Question

If

$\mathit{f}\left(x\right)\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathit{a}}_{\mathbf{0}}\mathbf{+}\mathbf{\sum }_{\mathbf{1}}^{\mathbf{\infty }}{\mathit{a}}_{\mathbf{n}}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathit{n}\mathit{x}\mathbf{}\mathbf{+}\mathbf{\sum }_{\mathbf{1}}^{\mathbf{\infty }}{\mathit{b}}_{\mathbf{n}}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathit{n}\mathit{x}\mathbf{=}\mathbf{\sum }_{\mathbf{-}\mathbf{\infty }}^{\mathbf{\infty }}{\mathit{c}}_{\mathbf{n}}{\mathit{e}}^{\mathbf{i}\mathbf{n}\mathbf{x}}$, use Euler's formula to find ${\mathit{a}}_{\mathbf{n}}$ and ${\mathit{b}}_{\mathbf{n}}$in terms of ${\mathit{c}}_{\mathbf{n}}$and $\mathbf{-}{\mathit{c}}_{\mathbf{n}}$, and to find ${\mathit{c}}_{\mathbf{n}}$and $\mathbf{-}{\mathit{c}}_{\mathbf{n}}$in terms of ${\mathit{a}}_{\mathbf{n}}$and${\mathit{b}}_{\mathbf{n}}$ a.

Short answer

The resultant expansion $c_n$and $c_{-n}$ are $c_n=\frac{1}{2i}\left(i{a}_n+b_n\right)$and $c_{-n}=\frac{1}{2}\left(a_n+i{b}_n\right)$.

Step-by-step solution

Given data

The given function is $f\left(x\right)=\frac{1}{2}{a}_0+\sum _1^{\infty }{a}_n\cos nx+\sum _1^{\infty }{b}_n\sin nx=\sum _{-\infty }^{\infty }{c}_n{e}^{inx}$.

Concept of Fourier series

An infinite sum of sines and cosines is used to represent the expansion of a periodic function f(x) into a Fourier series.

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

Simplify the expression

By Fourier series expansion, $f\left(x\right)=\sum _{n=-\infty }^{\infty }{c}_n{e}^{inx}$ and $f\left(x\right)=c_0+\sum _{n=-\infty }^{-1}{c}_n{e}^{inx}+\sum _{n=1}^{\infty }{c}_n{e}^{inx}$.

Check and simplify the expression

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

$$\begin{gathered} f\left(x\right)=c_0+\sum _{n=1}^{\infty }\left[c_{-n}{e}^{-inx}+c_n{e}^{inx}\right] \\ f\left(x\right)=c_0+\sum _{n=1}^{\infty }\left[\left(c_n+c_{-n}\right)\cos \left(nx\right)+i\left(c_n+c_{-n}\right)\sin nx\right] \end{gathered}$$

By comparison of above series:

$$\begin{gathered} a_0=2c_0 \\ {a}_n=c_n+c_{-n} \\ {b}_n=i\left(c_n+c_{-n}\right) \end{gathered}$$

This implies that $c_0=\frac{a_0}{2}$.

$$\begin{gathered} c_n=\frac{1}{2i}\left(i{a}_n+b_n\right) \\ {c}_{-n}=\frac{1}{2}\left(a_n+i{b}_n\right) \end{gathered}$$

Therefore, the $c_n$and $c_{-n}$ are $c_n=\frac{1}{2i}\left(i{a}_n+b_n\right)$and $c_{-n}=\frac{1}{2}\left(a_n+i{b}_n\right)$ respectively.