mm-202 === Subject: Even Functions? How do I tell if a function is even? I understand that a function is even if f(x) = f(-x), but how, for example, could I tell if (x^3 - x)^4(x^2 + 1)^3 or (x^2 + 1)^3 is even or not? === Subject: Re: Even Functions? >I understand that a function is even if f(x) = f(-x), but how, for example, >could I tell if (x^3 - x)^4(x^2 + 1)^3 or (x^2 + 1)^3 is even or not? Substitute -x in for x, and try to simplify to the original equation. Doug === Subject: Re: Even Functions? > How do I tell if a function is even? I understand that a function is even if f(x) = f(-x), but how, for > example, could I tell if (x^3 - x)^4(x^2 + 1)^3 or (x^2 + 1)^3 is even or > not? Replace x in the expression by -x and then move the minus signs around until you either get back to the original expression or you figure out that this is impossible to do. Also you can use properties of even and odd functions such as, (1) a product of two even functions or two odd functions is even and a product of an even and an odd function is odd, (2) A sum of even or odd functions is even or odd respectively and the sum of an even and an odd function is neither even nor odd unless one of them is identically 0, (3) constant functions are even, (4) the identity map (x) is odd. Have a tolerable existence. Eli === Subject: Re: Even Functions? Adjunct Assistant Professor at the University of Montana. >How do I tell if a function is even? I understand that a function is even if f(x) = f(-x), but how, for example, >could I tell if (x^3 - x)^4(x^2 + 1)^3 or (x^2 + 1)^3 is even or not? Check to see if f(-x) is the same as f(x). In the second case, f(x) = (x^2+1)^3, we have f(-x) = ((-x)^2 + 1)^3 = (x^2+1)^3 = f(x) for all x, because (-x)^2=x^2. So f(x) is even. In the former case, you have g(x) = (x^3-x)^4 (x^2+1)^3. g(-x) = ( (-x)^3 - (-x))^4 ( (-x)^2 + 1)^3 = (-x^3 + x)^4 (x^2+1)^3 = (-(x^3-x))^4 (x^2+1)^3 = (x^3-x)^4 (x^2+1)^3 (because (-a)^4 = a^4) = g(x) so for all x, g(-x)=g(x), and therefore the function is even. To show a function is not even (e.g., h(x) = (x^3-x)^3), then exhibit two points, a and -a, such that h(a) is not equal to h(-a). In the example h(x) = (x^3-x)^3, note that h(2) = (8-2)^3 = 6^3 and h(-2) = (-8+2)^3 = -6^3 so h(2) is not equal to h(-2), and therefore h is not even. Its not denial. Im just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Even Functions? >How do I tell if a function is even? I understand that a function is even if f(x) = f(-x), but how, for example, >could I tell if (x^3 - x)^4(x^2 + 1)^3 or (x^2 + 1)^3 is even or not? Check to see if f(-x) is the same as f(x). In the second case, f(x) = (x^2+1)^3, we have f(-x) = ((-x)^2 + 1)^3 = (x^2+1)^3 = f(x) for all x, because > (-x)^2=x^2. So f(x) is even. In the former case, you have g(x) = (x^3-x)^4 (x^2+1)^3. g(-x) = ( (-x)^3 - (-x))^4 ( (-x)^2 + 1)^3 > = (-x^3 + x)^4 (x^2+1)^3 How did you go from the above.... > = (-(x^3-x))^4 (x^2+1)^3 ... to the above....? Where did the extra brackets come from and why did + x become - x? > = (x^3-x)^4 (x^2+1)^3 (because (-a)^4 = a^4) ... then to the above...? I thought (-a)^3 = (-a)^3 so where did the negative go? > = g(x) so for all x, g(-x)=g(x), and therefore the function is even. correct, I just need some further clarification. === Subject: Re: Even Functions? >How do I tell if a function is even? I understand that a function is even if f(x) = f(-x), but how, for > example, >could I tell if (x^3 - x)^4(x^2 + 1)^3 or (x^2 + 1)^3 is even or not? Check to see if f(-x) is the same as f(x). In the second case, f(x) = (x^2+1)^3, we have f(-x) = ((-x)^2 + 1)^3 = (x^2+1)^3 = f(x) for all x, because > (-x)^2=x^2. So f(x) is even. In the former case, you have g(x) = (x^3-x)^4 (x^2+1)^3. g(-x) = ( (-x)^3 - (-x))^4 ( (-x)^2 + 1)^3 > = (-x^3 + x)^4 (x^2+1)^3 How did you go from the above.... = (-(x^3-x))^4 (x^2+1)^3 ... to the above....? Where did the extra brackets come from and why did + > x become - x? -x^3 + 3 = -(x^3 - x) He just used the distributive property. The minus sign out front can be considered a multiplication by -1. He factored out -1 from each term, which changes the sign of each. Thus: -(a+b) = -a - b -(a-b) = -a - (-b) = -a + b and -(x^3-x) = -x^3 - (-x) = -x^3 + x It looks like you need to review some elementary algebra. > = (x^3-x)^4 (x^2+1)^3 (because (-a)^4 = a^4) ... then to the above...? I thought (-a)^3 = (-a)^3 so where did the > negative go? Elementary algebra again. You had: [-(x^3 + 3)]^4 = (-1)^4 * (x^3+3)^4 and (-1)^4 = 1 Here the elementary fact is: (ab)^m = a^m * b^m > = g(x) so for all x, g(-x)=g(x), and therefore the function is even. correct, I just > need some further clarification. Before tackling this kind of thing, you need to be so comfortable with the elementary stuff that you can do it in your sleep. Review material on simplifying complicated algebraic expressions. - Randy === Subject: Re: Even Functions? Before tackling this kind of thing, you need to be so > comfortable with the elementary stuff that you can do it > in your sleep. Review material on simplifying complicated > algebraic expressions. after a few years off. === Subject: Re: Mean of a random variable >:>Is it possible that for a continuous random variable the following holds: >:>E[x]=integral_0^infty{1-F(x)} where F(x)=c.d.f. >: More than just possible. Use integration by parts. It should be true for >: any distribution (continuous or not) supported on the nonnegative reals, >: such that (1-F(x)) x -> 0 as x -> +infinity. That condition is necessary but not sufficient for the finiteness of the integral. The integral is infinite if and only if the expectation does not exist. If the product does not converge to 0, there exist c_n with c_n >= 2 c_{n-1} and also (1-F(c_n))c_n >= h > 0. Then (1-F(x)) >= h/c_n on the interval (c_{n-1}, c_n), so its integral on that interval is at least h/2. That it is not sufficient, take F(x) = 1 - e/[(e+x)ln(e+x)]. Then the indefinite integral of (1-F(x)) is ln(ln(e+x)), which diverges. -- 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: moded out (was: algebra index problem..... > which elements of this quotient need to be moded out to get >> Is that the usual spelling, moded? Dont know. Its really more slang than actual English anyway, > (we mod out by...), and I think Ive always used the present tense > in any written papers Ive ever submitted. I guess it should be mod-ed > or something... Below are counts of papers using tenses of {factor, mod, quotient} out. xy out | -s -ding -ing -ded -ed | Total --------+------------------------+----- factor | 40 139 67 | 246 mod | 17 30 2 15 0 | 64 quotient| 6 15 2 | 23 So it seems authors greatly favor the double-d spelling, i.e. modding out (30) vs. moding out (2) modded out (15) vs. moded out (0) Interestingly, most all of the matches occur in reviews on physics or closely related fields, perhaps re§ecting an innate bias towards an active vs. passive viewpoint. Such active views amount to < 12% of the 2854 reviews with passive terms quotient group or factor group. -Bill Dubuque === Subject: Re: moded out (was: algebra index problem..... > which elements of this quotient need to be moded out to get Is that the usual spelling, moded? >> Dont know. Its really more slang than actual English anyway, >> (we mod out by...), and I think Ive always used the present tense >> in any written papers Ive ever submitted. I guess it should be mod-ed >> or something... Below are counts of papers using tenses of {factor, mod, quotient} out. xy out | -s -ding -ing -ded -ed | Total > --------+------------------------+----- > factor | 40 139 67 | 246 > mod | 17 30 2 15 0 | 64 > quotient| 6 15 2 | 23 So it seems authors greatly favor the double-d spelling, i.e. modding out (30) vs. moding out (2) > modded out (15) vs. moded out (0) Interestingly, most all of the matches occur in reviews > on physics or closely related fields, perhaps re§ecting > an innate bias towards an active vs. passive viewpoint. > Such active views amount to < 12% of the 2854 reviews > with passive terms quotient group or factor group. It is standard in English to double the letter when appending a suffix to indicate that the short vowel sound is being retained e.g. bated vs batted. But there are exceptions -- happened for one. In this case, the only acceptable spelling is modded/modding. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === Subject: Re: moded out (was: algebra index problem..... <>sSHfTy;{Dhe&:+?b`9fUj5A~$gIYlYT0/$-asR-K~3S3[]q.R3YSmpR|$- GiZp>UN2a}!Fmw+%h }YL`!h_XXr5Q>_nGsY2_ > Below are counts of papers using tenses of {factor, mod, quotient} out. Heres another strange question. In England do they say factorize out ?? -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: moded out (was: algebra index problem..... > Below are counts of papers using tenses of {factor, mod, quotient} >> out. Heres another strange question. > In England do they say factorize out ?? I dont recall hearing that. Is that what they say in N. America? -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === Subject: Re: moded out (was: algebra index problem..... <>sSHfTy;{Dhe&:+?b`9fUj5A~$gIYlYT0/$-asR-K~3S3[]q.R3YSmpR|$- GiZp>UN2a}!Fmw+%h }YL`!h_XXr5Q>_nGsY2_ > It is standard in English to double the letter when appending > a suffix to indicate that the short vowel sound is being retained > e.g. bated vs batted. But there are exceptions -- happened for one. > In this case, the only acceptable spelling is modded/modding. Another unusual exception: busing, bussing (they dont mean the same thing!) Twenty years ago there were (perhaps mythological) newspaper headlines about forced bussing for schoolchildren. -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: moded out (was: algebra index problem..... Nntp-Posting-Host: apps.cwi.nl ... > It is standard in English to double the letter when appending > a suffix to indicate that the short vowel sound is being retained > e.g. bated vs batted. But there are exceptions -- happened for one. > In this case, the only acceptable spelling is modded/modding. Isnt the exception (just like a similar exception in Dutch) that it is (perhaps sometimes) not doubled if the syllable has no stress? In Dutch the no-double rule comes in play when the last syllable of the base word has an (unstressed) schwa. -- 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: Conditional Expectaion Question > [...] > Robert/Stephen, still > havent gotten it yet, though. Robert, the steps in your approach above look simple and direct. The > hard part would seem to be showing that E[XY|X] = XE[Y|X]. At least > this is what Im having a hard time showing. I can see intuitively that the composite RV XY will turn into a > simple constant times the RV Y if X is given. (A note on notation: > I think the notation should really read E[XY|x], where X is the > random variable (i.e., the *function* mapping the elements in the > sample space to the real numbers) and x is a real number in the > range of X. However, I cant express it mathematically. Can you > help me there? I assume that you are taking an applied stochastic processes/probability models course, so I will avoid appealing to the measure theoretic arguments here. (The measure theoretic approach to conditional expectation is quite delicate and sophisticated.) First note that E[X|Y] *is* correct notation: E[X|Y] is a random variable (which depends on Y). You are probably more comfortable with the notation E[X| Y=y] (which presents some technical difficulties when Y is continuous - you are conditioning on an event of probability 0). Think of this as a function f(y), so that E[X|Y]= f(Y), a random variable (a function of Y). Does that work for you? As I have pointed out previously, it is a general property that E[g(X)Y | X] = g(X) E[Y|X]. Your intuition is correct: When you are given X, X acts like a constant. Does the following argument convince you? E[g(X)Y | X=x] = E[g(x)Y| X=x] = g(x)E[Y| X=x] Now replace the scalar x with the random variable X. Or, if we let f(x) = E[g(X)Y | X=x] = g(x)E[Y| X=x], then E[g(X)Y | X] = f(X) = g(X) E[Y|X]. Stephen J. Herschkorn === Subject: Re: help with problem > can somone help me out with this econ problem... I would appreciate it: assume an industry with two firms facin an inverse market demand of > P=100-Q. The product is homogeneous, and each firm has a cost function of > 600+10q+0.25q^2 > Assume firms agree to equally share the market. > Try sci.econ -- Will Twentyman email: wtwentyman at copper dot net === Subject: Erdos and Selfridge paper Ive been looking at the paper by Erdos and Selfridge titled The Product of Consecutive Integers Is Never a Power and, like every other math paper, it has been cleaned up and made incomprehensible to the non-mathematician (me). So Im going to need some clarification. They define two theorems, the first stating The product of two or more consecutive positive integers is never a power and the second one stating Let k, l, n be integers k>=3, l>=2 and n+k>=p^k, where p^k is the least prime satisfying p^k>=k. Then there is a prime p>=k for which alpha sub p is not congruent to 0 (mod l), alpha sub p is the power of p dividing (n+1)...(n+k). As an example, suppose k=5, n =23, giving the sequence 24, 25, 26, 27, 28. In this case, would the least prime be 2 since 2^5 = 32? And what would alpha sub p be in this example? On the next page, they state a theorem of Sylvester and Schur that there is always a prime greater than k which divides (n+1)...(n+k), since n>k. Such a prime divides only one of the k factors, so n+k>>=(K+1)^l, whence n>k^l. In the example I gave above, k=5, n=23 and n+k>=(k+1)^l is true only if l=1. Is this a correct interpretation? === Subject: Re: Erdos and Selfridge paper > Ive been looking at the paper by Erdos and Selfridge titled The > Product of Consecutive Integers Is Never a Power and, like every other > math paper, it has been cleaned up and made incomprehensible to the > non-mathematician (me). So Im going to need some clarification. They define two theorems, the first stating The product of two or > more consecutive positive integers is never a power and the second > one stating Let k, l, n be integers k>=3, l>=2 and n+k>=p^k, where > p^k is the least prime satisfying p^k>=k. Then there is a prime p>=k > for which alpha sub p is not congruent to 0 (mod l), alpha sub p is > the power of p dividing (n+1)...