mm-2479 === Subject: differentiable structure on S^2 using spherical coordiates sphere in R^3 F(theta,phi) = (sin(phi)cos(theta), sin(phi)sin(theta), cos(phi)) I am trying to use this map to define a differentiable structure on S^2. I tried taking those subsets of R^2 that would map to some open hemispheres that cover S^2. The problem is that some of those maps will not be injective. In particular, I can't seem to cover the north and Can this map be used to define a differentiable structure on S^2? If so how can we cover the sphere with injective images of F? === Subject: Re: basic algebraic geometry question Yes, the definition of algebraic dependence or independence applies to any pair of elements (actually any set of elements) of the extension field. So in particular, the inclusions Q contained in Q[x,y] contained in Q(x,y) allows us to talk about the algebraic dependence or independence of a pair of elements of Q[x,y] over Q. To say that f,g in Q[x,y] are algebraically dependent over Q means precisely that there exists a nonzero polynomial p in Q[x,y] such that p(f,g) is identically 0. quasi === Subject: good finanical math/engineering forums? Hi all, Can anybody recommend some good financial math/engineering forums and === Subject: Re: good finanical math/engineering forums? === Subject: Re: Proof that 1 is the only Odd Perfect Number and thus the oldest math conjecture is conquered DON'T CALL IT A PERFECT NUMBER! CALL IT SOMETHING ELSE! Your definition is not standard, so you should not use standard terminology. --- Christopher Heckman === Subject: Re: Proof that 1 is the only Odd Perfect Number and thus the oldest math conjecture is conquered DON'T CALL IT A PERFECT NUMBER! CALL IT SOMETHING ELSE! Your definition is not standard, so you should not use standard terminology. So tell me, where do you have any room for trashcanning a definition found to have a flaw. And what counts as standard-- used for a thousand years, whether flawed or not. Seems to me, Chris, you do not look at math flexible enough, but rather a dogmatic closed minded attitude. I pointed out to you that the standard definition of Perfect Number is flawed in that it treats the number 1 as not perfect. Plus the fact that this standard definition is asymmetrical in that it treats numbers such as 1,4,9,16,25, etc etc as differently from all other numbers. A definition should treat all numbers uniformily. Divisor Pairs is a symmetrical definition for Perfect Number. So where is there any room in your career in mathematics for the idea that a definition can be flawed and needs to be trashcanned. You should not treat mathematics, Chris, as if it were religion. Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: Proof that 1 is the only Odd Perfect Number and thus the oldest math conjecture is conquered No language is perfect (no pun intended), but you can't change a language by simply redefining terms to your own taste, not if you want to be understood. The term perfect number is already in use, so you can't redefine it. If you think that a better concept is to allow duplicate factors for squares, and to include 1, then fine, but then you need to make up your own term. You can use any term that's not already being used. Be creative -- think of a good name, but you can't steal the existing name. Also, if you want to prove a conjecture about perfect numbers, you proof must somehow apply to the standard definition otherwise you haven't proved the existing conjecture. For example, if someone decides to redefine the term prime to include the standard primes (2,3,5,7,11,13, ...) as well as all the negative integers (-1,-2,-3,-4,...), then it's trivial to express any integer as a sum of 2 primes, but of course, it would not really qualify as a proof of the Goldbach conjecture since the Goldbach conjecture refers to standard primes. Bottom line -- do _not_ redefine existing standard terms -- it's a deal breaker. If you persist, people following the thread as well as those such as Chris who are working hard to clarify/verify/debunk your proof will probably just give up on you. quasi === Subject: Re: Proof that 1 is the only Odd Perfect Number and thus the oldest math conjecture is conquered The change is not for taste. A definition should measure every number to the same standards. The old perfect number definition does not do that. It measures these numbers of 1,4,9,16,etc etc differently from the measure of all other numbers. A definition should never do this. That is why I call it asymmetrical. For 10 all the divisor pairs are included, for 9 you omit the divisor of 3. This is important for those who study the deficiency or abundance in perfect numbers. None of the study of abundance or deficiency is of use with the old definition of perfect because it is random and not a uniform definition. To be a divisor of a number means it is a member of a divisor pair, so why count pairs for some numbers and neglect pair members in other numbers. Can you see how that is a randomness. Such a definition is random and ill-defined. Yes I can, because if something is flawed, intelligent people discontinue the use of a flawed definition. Surely you do not believe that definitions of mathematics are immutable. You are forgetting something. Randomness is a flaw. I am trying to tell you that the definition is flawed and you seem to not be cognizant of that. Redefining is a different act from that of pointing out a flaw in a definition. A week or so ago, you posted that you expected me to admit when I was wrong. Well I admitted that my Goldbach had a flaw. But it seems that you and Chris are now unable to reciprocate when you are found wrong. The definition is flawed and your reaction should not be to call me a redefiner but to inspect the claim of flaw. Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: Proof that 1 is the only Odd Perfect Number and thus the oldest math conjecture is conquered Trashcanning!?!?!? I've never said your definition was BAD; I just objected to you calling a number perfect if it satisfies that definition. Is 2 + 2 = 6 a true statement? Yes, if 2 has been redefined as three. If you didn't know that, you would say that 2 + 2 = 6 is false. Do you see the analogy? Think before you type. Yes. The fact is this: If ANY mathematician who speaks English reads the sentence: N is a perfect number, they will NOT be using your definition. On the other hand, if you say a number is AP-perfect if it satisfies your definition, then NO ONE will have trouble with the setence N is an AP-perfect number. The point is that the original definition of perfect, even if flawed, is still used. (You seem to be the only person who has though it was flawed, though, so no one's bothered changing the definition.) Did you get the point of my modern language that I used in your post? It is only flawed in your opinion. You're entitled to your own opinion, but you don't speak for the mathematics community as a whole. It's the difference between facts and opinion. It's good that you're calling them Divisor Pairs ... Now you just need to come up with a word other than perfect to describe when the sum of the Divisor Pairs of n is n. If you can come up with a term which is not used in mathematics, that will work fine. I've been suggesting AP-perfect (to show that it's based on your idea of Divisor Pairs), but you could also say that n is divisorly perfect or perfect in divisors if the sum of the Divisor Pairs of n is n. See? It's not hard; you could probably brainstorm a dozen more possiblities. And it will end all this pointless discussion. No, but you should treat it more like language, and the idea of language is to communicate. If you use words in a non-standard way, it will obstruct that communication. If I wanted to let you know I'm thinking Go work on your result, I wouldn't say, Go yourself, now, would I? No, not if I want to communicate an idea. --- Christopher Heckman === Subject: Re: Proof that 1 is the only Odd Perfect Number and thus the oldest math conjecture is conquered Good. Just make sure that when the Summation of the Divisor Pairs of n is n, that you don't call n a perfect number. Because when mathematicians see the phrase perfect number, they'll assume you're using the standard definition. True. And this is a statement you can include in your write-up. You shouldn't really say ill-defined in your write-up; a better thing to say would be that the old definition makes your proof harder. AP-Perfect, not perfect. Now, we get to the criticism of the main proof: It's not valid. If you add up an even number of fractions whose numerator and denominator are both odd, the sum will have an even numerator. For instance, 1/9 + 9/9 + 3/9 + 3/9 = (1/9 + 9/9) + (3/9 + 3/9) = (10/9) + (6/9) = 16/9, an even number being in the numerator despite all the original fractions being odd/odd. Can you provide some examples in the literature saying that 1 is a perfect number? If a large proportion says 1 is perfect, then you should be able to give me 2 or 3 examples. --- Christopher Heckman === Subject: Re: Proof that 1 is the only Odd Perfect Number and thus the oldest math conjecture is conquered I have more faith in that mathematicians will see the logic for a revamping of the definition. Of course some will resist, but most mathematicians are built of a stock of good logic and reasoning and will embrace the revised definition. So I have more faith than you do. Indeed it makes my proof harder and I would have to treat numbers of 4,9,16,25 etc etc as a special case. But it still renders the proof since this sequence is impossible to have a odd perfect number. But the important thing is to clean up the mess, rather than offer a proof with the mess remaining. I would rather fight to clean up the mess than to be credited with a proof with the mess still intact. Sorry Chris, but you seem to have not comprehended the argument. 1/9 + 9/9 + 3/9 + 3/9 is 16/9 and to be perfect requires 18/9 so there is missing a term of 2/9 and so a numerator has a even number which is impossible to divide into an odd number. All odd numbers (except 1) will have a missing term with an even number in the numerator in order to be perfect. Give me any odd number, I will add up the Divisor Pairs and at the end they will fall short of being perfect because there is missing a term with an even number in the numerator. Well I include every statement that says words to the effect Besides 1, no odd perfect number or statements such as 1 is the only odd perfect number I used to read a number of popularized math books in the early 1990s where I ran across such statements. So apparently quite a number of people believe 1 is perfect. I sold my math books when I moved out West. Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: Revisiting my (alleged) proof of the Goldbach Conjecture Yes. This is the normal sigma(n) function. If 9 is a divisor, then 3 and 1 will also be divisors automatically, so they will be listed. The standard definition is to look at the set of divisors, so repetition isn't an issue. --- Christopher Heckman === Subject: Re: Revisiting my (alleged) proof of the Goldbach Conjecture Yes. It involves finding the planar dual of the map. The dual of a map is a graph G, where each region is represented by one vertex in G. The edge uv between u and v exists if the regions corresponding to u and v share a non-trivial boundary (an actual line segment). Then k-coloring the map is the same as k-coloring the graph G. --- Christopher Heckman === Subject: Re: Revisiting my (alleged) proof of the Goldbach Conjecture Yes, he should have struck out the first sentence. Not necessarily. (Notice I did not say NO.) THE DIFFERENCE between whether there are borders or not IS VITAL TO THE ANSWER. If you went up to anyone who knows about the problem, even a superbeing like God, and asked, How many colors are needed to color the regions in a map?, you would get some questions asking you to specify whether the boundaries are to be colored, how to handle pie charts, etc. The Pythagorean Theorem is not true in spherical geometry. (Spherical geometry is used for navigation, so you can hardly call it psychology.) Consider the triangle running from the north pole south along the prime meridian to the equator, west 90 degrees, and north 90 degrees to the north pole. This is a right triangle, and the sides have lengths C, C, and C, for some constant C. But C^2 + C^2 is not C^2. The Pythagorean Theorem has over 50 proofs. Certainly not all of them use the same technique. You forgot