SOLUTION: Find Fourier series for 𝑓(𝑥) = 𝑥, −2 < 𝑥

SOLUTION: Find Fourier series for 𝑓(𝑥) = 𝑥, −2 < 𝑥 < 2, f(x + 4) = f(x).

Question 1165582: Find Fourier series for 𝑓(𝑥) = 𝑥, −2 < 𝑥 < 2, f(x + 4) = f(x).
Answer by CPhill(2138) (Show Source): You can put this solution on YOUR website!
The function $f(x) = x$ is defined over the interval $(-2, 2)$ and is periodic with period $T=4$.
The period is $2L = 4$, so $L=2$.
Since the function $f(x) = x$ is an **odd function** (because $f(-x) = -x = -f(x)$), its Fourier series will only contain sine terms. This means the coefficients $a_0$ and $a_n$ will be zero.
$$\mathbf{a_0 = 0}$$
$$\mathbf{a_n = 0}$$
The Fourier series is given by:
$$f(x) = \sum_{n=1}^{\infty} b_n \sin\left(\frac{n\pi x}{L}\right)$$
### 1. Calculate the $b_n$ Coefficients
For an odd function, the coefficient $b_n$ is calculated as:
$$b_n = \frac{2}{L} \int_{0}^{L} f(x) \sin\left(\frac{n\pi x}{L}\right) dx$$
Substitute $f(x) = x$ and $L=2$:
$$b_n = \frac{2}{2} \int_{0}^{2} x \sin\left(\frac{n\pi x}{2}\right) dx$$
$$b_n = \int_{0}^{2} x \sin\left(\frac{n\pi x}{2}\right) dx$$
We use **Integration by Parts (IBP)**: $\int u \, dv = uv - \int v \, du$.
* Let $u = x \implies du = dx$
* Let $dv = \sin\left(\frac{n\pi x}{2}\right) dx \implies v = -\frac{2}{n\pi} \cos\left(\frac{n\pi x}{2}\right)$
$$b_n = \left[ x \left(-\frac{2}{n\pi} \cos\left(\frac{n\pi x}{2}\right)\right) \right]_0^2 - \int_{0}^{2} \left(-\frac{2}{n\pi} \cos\left(\frac{n\pi x}{2}\right)\right) dx$$
#### A. Evaluate the $uv$ Term
$$\left[ -\frac{2x}{n\pi} \cos\left(\frac{n\pi x}{2}\right) \right]_0^2$$
$$= \left( -\frac{2(2)}{n\pi} \cos\left(\frac{n\pi (2)}{2}\right) \right) - \left( -\frac{2(0)}{n\pi} \cos(0) \right)$$
$$= -\frac{4}{n\pi} \cos(n\pi) - 0$$
Recall that $\cos(n\pi) = (-1)^n$.
$$\text{Term 1} = -\frac{4}{n\pi} (-1)^n = \frac{4}{n\pi} (-1)^{n+1}$$
#### B. Evaluate the $\int v \, du$ Term
$$\int_{0}^{2} \frac{2}{n\pi} \cos\left(\frac{n\pi x}{2}\right) dx$$
$$= \frac{2}{n\pi} \left[ \frac{2}{n\pi} \sin\left(\frac{n\pi x}{2}\right) \right]_0^2$$
$$= \frac{4}{n^2\pi^2} \left[ \sin\left(\frac{n\pi (2)}{2}\right) - \sin(0) \right]$$
$$= \frac{4}{n^2\pi^2} [\sin(n\pi) - 0]$$
Since $\sin(n\pi) = 0$ for all integers $n$, this entire term is **0**.
#### C. Final $b_n$ Coefficient
$$b_n = \text{Term 1} - 0$$
$$\mathbf{b_n = \frac{4}{n\pi} (-1)^{n+1}}$$
### 2. Form the Fourier Series
Substitute the $b_n$ coefficients back into the Fourier series expansion:
$$f(x) = \sum_{n=1}^{\infty} b_n \sin\left(\frac{n\pi x}{2}\right)$$
$$f(x) = \sum_{n=1}^{\infty} \frac{4}{n\pi} (-1)^{n+1} \sin\left(\frac{n\pi x}{2}\right)$$
---
### Expanded Series (First few terms)
$$f(x) = \frac{4}{\pi} \left[ \frac{(-1)^{1+1}}{1} \sin\left(\frac{1\pi x}{2}\right) + \frac{(-1)^{2+1}}{2} \sin\left(\frac{2\pi x}{2}\right) + \frac{(-1)^{3+1}}{3} \sin\left(\frac{3\pi x}{2}\right) + \ldots \right]$$
$$f(x) = \frac{4}{\pi} \left[ \frac{1}{1} \sin\left(\frac{\pi x}{2}\right) - \frac{1}{2} \sin(\pi x) + \frac{1}{3} \sin\left(\frac{3\pi x}{2}\right) - \frac{1}{4} \sin(2\pi x) + \ldots \right]$$

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