Convergence of Fouier Series and Evaluating Sums Using Fouier's Law

[dikmikkel]

Homework Statement

Consider the 2$\pi$-periodic function f(t) = t t in [-Pi;Pi]
a) show that the real fouier series for f(t) is:
$f(t) ~ \sum\limits_{n=1}^{\infty}\frac{2}{n}(-1)^{n+1}\sin nt$
b)
Use the answer to evaluate the following : $\sum\limits_{n=1}^{\infty}\dfrac{(-1)^{n+1}}{2n-1}$
Hint: Use Fouier's law with t = $\pi$/2

Homework Equations

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

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.

 

[ahm_11]

The answer should be pi/4 ...


from Fourier series we got ...

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

for t = $\frac{\pi}{2}$ ,

$\frac{\pi}{2}$ = 2 - 2/3 + 2/5 - 2/9 + ...

which followed ,

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

hence, [itex]\sum\frac{(-1)ⁿ⁺¹}{2n - 1}[/itex] = $\frac{\pi}{4}$

 

[dikmikkel]

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)?