Example 4. Expand the function $f(x)=10-x$, $5<x<15$ into a Fourier series.
Solution. We will construct the graph of the function $f(x)$ (Fig. 5.3). The dashed line represents the periodic extension with a period of $2l=10$ $(l=5)$. From the drawing, we conclude that the Fourier series expansion of this function coincides with the expansion of the function $f(x)=-x, x \in (-5; 5)$. This allows us to simplify the calculations somewhat.

![](https://cdn.mathpix.com/cropped/2024_05_22_446938f1e4808e7d6e07g-125.jpg?height=264&width=578&top_left_y=596&top_left_x=1902)

Fig. 5.3

From the oddness of $f(x)$, it follows that $a_{n}=0, n=0,1,2, \ldots$. It remains to calculate only the coefficients $b_{n}$; we will again use the oddness of $f(x)$ (replacing the interval $[-5; 5]$ with $[0; 5]$):

$b_{n}=-\frac{2}{5} \int_{0}^{5} x \sin \frac{n \pi x}{5} d x=$

$$
=-\frac{2}{5}\left[-\left.\frac{5 x \cos \frac{n \pi x}{3}}{n \pi}\right|_{0} ^{5}+\frac{5}{n \pi} \int_{0}^{5} \cos \frac{n \pi x}{5}\right]=\frac{10}{n \pi}(-1)^{n}
$$

Thus, the obtained Fourier series will converge to the function $f(x)=10-x$ in the interval $(5,15)$.

Answer. $10-x=\frac{10}{\pi} \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n} \sin \frac{n \pi x}{5}(5<x<15)$.