Physics (Week 203) >You can find some of my thoughts in these homework problem >sets for the quantum gravity seminar here at UCR. I hope to >say more later. on all of these links. The correct links are as follows: k-colorings as categorified coherent states answers: Euler's proof that 1 + 2 + 3 + ... = -1/12 answers: Bernoulli numbers answers: The Riemann zeta function answers: === === Subject: : Re: This Week's Finds in Mathematical Physics (Week 203) A random sampling of the links below indicates that they are broken. Igor k-colorings as categorified coherent states > http://math.ucr.edu/home/baez/coherent.pdf > answers: > http://math.ucr.edu/home/baez/coherent_miguel.pdf > http://math.ucr.edu/home/baez/coherent_erin.pdf > http://math.ucr.edu/home/baez/coherent_jeffrey.pdf > http://math.ucr.edu/home/baez/coherent_derek.pdf > Euler's proof that 1 + 2 + 3 + ... = -1/12 > http://math.ucr.edu/home/baez/zeta.pdf > answers: > http://math.ucr.edu/home/baez/zeta_erin.pdf > http://math.ucr.edu/home/baez/zeta_jeffrey.pdf > http://math.ucr.edu/home/baez/zeta_miguel.pdf > Bernoulli numbers > http://math.ucr.edu/home/baez/bernoulli.pdf > answers: > http://math.ucr.edu/home/baez/bernoulli_morton.pdf > http://math.ucr.edu/home/baez/bernoulli_miguel.pdf > http://math.ucr.edu/home/baez/bernoulli_erin.pdf > The Riemann zeta function > http://math.ucr.edu/home/baez/zeta2.pdf > answers: > http://math.ucr.edu/home/baez/zeta2_erin.pdf > http://math.ucr.edu/home/baez/zeta2_jeffrey.pdf > http://math.ucr.edu/home/baez/zeta2_miguel.pdf === === Subject: : probability of the linear combination of inequalities If for any vector A and its approximation A' we have Pr{|A'|^2 >= (1+c) |A|^2} <= p (1) Pr{|A'|^2 <= (1-c) |A|^2} <= p (2) Then we have Pr{|A'|^2 <= (1+c) |A|^2} >= 1-p (3) Pr{|A'|^2 >= (1-c) |A|^2} >= 1-p (4) If (3) and (4) are true, then by applying (3) and (4) to the vector h, x and h-x respectively, we have Pr{|h'|^2 >= (1-c) |h|^2} >= 1-p (5) Pr{|x'|^2 >= (1-c) |x|^2} >= 1-p (6) Pr{ -|h'-x'|^2 >= - (1+c) |h-x|^2} >= 1-p (7) Now plus the inequalities in (5), (6) and (7) together, the question is : Pr {|h'|^2 + |x'|^2 - |h'-x'|^2 >= (1-c)|h|^2 + (1-c)|x|^2 -(1+c)|h-x|^2} >= ? === === Subject: : integral help Can someone explain how to solve this integral analytically: Int( -tan(t) * sin(t) * dt ) === === Subject: : Re: integral help > Can someone explain how to solve this integral analytically: > Int( -tan(t) * sin(t) * dt ) Rewrite tan(t) = sin(t)/cos(t), and simplify the integrand. You should be able to get the whole expression in terms of a single trig function, which expression is not too difficult to integrate. Dale === === Subject: : Re: recursive to closed-form > thanks for your help. So the series for sqrt(x+1) could either be > (sum from n=0 to infinity) (-1)^(n+1) gamma(n-1/2)/(2sqrt(pi))*x^n or > (sum from n=0 to infinity) (-1)^(n+1) * (2n-3)!!/2^n *x^n ? > how were you able to find those closed form equations? thanks There is a book by Herbert S. Wilf called generatingfunctionology. It is available for free download from his web site. It was also published by Academic Press. I have read only a few pages of this book, but finding closed forms for recursively defined integer sequences is certainly one of the topics covered in this book, and the this sort of thing. === === Subject: : Re: recursive to closed-form > thanks for your help. So the series for sqrt(x+1) could either be > (sum from n=0 to infinity) (-1)^(n+1) gamma(n-1/2)/(2sqrt(pi))*x^n or > (sum from n=0 to infinity) (-1)^(n+1) * (2n-3)!!