анализ - Разложение в ряд Фурье функции $%1-2x$%

Воспользуйтесь стандартными формулами

$$f(x)=\frac{a_0}{2}+ \sum_{n=1}^{\infty}\!\left(a_n \cos\frac{\pi nx}{l}+ b_n\sin\frac{\pi nx}{l}\right)\!,\quad x\in(-l,l)$$

$$a_0= \frac{1}{l}\int\limits_{-l}^{l}f(x)\,dx= \frac{1}{1/2}\int\limits_{-1/2}^{1/2}(1-2x)\,dx=\ldots =2$$

$$a_n= \frac{1}{l}\int\limits_{-l}^{l}f(x)\cos\frac{\pi nx}{l}\,dx= \frac{1}{1/2}\int\limits_{-1/2}^{1/2}(1-2x)\cos\frac{\pi nx}{1/2}\,dx= \ldots = \frac{2\sin\pi n}{\pi n}=0$$

$$\begin{aligned}b_n&= \frac{1}{l}\int\limits_{-l}^{l}f(x)\sin\frac{\pi nx}{l}\,dx= \frac{1}{1/2}\int\limits_{-1/2}^{1/2}(1-2x)\sin\frac{\pi nx}{1/2}\,dx= \ldots=\\ &=\frac{2(\pi n\cos\pi n-\sin\pi n)}{\pi^2n^2}= \frac{2(-1)^n}{\pi n}\end{aligned}$$

$$f(x)=1+\frac{2}{\pi}\sum_{n=1}^{\infty}\frac{(-1)^n}{n}\sin(2\pi nx),\quad x\in\!\left(-\frac{1}{2};\,\frac{1}{2}\right)$$