to leave out the word not in that last sentence. --- Christopher Heckman === Subject: Re: Revisiting my (alleged) proof of the Goldbach Conjecture The statement of a problem can change what the final answer turns out to be. That's something I've mentioned before, and it can turn the 4CT into the 2CT (where you include borders), or a 6CT (if the regions don't behave nicely), or an n-CT (if you change what it means for two regions to share a boundary). Changing the basic assumptions in a problem will create a new problem, or change the answer. In Euclidean geometry, there is an axiom called the Parallel Postulate. This statement seems reasonable (For every line and every point not on that line, there is a line through the point parallel to the original line), but it is not provable using the other assumptions of geometry. This axiom can be used to show that the sum of the angles in a triangle is 180 degrees. If you change the assumption (to For every line and every point not on that line, there are no lines through the point parallel to the original line or For every line and every point not on that line, there is an infinite number of lines through the point parallel to the original line), then this affects what the sum of the angles in a triangle is. If you do geometry on a sphere, you're accepting the first version, and the sum is greater than 180 (consider the equator, prime meridian, and 90 degree W line on a globe; the 3 angles sum to 270). If you do geometry in another way, with the assumption that there are an infinite number of parallel lines through the point, you find out that the sum of the angles in a triangle is less than 180 degrees. Non-Euclidean geometry was philosophy (not psychology, as you stated), in that it had no real application ... until Einstein came along and said that the universe really isn't governed by Euclidean geometry, but one of the non-Euclidean geometries that mathematicians had been playing around with; do you thus ignore General Relativity for the same reason you refuse to see variations of problems as new problems? There are other situations where maybe you generalize a certain set of results and show that these results are all basically the same idea. Group Theory, for isntance, is based on modular arithmetic in different moduli. If the borderlines are thin enough, then the same result happens. It's the circular shape that's important. It depends on the problem. If the problem becomes ridiculously easy (like including the borders on a map, or saying the slices of a pie chart all have to be colored differently), the answer is no. If some new idea is needed to solve the new problem, the answer is yes. Generally the mathematics community as a whole determines which problems are the ones to go after. (Although NONE would accept a proof of a trivial result.) You could ask questions like: Does the proof of Theorem A really depend on Theorem B, or can it be done without Theorem B? Finding quick, clever proofs of lots of generalized results is often done this way. No, but in a proof where you don't use the JCT, something else will take its place. The house will still stand. Appel and Haken proved a different result than the one you did. The results are compatible, and mathematics only grows as a result of solving problems. The proving of more results, and the solving of new problems, still belongs in mathematical territory. Maybe you're thinking of mathematics as what's called Applied Mathematics, and psychology land as Pure Mathematics (without any immediate application). People go into Pure Mathematics to find out The Truth. And tomorrow, a Pure Mathematics result can How about looking at maps on other surfaces, like on the torus (the skin of a doughnut)? --- Christopher Heckman === Subject: Re: Revisiting my (alleged) proof of the Goldbach Conjecture That (parallel postulate in Euclidean axioms) is not an example of where a theorem of mathematics ignores Domain Elements and comes up with a new theorem. For those that believe in the 4CM such as Chris, then the burden is on these folk to provide us with a example of mathematics where a ignoring of domain elements leads to a new theorem of mathematics. The 4CM is the only alleged math theorem that is a theorem built on the ignoring of domain elements. It is unique. That that implies it is a phony theorem, a fakery. For mathematicians of the future to accept the 4CM, we will have to be shown that you can take a theorem, ignore domain elements and bring forth a new true theorem. You will have to provide some logical reasoning why this can be done. What the 4CM truly is, is a idea of Sensory Eye Perception, that we can more easily look at a map if there are four colors. Senses and perception is not mathematics. And the reason that the alleged proof of 4CM by Appel and Hakken is this dreadful computer proof is because it is not mathematics in the first place-- it is sensory eye perception. The 4CM is so vastly different from all other theorems of mathematics, is because, well, it is not mathematics. For staunch believers of the 4CM, the burden is on them to show us where in mathematics is another example of where you can ignore domain elements and have a mathematical statement worthy of a proof. And the burden is on them to provide us with some convincing argument that the deletion or ignoring of domain elements is logically reasonable. Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: Revisiting my (alleged) proof of the Goldbach Conjecture No, but the theorem that states the sum of the angles in a triangle IS. (You really need to read more carefully, you know.) I did, and you misunderstood it. (By the way, it's the Four Color _Theorem_, usually abbreviated 4CT.) Not true, either. See above. Changing the hypothesis is common in mathematics, and leads to proofs using other techniques. I guess AP would rather mathematics stay in the Dark Ages. Nope, not if you're colorblind. True, mathematics can be done without being able to see or perceive, because everything is a priori. Well, you've managed to come up with a new argument against the Appel-Haken proof, one which hasn't shown up in the last nearly-30 years, and I have to applaud you for that. Unfortunately, it's not a valid point. The basic issue (that you seem to keep ignoring) is: If you change the problem, you can change the answer. If x^2 = 0, then x = 0 (and no other values). If x^2 + 2x + 1 = 0, then IT IS NOT THE CASE that x = 0 (and no other values). AP would like you to think that by changing the problem, the answer stays the same, or that the new problem is not worth solving. In that case, you'd better tell Neil Robertson, Daniel Sanders, Paul Seymour, and Robin Thomas, because they've also been deluded, in addition to Appel & Haken. I have, above. Well, by your logic, your proof of the 2MT is incorrect because it doesn't use Chebychev Theorem, which is a true statement. --- Christopher Heckman === Subject: Re: Revisiting my (alleged) proof of the Goldbach Conjecture Just as easy as it is to adopt a definition, it is just as easy to trashcan it when found to have flaws. My new definition of Perfect Numbers counts divisor pairs and thus 1 is perfect. And my definition is completely symmetrical in that 9 has divisor pairs of 3,3 whereas the proper divisor definition counts only one 3 for 9. So at some point in time, people who study these things are going to say the old definition is seriously flawed and throw it away. And then they adopt the new definition of divisor pairs. Ken, definitions are not eternal unchanging entities. Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: Revisiting my (alleged) proof of the Goldbach Conjecture But no one who has worked with it has actually done this ... What does that suggest to you? Perhaps that you're the only one that considers it flawed? But why get rid of a definition when you can add one of your own: AP-perfectness, pefectness in divisors, divisorly perfect, etc.? --- Christopher Heckman === Subject: Re: Revisiting my (alleged) proof of the Goldbach Conjecture Why is it so important TO YOU that 1 be a perfect number? You seem to be the only person who worries so much about 1. I could argue that a different definition should be used so that 23 becomes a perfect number, but I would expect to be asked why I would want it that way, as well. Yes, because (1) {1, 2, 4} are the positive factors of 4, and (2) the sum of the elements in {1, 2, 4} is 7. Points (1) and (2) are consistent with the rest of mathematics. You could have shortened that last sentence to If n is not a perfect square, n is perfect exactly when n is AP-perfect (you can use this instead of my definition of perfect number). A perfect square may fail to be perfect and AP-perfect (1 + 3 = 4, 1 + 3 + 3 = 7, neither is equal to 9), or may be one and not the other. No, this is not true! 6 is a factor of 12, and is not less than or equal to sqrt(12). It is NOT ill-defined. The definition of perfect numbers is clear-cut, and it is straightforward to determine whether a number is perfect or not. Now, you can INTRODUCE new definitions in proofs, but don't use the same terminology that already exists. No one would have a problem if you introduced your new idea as AP-perfect numbers instead of a new definition of perfect numbers, and pointed out that this is done to make your proof go through. Once you've done this, and proven your result (actually the proof is flawed, but that's a discussion in another part of the thread), you've shown that (1) There are no odd AP-perfect numbers, except for 1. You can then relate your result to the traditional definition by pointing out that (2) For non-squares n, n is perfect iff it is AP-perfect, and (3) There are no perfect odd squares (which has been proved a long time ago). Then (1), (2), and (3) WOULD prove that there are no odd perfect numbers, according to the old definition, so you've taken care of BOTH definitions. But you need to take the extra step to show that what you've done actually relates to the problem as defined by the rest of the mathematical world. --- Christopher Heckman === Subject: Re: Revisiting my (alleged) proof of the Goldbach Conjecture It is NOT ill-defined. The definition of perfect numbers is clear-cut, and it is straightforward to determine whether a number is perfect or not. Come on, you have been in mathematics long enough to know that if a definition treats the number 10 differently then the number 9, then the definition is ill-defined. Your standard definition of Perfect Number says that 10 is 1 + 10 + 2 +5 because 1 is the divisor pair to 10 and 2 is the divisor pair to 5. Now your standard definition treats 9 differently for it says that 9 is 1 + 9 + 3. It neglects, ignores and deletes the divisor pair of 3 to that of 3. So your definition is phony and ill-defined. A definition must treat all numbers in the same manner. The standard definition is asymmetrical and thus ill-defined. It matters not how loud you scream that it is standard. Face facts, it is flawed. Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: Revisiting my (alleged) proof of the Goldbach Conjecture You've just shown that you know NOTHING about what it means for something to be ill-defined. (That is, your concept of ill-defined is ill-defined.) A function (for example) is well-defined if it has the property that if x = y, then f(x) = f(y). An ill-defined function would be one that does not have a consistent value for f(x). For instance, the function f(x) is a number y such that y^2 = x is ill-defined, because f(4) could be 2 or -2; the definition does not result in one unambigous value for f(4). --- not _my_ standard definition, _the_ standard definition (I didn't make it up, stop saying I did) --- No, it's consistent with itself. If there were any serious problems, the definition would have been removed a long time ago. (You never hear about aether in physics, for this reason.) The factors of 10 (other than 10) are {1, 2, 5}, and 10 is not perfect because the sum of the elements of this set is not 10. The factors of 9 (other than 9) are {1, 3}, and 9 is not perfect because the sum of the elements of this set is not 9. The standard definition of perfectness does: The factors of N (other than N) are [SOME SET], and N is or is not perfect, depending on whether the sum of the elements of this set is N. No special cases for N, so the definition treats all numbers the same. ... in your opinion. YOU DON'T HAVE TO USE IT, though, so why are you whining about it? It's not just me; it's