Question

find the fourier series of f(x) = e^-ax in (-π π)

Answer 1

Step-by-step explanation:

$\: All functions [math]f(x)[/math]between x=[[math]-\pi,\pi[/math]] can be expressed as:</p><p>[math]f(x)=a_0+\sum_{n=1}^{\infty} a_n \cos(n x)+b_n \sin(nx)[/math]</p><p></p><p>Using Fourier's formula gives us every term:</p><p></p><p>Constant:</p><p>[math]a_0=\frac{1}{2 \pi} \int_{-\pi}^{\pi} dx e^{a x} [/math]</p><p>[math]a_0=\frac{2 \sinh (\pi a)}{a}[/math]</p><p></p><p>Cosine terms:</p><p>[math]a_n=\frac{2}{2 \pi} \int_{-\pi}^{\pi} dx e^{a x} \cos(n x) [/math]</p><p>[math]a_n=\frac{(-1)^n 2 a \sinh (\pi a)}{\pi(a^2+n^2)}[/math]</p><p></p><p>Sine Terms:</p><p>[math]b_n=\frac{2}{2 \pi} \int_{-\pi}^{\pi} dx e^{a x} \sin(n x) [/math]</p><p>[math]b_n=\frac{(-1)^{n+1} 2 n \sinh (\pi a)}{\pi(a^2+n^2)}[/math]</p><p></p><p>Hope that helps!$