Infinite Series Involving PiDate: 12/16/97 at 01:00:59 From: Eric Tak Subject: Infinite series involving pi Sorry, I am just totally confused on this problem. Here it is: 1 + 1/2^2 + 1/3^2 + 1/4^2 + 1/5^2 + . . . . . = pi^2/6 Please, I need a clear, easy-to-understand reason or proof for WHY this is correct. If a simple solution is unavailable, a hard solution will also be greatly appreciated. Also, what does: 1 - 1/2^2 + 1/3^2 - 1/4^2 + 1/5^2 - . . . equal? Thank you! Date: 12/16/97 at 19:30:24 From: Doctor Anthony Subject: Re: Infinite series involving pi This can be proved by advanced trig. manipulations, but a more elegant method is with the aid of Fourier series. A brief note on Fourier series is necessary if you are to understand the method. For a great many functions within the range -pi < x < pi we can express f(x) as an infinite series of the form: f(x) = a(0)/2 + SUM(n=1 to inf.)[a(n)cos(nx) + b(n)sin(nx)] where the a(i), b(i) are constants. So f(x) = a(0)/2 + a(1)cos(x) + a(2)cos(2x) + a(3)cos(3x) + .... + b(1)sin(x) + b(2)sin(2x) + b(3)sin(3x) + .... This series is periodic with period 2.pi and the convention is to take the interval -pi < x < pi. The way to evaluate the constants a(i),is to multiply both sides by cos(nx) and then integrate between -pi and pi. INT(-pi to pi)[f(x)cos(nx)dx] = pi.a(n) (proof given below) So a(n) = (1/pi)INT(-pi to pi)[f(x)cos(nx)dx] Similarly b(n) = (1/pi)INT(-pi to pi)[f(x)sin(nx)dx] and a(0) = (1/pi)INT(-pi to pi)[f(x).dx] Proof of above results. INT(-pi to pi)[cos(mx)cos(nx)dx] m not equal n = (1/2)INT[cos(m+n)x + cos(m-n)x]dx between -pi and pi = 0 since sin(any multiple of pi) = 0 but INT(-pi to pi)[cos^2(nx)]dx = pi So all the terms cos(mx)cos(nx) (m not equal to n) disappear and we get only the term in cos^2(nx) This gives the result INT[f(x)cos(nx)dx] = a(n).pi and so a(n) = (1/pi)INT[f(x).cos(nx).dx] In the case n=0 we get INT[f(x).dx] = INT[a(0)/2)dx] + 0 + 0+ .. = a(0).pi So a(0) = (1/pi)INT[f(x).dx] Similarly b(n) = (1/pi)INT[f(x)sin(nx)dx] since all terms in sin(mx)sin(nx) (m not equal to n) disappear leaving only the term in sin^2(nx). Also all terms in sin(mx)cos(nx) whether or not m=n disappear on integration between -pi and pi. This means that we can evaluate all the a(i), b(i) and express f(x) as an infinite series in sines and cosines of multiples of x. There are a great many functions that can be expressed in this way in the interval -pi to pi. We can now turn our attention to finding the sum of the series 1 + 1/2^2 + 1/3^2 + 1/4^2 + .... Let f(x) = x^2 This is an even function of x so we know that NONE of the sine terms of the Fourier series can be present. We need only look at the a(i) terms. For all a(n) we get pi.a(n) = INT(-pi to pi)[x^2.cos(nx)dx] When n = 0 this gives a(0) = (2/3)pi^2 When n greater than 0 integration by parts in 3 steps gives a(n) = 4(-1)^n/n^2 The Fourier series is pi^2/3 + 4.SUM(1 to infinity)(-1)^n/n^2 cos(nx) With x=pi, we get pi^2 = pi^2/3 + 4.SUM(-1)^n/n^2 cos(n.pi) and since cos(n.pi)= (-1)^n this produces +'s for each term 2.pi^2/3 = 4.SUM[1 + 1/2^2 + 1/3^2 + ..... to infinity] pi^2/6 = 1 + 1/2^2 + 1/3^2 + 1/4^2 + ..... to infinity. As for the other series, first note that: 1 - 1/4 + 1/9 - 1/16 + 1/25 - ....... = pi^2/12 Then let f(x) = x^2. This is an even function of x so we know that NONE of the sine terms of the Fourier series can be present. We need only look at the a(i) terms. For all a(n) we get pi.a(n) = INT(-pi to pi)[x^2.cos(nx)dx] When n = 0 this gives a(0) = (2/3)pi^2 When n greater than 0 integration by parts in 3 steps gives: a(n) = 4(-1)^n/n^2 The Fourier series is pi^2/3 + 4.SUM(1 to infinity)(-1)^n/n^2 cos(nx) With x=0, we get: 0 = pi^2/3 + 4.SUM(-1)^n/n^2 cos(0) and since cos(0) = 1 this produces alternate + and - for the terms. -pi^2/3 = 4.SUM[-1 + 1/2^2 - 1/3^2 + ..... to infinity] -pi^2/12 = -1 + 1/2^2 - 1/3^2 + 1/4^2 + ..... to infinity. pi^2/12 = 1 - 1/4 + 1/9 - 1/16 + ... to infinity (Well you did ask for a 'hard' solution.) -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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