- Thread starter
- Banned
- #1

- Thread starter Poirot
- Start date

- Thread starter
- Banned
- #1

- Moderator
- #2

- Feb 7, 2012

- 2,765

Yes: Use the fact that $f(t)$ is the sum of its Fourier series at the point $t=0$.define f(t)=|t|, t between - pi and pi.

I have found the fourier co-efficents of f and am now charged with showing that the infinite series of 1/(2m+1)^2 is equal to (pi^2)/8. Can I use the fourier co-efficents?

- Thread starter
- Banned
- #3

- Moderator
- #4

- Feb 7, 2012

- 2,765

You said that you had found the Fourier coefficients of $f$, so you can write down its Fourier series. If the Fourier coefficients are $a_n$ and $b_n$ then the Fourier series is \(\displaystyle a_0 + \sum_{n=1}^\infty (a_n\cos nt + b_n\sin nt)\). There is a theorem which says that if the function $f$ is continuous then it is equal to the sum of its Fourier series.How do I find it's fourier series?

- Thread starter
- Banned
- #5

I will tell you what I have found: If g(n) is the fourier coeffient of f at n, then g(0)=pi/2You said that you had found the Fourier coefficients of $f$, so you can write down its Fourier series. If the Fourier coefficients are $a_n$ and $b_n$ then the Fourier series is \(\displaystyle a_0 + \sum_{n=1}^\infty (a_n\cos nt + b_n\sin nt)\). There is a theorem which says that if the function $f$ is continuous then it is equal to the sum of its Fourier series.

g(n)=-2/pi(n)^2, when n is odd, and g(n)=0 for all non-zero even n.

If I plug in zero in the fourier series, the sin coefficents vanish, and I don't get the right answer.

- Admin
- #6

- Mar 5, 2012

- 9,265

It's kind of similar... or (almost) the same.

- Thread starter
- Banned
- #7

Thanks.

It's kind of similar... or (almost) the same.

I get f(t)=pi/4 - (2cost/pi +2cos2t/4pi +.........). What do I do from here?

- Moderator
- #8

- Feb 7, 2012

- 2,765

That is mostly correct. The sine coefficients all vanish (because $|t|$ is an even function). For the cosine coefficients, notice that \(\displaystyle g(n) = \frac1\pi\int_{-\pi}^\pi |t|^n\cos nt\,dt = \frac2\pi\int_0^\pi t^n\cos nt\,dt\) (because the integral from $-\pi$ to $0$ is the same as the integral from $0$ to $\pi$). That gives answers twice what you found, namely $g(0) = \pi$ and $g(n) = -\frac4{\pi n^2}$ when $n$ is odd (and 0 for nonzero even $n$). If you write the odd number $n$ as $2m+1$ then the Fourier series becomes $$\tfrac12g(0) + \sum_{n=1}^\infty g(n)\cos nt = \frac\pi2 - \sum_{m=0}^\infty \frac4{(2m+1)^2\pi}\cos (2m+1)t.$$ Now see what that comes to when $t=0.$I will tell you what I have found: If g(n) is the fourier coeffient of f at n, then g(0)=pi/2

g(n)=-2/pi(n)^2, when n is odd, and g(n)=0 for all non-zero even n.

If I plug in zero in the fourier series, the sin coefficents vanish, and I don't get the right answer.

- Thread starter
- Banned
- #9

$f(t)=t^2$

fourier coefficents are g(0)=pi^2/3 and g(n)=2/(n^2) .(-1)^n otherwise.

Deduce that the infinite series of 1/n^4 is equal to pi^4/90.

Whatever t I input, I don't get, and indeed don't see how I am going to get the extra factor of 1/n^2 in the series. I don't know whether it's relevant but I was also asked to show that the norm of f squared is pi^4/5 (which I was able to do).

- Moderator
- #10

- Feb 7, 2012

- 2,765

Use Parseval's theorem (see equation (4) in that link).

$f(t)=t^2$

fourier coefficents are g(0)=pi^2/3 and g(n)=2/(n^2) .(-1)^n otherwise.

Deduce that the infinite series of 1/n^4 is equal to pi^4/90.

Whatever t I input, I don't get, and indeed don't see how I am going to get the extra factor of 1/n^2 in the series. I don't know whether it's relevant but I was also asked to show that the norm of f squared is pi^4/5 (which I was able to do).