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### Problem: 
We need to find the Fourier series for the function \( f(x) = x^2 \) in the interval \( (0, 2\pi) \), and then deduce the given series:

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

### Step 1: Fourier Series Setup
The Fourier series for a periodic function \( f(x) \) with period \( 2\pi \) is generally given by:

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

Where:
- \( a_0 \) is the average value of the function over the interval \( [0, 2\pi] \),
- \( a_n \) and \( b_n \) are the Fourier coefficients.

The Fourier coefficients are given by:

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

Since \( f(x) = x^2 \) is an even function, all the \( b_n \) terms will be zero, because:

\[
b_n = \frac{1}{\pi} \int_0^{2\pi} x^2 \sin(nx) \, dx = 0
\]

### Step 2: Calculating \( a_0 \)
The constant term \( a_0 \) is calculated as follows:

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

\[
\int_0^{2\pi} x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^{2\pi} = \frac{(2\pi)^3}{3} = \frac{8\pi^3}{3}
\]

\[
a_0 = \frac{1}{\pi} \cdot \frac{8\pi^3}{3} = \frac{8\pi^2}{3}
\]

Thus, the constant term is:

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

### Step 3: Calculating \( a_n \)
Now, we calculate the \( a_n \) terms:

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

This integral can be solved using integration by parts or a known result. The result is:

\[
a_n = \frac{4}{n^2}
\]

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

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

### Step 5: Deduction of the Given Series
To deduce the series involving \( \frac{\pi^2}{12} \), we evaluate the Fourier series at \( x = \pi \):

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

Since \( \cos(n\pi) = (-1)^n \), the equation becomes:

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

Rearranging the equation:

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

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

Dividing both sides by 4:

\[
\frac{\pi^2}{12} = \sum_{n=1}^{\infty} \frac{(-1)^n}{n^2}
\]

This gives the desired series:

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

### Conclusion:
The Fourier series for \( f(x) = x^2 \) on \( (0, 2\pi) \) is:

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

And the given series is deduced as:

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

---

Do you want more details on any step? Here are five questions for further exploration:

1. How is the Fourier series derived for other functions like \( e^x \)?
2. What role does symmetry (even/odd functions) play in Fourier series?
3. How can integration by parts be used to solve \( a_n \) in detail?
4. How does Parseval's theorem relate to Fourier series?
5. Can the Fourier series be applied to non-periodic functions?

**Tip:** For even functions, all sine terms in the Fourier series vanish due to symmetry, making the series easier to compute.

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 compute the Fourier series for f(x) = x^2 on the interval (0, 2π) and deduce the series π^2/12 = 1/1^2 - 1/2^2 + 1/3^2 - ... Detailed explanation includes calculation of Fourier coefficients and series representation.

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