The discussion centers on using Fourier coefficients to demonstrate that the infinite series of 1/(2m+1)^2 equals (π²)/8. The function defined is f(t) = |t| for t in the interval [-π, π]. The Fourier coefficients are identified as g(0) = π/2, g(n) = -2/πn² for odd n, and g(n) = 0 for even n. The Fourier series is constructed, and it is confirmed that the sine coefficients vanish, leading to the conclusion that the series converges to the desired result when evaluated at t = 0.
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Yes: Use the fact that $f(t)$ is the sum of its Fourier series at the point $t=0$.Poirot said: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?
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 $$a_0 + \sum_{n=1}^\infty (a_n\cos nt + b_n\sin nt)$$. There is a theorem which says that if the function $f$ is continuous then it is equal to the sum of its Fourier series.Poirot said:How do I find it's Fourier series?
Opalg said: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 $$a_0 + \sum_{n=1}^\infty (a_n\cos nt + b_n\sin nt)$$. There is a theorem which says that if the function $f$ is continuous then it is equal to the sum of its Fourier series.
I like Serena said:Take a look at this thread: http://www.mathhelpboards.com/f16/fourier-series-3924/
It's kind of similar... or (almost) the same. ;)
That is mostly correct. The sine coefficients all vanish (because $|t|$ is an even function). For the cosine coefficients, notice that $$g(n) = \frac1\pi\int_{-\pi}^\pi |t|^n\cos nt\,dt = \frac2\pi\int_0^\pi t^n\cos nt\,dt$$ (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 $t=0.$Poirot said: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.
Use Parseval's theorem (see equation (4) in that link).Poirot said:Thanks, solved it. Now I have a similar problem:
$f(t)=t^2$
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).