# Sum of infinite Fourier series

1. Jul 20, 2010

### bobred

1. The problem statement, all variables and given/known data
Show that

$$\sum_{r=0}^\infty\frac{1}{(2r+1)^2}=\frac{\pi^2}{8}$$

2. Relevant equations
The equation of the function is

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

3. The attempt at a solution
We are given the condition that t=0, so the cos terms are all 1 giving

$$\left(\cos t+\dfrac{\cos3t}{3^{2}}+\dfrac{\cos5t}{5^{2}}+\cdots\right)=\dfrac{1}{1^{2}}+\dfrac{1}{3^{2}}+\dfrac{1}{5^{2}}+\dfrac{1}{7^{2}}\cdots=\frac{\pi^2}{8}$$

Last edited: Jul 20, 2010
2. Jul 20, 2010

### xcvxcvvc

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:
$$\sum_{i=0}^\infty\frac{1}{(2r+1)^2}=\frac{1}{(2r+1)^2}\sum_{i=0}^\infty1=\frac{1}{(2r+1)^2}\infty = \infty$$

3. Jul 20, 2010

### hunt_mat

I think it's just a typo.

4. Jul 20, 2010

### bobred

Hi

The i should have been r.

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

Thanks

James

5. Jul 20, 2010

### hunt_mat

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. Jul 20, 2010

### bobred

The original function is piecewise

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

With a period of $$2\pi$$

James

7. Jul 20, 2010

### hunt_mat

You have the value of it's Fourier series at t=0, now calculate what f(0) is and equate the two values.

8. Jul 20, 2010

### bobred

Hi

Of course, duh.

James

9. Jul 22, 2010

### bobred

Hi

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

$$\sum_{r=0}^\infty\frac{(-1)^r}{2r+1}=\frac{1}{1}-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}$$

I know this converges to $$\frac{\pi}{4}$$, 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. Jul 22, 2010

### Gib Z

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. Jul 23, 2010

### hunt_mat

Don't integrate but differentiate.

12. Jul 23, 2010

### Gib Z

Whoops, my mistake.

13. Jul 25, 2010

### bobred

I differentiate the approximation and set $$t=-\frac{\pi}{2}$$ giving

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

$$\frac{\pi}{2}=\frac{2}{\pi}\sum_{r=0}^\infty\frac{(-1)^r}{2r+1}$$

To which I find

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

Any ideas where I'm going wrong?

Thanks

14. Jul 25, 2010

### Dickfore

$$\sum_{r = 0}^{\infty}{\frac{1}{(2 r + 1)^{2}}} = \sum_{r = 0}^{\infty}{\frac{1}{r^{2}}} - \sum_{r = 1}^{\infty}{\frac{1}{(2 r)^{2}}} = \left(1 - \frac{1}{4}\right) \, \sum_{r = 1}^{\infty}{\frac{1}{r^{2}}}$$

The sum:

$$\zeta(2) \equiv \sum_{r = 1}^{\infty}{\frac{1}{r^{2}}}$$

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

15. Jul 25, 2010

### bobred

Hi

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

Jamees

16. Jul 25, 2010

### Mute

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. Jul 25, 2010

### vela

Staff Emeritus
Why not just set t=pi/2 in the original function?

18. Jul 26, 2010

### bobred

Hi

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

$$\frac{\pi}{4}=\sum_{r=0}^\infty\frac{(-1)^r}{2r+1}$$