(n+k). As an example, suppose k=5, n =23, giving the sequence 24, 25, 26, 27, > 28. In this case, would the least prime be 2 since 2^5 = 32? And what > would alpha sub p be in this example? The phrase, p^k is the least prime..., makes no sense, as p^k isnt a prime (unless k = 1). Did you get the wording right? Or did it perhaps say p is the least prime...? Now no matter what k is, 2 is the least prime p satisfying p^k greater than or equal to k, so if youve quoted it correctly, p is always 2. That seems pretty silly. Can you check the paper again and see whether youve got it right? Also, if k = 5, and n = 23, and p = 2, then the condition n + k greater than or equal to p^k is not met. > On the next page, they state a theorem of Sylvester and Schur that > there is always a prime greater than k which divides (n+1)...(n+k), > since n>k. Such a prime divides only one of the k factors, so > n+k>>=(K+1)^l, whence n>k^l. What does >>= mean? -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Erdos and Selfridge paper They define two theorems, the first stating The product of two or > more consecutive positive integers is never a power and the second > one stating Let k, l, n be integers k>=3, l>=2 and n+k>=p^k, where > p^k is the least prime satisfying p^k>=k. Then there is a prime p>=k > for which alpha sub p is not congruent to 0 (mod l), alpha sub p is > the power of p dividing (n+1)...(n+k). The phrase, p^k is the least prime..., makes no sense, as p^k > isnt a prime (unless k = 1). Did you get the wording right? Or > did it perhaps say p is the least prime...? > Now no matter what k is, 2 is the least prime p satisfying > p^k greater than or equal to k, so if youve quoted it correctly, > p is always 2. That seems pretty silly. Can you check the paper > again and see whether youve got it right? Im looking at a Xerox copy of the paper as it appears in Illinois J. Math, 1975, and thats exactly what it says. Im thinking this might be a typo and he actually means p is the least prime satisfying p^k>= k. Wouldnt p^k ALWAYS be greater than k for any p>=2 and for any k>1? > Also, if k = 5, and n = 23, and p = 2, then the condition > n + k greater than or equal to p^k is not met. That is certainly true, which makes their paper even more confusing. > On the next page, they state a theorem of Sylvester and Schur that > there is always a prime greater than k which divides (n+1)...(n+k), > since n>k. Such a prime divides only one of the k factors, so > n+k>>=(K+1)^l, whence n>k^l. What does >>= mean? It means I made a typo. Delete one of the > signs. I dont know if I mentioned the title of this paper, namely The Product of Consecutive Integers Is Never a Power. Ive been looking for a proof of this theorem for a long time. Now that Ive found it, I dont understand it. (Im not a mathematician, but Ive had a long-time interest in number theory.) === Subject: Re: EVIL is real (was: Re: Simple principle in core error proof) : Now Im informing all of you that the people arguing against me are : EVIL, yes they are real, live EVIL people as mathematics is that : important, so its important enough for Evil itself to send minions : like them. Evil may be real, but that doesnt change the fact that math is abstract. PEople who point out mistakes in a chain of abstract reasoning do not become evil just because the person making the mistakes is a good person or means well. And even if the people pointing out the mistakes ARE evil, that doesnt mean theyre WRONG. -- --- Its difficult ... you need to be united to have any strength, but internal issues have to be addressed. --- E. Ray Lewis, on liberalism in America === Subject: Roots of equation If I have an equation x^4 - x^3 - 8x^2 + 12x and I know that (x - 2) is a root of the equation, how do I work out the other factor? Another example: e.g. x^2 - 6x^2 - 8x + 24 = (x - 2)(x^3 + 2x^2 - 2x -12) How do I get the (x^3 + 2x^2 - 2x -12) part from the knowing that 2 is a root? Is there a technique for finding this equation? === Subject: Re: Roots of equation >If I have an equation x^4 - x^3 - 8x^2 + 12x and I know that (x - 2) is a >root of the equation, how do I work out the other factor? Another example: e.g. x^2 - 6x^2 - 8x + 24 = (x - 2)(x^3 + 2x^2 - 2x -12) How do I get the (x^3 + 2x^2 - 2x -12) part from the knowing that 2 is a >root? Is there a technique for finding this equation? Here is a french poem explaining how to do do this. My (French) father-in-law once told me that this enabled him to solve problems of this type in about 1/4 the time of everyone else, much to the mystification of his teacher. Division dun polyn.99me par (x-a) LE premier terme du quotient A pour coefficient le premier terme DU dividende. UN terme quelconque du quotient A pour coefficient le coefficient du terme PR.83C.83DENT MULTIPLI.83 par a AUQUEL on ajoute ALG.83BRAIQUEMENT le coefficient du terme correspondant DU dividende. Derek Holt. === Subject: Re: Roots of equation >If I have an equation x^4 - x^3 - 8x^2 + 12x and I know that (x - 2) is a >root of the equation, how do I work out the other factor? You do a polynomial long division, which is exactly the same as for numbers, see http://www.karlscalculus.org/notes.html#plongdiv You should also immediately recognize ïx as another factor. Another example: e.g. x^2 - 6x^2 - 8x + 24 = (x - 2)(x^3 + 2x^2 - 2x -12) How do I get the (x^3 + 2x^2 - 2x -12) part from the knowing that 2 is a >root? Is there a technique for finding this equation? > Same thing here, polynomial long division of (I assume) x^4 - 6x^2 - 8x + 24 divided by (x-2) = (x^3 + 2x^2 - 2x -12). You divide by (x-2) because you know a root is x=2 which means x-2=0. === Subject: Re: Roots of equation > If I have an equation x^4 - x^3 - 8x^2 + 12x and I know that (x - 2) is a > root of the equation, how do I work out the other factor? long division > Another example: e.g. x^2 - 6x^2 - 8x + 24 You mean x^4 - 6x^2 - 8x + 24 ? > = (x - 2)(x^3 + 2x^2 - 2x -12) How do I get the (x^3 + 2x^2 - 2x -12) part from the knowing that 2 is a > root? Is there a technique for finding this equation? long division -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === Subject: Re: Roots of equation If I have an equation x^4 - x^3 - 8x^2 + 12x and I know that (x - 2) is a > root of the equation, how do I work out the other factor? long division Another example: e.g. x^2 - 6x^2 - 8x + 24 You mean x^4 - 6x^2 - 8x + 24 ? Yep I did. > = (x - 2)(x^3 + 2x^2 - 2x -12) How do I get the (x^3 + 2x^2 - 2x -12) part from the knowing that 2 is a > root? Is there a technique for finding this equation? long division === Subject: Re: JSH: Simple math facts >> P(m) = (5 a_1 + 7)(5 a_2 + 7)(5 a_3 + 7). >> [...] >> Therefore, >> P(m)/(49) = (5 a_1/7 + 1)(5 a_2/7 + 1)(5 a_3 + 7) >> follows from the *constant* terms. >> It looks to me like it follows from simple division. >> Ok, so P(m)/49 is that expression. Now what? Whats your point? >> V. >>(5 a_1/7 + 1)(5 a_2/7 + 1)(5 a_3 + 7) = >> 300125 m^3 - 18375 m^2 + 360 m + 22 >>where readers can see that the constant term is 22, which is coprime >>to 7. >> And.....? Why is this important? >>1 as their constant term, while the third of the as is 3 when m=0, >>which gives the last factor as 22 as required. >> I have no idea why you think this is significant. Im not talking about >> whether or not anyone agrees or disagrees, Im just wondering why your >> statement matters. So a_1(m) = 3, a_2(m) = a_3(m) = 0 when m = 0. So >> what? The constant terms are independent of m, right? So any changes to >them must occur without regard to the value of m. Heres a simpler example to illustrate the principle. Q(m) = (7m + 7)(7m + 7)(5m + 22) and notice that the constant term for Q(m) is 7(7)(22), and that Q(m) >has 49 as a factor. But dividing by 49 gives Q(m)/49 = (m+1)(m+1)(5m + 22) and the constant term is now 22. No kidding. You didnt answer my question. So the constant terms are independent of m, so what? Why does this *MATTER*? Has someone claimed that they do? Has someone claimed that they dont? What does this have to do with your fundamental issue of the algebraic integers not being complete? In a well written proof I can look at statements and tell what they *mean*. Not what they say (thats obvious) but why they are there in the proof and how they lead towards the solution. I accept that the constant term goes from 49*22 to 22 when dividing by 49, but Im really not sure what this shows. Alan -- Defendit numerus === Subject: Re: JSH: Simple math facts >> P(m) = (5 a_1 + 7)(5 a_2 + 7)(5 a_3 + 7). >> [...] >> Therefore, >> P(m)/(49) = (5 a_1/7 + 1)(5 a_2/7 + 1)(5 a_3 + 7) >> follows from the *constant* terms. >> It looks to me like it follows from simple division. >> Ok, so P(m)/49 is that expression. Now what? Whats your point? >> V. >>(5 a_1/7 + 1)(5 a_2/7 + 1)(5 a_3 + 7) = >> 300125 m^3 - 18375 m^2 + 360 m + 22 >>where readers can see that the constant term is 22, which is coprime >>to 7. >> And.....? Why is this important? >>1 as their constant term, while the third of the as is 3 when m=0, >>which gives the last factor as 22 as required. >> I have no idea why you think this is significant. Im not talking about >> whether or not anyone agrees or disagrees, Im just wondering why your >> statement matters. So a_1(m) = 3, a_2(m) = a_3(m) = 0 when m = 0. So >> what? The constant terms are independent of m, right? So any changes to >them must occur without regard to the value of m. Heres a simpler example to illustrate the principle. Q(m) = (7m + 7)(7m + 7)(5m + 22) and notice that the constant term for Q(m) is 7(7)(22), and that Q(m) >has 49 as a factor. But dividing by 49 gives Q(m)/49 = (m+1)(m+1)(5m + 22) and the constant term is now 22. No kidding. You didnt answer my question. So the constant terms are > independent of m, so what? Why does this *MATTER*? Has someone claimed > that they do? Has someone claimed that they dont? What does this > have to do with your fundamental issue of the algebraic integers not being > complete? What does it matter? Well it means that P(m)/49 factors as (5 a_1/7 + 1)(5 a_2/7 + 1)(5 a_3 + 7). However, 5a_1/7 and 5a_2/7 are not algebraic integers for all m. The point is that dividing by 49 is not an activity that is dependent on ms value, so you have that one factorization, which gives the correct constant terms of 1, 1, and 22. > In a well written proof I can look at statements and tell what they *mean*. > Not what they say (thats obvious) but why they are there in the proof and > how they lead towards the solution. I accept that the constant term goes > from 49*22 to 22 when dividing by 49, but Im really not sure what this shows. Alan Well you posters keep doing your best to *ignore* posts where I step out the proof, and instead spend a LOT of effort apparently trying to convince people that Im wrong, without bothering to consider the mathematical argument that I present step-by-step, which is telling. Youre playing a social game, dedicated to trying to hide the truth, while I keep saying, hey, I can step it out, step-by-step, talk about *each* and every aspect of the proof, and give context. My take on it all is that mathematicians are, besides being intellectually weak, running away from a truth they dont want to accept. But you know that you need to try and block others from accepting that truth, or theyll force you to stop running. So you keep posting in reply to me. You have to keep posting in reply to me, because of the social game. James Harris === Subject: Continuum Hypothesis Where can I found a G.9adels proof of the CH (hopefully the original translated into English)? I say proof because he didnt prove CH but rather proved that it couldnt dis/proved. === Subject: Re: Continuum Hypothesis <>sSHfTy;{Dhe&:+?b`9fUj5A~$gIYlYT0/$-asR-K~3S3[]q.R3YSmpR|$- GiZp>UN2a}!Fmw+%h }YL`!h_XXr5Q>_nGsY2_ > Where can I found a G.9adels proof of the CH (hopefully the original > translated into English)? > I say proof because he didnt prove CH but rather proved that it couldnt > dis/proved. He showed it cannot be disproved. Only later the other case was done by Cohen. I believe you can find both cases in: P. J. Cohen, Set Theory and the Continuum Hypothesis. New York: W. A. Benjamin, 1966. and if not, G.9adels own book is in English: K. G.9adel, The Consistency of the Continuum-Hypothesis. Princeton, NJ: Princeton University Press, 1940. -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: Transcendental numbers I know that rational and irrational numbers are everywhere dense in > the real number continuum, but what about transcendental numbers? Do > intervals exist in the real number continuum that contain no > transcendental numbers? Has this problem even been solved and is there > a proof? (Maybe this belongs in another math group?) There are more transcententals in any real interval than rational > or even algebraic numbers. Transcendentals are , if anything, more > dense, since there are uncontably many in any real interval whereas > there are only countably many algebraics (including the rationals). By what mechanism can you count the algebraics? Don Cool === Subject: Re: Transcendental numbers I know that rational and irrational numbers are everywhere dense in > the real number continuum, but what about transcendental numbers? Do > intervals exist in the real number continuum that contain no > transcendental numbers? Has this problem even been solved and is there > a proof? (Maybe this belongs in another math group?) There are more transcententals in any real interval than rational > or even algebraic numbers. Transcendentals are , if anything, more > dense, since there are uncontably many in any real interval whereas > there are only countably many algebraics (including the rationals). By what mechanism can you count the algebraics? How many polynomials (with rational coefficients) of degree one are there? They are of the form a_0 + a_1*x. There are countably many possibilities for the constant term a_0, and countably many possibilities for a_1, the coefficient of x. Countable times countable is countable, so there are countably many degree 1 polys. Each has 1 root, so there are countably many numbers that are roots of degree 1 polys. How many of degree two? Same logic applied to a_0 + a_1*x + a_2*x^2. Countable times countable times countable