the entire mathematics community. Since you come from the physics community, maybe you'll understand it this way: What would you do if someone came up with a new definition for the meter, saying the old one was flawed but not giving a good reason for the new one? Would you change over just because it's new -- which requires rewriting textbooks and creating different-sized rulers -- or say, We're going to stick with the old definition ? Then it is up to you to PROVE it's flawed. And, looking at the last three proofs you've posted, you're not up to the task. But like I say, Ludwig, YOU DON'T HAVE TO USE THE STANDARD DEFINTION. All you have to do is to SAY you're not using it, and USE SOME OTHER TERM other than the one used for the definition: AP-perfect P-perfect divisorly perfect perfect in divisors paired perfect square-perfect (since only squares are treated differently) 1-perfect (since it makes 1 perfect) n^2-perfect ... You could surely have come up with a suitable term in the time it took for you to bang out an un-thought-out reply. --- Christopher Heckman === Subject: How to evaluate this integral using pencil and paper? int(1/(cos(x)-cos(y), w.r.t x, x from 0 to y-epsilon) === Subject: Re: How to evaluate this integral using pencil and paper? use your pencil to write an order form for a computer. === Subject: Re: How to evaluate this integral using pencil and paper? Hint: 1/(cos(x) - cos(y)) = = 1/(2tan((x + y)/2)sin(y)) - 1/(2tan((x - y)/2)sin(y)). Jose Carlos Santos === Subject: Re: Invariant polynomials I'm very puzzled by that error message. Was your A not a Matrix? There were, however, typos in my code: I left off a ] in the lines for eqsA and eqsB. Also, in some cases you may need to expand dA and dB. And, it seems, for Maple 8 you need to make the list of coefficients into actual equations. Try the version below. I've made it into a procedure that accepts a positive integer n and a set or list of generators, each of which is a 4 x 4 Matrix. invariants:= proc(n::posint, S::{set(Matrix),list(Matrix)}) local b,d,A,f,fA,dA,eqsA,MA,R, i, j, k, l, IB; b:= [seq(seq(seq(seq(x^i*y^j*z^k*w^l, l=0..n-i-j-k),k=0..n),j=0..n),i=0..n)]: d:= nops(b): f:= unapply(add(c[i]*b[i],i=1..d),x,y,z,w): R:= NULL; for A in S do dA:= fA - f(x,y,z,w): MA:= LinearAlgebra:-GenerateMatrix(eqsA,[seq(c[i],i=1..d)])[1]: R:= R, LinearAlgebra:-NullSpace(MA); od: IB:= LinearAlgebra:-IntersectionBasis([R]); end; For example: invariants(3,[A,B]); 2 2 2 2 2 2 2 3 {-1, y w + y z + x y, w + z + x , y, y , y } Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Invariant polynomials I put with Maple 9.5. In any case your new precedure works perfectly. === Subject: Re: revisiting my old alleged proof of Infinitude of Perfect Numbers This one has mistakes in it, too. You are on thin ice here. In some sense, what you've said is true, but it's also true that the limit of N!/2^N gets arbitrarily large as N gets big. More obvious mistakes occur later. No, the sequence is not endless; it only runs from N!-N to N!-2. It can be _arbitrarily long_, though, but for each N, the sequence is finite. 2^n at n=infinity is a meaningless expression. For instance, what is the last digit in 2^infinity? (The Adics don't have prime numbers, so 2^n would have to be an integer, not an adic.) Also, you probably didn't know it then, but M has to be an integer, so it can't be infinity. This is where the camel's back gets broken. You've got the same problem as in your alleged proof of GC: The set S(N) = {N!-N, ..., N!} is not guaranteed to contain a prime number, unless there is some M such that {M, M+1, ..., 2M} is a subset of S(N); if such an M exists, then you CAN use Chebyshev's Theorem. But this would require that is, N! - N <= M <= N, so N! <= 2N. This is only true if N is 0, 1, 2, or 3, and N!-1 a prime only when N=3, a far cry from all N. S(0) = {1}, no M works S(1) = {0, 1}, no M works S(2) = {0, 1, 2}, M=1 works ({1,2} is a subset of S(2)) S(3) = {3, 4, 5, 6}, M=3 works ({3,...,2*3} is a subset of S(3)) This is another place where a warning flag should have gone up. If you do succeed in finding N and n such that N! - 1 = 2^n - 1, then you know that N! = 2^n as well. But N! is N <= 2. This means that (N,n) is one of (0,0), (1,0), or (2,1), only a finite number of solutions, as opposed to the infinite number that you claimed. (Even if you tack on (infinity, infinity), this equation still has a FINITE number of solutions.) I haven't read any further, since you claim the same method works for what follows. The IPN proof is busted, and you have to head back to the drawing board. --- Christopher Heckman === Subject: Re: revisiting my old alleged proof of Infinitude of Perfect Numbers Yes, I know I made alot of mistakes about that. Even believing for a moment that it would be difficult to find a number with a variance of 2025 when it can be found in a few minutes. But who cares how many mistakes are made, just as long as a few good ideas are discovered. Well it is sad, Dik, that you respond to a post I made earlier today for which I had abandoned by the time you responded to it. An active moving mind is the sign of a very healthy mind. Dik, I could use your help in tackling this Infinitude of Perfect Numbers. What I am going to try to do is set up the Factorial N! and then focus on that prime-free sequence of numbers that of N!-N to N! Numbers --- notice that in the construction that the number N!-1 is the only prime number candidate for the consecutive numbers of N!-N to N!, until you check what this number actually is, it is unknown whether N!-1 is prime or composite. This number N!-1 will prove the infinitude of perfect numbers. Numbers of the form N!-1 are infinite since the factorial sequence is infinite. B) Using the theorem that between M and 2M, there has to exist at least one prime number. Thus between 2n and 2n+1 there has to exist at least one prime number. C) The set of primes are infinite. 2^n (2^n) - 1 N! N-1 prime-free numbers 2^1=2 1 1!=1 2^2=4 3 2!=2 2^3=8 7 3!=6 2^4=16 4!=24 20,21,22 2^5=32 31 5!=120 115,116,117,118 --- end quoting --- Dik, I want to focus on those consecutive prime free numbers from N!