MHB Solve b) Using Parseval's Theorem

[goohu]

Hello good folks!

I'm stuck trying to solve the problem b). In the theory book examples they are skipping steps and shortly states 'use algebra' and parsevals theorem to rewrite the Fourier series into the answer that is given.

So I've tried to use parsevals theorem but I still can't rewrite the result into the sum we are looking for.

View attachment 9282

In case the picture is too blurry;

a)

f(t) = $$\frac{{pi}^{2}}{3} + \sum_{k=1}^{\infty} \frac{4*{-1}^{k}}{{k}^{2}} cos(kt)$$

b) Calculate the sum of $$\sum_{k=1}^{\infty} \frac{{-1}^{k+1}}{{k}^{2}}$$

 

[I like Serena]

Hello good folks!

I'm stuck trying to solve the problem b). In the theory book examples they are skipping steps and shortly states 'use algebra' and parsevals theorem to rewrite the Fourier series into the answer that is given.

So I've tried to use parsevals theorem but I still can't rewrite the result into the sum we are looking for.
In case the picture is too blurry;

a)

f(t) = $$\frac{{pi}^{2}}{3} + \sum_{k=1}^{\infty} \frac{4*{-1}^{k}}{{k}^{2}} cos(kt)$$

b) Calculate the sum of $$\sum_{k=1}^{\infty} \frac{{-1}^{k+1}}{{k}^{2}}$$

Hi goohu,

We also have that $f(t)=t^2$.
What do we get if we substitute $t=0$?

 

[goohu]

f(t)=t^2.
f(0) = 0.

If we plug in t=0 into the Fourier series the cos term simply becomes 1. But what do we do from here?

 

[I like Serena]

f(t)=t^2.
f(0) = 0.

If we plug in t=0 into the Fourier series the cos term simply becomes 1. But what do we do from here?

More specifically we get:
$$0 = \frac{{\pi}^{2}}{3} + \sum_{k=1}^{\infty} \frac{4\cdot{(-1)}^{k}}{{k}^{2}} \cdot 1$$
Can we rewrite that equation into the desired form?

 

[goohu]

Thanks Klaas! I solved the problem now.

Is this trick where you put t = 0 always applicable? IIRC musnt the function be continuous?

Before this I used the formula where you square both sides and it didnt work to rewrite it from there. How do know when to use which method?

 

[I like Serena]

Thanks Klaas! I solved the problem now.

Is this trick where you put t = 0 always applicable? IIRC musnt the function be continuous?

Before this I used the formula where you square both sides and it didnt work to rewrite it from there. How do know when to use which method?

The 'trick' here is that we write a function $f(t)$ as a Fourier Series.
Then we can substitute any value for $t$ that we want.
The condition to write a function as a Fourier Series is:

If f is continuous and the derivative of f(t) (which may not exist everywhere) is square integrable, then the Fourier series of f converges absolutely and uniformly to f(t).​

What do you mean by squaring both sides?
In this particular case the requested sum in (b) is simply an application of the result found in (a).
More generally, a Fourier Series is just one of the many tools that make some problems suddenly easy to solve.

 

[goohu]

Sorry for the late reply, but here's an example of "squaring both sides" its just something found in my formula sheet:

I also got stuck at the last step trying to find $$\sum_{k=1}^{\infty} \frac{1}{{k}^{2}+4}$$

View attachment 9307

 

[Opalg]

Sorry for the late reply, but here's an example of "squaring both sides" its just something found in my formula sheet:

I also got stuck at the last step trying to find $$\sum_{k=1}^{\infty} \frac{1}{{k}^{2}+4}$$

I have not tried to follow all of your calculations, but there is one obvious mistake. If $c_k$ is a complex number then $|c_k|^2$ is not the same as $c_k^2$. In fact, $|c_k|^2 = c_k\overline{c_k}$ (where the bar denotes the complex conjugate).

So if $c_k = \dfrac{e^{4\pi}-1}{2\pi(2-ik)}$ then $|c_k|^2 = \dfrac{(e^{4\pi}-1)^2}{4\pi^2(2-ik)(2+ik)} = \dfrac{(e^{4\pi}-1)^2}{4\pi^2(4+k^2)}$.

 

[goohu]

It turns out I almost got the right answer (same as the picture) and the left hand side becomes $$\sum_{k=1}^{\infty} \frac{1}{{k}^{2}+4}$$. However the right hand side misses a -1/8 term.

 

[Opalg]

It turns out I almost got the right answer (same as the picture) and the left hand side becomes $$\sum_{k=1}^{\infty} \frac{1}{{k}^{2}+4}$$. However the right hand side misses a 1/8 term.

I think that the missing 1/8 possibly comes from the fact that when you combine the terms indexed by $k$ and $-k$ in the sum $$\sum_{k=-\infty}^{\infty}|c_k|^2$$, in order to get a sum $$\sum_{k=0}^{\infty} |c_k|^2$$, you are in danger of counting the $k=0$ term twice.