Question

The symbol $\mathbf{\left[}\mathbf{x}\mathbf{\right]}$means the greatest integer less than or equal to x(for example,$\mathbf{\left[}\mathbf{3}\mathbf{\right]}\mathbf{=}\mathbf{3}\mathbf{,}\mathbf{[}\mathbf{2}\mathbf{.1}\mathbf{]}\mathbf{=}\mathbf{2}\mathbf{,}\mathbf{[}\mathbf{-}\mathbf{4}\mathbf{.5}\mathbf{]}\mathbf{=}\mathbf{-}\mathbf{5}$Expand $\mathit{x}\mathbf{-}\mathbf{\left[}\mathbf{x}\mathbf{\right]}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}$in an exponential Fourier series of period 1.

Short answer

The exponential Fourier series of period 1 of $x-\left[x\right]-\frac{1}{2}$is$f\left(x\right)=\frac{i}{2\pi }\sum _{n=-\infty }^{\infty }\frac{e^{i2\pi nx}}{n},n\ne 0$.

Step-by-step solution

Given information

It is given that the symbol $\left[x\right]$ means the greatest integer less than or equal to x. The function $x-\left[x\right]-\frac{1}{2}$is to be expanded in an exponential Fourier series of period 1.

Write complex series formulas

The following are the formulas for the complex Fourier series,

$\begin{array}{c}\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{\sum }_{\mathbf{-}\mathbf{\infty }}^{\mathbf{\infty }}{\mathbf{c}}_{\mathbf{n}}{\mathbf{e}}^{\mathbf{i}\mathbf{n}\mathbf{\pi }\mathbf{x}\mathbf{/}\mathbf{l}}\\ {\mathbf{c}}_{\mathbf{n}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}\mathbf{l}}\underset{-l}{\overset{l}{\int }}f\left(x\right)e^{-in\pi x/l}dx\end{array}$

Here the period of $f\left(x\right)$is $2l$and the frequencies of the terms in the series are $n/2l$.

Sketch the function

The function on the interval can be written as follows.

$f\left(x\right)=\left\{\begin{array}{ll}x+1/2,& x\in [-1/2,0]\\ x-1/2,& x\in [0,1/2]\end{array}\right.$

The graph of the given function is as follows.

Find exponential Fourier series

Use formula$c_n=\frac{1}{2l}\underset{-l}{\overset{l}{\int }}f\left(x\right)e^{-in\pi x/l}dx$to find coefficients.

$\begin{array}{c}{c}_n=\frac{1}{2l}\underset{-l}{\overset{l}{\int }}f\left(x\right)e^{-in\pi x/l}dx\\ =\underset{-1/2}{\overset{0}{\int }}(x+\frac{1}{2})e^{-i2n\pi x}dx+\underset{0}{\overset{1/2}{\int }}(x-\frac{1}{2})e^{-i2n\pi x}dx\\ ={[-\frac{x}{2\pi in}{e}^{-i2\pi nx}+\frac{1}{4{\pi }^2{n}^2}{e}^{-i2\pi nx}+\frac{1}{4\pi in}{e}^{-i2\pi nx}]|}_0^{1/2}+{[-\frac{x}{2\pi in}{e}^{-i2\pi nx}+\frac{1}{4{\pi }^2{n}^2}{e}^{-i2\pi nx}-\frac{1}{4\pi in}{e}^{-i2\pi nx}]|}_{1/2}^0\\ =-\frac{2}{4\pi in}\end{array}$

Solve further to get the coefficients.

$\begin{array}{c}{c}_n=-\frac{1}{2\pi in}\\ =\frac{i}{2\pi n}\end{array}$

Now use formula $f\left(x\right)=\sum _{-\infty }^{\infty }{c}_n{e}^{in\pi x/l}$to write the series.

$f\left(x\right)=\frac{i}{2\pi }\sum _{n=-\infty }^{\infty }\frac{e^{i2\pi nx}}{n},n\ne 0$