is countable, so there are countably many polys. Each poly has two roots, so thats countably many possible roots. Dot dot dot . . . there are countably many reals that are roots of polys of degree 1, of degree 2, . . ., of degree n, . . . Adding them all up, we get a countable sum of countable numbers, which is countable. === Subject: Re: Transcendental numbers > I know that rational and irrational numbers are everywhere dense in > the real number continuum, but what about transcendental numbers? Do > intervals exist in the real number continuum that contain no > transcendental numbers? Has this problem even been solved and is there > a proof? (Maybe this belongs in another math group?) There are more transcententals in any real interval than rational > or even algebraic numbers. Transcendentals are , if anything, more > dense, since there are uncontably many in any real interval whereas > there are only countably many algebraics (including the rationals). By what mechanism can you count the algebraics? How many polynomials (with rational coefficients) of degree one are > there? They are of the form a_0 + a_1*x. There are countably many > possibilities for the constant term a_0, and countably many > possibilities for a_1, the coefficient of x. Countable times countable > is countable, so there are countably many degree 1 polys. Each has 1 > root, so there are countably many numbers that are roots of degree 1 > polys. How many of degree two? Same logic applied to a_0 + a_1*x + a_2*x^2. > Countable times countable times countable is countable, so there are > countably many polys. Each poly has two roots, so thats countably many > possible roots. Dot dot dot . . . there are countably many reals that are roots of polys > of degree 1, of degree 2, . . ., of degree n, . . . Adding them all up, we get a countable sum of countable numbers, which > is countable. One tiny thing you left that makes a world of difference: a polynomial must be of finite degree. Otherwise you would be claiming the R is countable. === Subject: Re: Transcendental numbers > I know that rational and irrational numbers are everywhere dense in > the real number continuum, but what about transcendental numbers? Do > intervals exist in the real number continuum that contain no > transcendental numbers? Has this problem even been solved and is > there > a proof? (Maybe this belongs in another math group?) There are more transcententals in any real interval than rational > or even algebraic numbers. Transcendentals are , if anything, more > dense, since there are uncontably many in any real interval whereas > there are only countably many algebraics (including the rationals). By what mechanism can you count the algebraics? How many polynomials (with rational coefficients) of degree one are > there? They are of the form a_0 + a_1*x. There are countably many > possibilities for the constant term a_0, and countably many > possibilities for a_1, the coefficient of x. Countable times countable > is countable, so there are countably many degree 1 polys. Each has 1 > root, so there are countably many numbers that are roots of degree 1 > polys. How many of degree two? Same logic applied to a_0 + a_1*x + a_2*x^2. > Countable times countable times countable is countable, so there are > countably many polys. Each poly has two roots, so thats countably many > possible roots. Dot dot dot . . . there are countably many reals that are roots of polys > of degree 1, of degree 2, . . ., of degree n, . . . Adding them all up, we get a countable sum of countable numbers, which > is countable. > One tiny thing you left that makes a world of difference: a polynomial must > be of finite degree. Otherwise you would be claiming the R is countable. Well, there are infinitely many orders of finite-ordered polynomials. Then there are infinitely many coefficients. Then, what you are looking at there is N x N x N x .... There are those and then some. Its been discussed here that thats uncountable. That means here that there may be a bijection between N x N x N x ... x N x ... and R. Do you say the algebraics are countable? There exists a one-to-one function from them to the Cartesian product N and itself infinitely many times. The Cartesian product of N and itself infinitely many times is uncountable, just ask Ullrich or Virgil. What the hell, you might say. Consider the polynomials, in particular the polynomial with integer coefficients. a_0 Thats a constant, or a_0 x^0. There is one for each integer, Z(-oo,oo). Then lets consider another coefficient. Lets only worry about the positive integers, Z+. a_1 x^1 + a_0 x^0. Then we have the Cartesian product Z+ x Z+ for each possible value of a_1 and a_0, representing a subset of the polynomials of order 1 with integer coefficients. 1 x^1 + 1 x^0 1 x^1 + 2 x^0 1 x^1 + 3 x^0 1 x^1 + ... 2 x^1 + 1 x^0 2 x^1 + ... ... Keep in mind that any polynomial a_1 x_1 + a_0 x^0 has roots that are roots of (a_1 x_1 + a_0 x^0) *r for real r, eg for coefficients (a_1, a_2)=(1, 3) that polynomial has the same roots as for (2, 6), (3, 9), (4, 12), etc. Anyways, to represent a polynomial with two positive integer coefficients we have the Cartesian product of Z+ x Z+. Then, go on to the third coefficient, etcetera. You might immediately notice that that set of polynomials can be represented by Z+ x Z+ x Z+ ..., one coordinate for each coefficient. The polynomial has finite rank, or order, degree, and its rank can range from zero to infinity. Thats very similar to an integer, an integer is finite, but there are infinitely many of them. The polynomials rank is an integer. So what we get then to represent only all polynomials with positive integer coefficients is the Cartesian product Z+ x Z+ x Z+ x ..., which has been claimed to be uncountable. Whats up with that? Ross === Subject: Re: Transcendental numbers > I know that rational and irrational numbers are everywhere dense in > the real number continuum, but what about transcendental numbers? > Do > intervals exist in the real number continuum that contain no > transcendental numbers? Has this problem even been solved and is > there > a proof? (Maybe this belongs in another math group?) There are more transcententals in any real interval than rational > or even algebraic numbers. Transcendentals are , if anything, more > dense, since there are uncontably many in any real interval whereas > there are only countably many algebraics (including the rationals). By what mechanism can you count the algebraics? How many polynomials (with rational coefficients) of degree one are > there? They are of the form a_0 + a_1*x. There are countably many > possibilities for the constant term a_0, and countably many > possibilities for a_1, the coefficient of x. Countable times countable > is countable, so there are countably many degree 1 polys. Each has 1 > root, so there are countably many numbers that are roots of degree 1 > polys. How many of degree two? Same logic applied to a_0 + a_1*x + a_2*x^2. > Countable times countable times countable is countable, so there are > countably many polys. Each poly has two roots, so thats countably many > possible roots. Dot dot dot . . . there are countably many reals that are roots of polys > of degree 1, of degree 2, . . ., of degree n, . . . Adding them all up, we get a countable sum of countable numbers, which > is countable. > One tiny thing you left that makes a world of difference: a polynomial must > be of finite degree. Otherwise you would be claiming the R is countable. Well, there are infinitely many orders of finite-ordered polynomials. > Makes no sense. > Then there are infinitely many coefficients. Then, what you are looking at there is N x N x N x .... > NxNxN . . . is not the polynomials, its the ring of formal power series. The polynomials are N+N+N+... where ï+ means direct sum. In an infinite direct sum of rings, all but finitely many terms of any element are zero. Thats what makes polynomials polynomials. Otherwise theyd be power series. > There are those and then some. Its been discussed here that thats uncountable. That means here > that there may be a bijection between N x N x N x ... x N x ... and R. > Well yes there is, but that has nothing to do with algebraic numbers or polynomials. > Do you say the algebraics are countable? Yes. There exists a one-to-one > function from them to the Cartesian product N and itself infinitely > many times. No, thats not true. The direct sum of countably many copies of N is countable. The direct product is not. The Cartesian product of N and itself infinitely many > times is uncountable, just ask Ullrich or Virgil. > I agree. But the Cartesian product of countably many copies of N gives you the formal power series, not polynomials. > What the hell, you might say. > Yeah, you got me there. > Consider the polynomials, in particular the polynomial with integer > coefficients. a_0 Thats a constant, or a_0 x^0. There is one for each integer, > Z(-oo,oo). Then lets consider another coefficient. Lets only worry about the > positive integers, Z+. a_1 x^1 + a_0 x^0. Then we have the Cartesian product Z+ x Z+ for each possible value of > a_1 and a_0, representing a subset of the polynomials of order 1 with > integer coefficients. 1 x^1 + 1 x^0 > 1 x^1 + 2 x^0 > 1 x^1 + 3 x^0 > 1 x^1 + ... > 2 x^1 + 1 x^0 > 2 x^1 + ... > ... Keep in mind that any polynomial a_1 x_1 + a_0 x^0 has roots that are > roots of (a_1 x_1 + a_0 x^0) *r for real r, eg for coefficients > (a_1, a_2)=(1, 3) that polynomial has the same roots as for (2, 6), > (3, 9), (4, 12), etc. Anyways, to represent a polynomial with two positive integer > coefficients we have the Cartesian product of Z+ x Z+. Then, go on to the third coefficient, etcetera. You might immediately > notice that that set of polynomials can be represented by Z+ x Z+ x Z+ > ..., one coordinate for each coefficient. The polynomial has finite rank, or order, degree, and its rank can > range from zero to infinity. Thats very similar to an integer, an > integer is finite, but there are infinitely many of them. The > polynomials rank is an integer. So what we get then to represent only all polynomials with positive > integer coefficients is the Cartesian product Z+ x Z+ x Z+ x ..., > which has been claimed to be uncountable. Whats up with that? You are confused about the distinction between a direct product and a direct sum. In a direct sum, all but finitely terms are zero. So for example in the countable direct sum of N, (1,1,1,1,0,0,0,0,0,0,0...) is an element, but (1,1,1,1,1...) is not. === Subject: Re: Transcendental numbers > I know that rational and irrational numbers are everywhere dense in > the real number continuum, but what about transcendental numbers? Do > intervals exist in the real number continuum that contain no > transcendental numbers? Has this problem even been solved and is > there > a proof? (Maybe this belongs in another math group?) There are more transcententals in any real interval than rational > or even algebraic numbers. Transcendentals are , if anything, more > dense, since there are uncontably many in any real interval whereas > there are only countably many algebraics (including the rationals). By what mechanism can you count the algebraics? How many polynomials (with rational coefficients) of degree one are > there? They are of the form a_0 + a_1*x. There are countably many > possibilities for the constant term a_0, and countably many > possibilities for a_1, the coefficient of x. Countable times countable > is countable, so there are countably many degree 1 polys. Each has 1 > root, so there are countably many numbers that are roots of degree 1 > polys. How many of degree two? Same logic applied to a_0 + a_1*x + a_2*x^2. > Countable times countable times countable is countable, so there are > countably many polys. Each poly has two roots, so thats countably many > possible roots. Dot dot dot . . . there are countably many reals that are roots of polys > of degree 1, of degree 2, . . ., of degree n, . . . Adding them all up, we get a countable sum of countable numbers, which > is countable. > One tiny thing you left that makes a world of difference: a polynomial must > be of finite degree. Otherwise you would be claiming the R is countable. There is no such thing as an infinite degree polynomial. Youre thinking of formal power series. === Subject: Re: Transcendental numbers I know that rational and irrational numbers are everywhere dense in > the real number continuum, but what about transcendental numbers? Do > intervals exist in the real number continuum that contain no > transcendental numbers? Has this problem even been solved and is there > a proof? (Maybe this belongs in another math group?) There are more transcententals in any real interval than rational > or even algebraic numbers. Transcendentals are , if anything, more > dense, since there are uncontably many in any real interval whereas > there are only countably many algebraics (including the rationals). By what mechanism can you count the algebraics? By counting their polynomials. Any (rational) polynomial has a finite number of a_i and a countable number of choices for each a_i, so (N_0)^n=N_0. Don Cool === Subject: Re: Help rearranging equations > I was wondering if any charitable person might give me some pointers > on rearranging the following equations. (1) -Tsin(u)sin(v) = ML(ucos(u)sin(v) + vsin(u)cos(v) - > u^2sin(u)sin(v) - v^2sin(u)sin(v) + 2uvcos(u)cos(v)) (2) Tcos(u) - MG = ML(usin(u) + u^2cos(u)) (3) -Tsin(u)cos(v) = ML(ucos(u)cos(v) - vsin(u)sin(v) - > u^2sin(u)cos(v) - v^2sin(u)cos(v) - 2uvcos(u)sin(v)) I have got this far with my problem, but my mathematical juices have > run dry. T, M, L and G are constants, however T is unknown so I would like to > get rid of it. Similarly I expect M to disappear at some point in the > rearrangement. Multiply equation (1) by cos(u) and equation (2) by (sin(u) sin(v)) then add equations (left side to left side and right side to right side) to get an equation without T in it. Similarly, multiply equation (1) by cos(v) and equation (3) by -sin(v) and add equations to get a second equation without T in it. This system of two (new) equations does not contain T, but, together with any one of the original equations, is a system or equations equivalent to the original system of equations. A