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To find the Fourier series expansion of the periodic function \( f(x) = x \) on the interval \( -\pi < x < \pi \), we proceed by calculating the Fourier coefficients.

### Fourier Series:
The Fourier series of a function \( f(x) \) with period \( 2\pi \) is given by:

\[
f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} \left( a_n \cos(nx) + b_n \sin(nx) \right)
\]

Where the coefficients \( a_0 \), \( a_n \), and \( b_n \) are defined as:

\[
a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \, dx
\]
\[
a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos(nx) \, dx
\]
\[
b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin(nx) \, dx
\]

### Step 1: Find \( a_0 \)
We first calculate \( a_0 \):

\[
a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} x \, dx
\]

This is a straightforward integral:

\[
a_0 = \frac{1}{\pi} \left[ \frac{x^2}{2} \right]_{-\pi}^{\pi} = \frac{1}{\pi} \left( \frac{\pi^2}{2} - \frac{(-\pi)^2}{2} \right) = \frac{1}{\pi} \cdot \frac{2\pi^2}{2} = 0
\]

Thus, \( a_0 = 0 \).

### Step 2: Find \( a_n \)
Next, we calculate \( a_n \):

\[
a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} x \cos(nx) \, dx
\]

Since \( x \cos(nx) \) is an odd function (because \( x \) is odd and \( \cos(nx) \) is even), the integral of an odd function over a symmetric interval (like \( -\pi \) to \( \pi \)) is zero:

\[
a_n = 0 \quad \text{for all } n
\]

### Step 3: Find \( b_n \)
Now, we calculate \( b_n \):

\[
b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} x \sin(nx) \, dx
\]

This is an integral of an odd function (both \( x \) and \( \sin(nx) \) are odd, so their product is even), and we need to compute it:

\[
b_n = \frac{1}{\pi} \left[ -\frac{x \cos(nx)}{n} \Big|_{-\pi}^{\pi} + \frac{1}{n} \int_{-\pi}^{\pi} \cos(nx) \, dx \right]
\]

Evaluating the boundary terms \( x \cos(nx) \Big|_{-\pi}^{\pi} \), we find that these terms vanish because \( \cos(n\pi) \) alternates between \( \pm 1 \), and the integrals of \( \cos(nx) \) over \( -\pi \) to \( \pi \) result in:

\[
b_n = \frac{2 (-1)^{n+1}}{n}
\]

### Final Fourier Series:
Thus, the Fourier series expansion of \( f(x) = x \) on \( -\pi < x < \pi \) is given by:

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

This is the Fourier series for the function \( f(x) = x \), which consists only of sine terms (since it is an odd function).

---

Would you like more details on any of these steps or any related concept? Here are five questions you could explore further:

1. How do we handle Fourier series for even and odd functions?
2. What are the implications of the Fourier coefficients being zero for certain terms?
3. How would the Fourier series change if the function were not odd?
4. Can you derive a Fourier series for a piecewise function?
5. How does the convergence of the Fourier series behave for discontinuous functions?

**Tip**: When calculating Fourier coefficients, always check the symmetry of the function to simplify the integrals (odd/even properties can eliminate terms).

Find the Fourier series expansion of the periodic function f(x) = x on -π < x < π with f(x + 2π) = f(x).

Learn how to find the Fourier series expansion of the periodic function f(x) = x on the interval -π < x < π, using Fourier coefficients for sine and cosine terms. This detailed solution explores the mathematical concepts of harmonic analysis and orthogonality.

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