Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP Posting-Version: version B 2.10.2 9/5/84; site ism780c.UUCP Path: utzoo!watmath!clyde!burl!ulysses!bellcore!decvax!ittatc!dcdwest!sdcsvax!sdcrdcf!ism780c!tim From: tim@ism780c.UUCP (Tim Smith) Newsgroups: net.math,net.physics Subject: Re: value of an integral ( zeta(4) by elementary means ) Message-ID: <805@ism780c.UUCP> Date: Fri, 28-Feb-86 22:41:46 EST Article-I.D.: ism780c.805 Posted: Fri Feb 28 22:41:46 1986 Date-Received: Sun, 2-Mar-86 07:35:57 EST References: <823@drux2.UUCP> <896@yale.ARPA> <796@ism780c.UUCP> Reply-To: tim@ism780c.UUCP (Tim Smith) Organization: Interactive Systems Corp., Santa Monica, CA Lines: 163 Xref: watmath net.math:2918 net.physics:3895 makdisi@yale-cheops.UUCP (Kamal Khuri-Makdisi) writes: > > The only proof I know that the sum of 1/n^4, n = 1 to infinity, > is pi^2/90 involves Fourier series -- anyone know a more > elementary way of proving this? > That's pi^4/90. An elementary proof is in "Challenging Mathematical Problems with Elementary Solutions, Volume II", by A.M. Yaglom and I.M. Yaglom. This is problem 145. Here is what they do: ( in the following, things like n,m,j, etc, are integers ) First, we will establish that zeta(2) = pi^2/6. This is just to show how they approach this kind of thing, and besides, we might as well be complete. First, establish the formula sin nx = C(n,1) sin^1 (x) cos^(n-1) (x) - C(n,3) sin^3 (x) cos^(n-3) (x) + C(n,5) sin^5 (x) cos^(n-5) (x) - ... which you can do by induction. This can be rewritten as sin nx = sin^n (x) [ C(n,1) cot^(n-1)(x) - C(n,3) cot^(n-3)(x) + ... ] We let n = 2m+1, and let x = j pi / n, 1 <= j <= m. Then we have sin(x) != 0, and sin(nx) = 0. This gives us C(2m+1,1) cot^(2m)(x) - C(2m+1,3) cot^(2m-2) (x) + ... = 0 or, in other words, the polynomial C(2m+1,1) z^m - C(2m+1,3) z^(m-1) + ... = P(z) has the roots cot^2 ( pi / n ), cot^2 ( 2 pi /n ), ... , cot^2 ( m pi /n ). Now, the sum of the roots of a polynomial A0 y^n + A1 y^(n-1) + ... is -A1/A0. Thus, cot^2 (pi/n) + cot^2 (2 pi/n) + ... + cot^2 (m pi/n) = C(2m+1,3) / C(2m+1,1 ) = m(2m-1)/3 Noting that csc^2 u = cot^2 u + 1, we get that csc^2 (pi/n) + csc^2 (2 pi/n) + ... + csc^2 (m pi/n) = m(2m+2)/3 Now we are ready to go places! From our elementary trig classes we know that 0 < cot w < 1/w < csc w when 0 < w < pi/2 or cot^2 w < 1/w^2 < csc^2 w Using this, and the formulas above for sums of cot^2 and csc^2, we get m(2m-1)/3 < n^2/pi^2 * ( 1/1^2 + 1/2^2 + ... + 1/m^2 ) < m(2m+2)/3 or ( putting in 2m+1 for n), m(2m-1) pi^2 m(2m+2) pi^2 ------------ < 1/1^2 + 1/2^2 + ... + 1/m^2 < ------------ 3 ( 2m+1 )^2 3 ( 2m+1 )^2 As m -> oo, the things on the end both -> pi^2/6. Now for zeta(4)! We need to evaluate the sum cot^4 (pi/n) + cot^4 (2 pi/n) + ... + cot^4 (m pi/n) In other words, the sum of the squares of the roots of of P(z). If we consider a polynomial A0 y^n + A1 y^(n-1) + A2 y^(n-2) + ..., with roots R1,R2,...,Rn, then we have sum Ri = -A1/A0, and sum RiRj ( i oo, we get zeta(4) = pi^4/90 They point out that we may evaluate sum cot^6, sum cot^8, etc, in the same sort of way, and get zeta(6) = pi^6/945 zeta(8) = pi^8/9450 zeta(10) = pi^10/93555 zeta(12) = 691 pi^12 / 638512875 Some other formulas for pi that they get by playing with trig functions are Vieta's formula: Let R1 = (1/2)^(1/2), and R(n+1) = (1/2 + 1/2 Rn)^(1/2) Then 2/pi = R1 R2 R3 ... Leibniz's formula: pi/4 = 1 - 1/3 + 1/5 - 1/7 + ... and Wallis's formula: pi 2 2 4 4 6 6 8 8 -- = - - - - - - - - ... 2 1 3 3 5 5 7 7 9 This is a great book. Ubizmo, these computers sure suck for typing equations! -- Tim Smith sdcrdcf!ism780c!tim || ima!ism780!tim || ihnp4!cithep!tim