similar, but more complicated, method applied to the two new equations will give a single without M in it, since all the equations are linear (at least as equations in T and M). === Subject: Math dependency logic One of the more odd things that can happen when you deal with a discoverer is that you find that things you thought were simple, suddenly seem complicated and iffy. So Im now facing a problem where people are doubting algebra, which leads to the fun question of, just what is the logic here? The issue has to do with a factor like x+2 and how you know that 2 is constant. If the logic is expanded upon, my hope is that many of you will realize what is happening in my core error proof, and quit doubting, if you are doubting, as if youre not, and understand it, then thats another problem entirely. Ok, so how do you know that with even a polynomial factor, like x+2, that 2 is *constant* and not related to x? Whats the mathematical logic? Ive talked about setting x=0, which would reveal the constant factor which you can look at, and I think that for some of you, the ability to *physically* see constants in polynomial factors, has left you without ever knowing the *why* behind how it works. How about this? There are an infinity of number and letter combinations, of which x+2 is just one of that infinity. If you set x=2 though, you just have 2, so how do you know from whence it came? I mean, with just 2, well, its just 2, so theres no reference back to x, or y, or any other of that infinity of possibilities. The answer is that 2 is just a number that has no other information except itself. Logically, you can say that 2 is *independent* of any particular variable, even if its associated with it, like in an expression like x+2. Now, as 0 clears out variables set to 0, you can show independent terms by setting a particular value to 0, as that reveals everything that can be associated with an infinity of other things. Its like how it could be y+2, or z+2, or alpha + 2, as seeing just 2, doesnt tell you anything about an association. Possibly the problem for some of you is that you *remember* the previous association. That is, you know that you had x+2, then you set x=0, so in your mind you see a phantom x that holds the association. Thats ok, but it doesnt have a *mathematical* reality because your memory is irrelevant. The reality is that as far as the math is concerned 2 is a number that can be by itself or associated with x or an infinity of other things, and the math isnt going to pick or choose based on what YOU remember. It handles all possibilities at once. And with *just* a number, it sees independence. Thats why setting a variable to 0 is such a useful and powerful tool, as it reveals infinite possibilities by removing a particular association. My argument works on that principle, as by setting x or m, depending on what Im calling the variable, I find that I have 7, 7 and 22 as *constant* terms for factors of a polynomial. I also know that polynomial has a factor that is constant as that factor is 49. When I divide the polynomial by 49 I have constant terms of 1, 1, and 22, which tells me something. Now those numbers dont care about x or m. They dont *know* about x or m, but you may remember them, so you think your memory has some effect, but the math sees *infinite* possibility, while you may fixate on an association you remember. In the realm of infinite possibility, its not possible for 7, 7, and 22 to go to 1, 1, and 22 based on the value of x, where they check it first to decide how theyll behave, because theyre just NUMBERS, and they dont have enough data associated with them to be making those kinds of decisions. Its like, if human beings on earth constrain 7, 7 and 22 so that they have to worry about the value of x and m, then, wow, weve affected all reality in all dimensions and all possibilities, just with our MEMORY of there being an x or m associated in some interesting expression. But, of course, your thoughts dont constrain those numbers as they keep operating quite well throughout Totality, despite what you may believe, or remember. Just remember that when youre considering the issue of constants in a math argument. James Harris === Subject: Re: Math dependency logic > The issue has to do with a factor like x+2 and how you know that 2 is > constant. And does 2 + 2 = 5, for very large values of 2? === Subject: Re: Math dependency logic >One of the more odd things that can happen when you deal with a >discoverer is that you find that things you thought were simple, >suddenly seem complicated and iffy. So Im now facing a problem where >people are doubting me, what is the logic here? Excelent question. :) Jim === Subject: Re: Math dependency logic Nntp-Posting-Host: apps.cwi.nl > One of the more odd things that can happen when you deal with a > discoverer is that you find that things you thought were simple, > suddenly seem complicated and iffy. So Im now facing a problem where > people are doubting algebra, which leads to the fun question of, just > what is the logic here? The issue has to do with a factor like x+2 and how you know that 2 is > constant. Nope, the issue has to do with constant terms of non-polynomials. What is the constant term of sqrt(1 + x) + sqrt(1 - x) + 1? -- 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: Math dependency logic Adjunct Assistant Professor at the University of Montana. > One of the more odd things that can happen when you deal with a > discoverer is that you find that things you thought were simple, > suddenly seem complicated and iffy. So Im now facing a problem where > people are doubting algebra, which leads to the fun question of, just > what is the logic here? The issue has to do with a factor like x+2 and how you know that 2 is > constant. Nope, the issue has to do with constant terms of non-polynomials. >What is the constant term of sqrt(1 + x) + sqrt(1 - x) + 1? I think it has two things really: one is something James let slip recently: he thinks that if a function is 0 at x, then it has a factor of x somewhere. This is in analogy to a polynomial, of course, where if P(x)=0, then there exists a polynomial Q(x) such that P(x)=x*Q(x). He realizes this is not the case for general functions, but he figures there must be some y, factor of x, such that f(x)=y*q(x). His second error is thinking that the constant term constains everything that does not change as x changes. Your example has f(x) = sqrt(1+x) + sqrt(1-x) + 1; he wants to write it f(x) = (f(x)-f(0)) + f(0) and call f(0) the constant term and f(x)-f(0) the nonconstant term. In polynomials, of course, if we think of P(x) as a bunch of summands (the monomials), then P(x)-P(0) consists of all summands which vary as x varies, and P(0) of all summands which do not vary as x varies. James seems to think the same in general, even though what he does already contradicts it. In your example, for instance, f(0) = sqrt(1)+sqrt(1)+1 = 3, so f(x) = [ sqrt(1+x)+sqrt(1-x)+1-3] + 3 = [sqrt(1+x)+sqrt(1-x)-2] + 3 with 3 the constant term and sqrt(1+x)+sqrt(1-x)-2 the nonconstant term. Notably (or trivially, for the rest of us), there is a summand in the nonconstant term which does NOT vary as x varies. James has the same thing: his third term, 5a_3(x)+7 has a_3(0)=3, so the constant term is 22, which means g_3(x) = (5a_3(x)-15) + 22, and there is a term which does not vary in the nonconstant term. Nonetheless, he seems to be claiming that if g_1(x) = 5a_1(x) + 7, then since a_1(0)=0, that means that a_1(x) has a factor of x somewhere; so he probably thinks that all summands of a_1(x) will vary as x varies, and so that when you divide by w_1(x) (which is 7 at x=0), it must be that g_1(x) splits as the constant term 1, and the NON CONSTANT TERM 5a_1(x)/w_1(x). And thus, that 7/w_1(x) is constant. Which is of course a mistake, as you amply expanded elsewhere. Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A mans capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of figures few readers can criticize. A great many people are staggered to this extent, that they imagine there must be the indefinite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Math dependency logic > One of the more odd things that can happen when you deal with a > discoverer is that you find that things you thought were simple, > suddenly seem complicated and iffy. So Im now facing a problem where > people are doubting algebra, which leads to the fun question of, just > what is the logic here? The issue has to do with a factor like x+2 and how you know that 2 is > constant. Nope, the issue has to do with constant terms of non-polynomials. >What is the constant term of sqrt(1 + x) + sqrt(1 - x) + 1? I think it has two things really: one is something James let slip > recently: he thinks that if a function is 0 at x, then it has a > factor of x somewhere. This is in analogy to a polynomial, of course, > where if P(x)=0, then there exists a polynomial Q(x) such that > P(x)=x*Q(x). He realizes this is not the case for general functions, > but he figures there must be some y, factor of x, such that > f(x)=y*q(x). His second error is thinking that the constant term constains > everything that does not change as x changes. Your example has f(x) = sqrt(1+x) + sqrt(1-x) + 1; he wants to write it f(x) = (f(x)-f(0)) + f(0) and call f(0) the constant term and f(x)-f(0) the nonconstant > term. In polynomials, of course, if we think of P(x) as a bunch of > summands (the monomials), then P(x)-P(0) consists of all summands > which vary as x varies, and P(0) of all summands which do not vary as > x varies. James seems to think the same in general, even though what > he does already contradicts it. In your example, for instance, f(0) = sqrt(1)+sqrt(1)+1 = 3, so f(x) = [ sqrt(1+x)+sqrt(1-x)+1-3] + 3 > = [sqrt(1+x)+sqrt(1-x)-2] + 3 with 3 the constant term and sqrt(1+x)+sqrt(1-x)-2 the nonconstant > term. Notably (or trivially, for the rest of us), there is a summand > in the nonconstant term which does NOT vary as x varies. James has the same thing: his third term, 5a_3(x)+7 has a_3(0)=3, so > the constant term is 22, which means g_3(x) = (5a_3(x)-15) + 22, and there is a term which does not vary in the nonconstant term. Nonetheless, he seems to be claiming that if g_1(x) = 5a_1(x) + 7, then since a_1(0)=0, that means that a_1(x) has > a factor of x somewhere; so he probably thinks that all summands of > a_1(x) will vary as x varies, and so that when you divide by w_1(x) > (which is 7 at x=0), it must be that g_1(x) splits as the constant > term 1, and the NON CONSTANT TERM 5a_1(x)/w_1(x). And thus, that > 7/w_1(x) is constant. Which is of course a mistake, as you amply expanded elsewhere. I got scared there. I thought math was going to crumble at the seams and fall into the hands of JSH and his Core Errorian (not an attack on Koreans) agents, but Arturo has saved us. > Why do you take so much trouble to expose such a reasoner as > Mr. Smith? I answer as a deceased friend of mine used to answer > on like occasions - A mans capacity is no measure of his power > to do mischief. Mr. Smith has untiring energy, which does > something; self-evident honesty of conviction, which does more; > and a long purse, which does most of all. He has made at least > ten publications, full of figures few readers can criticize. A great > many people are staggered to this extent, that they imagine there > must be the indefinite something in the mysterious all this. > They are brought to the point of suspicion that the mathematicians > ought not to treat all this with such undisguised contempt, > at least. > -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan > Arturo Magidin > magidin@math.berkeley.edu === Subject: Re: Math dependency logic very good explanation; will James Harris pursue it? > In your example, for instance, f(0) = sqrt(1)+sqrt(1)+1 = 3, so f(x) = [ sqrt(1+x)+sqrt(1-x)+1-3] + 3 > = [sqrt(1+x)+sqrt(1-x)-2] + 3 with 3 the constant term and sqrt(1+x)+sqrt(1-x)-2 the nonconstant > term. Notably (or trivially, for the rest of us), there is a summand > in the nonconstant term which does NOT vary as x varies. James has the same thing: his third term, 5a_3(x)+7 has a_3(0)=3, so > the constant term is 22, which means g_3(x) = (5a_3(x)-15) + 22, and there is a term which does not vary in the nonconstant term. Nonetheless, he seems to be claiming that if g_1(x) = 5a_1(x) + 7, then since a_1(0)=0, that means that a_1(x) has > a factor of x somewhere; so he probably thinks that all summands of > a_1(x) will vary as x varies, and so that when you divide by w_1(x) > (which is 7 at x=0), it must be that g_1(x) splits as the constant > term 1, and the NON CONSTANT TERM 5a_1(x)/w_1(x). And thus, that > 7/w_1(x) is constant. --les ducs dEnron! === Subject: Re: Math dependency logic > One of the more odd things that can happen when you deal with a > discoverer is that you find that things you thought were simple, > suddenly seem complicated and iffy. So Im now facing a problem where > people are doubting algebra, which leads to the fun question of, just > what is the logic here? The issue has to do with a factor like x+2 and how you know that 2 is > constant. Nope, the issue has to do with constant terms of non-polynomials. >What is the constant term of sqrt(1 + x) + sqrt(1 - x) + 1? I think it has two things really: one is something James let slip > recently: he thinks that if a function is 0 at x, then it has a > factor of x somewhere. This is in analogy to a polynomial, of course, > where if P(x)=0, then there exists a polynomial Q(x) such that > P(x)=x*Q(x). He realizes this is not the case for general functions, > but he figures there must be some y, factor of x, such that > f(x)=y*q(x). There is a large amount of deja vu here. Years ago James argued that x^2 + y^2 could be factored in linear terms. -- 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: Math dependency logic > I think it has two things really: one is something James let slip > recently: he thinks that if a function is 0 at x, then it has a > factor of x somewhere. This is in analogy to a polynomial, of course, > where if P(x)=0, then there exists a polynomial Q(x) such that > P(x)=x*Q(x). He realizes this is not the case for general functions, > but he figures there must be some y, factor of x, such that > f(x)=y*q(x). Doesnt this