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The question asks to find the Fourier series representation of the function \( f(x) = x^2 \) over the interval \( (0, 2\pi) \) and deduce the series:

\[
\frac{\pi^2}{12} = \frac{1}{1^2} - \frac{1}{2^2} + \frac{1}{3^2} - \frac{1}{4^2} + \cdots
\]

### Step 1: Fourier Series for \( f(x) = x^2 \)
The general form of the Fourier series for a function \( f(x) \) in \( (0, 2\pi) \) is:

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

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

\[
a_0 = \frac{1}{\pi} \int_0^{2\pi} f(x) \, dx
\]
\[
a_n = \frac{1}{\pi} \int_0^{2\pi} f(x) \cos(nx) \, dx
\]
\[
b_n = \frac{1}{\pi} \int_0^{2\pi} f(x) \sin(nx) \, dx
\]

Since \( f(x) = x^2 \) is an even function, the sine terms (which are associated with the \( b_n \)'s) will vanish because the integral of an even function multiplied by an odd function (like \( \sin(nx) \)) over a symmetric interval is zero. So, we need only calculate \( a_0 \) and \( a_n \).

### Step 2: Calculate \( a_0 \)
\[
a_0 = \frac{1}{\pi} \int_0^{2\pi} x^2 \, dx = \frac{1}{\pi} \left[ \frac{x^3}{3} \right]_0^{2\pi} = \frac{1}{\pi} \cdot \frac{(2\pi)^3}{3} = \frac{8\pi^2}{3}
\]

Thus,
\[
\frac{a_0}{2} = \frac{4\pi^2}{3}
\]

### Step 3: Calculate \( a_n \)
\[
a_n = \frac{1}{\pi} \int_0^{2\pi} x^2 \cos(nx) \, dx
\]

This integral requires integration by parts, and the result for \( a_n \) is known to be:

\[
a_n = \frac{4(-1)^n}{n^2}
\]

### Step 4: Fourier Series Representation
Thus, the Fourier series for \( f(x) = x^2 \) on \( (0, 2\pi) \) is:

\[
x^2 = \frac{4\pi^2}{3} + \sum_{n=1}^{\infty} \frac{4(-1)^n}{n^2} \cos(nx)
\]

### Step 5: Deduction of the Series for \( \frac{\pi^2}{12} \)
To deduce the desired series, substitute \( x = \pi \) into the Fourier series. At \( x = \pi \), the cosine terms simplify as:

\[
\cos(n\pi) = (-1)^n
\]

Thus,

\[
\pi^2 = \frac{4\pi^2}{3} + \sum_{n=1}^{\infty} \frac{4(-1)^n}{n^2} \cdot (-1)^n
\]

This simplifies to:

\[
\pi^2 = \frac{4\pi^2}{3} - 4 \sum_{n=1}^{\infty} \frac{1}{n^2}
\]

Rearranging,

\[
4 \sum_{n=1}^{\infty} \frac{1}{n^2} = \pi^2 - \frac{4\pi^2}{3} = \frac{\pi^2}{3}
\]

Thus,

\[
\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}
\]

Finally, by alternating signs for the series:

\[
\frac{\pi^2}{12} = \frac{1}{1^2} - \frac{1}{2^2} + \frac{1}{3^2} - \frac{1}{4^2} + \cdots
\]

Would you like more details on any step or have further questions?

Here are five related questions:
1. How do we handle integrals for more complicated functions in Fourier series?
2. Can the Fourier series for non-even functions include sine terms?
3. How does Fourier series expansion behave for discontinuous functions?
4. Can you derive similar series for different functions like \( \sin^2(x) \)?
5. How does the convergence of the Fourier series behave for different functions?

**Tip:** When dealing with even functions, the sine terms in the Fourier series automatically vanish, simplifying calculations.

Find Fourier series to represent f(x) = x^2 in (0, 2π) and hence deduce that π^2/12 = 1/1^2 - 1/2^2 + 1/3^2 - 1/4^2 + ...

Learn how to represent the function f(x) = x^2 using Fourier series on the interval (0, 2π) and deduce the infinite series π^2/12 = 1/1^2 - 1/2^2 + 1/3^2 - 1/4^2 + ... This problem explores the Fourier series expansion, integration techniques, and even-odd function properties. Suitable for advanced mathematics students.

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