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To solve the problem, we need to find the Fourier series representation of the function \( f(x) = x^2 \) on the interval \( (0, 2\pi) \). From this, we will deduce that:

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

### Step 1: General form of Fourier Series
The Fourier series of a function \( f(x) \) defined on the interval \( [0, 2\pi] \) 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 Fourier coefficients \( a_n \) and \( b_n \) are calculated as follows:

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

### Step 2: Calculating the coefficients for \( f(x) = x^2 \)
#### 1. \( a_0 \) (Constant term):
\[
a_0 = \frac{1}{\pi} \int_0^{2\pi} x^2 \, dx = \frac{1}{\pi} \left[ \frac{x^3}{3} \right]_0^{2\pi} = \frac{1}{\pi} \cdot \frac{(2\pi)^3}{3} = \frac{8\pi^2}{3}
\]

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

#### 2. \( a_n \) (Cosine terms):
\[
a_n = \frac{1}{\pi} \int_0^{2\pi} x^2 \cos(nx) \, dx
\]
We will use integration by parts here. This integral evaluates to:

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

#### 3. \( b_n \) (Sine terms):
\[
b_n = \frac{1}{\pi} \int_0^{2\pi} x^2 \sin(nx) \, dx
\]
Since \( x^2 \sin(nx) \) is an odd function over \( [0, 2\pi] \), the integral evaluates to zero, so:

\[
b_n = 0
\]

### Step 3: Fourier series representation
Now, the Fourier series representation of \( f(x) = x^2 \) is:

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

### Step 4: Deducing the series
To deduce the series for \( \frac{\pi^2}{12} \), set \( x = \pi \) in the Fourier series expansion:

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

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

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

Rearranging gives:

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

Now subtract \( \frac{1}{1^2} \) to form the alternating series given in the question:

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

This is the desired result.

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Would you like more details on any step? Here are five related questions:

1. How does integration by parts work for calculating Fourier coefficients?
2. What is the significance of \( \cos(nx) \) and \( \sin(nx) \) terms in Fourier series?
3. Can the Fourier series be used to approximate discontinuous functions?
4. How would the Fourier series change if \( f(x) \) was periodic with a different period?
5. What other mathematical results can be derived from the Fourier series of basic functions?

**Tip:** Always check for symmetry in functions when calculating Fourier coefficients, as this can simplify calculations (e.g., odd or even functions leading to zero sine or cosine terms).

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

Learn how to find the Fourier series representation of the function f(x) = x^2 in the interval (0, 2π) and deduce the result π^2 / 12 = 1 / 12 - 1 / 2^2 + 1 / 3^2 - .... This problem involves key concepts like Fourier coefficients, trigonometric series, and series expansion.

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