/2^n *x^n ? > how were you able to find those closed form equations? thanks Here's an approach. Start by removing the '-' in front of the n by defining g(n) = (-1)^(n+1) f(n). That should give you g(n) = g(n-1) * (n - 3/2). Then compare with the Gamma function, gamma(n) = (n-1) * gamma(n-1) you should see something like g(n) / g(0) = gamma(n-1/2) / gamma(-1/2). That brings you close to Robert's answer. Going back to your original problem (1+x)^(1/2) you can just expand using the binomial theorem to get sum x^n * binom(1/2, n). === === Subject: : International Journal of Mathematics - Vol 15 No 1 International Journal of Mathematics View table-of-contents and abstracts at http://www.worldscinet.com/ijm.html Contents: Compactness Of Certain Families Of Pseudo-Holomorphic mappings into ${mathbb C}^n$ Herve Gaussier And Kang-Tae Kim Rational Families Of Vector Bundles On Curves Ana-Maria Castravet Maximal Coactions Siegfried Echterhoff Covering Dimension And Quasidiagonality Eberhard Kirchberg And Wilhelm Winter Hiroshi Yamauchi For more information, go to http://www.worldscinet.com/ijm.html === === Subject: : Re: GCD for Rational Numbers > So, yes, it is pretty well known, but calling it a GCD is a rather > breathtaking abuse of language. Let define GCD(a/b, c/d) = GCD(a,c)/LCM(b,d) LCM(a/b, c/d) = LCM(a,c)/GCD(b,d) It's easy to verify that they are distributive GCD(m/n * a/b , m/n * c/d) = m/n * GCD(a/b, c/d) associate, commutative and satisfy absorption law LCM(a/b, GCD(a/b, c/d)) = a/b (These are properties from Mathworld page). I seem to have trouble with Knuth identity, however, GCD(2^a/b-1, 2^c/d-1) = 2^GCD(a/b,c/d) -1 as it seem to require extending GCD definition to radicals! === === Subject: : Re: GCD for Rational Numbers Adjunct Assistant Professor at the University of Montana. [.snip.] >Looks to me like you are finding the common measure of the two >rationals of smallest height (the largest common measure, perhaps): >the denominator is the lcm of the denominators of a and b (when >written in lowest terms), and the numerator is the gcd of the >numerators of a and b (when written in lowest terms). In particular, >it measures both a and b (that is, there is an integer multiple of it >which equals a and equals b), and has the smallest of max{|p|,|q|} >(when written as p/q) of all numbers that measure both a and b. In fact, it is simply the largest common measure of teh two rationals; writing a = p/q and b=r/s with gcd(p,q)=gcd(r,s)=1, then any common measure will be of the form a/k=b/m for some integers k and m. Thus, its reduced numerator will be a divisor of both p and r, hence of the gcd of p and r; and the reduced denominator will be a multiple of both q and s, hence of their lcm. So any common measure of the form u/v, with gcd(u,v)=1 will have |u|<= gcd(p,r) and |v|>=lcm(q,s), and so will be smaller than the common measure found. >So, yes, it is pretty well known, but calling it a GCD is a rather >breathtaking abuse of language. Try searching for common measures, >or look up however it was that Euclid called it. >Thank you, Arturo, googling common measures indeed revealed pretty >elementary topic behind it. But what if I change min into max in the >above definition? (Naturally, I'm thinking of LCM as dual to GCD;-) Does >this concept also have a common name that I'm not aware of? If you change to lcms, then from p/q and r/s you get the rational that has gcd(q,s) as denominator and lcm(p,r) as numerator. This would be the smallest rational that has both a and b as measures (the smallest rational measured by both a and b). For suppose that the rational u/v is measured by both a=p/q and b=r/s. Then ka = mb = u/v for some integers k and m. The reduced form of ka will have a denominator that divides q and numerator which is a multiple of p; and the reduced form of mb will have a denominator that divides s and a numerator that is a multiple of r. Thence u/v will have u a multiple of lcm(p,r) and a dneominator a divisor of gcd(q,s). So any rational measured by both a and b will be of the form u/v with lcm(p,r)<=u and gcd(q,s)>=v, hence