work for all analytic functions? Have a tolerable existence. Eli === Subject: Re: Math dependency logic I think it has two things really: one is something James let slip > recently: he thinks that if a function is 0 at x, then it has a > factor of x somewhere. This is in analogy to a polynomial, of course, > where if P(x)=0, then there exists a polynomial Q(x) such that > P(x)=x*Q(x). He realizes this is not the case for general functions, > but he figures there must be some y, factor of x, such that > f(x)=y*q(x). Doesnt this work for all analytic functions? Perhaps so. But the functions he uses are not analytic. They involve roots (cube and square roots in this case). -- 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: Math dependency logic > One of the more odd things that can happen when you deal with a > discoverer is that you find that things you thought were simple, > suddenly seem complicated and iffy. So Im now facing a problem where > people are doubting algebra, which leads to the fun question of, just > what is the logic here? > More nonsequiters than usual today. > The issue has to do with a factor like x+2 and how you know that 2 is > constant. > Today is Monday, so 2 is constant. > If the logic is expanded upon, my hope is that many of you will > realize what is happening in my core error proof, and quit doubting, > if you are doubting, as if youre not, and understand it, then thats > another problem entirely. Ok, so how do you know that with even a polynomial factor, like x+2, > that 2 is *constant* and not related to x? > Tomorrow is Tuesday, 2 will still be constant. > Whats the mathematical logic? > Constants are constant. > Ive talked about setting x=0, which would reveal the constant factor > which you can look at, and I think that for some of you, the ability > to *physically* see constants in polynomial factors, has left you > without ever knowing the *why* behind how it works. How about this? There are an infinity of number and letter combinations, of which x+2 > is just one of that infinity. If you set x=2 though, you just have 2, > so how do you know from whence it came? > If you set x = 2 in the expression x+2, you get 4. Dont you? > I mean, with just 2, well, its just 2, so theres no reference back > to x, or y, or any other of that infinity of possibilities. The answer is that 2 is just a number that has no other information > except itself. > 2 is 2, yes. > Logically, you can say that 2 is *independent* of any particular > variable, even if its associated with it, like in an expression like > x+2. Now, as 0 clears out variables set to 0, you can show independent > terms by setting a particular value to 0, as that reveals everything > that can be associated with an infinity of other things. > You are really off the deep end today. I mean, even though your mathematical reasoning is wrong, it is usually interesting and capable of being either true or false. This latest stuff makes no sense at all. > Its like how it could be y+2, or z+2, or alpha + 2, as seeing just 2, > doesnt tell you anything about an association. Possibly the problem for some of you is that you *remember* the > previous association. That is, you know that you had x+2, then you > set x=0, so in your mind you see a phantom x that holds the > association. > Oh I see where this is going. All along you have clearly not understood the distinction among polynomial expressions, polynomial functions, and polynomial expressions evaluated at a particular value. I believe you may yourself be dimly starting to understand what it is you are confused about. === Subject: Re: Math dependency logic James, you are becoming less and less coherent. If this error is so CORE to mathematics, cant you just give a few simple quadratics with integral coefficients between -100 and 100 to prove your point? > One of the more odd things that can happen when you deal with a > discoverer is that you find that things you thought were simple, > suddenly seem complicated and iffy. So Im now facing a problem where > people are doubting algebra, which leads to the fun question of, just > what is the logic here? The issue has to do with a factor like x+2 and how you know that 2 is > constant. If the logic is expanded upon, my hope is that many of you will > realize what is happening in my core error proof, and quit doubting, > if you are doubting, as if youre not, and understand it, then thats > another problem entirely. Ok, so how do you know that with even a polynomial factor, like x+2, > that 2 is *constant* and not related to x? Whats the mathematical logic? Ive talked about setting x=0, which would reveal the constant factor > which you can look at, and I think that for some of you, the ability > to *physically* see constants in polynomial factors, has left you > without ever knowing the *why* behind how it works. How about this? There are an infinity of number and letter combinations, of which x+2 > is just one of that infinity. If you set x=2 though, you just have 2, > so how do you know from whence it came? I mean, with just 2, well, its just 2, so theres no reference back > to x, or y, or any other of that infinity of possibilities. The answer is that 2 is just a number that has no other information > except itself. Logically, you can say that 2 is *independent* of any particular > variable, even if its associated with it, like in an expression like > x+2. Now, as 0 clears out variables set to 0, you can show independent > terms by setting a particular value to 0, as that reveals everything > that can be associated with an infinity of other things. Its like how it could be y+2, or z+2, or alpha + 2, as seeing just 2, > doesnt tell you anything about an association. Possibly the problem for some of you is that you *remember* the > previous association. That is, you know that you had x+2, then you > set x=0, so in your mind you see a phantom x that holds the > association. Thats ok, but it doesnt have a *mathematical* reality because your > memory is irrelevant. The reality is that as far as the math is concerned 2 is a number that > can be by itself or associated with x or an infinity of other things, > and the math isnt going to pick or choose based on what YOU remember. It handles all possibilities at once. And with *just* a number, it sees independence. Thats why setting a variable to 0 is such a useful and powerful tool, > as it reveals infinite possibilities by removing a particular > association. My argument works on that principle, as by setting x or m, depending > on what Im calling the variable, I find that I have 7, 7 and 22 as > *constant* terms for factors of a polynomial. I also know that polynomial has a factor that is constant as that > factor is 49. When I divide the polynomial by 49 I have constant terms of 1, 1, and > 22, which tells me something. Now those numbers dont care about x or m. They dont *know* about x > or m, but you may remember them, so you think your memory has some > effect, but the math sees *infinite* possibility, while you may fixate > on an association you remember. In the realm of infinite possibility, its not possible for 7, 7, and > 22 to go to 1, 1, and 22 based on the value of x, where they check it > first to decide how theyll behave, because theyre just NUMBERS, and > they dont have enough data associated with them to be making those > kinds of decisions. Its like, if human beings on earth constrain 7, 7 and 22 so that they > have to worry about the value of x and m, then, wow, weve affected > all reality in all dimensions and all possibilities, just with our > MEMORY of there being an x or m associated in some interesting > expression. But, of course, your thoughts dont constrain those numbers as they > keep operating quite well throughout Totality, despite what you may > believe, or remember. Just remember that when youre considering the issue of constants in a > math argument. > James Harris === Subject: Integral How do I integrate this? f(x) = sin(x)/x === Subject: Re: Integral > How do I integrate this? f(x) = sin(x)/x > I havent tried it, but cant you just take sin(x) * (1/x) and use integration by parts. Lurch === Subject: Re: Integral > How do I integrate this? f(x) = sin(x)/x > You could write sin(x)/x = 1- x^2/3! + x^4/5! - ... [using the expansion of sin(x)] and integrate it. You cant really do better than that. === Subject: Re: Integral You cant. > How do I integrate this? f(x) = sin(x)/x > === Subject: Re: Integral > You cant. You can, you just dont know a name for it. I hate it when people say you cant integrate something, just because the result is beyond the reach of elementary functions. === Subject: Re: Integral You cant. You can, you just dont know a name for it. I certainly cant, and I dont think you can either. (This is semantics) > I hate it when people say you cant integrate something, just because the > result is beyond the reach of elementary functions. === Subject: Re: Integral > You cant. You can, you just dont know a name for it. > I certainly cant, and I dont think you can either. > (This is semantics) Saying one cannot integrate that function is equivalent to not being able to integrate that function. So you would dare to say that sin(x)/x is not integrable? Wrong. And even if it is semantics, it still gives me the creeps when someone says something like that. === Subject: Re: Integral >You cant. > You can, you just dont know a name for it. Its called a Sine Integral : Si(z)= int_0^z sin(x)/x dx We may add : lim_{x->oo} Si(x)= pi/2 Si(z)= sum_{n=0}^oo (-1)^n*z^(2n+1)/((2n+1)*(2n+1)!) Si(z)= (E1(i*z)-E1(-i*z))/(2*i)+pi/2 with E1(z) the exponential integral For further information see http://functions.wolfram.com/GammaBetaErf/SinIntegral/ Raymond > I hate it when people say you cant integrate something, just because the > result is beyond the reach of elementary functions. === Subject: Re: Integral >You cant. >> You can, you just dont know a name for it. Its called a Sine Integral : > Si(z)= int_0^z sin(x)/x dx > Give it a name and its solved, yup. It doesnt really address the OPs question though, does it. He obviously wants to know how to work out the integral. He has probably found that the methods he has learned dont seem to work and wants to know if there is a clever method he hasnt figured out yet. And the answer for him is no, there isnt any way to work it. Of course, giving the antiderivative a name and using that for the answer has its uses, somewhat like in the medical profession when some mysterious disease shows up. They may not know what causes it, how it spreads, how to treat it, or anything else about it, but they can give it a name. Our tests indicate that you have di-§ukus. Feel better now? --Lynn === Subject: Re: Integral > You can, you just dont know a name for it. >> Its called a Sine Integral : >> Si(z)= int_0^z sin(x)/x dx >Give it a name and its solved, yup. It doesnt really address the >OPs question though, does it. He obviously wants to know how to work >out the integral. He has probably found that the methods he has >learned dont seem to work and wants to know if there is a clever >method he hasnt figured out yet. And the answer for him is no, there >isnt any way to work it. >Of course, giving the antiderivative a name and using that for the >answer has its uses, somewhat like in the medical profession when some >mysterious disease shows up. They may not know what causes it, how it >spreads, how to treat it, or anything else about it, but they can give >it a name. Our tests indicate that you have di-§ukus. Feel better >now? But once you have the name, you can look it up and see what is known about it. And for functions such as Si, there is in fact a lot of information that can be found in standard reference works or in computer algebra systems such as Maple or Mathematica. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: Integral You cant. >You can, you just dont know a name for it. >> Its called a Sine Integral : >> Si(z)= int_0^z sin(x)/x dx > Give it a name and its solved, yup. It doesnt really address the > OPs question though, does it. He obviously wants to know how to work > out the integral. He has probably found that the methods he has > learned dont seem to work and wants to know if there is a clever > method he hasnt figured out yet. And the answer for him is no, there > isnt any way to work it. Of course, giving the antiderivative a name and using that for the > answer has its uses, somewhat like in the medical profession when some > mysterious disease shows up. They may not know what causes it, how it > spreads, how to treat it, or anything else about it, but they can give > it a name. Our tests indicate that you have di-§ukus. Feel better > now? --Lynn Of course I only gave it a name ;-) (hint 1: nothing snipped?? hint 2:the name (and the link) indicates that no closed form using only a finite number of ïelementary functions is known (this may be proved but Ill stop here...)). R. === Subject: Re: Rational function My thoughts exactly! > How about the guy with the old chestnut comment... I suppose you mean Bob Kolker. But he might not be reading _this_ thread; > after all, that comment was made by him in a different thread. You mean he does something other than read sci.math? Like he has a life? I am having trouble finding the equation of a rational function where > y^x = x^y and goes through (2,4) (4,2) (2.48832, 2.985984) > (2.985984,2.48832) (2.25,3.375) (3.375,2.25) (2.44140625,3.0517578) > (3.0517578, 2.44140625). The > numbers are rounded off ofcourse, but they all are true for y^x = x^y. Perhaps youre annoyed that nobody has answered your question yet. But > were not compelled to answer questions, and your particular question, > taken at face value, doesnt seem interesting. Sorry. OTOH, if you could > show us _why_ we should be interested in your question, then you would > probably get some useful responses. > Meanyheads. You dont have anything useful to do. Just pull the answer out > of your overgrown brains sitting on your spindly useless bodies and tell me > the answer. Una Grajjit === Subject: Re: Rational function > I am having trouble finding the equation of a rational function where y^x > = > x^y and goes through (2,4) (4,2) (2.48832, 2.985984) (2.985984,2.48832) > (2.25,3.375) (3.375,2.25) (2.44140625,3.0517578) (3.0517578,2.44140625). > The > numbers are rounded off ofcourse, but they all are true for y^x = x^y. Any > See, for example: http://mathworld.wolfram.com/LagrangePolynomial.html === Subject: Re: Rational function I certainly didnt mean to get off the subject. I like your excuse David, but again I think its because you dont know an answer. You seem to have answered many other un-interesting questions as you put it, on the forum. You and Mr. Old Chestnut seem to be alike. If you dont have an answer, please feel free not to reply. :) (I dont want to get off of the subject too much) > How about the guy with the old chestnut comment... > I am having trouble finding the equation of a rational function where y^x > = > x^y and goes through (2,4) (4,2) (2.48832, 2.985984) (2.985984,2.48832) > (2.25,3.375) (3.375,2.25) (2.44140625,3.0517578) (3.0517578,2.44140625). > The > numbers are rounded off ofcourse, but they all are true for y^x = x^y. Any === Subject: Re: Rational function > I certainly didnt mean to get off the subject. I like your excuse David, > but again I think its because you dont know an answer. You seem to have > answered many other un-interesting questions as you put it, on the forum. Interesting, like beauty, is in the eye of the beholder. > You and Mr. Old Chestnut seem to be alike. If you dont have an answer, > please feel free not to reply. :) (I dont want to get off of the subject > too much) They both know the answers. I think nobody realized that you dont have a clue. x^y = y^x implicitly defines y as a function of x. This function is not a rational function (surprise!). Any rational function will only approximate the function. The study of approximating functions is interpolation. It is well-studied interpolation and polynomials or even on interpolating polynomials, and you probably want -linear. I think all calculus books still at least mention LaGrange interpolation, but youre probably better off to look at a numerical analysis book. Cubic splines might work very nicely. Another place to look might be graduation theory. The choice of interpolation methods is almost always made on non-mathematical grounds. What are you willing to give up in order to get the data in the form you like? Jon Miller === Subject: Re: Rational function The reason I think it is a rational function is that y = 8/x comes close. What is it with you people and rude comments. Im not offended, but I think its imature, and Im probably younger than most of you. > I certainly didnt mean to get off the subject. I like your excuse David, > but again I think its because you dont know an answer. You seem to have > answered many other un-interesting questions as you put it, on the > forum. Interesting, like beauty, is in the eye of the beholder. You and Mr. Old Chestnut seem to be alike. If you dont have an answer, > please feel free not to reply. :) (I dont want to get off of the subject > too much) They both know the answers. I think nobody realized that you dont have a > clue. x^y = y^x implicitly defines y as a function of x. This function is not a > rational function (surprise!). Any rational function will only approximate > the function. The study of approximating functions is interpolation. It is well-studied interpolation > and polynomials or even on interpolating polynomials, and you probably > want -linear. I think all calculus books still at least mention LaGrange > interpolation, but youre probably better off to look at a numerical > analysis book. Cubic splines might work very nicely. Another place to look > might be graduation theory. The choice of interpolation methods is almost always made on > non-mathematical grounds. What are you willing to give up in order to get > the data in the form you like? Jon Miller === Subject: bug in the faq section on CH constitute permission for an emailed response. 26 4E BF 1A 92 The discussion of Freilings Axiom of Symmetry, in the excerpt from Bill Allen, seems to have a typo. The complete bit is: Let A be the set of functions mapping Real Numbers into countable sets of Real Numbers. Given a function f in A , and some arbitrary real numbers x and y , we see that x is in f(y) with probability 0, i.e. x is not in f(y) with probability 1. Similarly, y is not in f(x) with probability 1. Let AX be the axiom which states ``for every f in A , there exist x and y such that x is not in f(y) and y is not in f(x) The intuitive justification for AX is that we can find the x and y by choosing them at random. In ZFC, AX = not CH. proof: If CH holds, then well-order R as r_0, r_1, .... , r_x, ... with x < aleph_1 . Define f(r_x) as { r_y : y >= x } . Then f is a function which witnesses the falsity of AX. If CH fails, then let f be some member of A . Let Y be a subset of R of cardinality aleph_1 . Then Y is a proper subset. Let X be the union of all the sets f(y) with y in Y , together with Y . Then, as X is an aleph_1 union of countable sets, together with a single aleph_1 size set Y , the cardinality of X is also aleph_1 , so X is not all of R . Let a be in R X , so that a is not in f(y) for any y in Y . Since f(a) is countable, there has to be some b in Y such that b is not in f(a) . Thus we have shown that there must exist a and b such that a is not in f(b) and b is not in f(a) . So AX holds. I focus on the fourth paragraph, in the proof that CH -> ~AX. We have: f(r_x) := { r_y : y >= x} Surely that should be f(r_x) := { r_y : y <= x} After all, given the first definition, the value of f would always be uncountable! Thomas === Subject: Proof for (x+y)^n = ... I have to prove that for all n from |N and for arbitrary x,y from |R the following is true: (x+y)^n = sum(comb(n,k)*x^k*y^(n-k), k, 0, n) (comb(n,k) - combination n above k) ( sum([..]) - comb(n,0)*x^0*y^(n-0)+ ... + comb(n,n)*x^n*y^(n-n) ) This is what I have done: ### Proof through complete induction toward n: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 1.) n = 0 => (x+y)^0 = sum(comb(0,k)*x^k*y^(0-k), k, 0, 0) = 1 2.) Let us suppose that the sentence is right for n. Then we have to prove it for n+1. 3.) (x+y)^(n+1) = (x+y)(x+y)^n = 2.) = (x+y)sum(comb(n,k)*x^k*y^(n-k), k, 0, n) = x*sum(comb(n,k)*x^k*y^(n-k), k, 0, n)+y*sum(comb(n,k)*x^k*y^(n-k), k, 0, n) = sum(comb(n,k)*x^(k+1)*y^(n-k), k, 0, n)+sum(comb(n,k)*x^k*y^(n-k+1), k, 0, n) ### And what do I have to do now? Can you give me a hint? Please, I beg you! :-) Karl. === Subject: Re: Proof for (x+y)^n = ... > I have to prove that for all n from |N and for arbitrary x,y > from |R the following is true: (x+y)^n = sum(comb(n,k)*x^k*y^(n-k), k, 0, n) (comb(n,k) - combination n above k) > ( sum([..]) - comb(n,0)*x^0*y^(n-0)+ ... + comb(n,n)*x^n*y^(n-n) ) This is what I have done: > ### > Proof through complete induction toward n: > ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ > 1.) n = 0 => (x+y)^0 = sum(comb(0,k)*x^k*y^(0-k), k, 0, 0) = 1 > 2.) Let us suppose that the sentence is right for n. Then we have > to prove it for n+1. 3.) > (x+y)^(n+1) = (x+y)(x+y)^n = 2.) = (x+y)sum(comb(n,k)*x^k*y^(n-k), k, 0, n) > = x*sum(comb(n,k)*x^k*y^(n-k), k, 0, n)+y*sum(comb(n,k)*x^k*y^(n-k), k, 0, n) > = sum(comb(n,k)*x^(k+1)*y^(n-k), k, 0, n)+sum(comb(n,k)*x^k*y^(n-k+1), k, 0, n) > ### And what do I have to do now? Can you give me a hint? Hints: 1) Collect terms with equal powers in ïx and ïy. 2) To do #1, it may be useful to change the index, i.e., summing f(k) for k=1 to 6 is the same as summing f(k+1) for k=0 to 5. 3) You may need to first prove a combinatorial identity. This identity will be the obvious one that you need once you collect terms. 4) Following Polya, if there is a problem you cant solve, there is a simpler problem you cant solve. Write out the above steps *explicitly* for cases of small n (n=1,2,3,4). 5) Look up Pascals triangle. Best wishes, Mike > Please, I beg you! :-) No, no, none of that. A simple cash deposit in my Swiss bank account will be quite sufficient :-). Best wishes, Mike === Subject: can anyone help me solve this problem? How many ways are there to distribute B non-distinct Bones between D distinct dogs? if you know the answer to this, could you let me know how you came up with it? === Subject: Re: can anyone help me solve this problem? Adjunct Assistant Professor at the University of Montana. >How many ways are there to distribute B non-distinct Bones between D distinct dogs? Number the dogs, 1,...,D. Since the bones are indistinguishable, just choose B dogs, and give them bones. If you do not allow giving more than one bone to a dog, you are just asking about the number of ways to choose a subset of size B out of {1,...,D}, i.e., D choose B. If you allow to give more than one bone to a dog, you have what are called combinations with repetitions. Its the same problem as counting how many different throws of B dice, each with D faces, there are; the solution then is (B+D-1) choose B. Its not denial. Im just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: can anyone help me solve this problem? > How many ways are there to distribute B non-distinct Bones between D > distinct dogs? > if you know the answer to this, could you let me know how you came up > with it? Sounds like you want each bone to pick a dog. Or mathematically, you want functions from the set of bones into the set of dogs--each dog then gets the pre-image of this function. Have you tried doing this explicitly for a small number of dogs and bones? If I had one bone and 10 dogs, what could I do? If I had two bones and ten dogs what could I do? Remember the joke about the mathematician: A mathematician is given an empty kettle and asked to boil some water, so he (yes, he, no woman would be silly enough to do what this mathematician is about to do) ... so he fills the kettle with water, lights the stove, etc. A while later he is given a kettle filled with water and asked to boil some water. The mathematician spills out the water, then presents the empty kettle and state I have now reduced this to the previously solved problem. So, if I have two bones, how can I reduce that to the previously solved problem. Best wishes, Mike === Subject: Leibniz & Newton Can anyone help me with this question? Why is Newton more widely accepted as the man who developed calculus, instead of Leibniz? Kavon kavon@mathisradical.com === Subject: Re: Leibniz & Newton In , on at 11:41 PM, Kavon Rueter said: >Can anyone help me with this question? Why is Newton more widely >accepted as the man who developed calculus, instead of Leibniz? And why isnt proper credit given to earlier work by, e.g., Archimedes, Fermat? FWIW, our current notation stems from both Liebniz and Newton. Physicist are more prone to use Newtons x dot notation, but both mathematicians and physicists use both when appropriate. The notation used for partial derivatives also stems from Liebnizs notation. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail will be subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do === Subject: Re: Leibniz & Newton > Can anyone help me with this question? Why is Newton more widely accepted > as the man who developed calculus, instead of Leibniz? Who invented calculus? In England, they say Newton; on the Contenant, they say Leibniz. In former centuries, this was a heated dispute! Kavon > kavon@mathisradical.com === Subject: Re: Leibniz & Newton Hi A N Niel. Didnt the Bernoullis get involved in the dispute and take sides? === Subject: Re: Leibniz & Newton > Can anyone help me with this question? Why is Newton more widely accepted > as the man who developed calculus, instead of Leibniz? > Its Leibnizs dy/dx notation that survives, rather than Newtons dot notation for derivatives. Perhaps Newton gets more credit because his mathematical discoveries in general were much more extensive than Leibnizs, and because Newton used his calculus to solve the problem of gravitation. === Subject: Re: Leibniz & Newton > Can anyone help me with this question? Why is Newton more widely >> accepted as the man who developed calculus, instead of Leibniz? Its Leibnizs dy/dx notation that survives, rather than Newtons dot > notation for derivatives. Perhaps Newton gets more credit because his > mathematical discoveries in general were much more extensive than > Leibnizs, and because Newton used his calculus to solve the problem of > gravitation. Actually, there is evidence that Newton created calculus before Leibnitz even though Leibnitz published first. I read this in a book on math history that I cant cite right now. However, this was not established until a long while after fact and the authors opinion is that the real reason that Newton gets more of the credit is because England had more power in the international community at the time than Germany had. Have a tolerable existence. Eli === Subject: Re: Leibniz & Newton : Actually, there is evidence that Newton created calculus before Leibnitz : even though Leibnitz published first. I read this in a book on math : history that I cant cite right now. However, this was not established : until a long while after fact and the authors opinion is that the real : reason that Newton gets more of the credit is because England had more : power in the international community at the time than Germany had. I have read that the Brittish Acience Academy was set to solve the dispute on who invented calculus first, and the Brittish Academy was at that time led by Newton... === Subject: Re: Leibniz & Newton Mikko, I know at the time Germany followed with Leibniz. The mathematicians were divided, often taking sides. Today, the dispute is all but forgotten. Well, until I had to go and post this original message, huh!! Kavon : Actually, there is evidence that Newton created calculus before Leibnitz > : even though Leibnitz published first. I read this in a book on math > : history that I cant cite right now. However, this was not established > : until a long while after fact and the authors opinion is that the real > : reason that Newton gets more of the credit is because England had more > : power in the international community at the time than Germany had. I have read that the Brittish Acience Academy was set to solve the dispute > on who invented calculus first, and the Brittish Academy was at that time > led by Newton... === Subject: Re: Leibniz & Newton hiding (from the plague) and developed calculus at his farm, while Leibnitz was developing his calculus at the same time, only doing so in Germany. The thought has made me wonder, about English speaking peoples preferring Newtons over Leibnizs. I wonder if Germans consider Leibniz the authority?? === Subject: Strange Automorphism Group Let R denote the field of real numbers. Then let R+ denote the additive group of the reals. Then my question concerns Aut(R+). In ZF, it can be proved that this group contains a subgroup isomorphic to R*, the multiplicative group of the reals. On the other hand, in ZFC, this group is huge and nonabelian and isomorphic to a huge |R|-dimensional general linear group. Is there a model or axiomatic extension of set theory in which the subgroup isomorphic to R* is all that is in Aut(R+)? Does ZF + Determinacy work? ---- David === Subject: Re: Strange Automorphism Group >Let R denote the field of real numbers. Then let R+ denote the >additive group of the reals. Then my question concerns Aut(R+). In ZF, >it can be proved that this group contains a subgroup isomorphic to R*, >the multiplicative group of the reals. On the other hand, in ZFC, this >group is huge and nonabelian and isomorphic to a huge |R|-dimensional >general linear group. >Is there a model or axiomatic extension of set theory in which the >subgroup isomorphic to R* is all that is in Aut(R+)? Does ZF + >Determinacy work? An automorphism of R+ is an additive function on R. Its well-known that all measurable additive functions on R are of the form x -> cx. So any model of ZF in which all functions are measurable will do. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Math dependency logic REVISED One of the more odd things that can happen when you deal with a discoverer is that you find that things you thought were simple, suddenly seem complicated and iffy. So Im now facing a problem where people are doubting algebra, which leads to the fun question of, just what is the logic here? The issue has to do with a factor like x+2 and how you know that 2 is constant. If the logic is expanded upon, my hope is that many of you will realize what is happening in my core error proof, and quit doubting, if you are doubting, as if youre not, and understand it, then thats another problem entirely. Ok, so how do you know that with even a polynomial factor, like x+2, that 2 is *constant* and not related to x? Whats the mathematical logic? Ive talked about setting x=0, which would reveal the constant factor which you can look at, and I think that for some of you, the ability to *physically* see constants in polynomial factors, has left you without ever knowing the *why* behind how it works. How about this? There are an infinity of number and letter combinations, of which x+2 is just one of that infinity. If you set x=0 though, you just have 2, so how do you know from whence it came? I mean, with just 2, well, its just 2, so theres no reference back to x, or y, or any other of that infinity of possibilities. The answer is that 2 is just a number that has no other information except itself. Logically, you can say that 2 is *independent* of any particular variable, even if its associated with it, like in an expression like x+2. Now, as 0 clears out variables set to 0, you can show independent terms by setting a particular value to 0, as that reveals everything that can be associated with an infinity of other things. Its like how it could be y+2, or z+2, or alpha + 2, as seeing just 2, doesnt tell you anything about an association. Possibly the problem for some of you is that you *remember* the previous association. That is, you know that you had x+2, then you set x=0, so in your mind you see a phantom x that holds the association. Thats ok, but it doesnt have a *mathematical* reality because your memory is irrelevant. The reality is that as far as the math is concerned 2 is a number that can be by itself or associated with x or an infinity of other things, and the math isnt going to pick or choose based on what YOU remember. It handles all possibilities at once. And with *just* a number, it sees independence. Thats why setting a variable to 0 is such a useful and powerful tool, as it reveals infinite possibilities by removing a particular association. My argument works on that principle, as by setting x or m, depending on what Im calling the variable, I find that I have 7, 7 and 22 as *constant* terms for factors of a polynomial. I also know that polynomial has a factor that is constant as that factor is 49. When I divide the polynomial by 49 I have constant terms of 1, 1, and 22, which tells me something. Now those numbers dont care about x or m. They dont *know* about x or m, but you may remember them, so you think your memory has some effect, but the math sees *infinite* possibility, while you may fixate on an association you remember. In the realm of infinite possibility, its not possible for 7, 7, and 22 to go to 1, 1, and 22 based on the value of x, where they check it first to decide how theyll behave, because theyre just NUMBERS, and they dont have enough data associated with them to be making those kinds of decisions. Its like, if human beings on earth constrain 7, 7 and 22 so that they have to worry about the value of x and m, then, wow, weve affected all reality in all dimensions and all possibilities, just with our MEMORY of there being an x or m associated in some interesting expression. But, of course, your thoughts dont constrain those numbers as they keep operating quite well throughout Totality, despite what you may believe, or remember. Just remember that when youre considering the issue of constants in a math argument. James Harris === Subject: Re: Math dependency logic REVISED > One of the more odd things that can happen when you deal with a > discoverer is that you find that things you thought were simple, > suddenly seem complicated and iffy. No, actually, a person whos a discover in math, makes what sense out of complicated things, and lays them on a firm foundation, so there is no iffiness. Its when you dont know what youre doing that what is simple seems complicated. MB === Subject: Re: Math dependency logic REVISED > One of the more odd things that can happen when you deal with a > discoverer is that you find that things you thought were simple, > suddenly seem complicated and iffy. So Im now facing a problem where > people are doubting algebra, which leads to the fun question of, just > what is the logic here? No one is doubting algebra. Were just certain that you arent using it properly. -- rs, Silverlock === Subject: Re: Math dependency logic REVISED > One of the more odd things that can happen when you deal with a > discoverer is that you find that things you thought were simple, > suddenly seem complicated and iffy. So Im now facing a problem where > people are doubting algebra, which leads to the fun question of, just > what is the logic here? The logic? Lets see: A person who proudly declares he is uneducated in mathematics does some algebra. He comes up with a result. People with training in mathematics universally disagree with the result. Therefore we conclude that it is ______ they are doubting. (a) the persons ability to do algebra (b) algebra itself (c) life, the universe, and everything (d) other , which leads to the fun question of, just what is the logic here? Apparently, the question has and still escapes you and it has been over seven years! Megalomaniac, go see a shrink so you can receive various psychotropic drugs! === Subject: Re: Math dependency logic REVISED > ? Please, take a math class. Learn some math before you revolutionize mathematics, it would be more useful (and necessary) than you probably think. Look at some real proofs, for example. Work some stuff out that is simpler than, say, fermats last theorem. Even at with my humble mathematical background, youre ïmusings are clearly the work of someone who has not been introduced properly to mathematics. All your proofs seem to be a sort of juggling of equations at (or about) a high school level. You might have a great intuition which would be a terrible thing to waste, which is what you are doing right now. Justin Van Winkle === Subject: Re: Math dependency logic REVISED I made. They were made under the assumption that James Harris was about 15. This is apparently incorrect. James, give it up. Your math isnt coherent. Its nonsense. Utter nonsense. It is entertaining to read the arguments though. (The arguments between you and the other posters to the group. Reading your mathematical ïarguments is painful, at best.) Justin Van Winkle > ? Please, take a math class. Learn some math before you revolutionize > mathematics, it would be more useful (and necessary) than you probably > think. Look at some real proofs, for example. Work some stuff out that is > simpler than, say, fermats last theorem. Even at with my humble mathematical background, youre ïmusings are clearly > the work of someone who has not been introduced properly to mathematics. > All your proofs seem to be a sort of juggling of equations at (or about) a > high school level. You might have a great intuition which would be a > terrible thing to waste, which is what you are doing right now. Justin Van Winkle === Subject: Re: Math dependency logic REVISED > I made. They were made under the assumption that James Harris was about 15. > This is apparently incorrect. James, give it up. Your math isnt coherent. Its nonsense. Utter > nonsense. It is entertaining to read the arguments though. (The arguments > between you and the other posters to the group. Reading your mathematical > ïarguments is painful, at best.) Justin Van Winkle And now readers should notice the psych out game. Usenet is easy, if you dont like a poster, dont read them. However, math society is STUCK. If it completely ignores me, then I can explain without hecklers distracting or lying, and people might see how simple my argument is, and realize that mathematicians are engaging in an amazing fraud. If they reply to me with mathematics, either they have to admit Im right, or obfuscate. But I SIMPLIFY, as Ive been doing lately, which eventually puts them in a box. Math IS wonderful in that way. So some of them just switch to insulting posts meant to convince other readers that I must be wrong because people SAY Im wrong, which is kind of interesting. Math society has gone rogue. Given a very simple and basic proof of a problem in the discipline, members of math society are choosing a fantasy world, where their discipline is perfect. Itd be like if Einstein introduced the general theory of relativity, and physicists spent a lot of time calling him a crank, refused to acknowledge experimental results verifying his theories, and spent a lot of effort trying to make sure that the rest of the world, never knew who he was. Mathematicians are basically behaving like sorry excuses for intellectuals. James Harris === Subject: Re: Math dependency logic REVISED readers -- and just who are these ghostly entities? have you gotten recruits from the Sokal School of Numbertheory, or what? anyway, you dont think that Einstein made any mistakes in his math?... how about his hypotheses? and what about Ienstien? > And now readers should notice the psych out game. --Dec.2000 ïWAND Chairman Paul ONeill, reelected to Board. Newsish? http://www.rand.org/publications/randreview/issues/rr.12.00/ http://members.tripod.com/~american_almanac === Subject: Re: Math dependency logic REVISED comments > I made. They were made under the assumption that James Harris was about 15. > This is apparently incorrect. James, give it up. Your math isnt coherent. Its nonsense. Utter > nonsense. It is entertaining to read the arguments though. (The arguments > between you and the other posters to the group. Reading your mathematical > ïarguments is painful, at best.) Justin Van Winkle And now readers should notice the psych out game. Usenet is easy, if you dont like a poster, dont read them. However, math society is STUCK. If it completely ignores me, then I > can explain without hecklers distracting or lying, and people might > see how simple my argument is, and realize that mathematicians are > engaging in an amazing fraud. If they reply to me with mathematics, either they have to admit Im > right, or obfuscate. But I SIMPLIFY, as Ive been doing lately, which > eventually puts them in a box. Math IS wonderful in that way. So some of them just switch to insulting posts meant to convince other > readers that I must be wrong because people SAY Im wrong, which is > kind of interesting. Math society has gone rogue. Given a very simple and basic proof of a > problem in the discipline, members of math society are choosing a > fantasy world, where their discipline is perfect. Itd be like if Einstein introduced the general theory of relativity, > and physicists spent a lot of time calling him a crank, refused to > acknowledge experimental results verifying his theories, and spent a > lot of effort trying to make sure that the rest of the world, never > knew who he was. Mathematicians are basically behaving like sorry excuses for > intellectuals. > James Harris But see, most everyone here has formal training in mathematics. I could argue with you all day long about a physics idea (since you have a physics degree), but if you disagree with me, does that entitle me to call you a liar? Why or why not? David Moran === Subject: Re: Math dependency logic REVISED Do you just have a lot of time on your hand (and no life), or do you have a *machine* write this for you? > One of the more odd things that can happen when you deal with a > discoverer is that you find that things you thought were simple, > suddenly seem complicated and iffy. So Im now facing a problem where > people are doubting algebra, which leads to the fun question of, just > what is the logic here? The issue has to do with a factor like x+2 and how you know that 2 is > constant. If the logic is expanded upon, my hope is that many of you will > realize what is happening in my core error proof, and quit doubting, > if you are doubting, as if youre not, and understand it, then thats > another problem entirely. Ok, so how do you know that with even a polynomial factor, like x+2, > that 2 is *constant* and not related to x? Whats the mathematical logic? Ive talked about setting x=0, which would reveal the constant factor > which you can look at, and