Prove the series has the value
Kenshin
1/(1^3) - 1/(3^3) + 1/(5^3) - 1/(7^3)+...=pi^3 / 8
The hint that my professor said is to use zeta function in some way,
but I can't use it...
Any idea? You don't need to use it, surely.
James Van Buskirk
news:753afc26-f57a-4c09...@k15g2000prk.googlegroups.com...
> Prove that
I like the approach taken by Murray Spiegel in his Schaum's Outline
on Fourier Analysis. Since you want the coefficients to decay as
1/n**3, we would like a function continuous through the first
derivative but with a discontinuity in the second derivative. Let
f(0) = f(1) = 0 and f'(0) = f'(1) = 1. Since there are four
conditions we need at least a cubic polynomial to satisfy them.
The first two conditions require that f(x) = x*(x-1)*(A*x+B).
f'(0) = -B = 1. f'(1) = A+B = A-1 = 1, so
f(x) = x*(x-1)*(2*x-1) = 2*x**3-3*x**2+x. Now take the periodic
extension of f from the interval [0,1] to the real numbers.
f(x) = a(0)+sum([(a(n)*cos(2*pi*n*x)+b(n)*sin(2*pi*n*x),n=1,infinity)])
Since f(1-x) = -f(x) the periodic extension is an odd function of
x so a(n) = 0 for all n.
b(n)/2 = integral from 0 to 1 of f(x)*sin(2*pi*n*x)*dx = 3/(2*n**3*pi**3)
So f(x) = 2*x**3-3*x**2+x
= sum([(3*sin(2*pi*n*x)/(n**3*pi**3),n=1,infinity)])
Now that we have the answer, what was the question? Oh yes, you
would like an alternating series. What happens if x = 1/4? We
get f(1/4) = 3/32 = sum([(3*sin(pi*n/2)/(n**3*pi**3),n=1,infinity)])
= sum([(3*sin(pi*(2*n+1)/2)/((2*n+1)**3*pi**3),n=0,infinity)])
= sum([(3*(-1)**n/((2*n+1)**3*pi**3),n=0,infinity)])
So sum([((-1)**n/(2*n+1)**3,n=0,infinity)]) = pi**3/32.
You might also find many ideas in chapter 1 or 2 of Titchmarsh.
--
write(*,*) transfer((/17.392111325966148d0,6.5794487871554595D-85, &
6.0134700243160014d-154/),(/'x'/)); end
Gerry Myerson
<753afc26-f57a-4c09...@k15g2000prk.googlegroups.com>,
Kenshin <ksa...@gmail.com> wrote:
> Prove that
>
> 1/(1^3) - 1/(3^3) + 1/(5^3) - 1/(7^3)+...=pi^3 / 8
Surely that 8 should be 32. Formula originally due
to Euler, 1748.
--
Gerry Myerson (ge...@maths.mq.edi.ai) (i -> u for email)
Han de Bruijn
Rob Johnson
I don't know about using the zeta function, but you should look at the
Dirichlet beta function:
<http://mathworld.wolfram.com/DirichletBetaFunction.html>
+oo
--- k 1
beta(x) = > (-1) -------- [1]
--- (2k+1)^x
k=0
These have nice representations as rational multples of pi^x when x
is a positive odd integer; e.g.
1 1 1 1 pi
--- - --- + --- - --- + ... = -- [2a]
1^1 3^1 5^1 7^1 4
1 1 1 1 pi^3
--- - --- + --- - --- + ... = ---- [2b]
1^3 3^3 5^3 7^3 32
1 1 1 1 5 pi^5
--- - --- + --- - --- + ... = ------ [2c]
1^5 3^5 5^5 7^5 1536
One way to evaluate these sums is by using the following identity
+oo
--- 1
> --- = pi cot(pi x) [3]
--- k+x
k=-oo
See <http://www.whim.org/nebula/math/infharmseries.html>.
Evaluating [3] at x = 1/4, we get [2a].
Taking two derivatives of [3], we get
+oo
--- 1 3 2
> ------- = pi cot(pi x) csc (pi x) [4]
--- (k+x)^3
k=-oo
Taking two derivatives of [4], we get
+oo
--- 1
> ------- [5]
--- (k+x)^5
k=-oo
1 5 3 2 4
= - pi (cot (pi x) csc (pi x) + 2 cot(pi x) csc (pi x))
3
Evaluating [4] and [5] at x = 1/4, we get [2b] and [2c].
Another way to look at these types of series is through generating
functions. The generating function of beta(2n+1) is
g(x)
+oo
--- 2n+1
= > beta(2n+1) x
---
n=1
+oo +oo
--- --- k x^{2n+1}
= > > (-1) -------------
--- --- (2k+1)^{2n+1}
n=1 k=0
+oo
--- k x/(2k+1)
= > (-1) ----------------
--- 1 - (x/(2k+1))^2
k=0
+oo
x --- k 1 1
= - > (-1) ( ------ + ------ )
2 --- 2k+1+x 2k+1-x
k=0
+oo
x --- k 1
= - > (-1) ------
2 --- 2k+1+x
k=-oo
+oo
x --- k 1
= - > (-1) ---------
4 --- k+(1+x)/2
k=-oo
x
= - pi csc(pi(1+x)/2)
4
= pi/4 x sec(pi/2 x) [6]
Indeed, the power series for pi/4 x sec(pi/2 x) is
pi pi^3 3 5 pi^5 5 61 pi^7 7 277 pi^9 9
-- x + ---- x + ------- x + ------- x + -------- x + ...
4 32 1536 184320 8257536
which agrees with [2].
Since cos(x) has a nice power series, we can derive a nice recurrence
for beta(2n+1):
For n = 0,
beta(1) = pi/4
and for n > 0,
n-1
--- (-pi^2/4)^{n-k}
beta(2n+1) = - > beta(2k+1) ---------------
--- (2n-2k)!
k=0
Rob Johnson <r...@trash.whim.org>
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Rob Johnson
I have written up a short paper detailing the computation of the
generating function and its use to derive a recurrence relation
for both zeta(2k) and beta(2k+1).
<http://www.whim.org/nebula/math/pdf/zeta-beta.pdf>