フーリエ級数の問題

${\displaystyle f\left(x\right)\sim \frac{3}{2}+\frac{4}{{\pi }^2}\left(\cos \frac{\pi }{2}x+\frac{1}{3^2}\cos \frac{3\pi }{2}x+\frac{1}{5^2}\cos \frac{5\pi }{2}x+\cdots \cdots \right)}$${\displaystyle +\frac{2}{\pi }\left(-\sin \frac{\pi }{2}x+\frac{1}{2}\sin \frac{2\pi }{2}x-\frac{1}{3}\sin \frac{3\pi }{2}x+\cdots \cdots \right)}$

${\displaystyle =\frac{1}{2}\left\{{\displaystyle {\int }_{-2}^{0}2\cos \frac{n\pi }{2}}xdx+{\displaystyle {\int }_0^2\left(2-x\right)}\cos \frac{nt}{2}xdx\right\}}$

${\displaystyle =\frac{1}{2}\left\{2{\left[\frac{2}{n\pi }\sin \frac{n\pi }{2}x\right]}_{-2}^0+{\left[\left(2-x\right)\frac{2}{n\pi }\sin \frac{n\pi }{2}x\right]}_0^2-{\displaystyle {\int }_0^2\left(-1\right)\frac{2}{n\pi }\sin \frac{n\pi }{2}x}\right\}}$

${\displaystyle =\frac{1}{2}\left\{\frac{4}{n\pi }\left(0-0\right)+\left(0-0\right)+\frac{2}{n\pi }{\left[-\frac{2}{n\pi }\cos \frac{n\pi }{2}x\right]}_0^2\right\}}$

${\displaystyle =\frac{1}{2}\left\{{\displaystyle {\int }_{-2}^{0}2\sin \frac{n\pi }{2}xdx+{\displaystyle {\int }_0^2\left(2-x\right)\sin \frac{n\pi }{2}xdx}}\right\}}$

${\displaystyle =\frac{1}{2}\left\{2{\left[-\frac{2}{n\pi }\cos \frac{n\pi }{2}x\right]}_{-2}^0+{\left[\left(2-x\right)\left(-\frac{2}{n\pi }\cos \frac{n\pi }{2}\right)\right]}_0^2-{\displaystyle {\int }_0^2\left(-1\right)\left(-\frac{2}{n\pi }\cos \frac{n\pi }{2}\right)dx}\right\}}$

${\displaystyle =\frac{1}{2}\left\{2\left(-\frac{2}{n\pi }\right)\left(1-\cos \left(-n\pi \right)\right)-2\left(-\frac{2}{n\pi }\right)-\frac{2}{n\pi }{\left[\frac{2}{n\pi }\sin \frac{n\pi }{2}\right]}_0^2\right\}}$

${\displaystyle f\left(x\right)\sim a_0+{\displaystyle \sum _{n=1}^{\infty }\left(a_n\cos \frac{n\pi }{2}+b_n\sin \frac{n\pi }{2}x\right)}}$

${\displaystyle \sim \frac{3}{2}+{\displaystyle \sum _{n=1}^{\infty }\left(-\frac{2}{n^2{\pi }^2}\left\{{\left(-1\right)}^n-1\right\}\cos \frac{n\pi }{2}x+\frac{2}{n\pi }{\left(-1\right)}^n\sin \frac{n\pi }{2}x\right)}}$

${\displaystyle \sim \frac{3}{2}+\frac{4}{{\pi }^2}\left(\cos \frac{\pi }{2}x+\frac{1}{3^2}\cos \frac{3\pi }{2}x+\frac{1}{5^2}\cos \frac{5\pi }{2}x+\cdots \cdots \right)}$${\displaystyle +\frac{2}{\pi }\left(-\sin \frac{\pi }{2}x+\frac{1}{2}\sin \frac{2\pi }{2}x-\frac{1}{3}\sin \frac{3\pi }{2}x+\cdots \cdots \right)}$