-N to N! so as the number N gets larger so does the number of consecutive prime free numbers. So I have composite consecutive numbers of a,b,c,d,e,f,g,h,i,j,k for a specific N and the only prime candidate is k, and k has the form of the (2^n)-1. So then I would like to slap onto this sequence the Chebyshev Theorem forcing me to say that k is prime because it is between M and 2M. Dik, do you see any hardship in this scheme of things? Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: revisiting my old alleged proof of Infinitude of Perfect Numbers If the first number of your sequence is N!-N, and the last one is N!, then Chebyshev provably won't say that one of the numbers in that sequence is prime, because there's no way to fit the interval [M,2M] inside of the interval [N!-N,N!] unless N is small (3 or less -- see the thread with your alleged GC proof). In fact, N!-1 can even be composite: 5!-1 = 119 = 7 * 17. --- Christopher Heckman === Subject: Re: revisiting my old alleged proof of Infinitude of Perfect Numbers If the first number of your sequence is N!-N, and the last one is N!, then Chebyshev provably won't say that one of the numbers in that sequence is prime, because there's no way to fit the interval [M,2M] inside of the interval [N!-N,N!] unless N is small (3 or less -- see the thread with your alleged GC proof). In fact, N!-1 can even be composite: 5!-1 = 119 = 7 * 17. on the advice of others. Because I never had experience with sequence proving. Some people are really good at this area of math, I am not. If I had 2 wishes for this proof attempt I would wish that the 2^n sequence lined up exactly with the N! sequence and so the 2^n-1 lined up exaclty with the N!-1 and thus a breeze to prove 2^n-1 is prime. And the second wish is that this method of factorial and Chebyshev is a universal tester for all other prime forms whether they are infinite or not. I think this second wish will come true if I can show this scheme works for Infinitude of Perfect Numbers. Chris, what I did in 1991, since 2^n does not line up with N! is that I did a morphed-Mathematical Induction of where it lines up at the case of 2, then saw that it lines up at infinity, ah but this is not satisfying. I feel confident that the proof is in this scheme but am unable to assemble the argument into a flowing proof. I feel that someone skilled with sequences can custom fit that N! to line up with the 2^n or vice versa. Dik is good with p-adics and probably good at these sequence manipulations. I smell the proof here, but just am not able to make it flow. Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: revisiting my old alleged proof of Infinitude of Perfect Numbers Now one idea that may work is to use the 31! for the case of 2^5 rather than trying to line up 5!. So can I line up 31! which has a heap of consecutive composites and then tie into the 2^5-1 = 31? I smell the proof is in this scheme but it requires some brilliant manipulation of these sequences. And if it works, I see it can be generalized as a tool to prove a large class of forms such as 2^n +1 as to the question of whether these primes are infinite or not infinite. So if this method works for Infinitude of Perfect Numbers it would be a widespread tool to prove the infinitude of other prime forms. Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: Rational vs irrational ... but irrelevant to this particular discussion. Or any positive real. So what? We were talking about reals. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Rational vs irrational If x is expressible, then consider the sequence of rationals 10^(-n) floor(10^n x) for positive integers n. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada ... unless f'(f(f(x))) = 0 ??? I don't follow this at all. Let's do the same thing with some other equation, say f(x) = g(x). Take f of both sides: f(f(x)) = f(g(x)) Take derivative: f'(f(x)) f'(x) = f'(g(x)) g'(x) Now use f(x) = g(x), and you get: f'(x) = g'(x) but note that this is just the der. of f(x) = g(x) Hence, what? Not at all. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Theorems with short proofs in analysis and long proofs in PA I'm seeking info about what we might call real world speed-up. Could non-specialists (e.g. Am. Math Monthly style) on number-theoretic claims (ones expressible in the language of first-order Peano Arithmetic) which have really neat short proofs in e.g. complex analysis and whose only known PA proofs are horribly long and nasty? [I'm talking here of ordinary theorems that you might enounter in a first course on mumber theory.] === Subject: Re: Theorems with short proofs in analysis and long proofs in PA You mean like the Prime Number Theorem, where there are shorter analytic proofs, and longer elementary proofs...? http://mathworld.wolfram.com/PrimeNumberTheorem.html http://www.answers.com/topic/prime-number-theorem Perhaps another example is Dirichlet's Theorem http://mathworld.wolfram.com/DirichletsTheorem.html that any arithmetic progression {an+b} with a,b relatively prime, contains infinitely many primes. Now in both of these cases, even the analytic proof is not really neat short... === Subject: Re: Theorems with short proofs in analysis and long proofs in PA The ultimate of speed up is Goodstein's Theorem. Provable in ZFC, not provable in FOL with Peano's Axioms. === Subject: Re: Cylinder plane intersection ellipse | It is clear as daylight. Excellent explanation. I tend do be suspicious | of my own shadow lest I commit a grave mathematical mistake to code. | Most of my code deals with impacting shapes and objects in 3d for DEM | simulations. | | Arjun For that type of work you need two essential subroutines through which to pass every vertex in the object. For example, suppose you want to display a cube. You can draw a square, but that has no perspective. This is a better cube (fixed font needed): _____________ | _________ /| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | |_________| | |/___________| And this has rotation as well: __________ / /| / / | / / | /_________/ | | | | | | | | | | | | | | | / | | / |_________|/ Can you write the code to draw these? Androcles. === Subject: Re: Cylinder plane intersection ellipse Since it is a circularly symmetric cylinder a simple parametrization is adequate.