lcm(p,r)/gcd(q,s) will be the smallest such rational. ======== === === Subject: : Re: GCD for Rational Numbers > In fact, it is simply the largest common measure of teh two rationals; > writing a = p/q and b=r/s with gcd(p,q)=gcd(r,s)=1, then any common > measure will be of the form a/k=b/m for some integers k and m. Thus, > its reduced numerator will be a divisor of both p and r, hence of the > gcd of p and r; and the reduced denominator will be a multiple of both > q and s, hence of their lcm. So any common measure of the form u/v, > with gcd(u,v)=1 will have |u|<= gcd(p,r) and |v|>=lcm(q,s), and so > will be smaller than the common measure found. ..... > If you change to lcms, then from p/q and r/s you get the rational that > has gcd(q,s) as denominator and lcm(p,r) as numerator. This would be > the smallest rational that has both a and b as measures (the smallest > rational measured by both a and b). > For suppose that the rational u/v is measured by both a=p/q and > b=r/s. Then ka = mb = u/v for some integers k and m. The reduced form > of ka will have a denominator that divides q and numerator which is a > multiple of p; and the reduced form of mb will have a denominator that > divides s and a numerator that is a multiple of r. Thence u/v will > have u a multiple of lcm(p,r) and a dneominator a divisor of > gcd(q,s). So any rational measured by both a and b will be of the form > u/v with lcm(p,r)<=u and gcd(q,s)>=v, hence lcm(p,r)/gcd(q,s) will be > the smallest such rational. Let's call proposed extension for rational numbers as GCDR (LCM, correspondingly). Are you suggesting that GCDR(a/b, c/d) = GCD(a,c)/LCM(b,d) GLCM(a/b, c/d) = LCM(a,c)/GCD(b,d) ? This clearly doesn't fit my definition. Consider GCDR(87/22,9/5)= =GCDR(2^-1*3^1*11^-1*29^1,3^2*5^-1)= =2^min(-1,0)*3^min(1,2)*5^min(-1,0)*11^min(-1,0)*29^min(0,1)= =2^-1*3^1*5^-1*11^-1=3/110 whereas GCD(87,9) = 3*29 LCM(22,5) = 110 === === Subject: : Re: GCD for Rational Numbers > GCD(87,9) = 3*29 ---------------^^^^^^^^^^ typo > LCM(22,5) = 110 Your definition is correct. Sorry about that. === === Subject: : Communications in Contemporary Mathematics - Vol 6 No 1 Communications in Contemporary Mathematics View table-of-contents and abstracts at http://www.worldscinet.com/ccm.html Contents: Geometric-Optics For Nonlinear Concentrating Waves In Focusing And Non-Focusing Two Geometries Slim Ibrahim Special Lagrangian Cycles And Hermitian Yang-Mills Connections Jingyi Chen Vertex Algebras And Vertex Poisson Algebras Haisheng Li Critical Points Of Master Functions And Flag Varieties E. Mukhin and A. Varchenko Global Regularity For A Singular Equation And Local H1 Minimizers Of A Nondifferentiable Functional Juan Davila Vertex Operator Representations Of Quantum Tori At Roots Of Unity Y. ig and K. Zhao For more information, go to http://www.worldscinet.com/ccm.html === === Subject: : Form of Lagrangian in the calculs of variations. I'm reading the first edition of Mechanics by Landau et al, published in 1960. Just before equation 3.1 on page 5 it says exactly this: Since space is isotropic, the Lagrangian must also be independent of the direction of v, and is therefore a function only of its magnitude, i.e. of v(bold)^2 = v(italic)^2: L = L(v(italic)^2) (3.1) He seems to be saying that L(sqrt(v(bold)^2) implies that L'(v(bold)^2) must also exist. Which seems reasonable since sqrt(x) is monotonic. Is this correct? === === Subject: : Re: Least square soln of affine transform, t is given by centroid difference ? >It is intuitive to say that, for 2D affine transforms, >X' = AX + t >where X' and X are both N by 2 vectors, N being the number of 2 >points, >the best estimate of t is given by the difference vector between the >centroid of the two point sets X and X'. >But I cannot see this from the least square