I think that for some of you, the ability > to *physically* see constants in polynomial factors, has left you > without ever knowing the *why* behind how it works. How about this? There are an infinity of number and letter combinations, of which x+2 > is just one of that infinity. If you set x=0 though, you just have 2, > so how do you know from whence it came? I mean, with just 2, well, its just 2, so theres no reference back > to x, or y, or any other of that infinity of possibilities. The answer is that 2 is just a number that has no other information > except itself. Logically, you can say that 2 is *independent* of any particular > variable, even if its associated with it, like in an expression like > x+2. Now, as 0 clears out variables set to 0, you can show independent > terms by setting a particular value to 0, as that reveals everything > that can be associated with an infinity of other things. Its like how it could be y+2, or z+2, or alpha + 2, as seeing just 2, > doesnt tell you anything about an association. Possibly the problem for some of you is that you *remember* the > previous association. That is, you know that you had x+2, then you > set x=0, so in your mind you see a phantom x that holds the > association. Thats ok, but it doesnt have a *mathematical* reality because your > memory is irrelevant. The reality is that as far as the math is concerned 2 is a number that > can be by itself or associated with x or an infinity of other things, > and the math isnt going to pick or choose based on what YOU remember. It handles all possibilities at once. And with *just* a number, it sees independence. Thats why setting a variable to 0 is such a useful and powerful tool, > as it reveals infinite possibilities by removing a particular > association. My argument works on that principle, as by setting x or m, depending > on what Im calling the variable, I find that I have 7, 7 and 22 as > *constant* terms for factors of a polynomial. I also know that polynomial has a factor that is constant as that > factor is 49. When I divide the polynomial by 49 I have constant terms of 1, 1, and > 22, which tells me something. Now those numbers dont care about x or m. They dont *know* about x > or m, but you may remember them, so you think your memory has some > effect, but the math sees *infinite* possibility, while you may fixate > on an association you remember. In the realm of infinite possibility, its not possible for 7, 7, and > 22 to go to 1, 1, and 22 based on the value of x, where they check it > first to decide how theyll behave, because theyre just NUMBERS, and > they dont have enough data associated with them to be making those > kinds of decisions. Its like, if human beings on earth constrain 7, 7 and 22 so that they > have to worry about the value of x and m, then, wow, weve affected > all reality in all dimensions and all possibilities, just with our > MEMORY of there being an x or m associated in some interesting > expression. But, of course, your thoughts dont constrain those numbers as they > keep operating quite well throughout Totality, despite what you may > believe, or remember. Just remember that when youre considering the issue of constants in a > math argument. > James Harris === Subject: Re: Math dependency logic REVISED > One of the more odd things that can happen when you deal with a > discoverer is that you find that things you thought were simple, > suddenly seem complicated and iffy. So Im now facing a problem where > people are doubting algebra, which leads to the fun question of, just > what is the logic here? The issue has to do with a factor like x+2 and how you know that 2 is > constant. > James, revising this post is not going to help. What on earth do you mean how do you know 2 is constant? Do you mean that sometimes its 3? Technical folk sometimes make a joke like 2+1 = 4, for suitable values of 2. But its a joke, not a philosophical argument. What are you talking about? === Subject: Re: Math dependency logic REVISED One of the more odd things that can happen when you deal with a > discoverer is that you find that things you thought were simple, > suddenly seem complicated and iffy. So Im now facing a problem where > people are doubting algebra, which leads to the fun question of, just > what is the logic here? The issue has to do with a factor like x+2 and how you know that 2 is > constant. > James, revising this post is not going to help. What on earth do you > mean how do you know 2 is constant? Do you mean that sometimes its 3? > Technical folk sometimes make a joke like 2+1 = 4, for suitable values > of 2. But its a joke, not a philosophical argument. What are you talking about? Mathematicians have been fighting me on the issue of constant terms being constant. Its crazy, its wacky, and I find it infuriating, so I set up a stage for context for readers on the other newsgroups. Notice the replies I got, and now pay attention carefully to the mathematics these people are fighting. Notice how Ill be strongly emphasizing constant terms all the way down. P(x) = 14706125 x^3 - 900375 x^2 + 17640 x + 1078 which has a constant term that is 1078. Well P(x) can also be written out as P(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3 so I can factor to get P(x) = (5 a_1 + 7)(5 a_2 + 7)(5 a_3 + 7) where the as are roots of a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x). Notice it *appears* that the constant terms for the three factors are all 7, which cant be right, as the constant term of P(x) is 1078, so setting x=0, reveals P(0) = (5(0) + 7)(5(0) + 7)(5(3) + 7) = 7(7)(22) as the cubic defining the as with x=0 is a^3 - 3a^2, which has roots, 0, 0 and 3, and Ive picked a_1 and a_2 for 0, so that leaves a_3 with a value of 3 when x=0. So let a_3 = b_3 + 3, where I keep indices matched. Then I have P(x) = (5 a_1 + 7)(5 a_2 + 7)(5 b_3 + 5(3) + 7) P(x) = (5 a_1 + 7)(5 a_2 + 7)(5 b_3 + 22) and now my constant terms work out correctly. But P(x) has 49 as a factor as every term in P(x) = 14706125 x^3 - 900375 x^2 + 17640 x + 1078 has 49 as a factor, so I can divide by 49, and dividing 1078 by 49 gives me 22, as the new constant term. Well that means that P(x)/49 = (5 a_1/7 + 1)(5 a_2/7 + 1)(5 b_3 + 22) is the only way that the constant terms keep matching. And if you let mathematicians debate that conclusion or cast doubt on it, while calling me names and acting like Im the one whos crazy, then you piss on the intellectual basis for civilization itself. Why let mathematicians get away with it? Whats in it for you? Why let them trample on civilization, as if honesty and integrity in a discipline dont mean anything if enough people in that discipline dont like the truth? Whats wrong with you people? Dont ANY of you believe in anything? James Harris === Subject: Re: Math dependency logic REVISED In sci.physics, James Harris One of the more odd things that can happen when you deal with a >> discoverer is that you find that things you thought were simple, >> suddenly seem complicated and iffy. So Im now facing a problem where >> people are doubting algebra, which leads to the fun question of, just >> what is the logic here? >> The issue has to do with a factor like x+2 and how you know that 2 is >> constant. >> James, revising this post is not going to help. What on earth do you >> mean how do you know 2 is constant? Do you mean that sometimes its 3? >> Technical folk sometimes make a joke like 2+1 = 4, for suitable values >> of 2. But its a joke, not a philosophical argument. >> What are you talking about? Mathematicians have been fighting me on the issue of constant terms > being constant. Its crazy, its wacky, and I find it infuriating, so > I set up a stage for context for readers on the other newsgroups. > Notice the replies I got, and now pay attention carefully to the > mathematics these people are fighting. Notice how Ill be strongly emphasizing constant terms all the way > down. P(x) = 14706125 x^3 - 900375 x^2 + 17640 x + 1078 which has a constant term that is 1078. Well P(x) can also be written out as P(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3 so I can factor to get P(x) = (5 a_1 + 7)(5 a_2 + 7)(5 a_3 + 7) where the as are roots of a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x). Why is x still in this cubic? Are we expected to solve for a as a function of x, while (before? after?) solving for x in the original equation? [rest snipped] -- #191, ewill3@earthlink.net Its still legal to go .sigless. === Subject: Re: Math dependency logic REVISED > Mathematicians have been fighting me on the issue of constant terms > being constant. Step 1. Misrepresent the opposition. [snip other steps, since 1st step is false.] -- A man with integrity identifies, acknowledges and corrects his errors. One without it ignores, denies or defends them. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: Math dependency logic REVISED >> Whats wrong with you people? Dont ANY of you believe in anything? I believe Ill have another beer. === Subject: Re: Math dependency logic REVISED >> Whats wrong with you people? Dont ANY of you believe in anything? I believe Ill have another beer. Im *tired* of the games. Mathematicians are running and there are posters trying to help them run, but Im giving notice, Ill keep putting the information out there. Notice how Ill be strongly emphasizing constant terms all the way down. P(x) = 14706125 x^3 - 900375 x^2 + 17640 x + 1078 which has a constant term that is 1078. Well P(x) can also be written out as P(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3 so I can factor to get P(x) = (5 a_1 + 7)(5 a_2 + 7)(5 a_3 + 7) where the as are roots of a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x). Notice it *appears* that the constant terms for the three factors are all 7, which cant be right, as the constant term of P(x) is 1078, so setting x=0, reveals P(0) = (5(0) + 7)(5(0) + 7)(5(3) + 7) = 7(7)(22) as the cubic defining the as with x=0 is a^3 - 3a^2, which has roots, 0, 0 and 3, and Ive picked a_1 and a_2 for 0, so that leaves a_3 with a value of 3 when x=0. So let a_3 = b_3 + 3, where I keep indices matched. Then I have P(x) = (5 a_1 + 7)(5 a_2 + 7)(5 b_3 + 5(3) + 7) P(x) = (5 a_1 + 7)(5 a_2 + 7)(5 b_3 + 22) and now my constant terms work out correctly. But P(x) has 49 as a factor as every term in P(x) = 14706125 x^3 - 900375 x^2 + 17640 x + 1078 has 49 as a factor, so I can divide by 49, and dividing 1078 by 49 gives me 22, as the new constant term. Well that means that P(x)/49 = (5 a_1/7 + 1)(5 a_2/7 + 1)(5 b_3 + 22) is the only way that the constant terms keep matching. And if you let mathematicians debate that conclusion or cast doubt on it, while calling me names and acting like Im the one whos crazy, then you piss on the intellectual basis for civilization itself. Why let mathematicians get away with it? Whats in it for you? Why let them trample on civilization, as if honesty and integrity in a discipline dont mean anything if enough people in that discipline dont like the truth? Whats wrong with you people? Dont ANY of you believe in anything? James Harris === Subject: Re: Math dependency logic REVISED One of the interesting things Ive noticed about mathematics is that the truth or falsity of a mathematical statement is not affected by the number of times it is repeated. Gib === Subject: Re: Math dependency logic REVISED >One of the interesting things Ive noticed about mathematics is that the >truth or falsity of a mathematical statement is not affected by the >number of times it is repeated. You can say that again. >Gib ************************ === Subject: Re: Math dependency logic REVISED [.snip.] >Notice how Ill be strongly emphasizing constant terms all the way >down. P(x) = 14706125 x^3 - 900375 x^2 + 17640 x + 1078 which has a constant term that is 1078. Well P(x) can also be written out as P(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3 so I can factor to get P(x) = (5 a_1 + 7)(5 a_2 + 7)(5 a_3 + 7) where the as are roots of a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x). Notice it *appears* that the constant terms for the three factors are >all 7, which cant be right, as the constant term of P(x) is 1078, so >setting x=0, reveals P(0) = (5(0) + 7)(5(0) + 7)(5(3) + 7) = 7(7)(22) as the cubic defining the as with x=0 is a^3 - 3a^2, which has roots, 0, 0 and 3, and Ive picked a_1 and a_2 >for 0, so that leaves a_3 with a value of 3 when x=0. So let a_3 = b_3 + 3, where I keep indices matched. Then I have P(x) = (5 a_1 + 7)(5 a_2 + 7)(5 b_3 + 5(3) + 7) P(x) = (5 a_1 + 7)(5 a_2 + 7)(5 b_3 + 22) and now my constant terms work out correctly. But P(x) has 49 as a factor as every term in P(x) = 14706125 x^3 - 900375 x^2 + 17640 x + 1078 has 49 as a factor, so I can divide by 49, and dividing 1078 by 49 >gives me 22, as the new constant term. Well that means that P(x)/49 = (5 a_1/7 + 1)(5 a_2/7 + 1)(5 b_3 + 22) is the only way that the constant terms keep matching. No. Heres another way: P(x)/49 = [ (5a_1(x)/w_1(x) + (7/w_1(x)) -1) + 1 ] * [ (5a_2(x)/w_2(x) + (7/w_2(x)) -1) + 1 ] * [ ( ( (5b_3(x)+22)/w_3(x) ) -22 ) + 22 ] where w_1(x) is the common factor of a_1(x) and 7 (which happens to equal 7 at x=0); w_2(x) is the common factor of a_2(x) and 7 (which happens to equal 7 at x=0); and w_3(x) is the common factor of a_3(x) and 7 (which happens to equal 1 at x=0). So long as w_1(x)*w_2(x)*w_3(x) = 49, the equality is true, the constant terms still match. You are ASSUMING that 7/w_1(x) = 1 for all values of x. But that is what you want to PROVE. So you are engaging in a circular argument. Remember: your definitions are: the constant term of f(x) is f(0); and we write f(x) as f(x) = g(x) + f(0), where g(x) = f(x)-f(0). That is, all you are doing is adding and subtracting f(0) in your expression, just as I did above. In order to conclude that (5a_1(x)/w_1(x) + (7/w_1(x)) -1) = 5a_1(x)/7 you must be assuming that w_1(x) = 7 for all x. But that is your claimed CONCLUSION, so you cannot assume it. To conclude that (5a_2(x)/w_2(x) + (7/w_2(x)) -1) = 5a_2(x)/7 for all x, you must be assuming that w_2(x)=7 for all x, but that is your claimed CONCLUSION, so you cannot assume it. To conclude that (5b_3(x)+22)/w_3(x) ) -22 = 5b_3(x)+22 you must be assuming that w_3(x) = 1 for all x; but that is your claimed CONCLUSION, so you cannot assume it.