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We are given the piecewise function \( f(x) \) defined as:

\[
f(x) = 
\begin{cases} 
a, & 0 \leq x \leq \pi \\
-a, & \pi \leq x \leq 2\pi
\end{cases}
\]

The problem is asking to find the Fourier coefficients, specifically the \( a_0 \) and \( a_n \) terms, for this function.

### Fourier Series Overview:

For a function defined on the interval \( [0, 2\pi] \), the Fourier series expansion is given by:

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

The coefficients \( a_0 \) and \( a_n \) are calculated using the following formulas:

- \( a_0 \) (the average value) is given by:

\[
a_0 = \frac{1}{\pi} \int_0^{2\pi} f(x) \, dx
\]

- \( a_n \) is given by:

\[
a_n = \frac{1}{\pi} \int_0^{2\pi} f(x) \cos(nx) \, dx
\]

- \( b_n \) is given by:

\[
b_n = \frac{1}{\pi} \int_0^{2\pi} f(x) \sin(nx) \, dx
\]

However, since \( f(x) \) is an even function, \( b_n = 0 \) for all \( n \). This leaves us with only \( a_0 \) and \( a_n \).

### Step 1: Calculate \( a_0 \)

\[
a_0 = \frac{1}{\pi} \int_0^{2\pi} f(x) \, dx
\]

We split the integral at \( x = \pi \):

\[
a_0 = \frac{1}{\pi} \left( \int_0^{\pi} a \, dx + \int_{\pi}^{2\pi} (-a) \, dx \right)
\]

\[
a_0 = \frac{1}{\pi} \left( a[\pi - 0] + (-a)[2\pi - \pi] \right) = \frac{1}{\pi} \left( a\pi - a\pi \right) = 0
\]

Thus, \( a_0 = 0 \).

### Step 2: Calculate \( a_n \)

\[
a_n = \frac{1}{\pi} \int_0^{2\pi} f(x) \cos(nx) \, dx
\]

We split the integral again:

\[
a_n = \frac{1}{\pi} \left( \int_0^{\pi} a \cos(nx) \, dx + \int_{\pi}^{2\pi} (-a) \cos(nx) \, dx \right)
\]

Using the substitution \( u = x - \pi \) in the second integral, the two integrals will combine, yielding:

\[
a_n = \frac{2a}{\pi} \int_0^{\pi} \cos(nx) \, dx
\]

The integral of \( \cos(nx) \) is \( \frac{\sin(nx)}{n} \), so we get:

\[
a_n = \frac{2a}{\pi} \left[ \frac{\sin(n\pi)}{n} - \frac{\sin(0)}{n} \right]
\]

Since \( \sin(n\pi) = 0 \), the result simplifies to:

\[
a_n = \frac{2a}{\pi} \cdot 0 = 0 \quad \text{for all odd} \ n.
\]

However, for even \( n \), we get non-zero results, depending on the specific values of \( n \).

Based on this information, the correct choice is \( a_0 = 0 \), which corresponds to option (c).

---

Do you have any questions or would you like further details?

### 5 Related Questions:
1. How do we calculate the \( b_n \) terms in a Fourier series?
2. What is the physical interpretation of the Fourier coefficients?
3. Can the Fourier series of a discontinuous function converge to the original function?
4. How do Fourier series apply to signal processing?
5. What is the difference between Fourier series and Fourier transform?

### Tip:
Always check the symmetry of the function before calculating the Fourier coefficients, as it can simplify the problem significantly (e.g., odd/even function).

Given a piecewise function f(x) = a for 0 ≤ x ≤ π and f(x) = -a for π ≤ x ≤ 2π, find the Fourier coefficients a_0 and a_n.

Learn how to find the Fourier coefficients a_0 and a_n for the piecewise function f(x), where f(x) = a for 0 ≤ x ≤ π and f(x) = -a for π ≤ x ≤ 2π. This solution involves calculating the integrals using Fourier series formulas and recognizing function symmetry.

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