solution, although I think >it can be proven. I handle a generalization of this regression at http://www.whim.org/nebula/math/affineregress.html Perhaps you might find what you are looking for there. Rob Johnson It is intuitive to say that, for 2D affine transforms, > X' = AX + t > where X' and X are both N by 2 vectors, N being the number of 2D points, > Your equation is, though understandable, not quite correct, > because X' and AX are N by 2 matrices while t is just an 1 by 2 vector. > You'd better write > X' = A X + eT t > where eT is the N by 1 vector whose entries are all equal to 1. > the best estimate of t is given by the difference vector between the > centroid of the two point sets X and X'. > Unfortunately, this statement is wrong. > What is true is that the best estimate of t is given by > the difference vector between the centroid of the point set X' > and the image of the centroid of the point set X under the linear map A: > t_opt = 1/N * e * (X' - A X) > where e = [1 1 1 ... 1] is the transpose of eT. I see what you are saying. This is correct. But how do you show my statement is wrong? If my statement is correct, it will be a further step of yours. It seems that your statement is a first step toward my statement. === === Subject: : Re: Least square soln of affine transform, t is given by centroid difference ? >It is intuitive to say that, for 2D affine transforms, >X' = AX + t >where X' and X are both N by 2 vectors, N being the number of 2D points, >Your equation is, though understandable, not quite correct, >because X' and AX are N by 2 matrices while t is just an 1 by 2 vector. >You'd better write > X' = A X + eT t >where eT is the N by 1 vector whose entries are all equal to 1. >the best estimate of t is given by the difference vector between the >centroid of the two point sets X and X'. >Unfortunately, this statement is wrong. >What is true is that the best estimate of t is given by >the difference vector between the centroid of the point set X' >and the image of the centroid of the point set X under the linear map A: > t_opt = 1/N * e * (X' - A X) >where e = [1 1 1 ... 1] is the transpose of eT. > I see what you are saying. This is correct. But how do you show my > statement is wrong? If my statement is correct, it will be a further > step of yours. It seems that your statement is a first step toward my > statement. Let me try to disprove your statement: Let's givt names to the centroids: c for the centroid of X and c' for the centroid of X' My statement is that t = c' - A c while you claim t = c' - c which would only be compatible with my statement in case A c = c. However, the matrix A can be computed as the matrix which minimizes the (length of the) difference Y' - A Y, where Y and Y' denote the point sets X and X' translated by c and c', so that the centroids of Y and Y' both become the origin 0. So, if we start with an arbitrary matrix A that is not the identity matrix and a point sets Y with centroid lying at the origin, then look at the image of this point set under A: Y' := A Y (for which the origin is also the centroid), then choose a vector c such that A c is not equal to c and set X := Y + eT c X':= Y', we obtain c' = 0 t = c' - A c not equal to c' - c which disproves your claim. By the way, I just realized that I partly mixed up column and row vectors in my arguments, but this shouldn't matter... === === Subject: : Euclid Algorithm for GCD(a/b,c/d) GCD(87/22-2*9/5) = ? 87/22-1*9/5 > min (87/22, 9/5) 87/22-2*9/5 = 39/110 < min (87/22, 9/5) 9/5-1*39/110 > min (9/5, 39/110) 9/5-2*39/110 > min (9/5, 39/110) 9/5-3*39/110 > min (9/5, 39/110) 9/5-4*39/110 > min (9/5, 39/110) 9/5-5*39/110 = 3/110 < min (87/22, 9/5) 39/110-1*3/110 > min (39/110, 3/110) 39/110-2*3/110 > min (39/110, 3/110) ... 39/110-13*3/110 = 0 -- stop GCD(87/22-2*9/5) = 3/110