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Fouiers series Calculate sum

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dikmikkel
#1
Nov3-11, 09:24 AM
P: 169
1. The problem statement, all variables and given/known data
Consider the 2[itex]\pi[/itex]-periodic function f(t) = t t in [-Pi;Pi]
a) show that the real fouier series for f(t) is:
[itex]f(t) ~ \sum\limits_{n=1}^{\infty}\frac{2}{n}(-1)^{n+1}\sin nt[/itex]
b)
Use the answer to evaluate the following : [itex]\sum\limits_{n=1}^{\infty}\dfrac{(-1)^{n+1}}{2n-1}[/itex]
Hint: Use Fouier's law with t = [itex]\pi[/itex]/2

2. Relevant equations
Fouiers Law? I'm danish, and therefore i'm not really sure what it's called.


3. The attempt at a solution
Part a i have done by finding the coefficients.
Part b) I can't see where the problem in part b and the answer to a relates. I've tried with Maple 15 to calculate the value and i'm getting Pi/4, but i keep getting something different for the series from a)
Please Help me, i would really like to understand this as i'm studying physics.
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ahm_11
#2
Nov3-11, 10:57 AM
P: 6
The answer should be pi/4 ...


from fourier series we got ...

t = [itex]\sum\frac{2}{n}(-1)n+1sinnt[/itex]

for t = [itex]\frac{\pi}{2}[/itex] ,

[itex]\frac{\pi}{2}[/itex] = 2 - 2/3 + 2/5 - 2/9 + ......

which followed ,

[itex]\frac{\pi}{2}[/itex] = 2 * [itex]\sum\frac{(-1)n+1}{2n - 1}[/itex]

hence, [itex]\sum\frac{(-1)n+1}{2n - 1}[/itex] = [itex]\frac{\pi}{4}[/itex]
dikmikkel
#3
Nov3-11, 11:41 AM
P: 169
Hmm, isn't that a backwards proof? Can't i use something like the even terms of sin(n*Pi/2) = 0 and the odd ones alternates between -1 and 1

And how is t = ... isn't it f(t)?


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