Chapter 10.2, Problem 23P

a.

To determine

To graph: The function defined by $f\left(x\right)=\{\begin{array}{l}-\frac{1}{2}x,-2\le x<0\\ 2x-\frac{1}{2}{x}^2,\text{ 0}\le x<2\end{array}$ with $f\left(x+4\right)=f\left(x\right)$ for three different periods.

b.

To determine

The Fourier series for the function $f\left(x\right)=\{\begin{array}{l}-\frac{1}{2}x,-2\le x<0\\ 2x-\frac{1}{2}{x}^2,\text{ 0}\le x<2\end{array}$.

c.

To determine

To plot: The partial sums corresponding to the Fourier series $f\left(x\right)=\frac{11}{12}+\frac{1}{{\pi }^2}{\displaystyle \sum _{n=1}^{\infty }\frac{{\left(-1\right)}^n-5}{n^2}\cos \frac{n\pi x}{2}}+{\displaystyle \sum _{n=1}^{\infty }\left(\frac{4\left(1-{\left(-1\right)}^n\right)}{n^3{\pi }^3}-\frac{{\left(-1\right)}^n}{n\pi }\right)\sin \frac{n\pi x}{2}}$ taking the partial sum $s_m\left(x\right)$ along the y-axis for $m=5$, $m=10$ and $m=20$.

d.

To determine

To describe: The convergence of the Fourier series $f\left(x\right)=\frac{11}{12}+\frac{1}{{\pi }^2}{\displaystyle \sum _{n=1}^{\infty }\frac{{\left(-1\right)}^n-5}{n^2}\cos \frac{n\pi x}{2}}+{\displaystyle \sum _{n=1}^{\infty }\left(\frac{4\left(1-{\left(-1\right)}^n\right)}{n^3{\pi }^3}-\frac{{\left(-1\right)}^n}{n\pi }\right)\sin \frac{n\pi x}{2}}$.