(x,y,z)= R(cos(th),sin(th),cos(th)*tan(alpha)). If it helps, try an experiment. Take a thin hollow cardboard cylinder and make two cuts.One at an angle as you showed sectioning a carrot and one along the one smallest generator. Spread the development flat out and observe a sine curve. Now recognize (x,y,z). === Subject: Topology Question This is a homework exercise, but I'm not sure if I've done it correctly. The problem is from Munkres text, 2.13.1. Let X be a topological space; let A be a subset of X. Suppose that for each x in A there is an open set U containing x such that A contains U. Show that A is open in X. Proof. Since X is a topological space and U is open there is a basis B for the topology on X such that U is in B. Then by definition A is open in X because for each x in A there is a basis element U in B such that x is in U and A contains U. [] Does this look right? If not, why? Kyle === Subject: Re: Topology Question Namely the trivial base of the topology itself, so why even use it? Your use of variables is vague unto confusing. That is not the definition of open. Maybe it is in metric space theory, but it's not the definition of open for a topological space. You seem to be treading water, using what you're supposed to prove, for the proof. No. Perhaps you're suffering a hang over from taking too much metric space theory. ;-) Hint: a union of open sets is open, by definition of the topology. === Subject: Re: Topology Question By the trivial base do you mean the collection {ø, X}? That's how the text defined open for a basis generated topology. The other definition was that a set U is open in X if U is in T where T is the topology on X. What is the other definition? Heh, prehaps I shouldn't have read anything prior to this class. P.S. What was that topology related URL I've seen you give out before? === Subject: Re: Topology Question What is U? What do you mean by U is open there? It's the there that bugs me. What's a basis element? I think I've already answered that. Hint: you don't need basis *at all* to solve this problem. Jose Carlos Santos === Subject: Re: Topology Question What do you mean I hate to press the ENTER key? Are my posts showing up strange or something? I post to sci.math via the Math Forum at Drexel. Someone else has commented to me on this as well. I appologize, but I'm not intentionally doing anything. U is an open set in X containing x such A contains U. I meant that since X is a topological space, and U is open in X, there exists a basis B.... I should have used a comma to separate. === Subject: Re: infinity Either an equivalence class of ordered pairs or a representative of that class, right? I don't see much difference between saying that natural numbers are sets and rational numbers are sets. The construction in the latter case is more complicated, but so what? But I'm not too keen on saying natural numbers are sets in the first place. Surely, there are certain sets that naturally model the naturals, but the natural numbers are not sets. I'm showing my structuralist biases here, but this is rather beside the point anyway. What I want to say is: if you want to say that natural numbers *are* sets, then I don't see why you want to reject the claim that rational numbers are sets too. -- Jesse F. Hughes You know that view most people have of mathematicians as brilliant people? What if they're not? -- James S. Harris === Subject: Re: infinity You don't think of it as an equivalence class of ordered pairs, where (m,n) ~ (p,q) if mq=np? === Subject: Re: infinity There's an obvious canonical representative of such classes, namely (m,n) with no common divisors. -- If .999... = 1 then (.999...)/1 should equal 1 let's see (.999...)/1 = .999... [Therefore] .999... still=/= 1 -- An astonishing proof by S. Enterprize === Subject: Re: math is more subjective than art appreciation Yes, I do think that this opinion is self-serving and closed-minded. Stupid, too. -- Jesse F. Hughes What you call reasonable is suspect since you've proven yourself to be an enemy of mathematics. -- James S. Harris defends the cause. === Subject: Re: math is more subjective than art appreciation Jesse, you're brilliant!! I agree with you completely!!1! Lee Rudolph f being a continuous positive function , f(x+y)=f(x).f(y) may be written: ln(f(x+y) ) / ln(f(y)) = ln(f(x))/ln(f(y)) + 1 , blocking y , y and ln(f(y)) become constant values . In this form with increment + 1 ,ln(f(x))/ln(f(y)) is an Abel function (counting iterations) of { see (x+y) / y = x/y +1 } , so ln(f(x))/ln(f(y)) = x/y or c*x/(c*y) isolating the variable x gives ln(f(x))= c*x and f(x)=exp(c*x) Alain. === Subject: Re: A Simple Functional Equation One more note. I do sometimes see z/|z| called the sign of z. Yes, but can someone just tell me if this reasoning is correct, especially f(x+y)=f(x)f(y) (1) derivate with respect to x f'(x+y)=f'(x)f(y) (2) let x = 0 f'(y)=f'(0)f(y) (3) but since y is an independent variable let's call it x f'(x)=f'(0)f(x) (4) f'(x)/f(x)=f'(0) (5) / Henrik === Subject: Re: Area of X^2*n+Y^2*n=a^2*n (((Assuming))) triangle of equal sides the area shall be a^2*k^2*3^(1/2) where k is a real root of k^(2*n)+(3^(1/2)*k-1)^(2*n)=1 === Subject: Re: Area of X^2*n+Y^2*n=a^2*n I don't see how the largest triangle can, in general, be equilateral. As n gets large the shape approaches a square, and the largest triangle inside a square is not equilateral. Actually, I think my first guess was incorrect. I'm wondering now if the answer is actually an isosceles triangle with apex at the intersection of the curve with the line y = x. === Subject: Re: Complex conjugate of an integral If that's what he means, why did he put the complex conjugates of a and b into his formula? === Subject: Celestial Mechanics I read an interesting story: the English astronomer and mathematician J. C. Adams (1819-1892), while still an undergraduate, performed an investigation to try to explain the reason for the irregularities in the motion of the planet Uranus. Adams theorized that the unexpected planetary orbit could be due to the presence of an as yet undiscovered planet in the vicinity.Once Adams did put his work on paper, he submitted it to the director of the Cambridge Observatory. The observatory took no action on his work. Several months later, Urbain Le Verrier submitted similar work to Johann Gottfried Galle, the