Question on Fourier series $\sum_{n=1}^\infty a_n \sin ( \pi nx) = f(x)$

Question

Assume $\sum_{n=1}^\infty a_n \sin ( \pi nx) = f(x)$ where $f: [0,1] \rightarrow \mathbb R$ continuous is and $f(0) = f(1)$. Can I then recorver the $a_n$ by using somehow the Fourier series of $f$ ? This question arises from some differential equation problem which is as follows:

Let $\Omega := \{0 < x < 1, 0<y\} \subseteq \mathbb R^2$. We then are looking for a function $u: \overline \Omega \rightarrow \mathbb C$ s.t.

$u \in C(\overline \Omega), u \in C^2(\Omega)$
$\Delta u = 0$ on $\Omega$
$\lim_{y \rightarrow \infty} u(x,y) = 0$ for all $x \in [0,1]$.
$u(0,y) = u(1,y) = 0$ for all $y \geq 0$
$u(x,0) = f(x)$ for all $x \in [0,1]$.

This lead me to $$ u(x,y) = \sum_{n \geq 1} a_n \sin (n \pi x) e^{-\pi n y} $$ But now I have to calculate the $a_n$ with my last properity.

What are the known things in your problem? – Mhenni Benghorbal Jul 01 '13 at 14:42 — Mhenni Benghorbal, Jul 01 '13 at 14:42
I have updated from where the question arises. – Jul 01 '13 at 14:58 — , Jul 01 '13 at 14:58

Ron Gordon · Accepted Answer · 2013-07-01T18:13:16.730

0

Use $u(x,0)=f(x)$. Then because the functions $\sin{(n \pi x)}$ form an orthogonal basis set, you may simply write that, for $n \gt 0$,

$$a_n = \frac{\displaystyle \int_0^1 dx \, f(x) \, \sin{(n \pi x)}}{\displaystyle\int_0^1 dx \, \sin^2{(n \pi x)}} = 2 \int_0^1 dx \, f(x) \, \sin{(n \pi x)}$$

edited Jul 01 '13 at 18:13

answered Jul 01 '13 at 15:13

Ron Gordon

141,538

score 0 · Answer 2 · answered Jul 01 '13 at 15:17

0

Just note that, you can evaluate $a_n$ as

$$ a_n = \int_{0}^{1}f(x)\sin(n\pi x)dx. $$

answered Jul 01 '13 at 15:17

Mhenni Benghorbal

48,099

Is this because of uniform convergence of the Fourierseries of $f$ ? – Jul 01 '13 at 15:17
@André: This is how we find the Fourier coefficients, See here at the end of my answer. – Mhenni Benghorbal Jul 01 '13 at 15:20

Question on Fourier series $\sum_{n=1}^\infty a_n \sin ( \pi nx) = f(x)$

2 Answers2