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To find the Fourier expansion of the given function \( f(x) \), we need to express the function as a Fourier series on the interval \( [-1, 1] \).

The given piecewise function is:
\[
f(x) = 
\begin{cases} 
-x, & -1 < x < 0 \\
x, & 0 < x < 1 
\end{cases}
\]

Since the function is odd (because \( f(-x) = -f(x) \)), we only need to find the sine terms of the Fourier series (Fourier sine series).

The general Fourier sine series for an odd function is given by:
\[
f(x) = \sum_{n=1}^{\infty} b_n \sin\left(n\pi x\right)
\]
where the Fourier coefficients \( b_n \) are computed as:
\[
b_n = \frac{2}{L} \int_0^L f(x) \sin\left(\frac{n\pi x}{L}\right) dx
\]
Here, \( L = 1 \) because the interval is \( [-1, 1] \).

### Step 1: Compute the Fourier Coefficients \( b_n \)

Since the function is odd, we only need to calculate the sine coefficients:

\[
b_n = 2 \int_0^1 f(x) \sin(n\pi x) dx
\]

For \( f(x) = x \) on \( [0, 1] \), we compute \( b_n \) as:
\[
b_n = 2 \int_0^1 x \sin(n\pi x) dx
\]

This integral can be solved using integration by parts. Let:
- \( u = x \), so \( du = dx \)
- \( dv = \sin(n\pi x) dx \), so \( v = -\frac{1}{n\pi} \cos(n\pi x) \)

Now applying integration by parts:
\[
b_n = 2 \left[ -\frac{x \cos(n\pi x)}{n\pi} \Big|_0^1 + \frac{1}{n\pi} \int_0^1 \cos(n\pi x) dx \right]
\]

Evaluating the first term at the limits:
\[
-\frac{x \cos(n\pi x)}{n\pi} \Big|_0^1 = -\frac{1 \cdot \cos(n\pi)}{n\pi} + 0 = -\frac{(-1)^n}{n\pi}
\]

Now for the second term:
\[
\frac{1}{n\pi} \int_0^1 \cos(n\pi x) dx = \frac{1}{n\pi} \left[ \frac{\sin(n\pi x)}{n\pi} \Big|_0^1 \right] = 0
\]

Thus, we have:
\[
b_n = \frac{2(-1)^{n+1}}{n\pi}
\]

### Step 2: Write the Fourier Series

Now that we have the coefficients \( b_n = \frac{2(-1)^{n+1}}{n\pi} \), the Fourier sine series expansion for \( f(x) \) is:

\[
f(x) = \sum_{n=1}^{\infty} \frac{2(-1)^{n+1}}{n\pi} \sin(n\pi x)
\]

This is the Fourier series for the given piecewise function.

---

Would you like a detailed breakdown of each calculation step or need further clarifications? Here are some related questions for exploration:

1. How does the parity of a function affect its Fourier series expansion?
2. What are the conditions for using only the sine terms in the Fourier series?
3. How can you verify that a function is odd or even?
4. How do you apply integration by parts for solving Fourier coefficients?
5. What is the physical significance of Fourier series in signal processing?

**Tip:** For odd functions defined on symmetric intervals like \( [-L, L] \), the Fourier series will only involve sine terms, simplifying the expansion process.

Find the Fourier expansion of f(x) = {-x, -1 < x < 0; x, 0 < x < 1}.

This solution details how to find the Fourier expansion of a piecewise function defined on the interval [-1, 1]. Using the Fourier sine series for odd functions, the problem demonstrates integral techniques like integration by parts to compute the Fourier coefficients.

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