Question

show that the Fourier series of
f(x)=e^x (-π,π)
Is
1/π sin hπ+∑_(n=1)^∞▒〖(2 sin hπ)/π(1+n^2 ) (-1)^n (cos nx-nsin nx)〗

Answer 1

$\displaystyle\textsf{The Fourier series of a continuous function}\, f(x) \, \textsf{is defiined as}\\ f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} (a_n \cos(nx) + b_n \sin(nx)) \\ \textsf{The period of of the function is}\, 2\pi, \\ \begin{aligned} \textsf{Hence}, \, a_0 &= \frac{1}{\pi} \int_{-\pi}^{\pi} e^x \, \mathrm{d} x\\ a_n &= \frac{1}{\pi} \int_{-\pi}^{\pi} e^x \cos(nx) \, \mathrm{d} x\\ b_n &= \frac{1}{\pi} \int_{-\pi}^{\pi} e^x \sin(nx) \, \mathrm{d} x \end{aligned} \\ \textrm{Calculating each term, we have}\\ \begin{aligned} a_0 &= \frac{1}{\pi} \int_{-\pi}^{\pi} e^x \, \mathrm{d} x \\&= \frac{1}{\pi}(e^x \vert_{-\pi}^{\pi}) \\&=\frac{1}{\pi} (e^{\pi} - e^{-\pi}) \\&=\frac{2\sinh(\pi)}{\pi} \end{aligned} \\ \begin{aligned} a_n &= \frac{1}{\pi} \int_{-\pi}^{\pi} e^x \cos(nx) \, \mathrm{d} x \\&= \frac{1}{n\pi} \int_{-\pi}^{\pi} e^x \, \mathrm{d}(\sin(nx)) \\&= \frac{1}{n\pi} \left(e^x \sin(nx) \vert_{-\pi}^{\pi} - \int_{-\pi}^{\pi} e^x \sin(nx) \, \mathrm{d} x\right) \\&= \frac{1}{n\pi} \left(e^{\pi} \sin(n\pi) - e^{\pi} \sin(-n\pi) + \frac{1}{n}\int_{-\pi}^{\pi} e^x\, \mathrm{d}(\cos(nx))\right) \end{aligned} \\ \sin(n\pi) = 0 \hspace{0.3cm} \forall \, n \in \mathbb{Z} \\ \begin{aligned} \therefore a_n &= \frac{1}{n\pi} \left(\frac{1}{n}\int_{-\pi}^{\pi} e^x\, \mathrm{d}(\cos(nx))\right) \\&= \frac{1}{n^2\pi} \left(e^x \cos(nx) \vert_{-\pi}^{\pi} - \int_{-\pi}^{\pi} e^x\cos(nx) \, \mathrm{d} x\right) \\&=\frac{1}{n^2\pi} \left(e^{\pi} \cos(n\pi) - e^{\pi} \cos(-n\pi) - \pi a_n\right) \end{aligned} \\ \cos(n\pi) = (-1)^n \hspace{0.3cm}\forall \, n \in \mathbb{Z} \\ \begin{aligned} \therefore a_n &= \frac{1}{n^2\pi} \left((e^{\pi} - e^{\pi})(-1)^n - \pi a_n\right) \\&=\frac{1}{n^2\pi} \left(2\sinh(\pi)(-1)^n - \pi a_n\right) \\ a_n + \frac{a_n }{n^2} &= \frac{2\sinh(\pi)(-1)^n}{n^2\pi} \\ a_n\left(1 + \frac{1}{n^2}\right) &= \frac{2\sinh(\pi)(-1)^n}{n^2\pi}\\ a_n \left(\frac{1 + n^2}{n^2}\right) &= \frac{2\sinh(\pi)(-1)^n}{n^2\pi}\\ a_n &= \frac{2\sinh(\pi)(-1)^n}{(n^2 + 1)\pi} \end{aligned} \\ \begin{aligned} b_n &= \frac{1}{\pi} \int_{-\pi}^{\pi} e^x \sin(nx) \, \mathrm{d} x \\&=-\frac{1}{n\pi} \int_{-\pi}^{\pi} e^x \, \mathrm{d}(\cos(nx)) \\&=-\frac{1}{n\pi} \left(e^x \cos(nx) \vert_{-\pi}^{\pi} - \int_{-\pi}^{\pi} e^x \cos(nx) \, \mathrm{d} x\right) \\&=-\frac{1}{n\pi} \left(e^{\pi} \cos(n\pi) - e^{\pi} \cos(-n\pi) - \int_{-\pi}^{\pi} e^x \cos(nx) \, \mathrm{d} x\right) \\&= -\frac{1}{n\pi} \left(2\sinh(\pi)(-1)^n - \int_{-\pi}^{\pi} e^x \cos(nx) \, \mathrm{d} x\right) \\&= -\frac{1}{n\pi} \left(2\sinh(\pi)(-1)^n - \pi a_n\right) \\&= -\frac{2\sinh(\pi)(-1)^n}{n\pi} + \frac{a_n}{n} \\&= -\frac{2\sinh(\pi)(-1)^n}{n\pi} + \frac{2\sinh(\pi)(-1)^n}{n(n^2 + 1)\pi} \\&=\frac{2\sinh(\pi)(-1)^n}{n\pi}\left(-1 + \frac{1}{n^2 + 1}\right) \\&= \frac{2n(-1)^{n + 1}\sinh(\pi)}{(n^2 + 1)\pi} \end{aligned} \\ \therefore \textsf{The Fourier series of the continuous function is} \\ f(x) = \frac{\sinh(\pi)}{\pi}+ 2\sum_{n=1}^{\infty} \frac{(-1)^n\sinh(\pi)}{(n^2 + 1)\pi} \left(\cos(nx) - n\sin(nx)\right)$