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The strategy I have been asked to take, is to show that Fourier coefficients of the function $f(x) = x$ on $[0, 1]$ are up to a constant equal to $1/n^2$.Then I should apply the norm

\begin{equation*} \lVert f \rVert_2 = (\int_0^1 \lvert f(x)\rvert^2 dx)^{1/2} \end{equation*}

in terms of Fourier coefficients.

So I know that for an inner product space we have that

\begin{equation*} x=\sum_{n=1}^\infty \langle x, e_n \rangle e_n \end{equation*}

and the space we seem to be working in is $L^2 [0,1]$.

The first difficulty I am having is finding an orthonormal sequence in this space. Without some$(e_n)_{n \in \mathbb N}$ to work with. I cannot see how I can progress with this solution and I am not sure how to come up with an infinite linearly independent set of functions.

Martin Sleziak
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user108896
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2 Answers2

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Let we consider $f(x)=x$ on $I=(-\pi,\pi)$.

$f$ is an odd function in $L^2(-\pi,\pi)$, hence its Fourier series is given by: $$ f(x) = \sum_{n\geq 1} c_n \sin(nx) \tag{1}$$ where integration by parts gives: $$ c_n = \frac{1}{\pi}\int_{-\pi}^{+\pi}f(x)\sin(nx)\,dx = \frac{2(-1)^{n+1}}{n}\tag{2}$$ hence, by Parseval's identity, $$ \frac{2\pi^3}{3}=\int_{-\pi}^{\pi}f(x)^2\,dx = \pi\sum_{n\geq 1}c_n^2 = 4\pi\sum_{n\geq 1}\frac{1}{n^2}\tag{3}$$ and simplifying: $$ \zeta(2)=\sum_{n\geq 1}\frac{1}{n^2}=\color{red}{\frac{\pi^2}{6}}.\tag{4}$$

Have also a look at one of the greatest question of all time: Different methods to compute $\sum\limits_{k=1}^\infty \frac{1}{k^2}$.

Community
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Jack D'Aurizio
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Using Omnomnomnom's choice:

Complete orthonormal set $e_n(x):=e^{2\pi i n x}$ on $[0,1]$. For function $f(x) := x$, the Fourier coefficients are $$ c_n = \langle f,e_n\rangle = \int_0^1 x e^{-2\pi i n x}\;dx = \frac{i}{2\pi n} ,\qquad{n \ne 0} \\ c_0 = \langle f,e_0\rangle = \int_0^1 x\;dx = \frac{1}{2} $$ Then compute $$ \|f\|^2 = \int_0^1 |x|^2\,dx = \frac{1}{3} $$ So by Parseval's identity $$ \|f\|^2 = \sum_{n=-\infty}^{+\infty} |c_n|^2 \\ \frac{1}{3} = \frac{1}{4} + 2\sum_{n=1}^\infty \frac{1}{4\pi^2n^2} \\ \frac{\pi^2}{6} = \sum_{n=1}^\infty \frac{1}{n^2} $$

GEdgar
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