Let's solve this problem step by step.

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.