Question

Expand the same functions as in Problems 5.1 to 5.11 in Fourier series of complex exponentials ${\mathit{e}}^{\mathbf{inx}}$ on the interval$\mathbf{(}\mathbf{-}\mathit{\pi }\mathbf{,}\mathit{\pi }\mathbf{)}$ and verify in each case that the answer is equivalent to the one found in Section 5.

Short answer

The resultantexpansionis $f\left(x\right)=1+i\left(\sum _{n=-\infty }^{\infty }{(-1)}^n\frac{e^{inx}}{n}\right)n\ne 0$.

Step-by-step solution

Given data

The given function is complex exponentials $e^{inx}$.

Concept of Fourier series

In terms of an infinite sum of sines and cosines, a Fourier series is an expansion of a periodic function.

The orthogonality relationships of the sine and cosine functions are used in Fourier series.

Find the coefficients

The $c_0$ coefficient is shown below.

$$\begin{gathered} c_0=\frac{1}{2\pi }{\int }_{-\pi }^{\pi }xdx \\ {c}_0=0 \end{gathered}$$

The other coefficients are as follows:

$$\begin{gathered} c_n=\frac{1}{2\pi }{\int }_{-\pi }^{\pi }x{e}^{-inx}dx \\ {c}_n=\frac{1}{2\pi }{\left[-\frac{x}{in}{e}^{-inx}+\frac{1}{n^2}{e}^{-inx}\right]}_{-\pi }^{\pi } \\ {c}_n=\frac{1}{2\pi }\left(-\frac{\pi }{in}{e}^{-in\pi }-\frac{\pi }{in}{e}^{in\pi }+\frac{1}{n^2}\left(e^{-in\pi }-e^{in\pi }\right)\right) \\ {c}_n=\frac{1}{2\pi }\left[-\frac{\pi }{in}2\cos \left(n\pi \right)-\frac{1}{n^2}2i\sin \left(n\pi \right)\right] \end{gathered}$$

So, $c_n=\frac{i}{n}(-1{)}^n$.

Then the function is, $f\left(x\right)=1+i\left(\sum _{n=-\infty }^{\infty }{(-1)}^n\frac{e^{inx}}{n}\right)n\ne 0$.

Check and simplify the expression

For check, change $n\to -n$.

$$\begin{gathered} f\left(x\right)=1+i\sum _{n=-\infty }^{-1}\frac{e^{inx}}{n}(-1{)}^n+i\sum _{n=1}^{\infty }{(-1)}^n \\ f\left(x\right)=1+ii\sum _{n=1}^{\infty }\frac{{(-1)}^n}{n}\left(e^{inx}-e^{-inx}\right) \\ f\left(x\right)=1-2i\sum _{n=1}^{\infty }{(-1)}^n\frac{\sin \left(nx\right)}{n} \end{gathered}$$

Therefore, the expansion is $f\left(x\right)=1+i\left(\sum _{n=-\infty }^{\infty }{(-1)}^n\frac{e^{inx}}{n}\right)n\ne 0$.