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

The following are the formulas for the complex Fourier series,

$\begin{array}{c}\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\underset{\mathbf{鈭�}\mathbf{鈭�}}{\overset{\mathbf{鈭�}}{\mathbf{鈭�}}}{\mathbf{c}}_{\mathbf{n}}{\mathbf{e}}^{\mathbf{i}\mathbf{n}\mathbf{蟺}\mathbf{x}\mathbf{/}\mathbf{l}}\\ {\mathbf{c}}_{\mathbf{n}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}\mathbf{l}}\underset{鈭�l}{\overset{l}{鈭�}}f\left(x\right)e^{鈭�in蟺x/l}dx\end{array}$

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

The function on the interval can be written as follows.

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

The graph of the given function is as follows.

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

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

Solve further to get the coefficients.

$\begin{array}{c}{c}_n=鈭�\frac{1}{2蟺in}\\ =\frac{i}{2蟺n}\end{array}$

Now use formula $f\left(x\right)=\underset{鈭�\mathrm{鈭�}}{\overset{\mathrm{鈭�}}{鈭�}}{c}_n{e}^{in蟺x/l}$to write the series.

$f\left(x\right)=\frac{i}{2蟺}\underset{n=鈭�\mathrm{鈭�}}{\overset{\mathrm{鈭�}}{鈭�}}\frac{e^{i2蟺nx}}{n},n鈮�0$