Find the Fourier series of f(x) = x^2
, x ∈ (0,2L). Hence deduce that

1/1^2
+
1/3^2
+
1/5^2
+ ⋯ ∞ =
π^2/8

### Finding the Fourier Series of \( f(x) = x^2 \) on \( (0, 2L) \)

The Fourier series of a function \( f(x) \) over the interval \( (0, 2L) \) is given by:

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

Where the coefficients \( a_0 \), \( a_n \), and \( b_n \) are given by:

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

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

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

For the function \( f(x) = x^2 \):

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

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

\[
a_0 = \frac{1}{L} \left[ \frac{x^3}{3} \right]_0^{2L} = \frac{1}{L} \cdot \frac{(2L)^3}{3} = \frac{8L^2}{3}
\]

Thus, \( \frac{a_0}{2} = \frac{4L^2}{3} \).

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

\[
a_n = \frac{1}{L} \int_0^{2L} x^2 \cos\left(\frac{n\pi x}{L}\right) \, dx
\]

This integral can be solved by parts, or using a known formula:

\[
a_n = \frac{2L^2}{n^2\pi^2} \left( 2 - \cos(n\pi) \right) = \frac{4L^2}{n^2\pi^2}
\]

#### 3. Calculate \( b_n \):

Since \( x^2 \) is an even function, \( b_n = 0 \).

### Fourier Series of \( x^2 \) on \( (0, 2L) \):

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

### Deduction of the Series Sum

Now, consider \( L = \frac{\pi}{2} \), and \( f(x) = x^2 \) on \( (0, \pi) \). The Fourier series becomes:

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

Evaluate at \( x = 0 \):

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

So:

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

Now, considering only the odd terms \( n = 1, 3, 5, \ldots \):

\[
\frac{1}{1^2} + \frac{1}{3^2} + \frac{1}{5^2} + \ldots = \frac{\pi^2}{8}
\]

This is because the series sum for only odd terms is half the sum of all terms for \( n \):

\[
\sum_{k=0}^{\infty} \frac{1}{(2k+1)^2} = \frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{n^2}
\]

Thus, the required series sum:

\[
\frac{1}{1^2} + \frac{1}{3^2} + \frac{1}{5^2} + \ldots = \frac{\pi^2}{8}
\]

---

Would you like to explore the integration details, or have any questions? Here are some further topics to consider:

1. Deriving the general formula for Fourier coefficients for any \( f(x) \).
2. Exploring the convergence properties of Fourier series.
3. Understanding the difference between even and odd extensions.
4. Investigating the Gibbs phenomenon in Fourier series.
5. Applying Fourier series to solve differential equations.

**Tip:** The Fourier series provides powerful insights into periodic functions, with applications ranging from signal processing to heat conduction.

Explore the Fourier series of f(x) = x^2 on the interval (0, 2L) and deduce the series sum 1/1^2 + 1/3^2 + 1/5^2 + ... = π^2/8. Understand the mathematical calculations involved in Fourier series coefficients and series summation.

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