Sum of infinite Fourier series

  • Thread starter bobred
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  • #1
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Homework Statement


Show that

[tex]\sum_{r=0}^\infty\frac{1}{(2r+1)^2}=\frac{\pi^2}{8}[/tex]


Homework Equations


The equation of the function is

[tex]F(t)&=&\dfrac{\pi}{4}-\dfrac{2}{\pi}\left(\cos t+\dfrac{\cos3t}{3^{2}}+\dfrac{\cos5t}{5^{2}}+\cdots\right)-\left(\sin t-\dfrac{\sin2t}{2}+\dfrac{\sin3t}{3}-\cdots\right)[/tex]


The Attempt at a Solution


We are given the condition that t=0, so the cos terms are all 1 giving

[tex]\left(\cos t+\dfrac{\cos3t}{3^{2}}+\dfrac{\cos5t}{5^{2}}+\cdots\right)=\dfrac{1}{1^{2}}+\dfrac{1}{3^{2}}+\dfrac{1}{5^{2}}+\dfrac{1}{7^{2}}\cdots=\frac{\pi^2}{8}[/tex]
 
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Answers and Replies

  • #2
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Homework Statement


Show that

[tex]\sum_{i=0}^\infty\frac{1}{(2r+1)^2}=\frac{1}{8}[/tex]


Homework Equations





The Attempt at a Solution


Did you write the question incorrectly? The expression that you're summing doesn't have the indexing variable in it anywhere, so it diverges toward infinity:
[tex]\sum_{i=0}^\infty\frac{1}{(2r+1)^2}=\frac{1}{(2r+1)^2}\sum_{i=0}^\infty1=\frac{1}{(2r+1)^2}\infty = \infty[/tex]
 
  • #3
hunt_mat
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I think it's just a typo.
 
  • #4
173
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Hi

The i should have been r.

I know from elsewhere that the sum is [tex]\frac{\pi^{2}}{8}[/tex], but I haven't really shown that, any hints?

Thanks

James
 
  • #5
hunt_mat
Homework Helper
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You have to know the original function which the series you have written down represents. Evaluate this function at t=0 and do a little algebra.

Mat
 
  • #6
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The original function is piecewise

[tex]f(t)=\begin{cases}
-t & \left(-\pi<t\leq0\right)\\
0 & \left(0<t\leq\pi\right)\end{cases}[/tex]

With a period of [tex]2\pi[/tex]

James
 
  • #7
hunt_mat
Homework Helper
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You have the value of it's Fourier series at t=0, now calculate what f(0) is and equate the two values.
 
  • #8
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Hi

Of course, duh.

Thanks for your help

James
 
  • #9
173
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Hi

I am next asked to choose an appropriate value for t and find the value of convergence of the following,

[tex]
\sum_{r=0}^\infty\frac{(-1)^r}{2r+1}=\frac{1}{1}-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}
[/tex]

I know this converges to [tex]\frac{\pi}{4}[/tex], but I am not sure where to start, the function is still the same as at the beginning of the thread

Any ideas where to start?

Thanks, James
 
  • #10
Gib Z
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Integrate both the original function and its Fourier Series term by term. Then try to substitute some value of x in to get that result.
 
  • #11
hunt_mat
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Don't integrate but differentiate.
 
  • #12
Gib Z
Homework Helper
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Don't integrate but differentiate.

Whoops, my mistake.
 
  • #13
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I differentiate the approximation and set [tex]t=-\frac{\pi}{2}[/tex] giving

[tex]\frac{\pi}{2}=\frac{2}{\pi}\sum_{r=0}^\infty\frac{1}{2r+1}\sin((2r+1)t)[/tex] as sin alternates sign we get

[tex]\frac{\pi}{2}=\frac{2}{\pi}\sum_{r=0}^\infty\frac{(-1)^r}{2r+1}[/tex]

To which I find

[tex]\sum_{r=0}^\infty\frac{(-1)^r}{2r+1}=\frac{\pi^2}{4}[/tex] and not [tex]\frac{\pi}{4}[/tex]

Any ideas where I'm going wrong?

Thanks
 
  • #14
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5
[tex]
\sum_{r = 0}^{\infty}{\frac{1}{(2 r + 1)^{2}}} = \sum_{r = 0}^{\infty}{\frac{1}{r^{2}}} - \sum_{r = 1}^{\infty}{\frac{1}{(2 r)^{2}}} = \left(1 - \frac{1}{4}\right) \, \sum_{r = 1}^{\infty}{\frac{1}{r^{2}}}
[/tex]

The sum:

[tex]
\zeta(2) \equiv \sum_{r = 1}^{\infty}{\frac{1}{r^{2}}}
[/tex]

can be found by expanding the function [itex]f(x) = x^{2}[/itex] in Fourier series in the region [itex][-\pi, \pi][/itex] and taking its value for [itex]x = \pi[/itex].
 
  • #15
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Hi

Sorry for my ignorrance, but I can't see how this helps with the question (see post#9)

Jamees
 
  • #16
Mute
Homework Helper
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I differentiate the approximation and set [tex]t=-\frac{\pi}{2}[/tex] giving

[tex]\frac{\pi}{2}=\frac{2}{\pi}\sum_{r=0}^\infty\frac{1}{2r+1}\sin((2r+1)t)[/tex] as sin alternates sign we get

[tex]\frac{\pi}{2}=\frac{2}{\pi}\sum_{r=0}^\infty\frac{(-1)^r}{2r+1}[/tex]

To which I find

[tex]\sum_{r=0}^\infty\frac{(-1)^r}{2r+1}=\frac{\pi^2}{4}[/tex] and not [tex]\frac{\pi}{4}[/tex]

Any ideas where I'm going wrong?

Thanks

The problem is that you differentiated f(t), so your series converges to f'(t), which is either equal to -1 or 0, but you set to it f'(t=-pi/2) = pi/2 and equated the pi/2 to the series instead of the actual value of f(t=-pi/2) = -1.
 
  • #17
vela
Staff Emeritus
Science Advisor
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Why not just set t=pi/2 in the original function?
 
  • #18
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Hi

If I put [tex]t=\pi/2[/tex] the [tex]f(\pi/2)=0[/tex] and we get

[tex]\frac{\pi}{4}=\sum_{r=0}^\infty\frac{(-1)^r}{2r+1}[/tex]

Thanks everyone for your help.

James
 

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