director of the Berlin Observatory. Galle acted on Le Verrier's work and became the first person to observe Neptune. Adams and Le Verrier conjectured,in order to make their predictions,that the new planet was twice as far from the Sun as Uranus (as the empirical Bode's Law required). This story suggested to me the following problem: let us consider n point masses wich interact following Newton's laws.If you know the value of the first n masses, m_1,...,m_(n-1) and their position at every instant, r_1(t),...,r_(n-1)(t), can you know m_n and r_n(t)? You can easily calculate, using Newton's laws, the force acting on the n-th point mass. So, if you knew also m_n, the problem would be trivial.But you don't know m_n! Maury === Subject: Re: Continuous Injections To the World Wide Wade: can you at least tell me what WLOG f mean? Maury === Subject: Re: operations between different numbers snip This is back to front, analytic rather than constructive. The integers and the unsigned rationals are constructed from the naturals, the half line constructed from the unsigned rationals, loops from them, algebras from loops. The subset approach works down from one branch of a tree, and fails to see the other branches. He previously said This only leads to fields with negative numbers and to algebras that only conserve one property. There are many other algebras. See http://library.wolfram.co/infocenter/MathSource/4894 . Start with sets and develop natural numbers N. An equivalence relation on ORDERED sets gives unsigned rationals Q+ [Landau 1930], and a continuity axiom completes these to the half line of unsigned continuous numbers (I call them Primals, P). Conserved properties are introduced by multiplying generalized vectors A={a1,a2,...,am}, B,... (indexed sets of primals) using Moufang Loop multiplication tables (groups + octonions). These have the conservation property Det[A] Det[B] = Det[AB] (found for groups by Frobenius in 1895). The individual factors of the symbolic determinant (I call them sizes) are conserved. Only now is it permissible to introduce signs, because an mxm multiplication table with r-fold symmetry can be collapsed to an (m/r)x(m/r) r-signed table via equivalence relations on sets or r primals. If r=2, real numbers are created, but other r's create other algebras. (In the 1860's Hamilton identified the r=2 relationship between the 8 & 16 elements of the quaternion and octonion tables and the 4 & 8 signed (real) elements of the corresponding algebras.) Applying r=2 to the C2 and C4 cyclic groups (of 2 and 4 elements) gives the rules of Real and Complex algebra:- C2 real C4 complex C4C2r 1 2 + - 1 2 3 4 1 i -1 -i a b c d 2 1 - + 2 3 4 1 i -1 -i 1 b c -d -a 3 4 1 2 -1 -i 1 i c -d a -b 4 1 2 3 -i 1 i -1 d -a -b c The C2 & C4 tables (and those of quaternions and octonions) only conserve one property, because their determinant factors are powers of the sum of the elements. Many other group tables define algebras with several conserved properties (or symmetries); C4C2 gives the table shown above as C4C2r, which conserves {a+b+c-d, a-b+c+d, (a-c)^2+(b+d)^2} (Calculate the determinant and factorise it yourself!) Monosized real and complex algebras are degenerate, losing many important properties of the multi-sized algebras. I am investigating Dozal, the algebra of 12-element vectors; it has operators that are neither real nor complex, deBroglie-like orbits with Planck areas, renormalization, supersymmetry, 3-phase quark waves in 3 families, etc.) So, in reply to the original question the answer is YES, but it extends to Moufang loops, which have the Frobenius conservation property. The associative property that defines groups is irrelevant. Roger Beresford (Loughborough) .. we must be willing to question basic, seemingly self-evident, p426.) === Subject: Re: Euler Elemens d'algebre in PDF version A scan (page by page and slow..) of volume II in french is here : http://siba3.unile.it/archives/images/euler.html The german books 'Vollst.8andige Anleitung zur Algebra' volume I and II are both available in djvu format here : http://www.mathematik.uni-bielefeld.de/~rehmann/DML/BOOKS/index.html (free reader for this format available at http://www.numdam.org/) The french books seem to be sold too : http://www.alibris.com/search/search.cfm?qwork=1996506&wauth=Euler%2C%20Leon hard&ptit=El%E9mens%20d%27algebre&pauth=Euler%2C%20Leonhard&pisbn=&pqty=2&pq t ynew=0&pbest=24%2E90&matches=2&qsort=r&cm_re=works*listing*title Hoping it helped, Raymond === Subject: Re: Linguistic ambiguity vs. mathematical precision Hi there As I am not a mathematician I would like to know from you what is the language of mathematics if it is not digital? Of course all the languages, natural or atificial, that exist satisfy needs and are justified. I was only inquiring about the future of human language in view of information growth. In addition I thought anything which is precise or imprecise has advantages and disadvantages. The imprecise nature of human language inspite of some disadvantages of ambiguity is superior for the same reason namely being ambiguous. What do you think? But I wonder why you started a new thread with my title? Jamshid === Subject: Re: Linguistic ambiguity vs. mathematical precision Hi Jim I am a linguist and the first time in your group. The latest developments in science and technology make me see language from a different perspective. One of the questions which interests me is: how human language can cope with knowledge growth say in a hundred years or so.In addition, analogue and digital modes of communication, as I understand them, have become the cenntre of my interest. I thought analogue mode is superior to digital mode because it gives room to (ambiguity) more messages (some implied and some literal, some metaphorical...) than digital mode. Thus Precision of mathematics in my understanding is a reduction and loss of information because it is based on digits. In digital mode there is a numerical transition. Is there nothing lost if you move by digits say from one to two? You said you don't agree analogue is imprecise. Perhaps your understanding is different or I got it wrong. I just wanted to find out because I can't argue mathematically. In what way have I done you a power of good? Jamshid