The image asks for the Fourier series of the function $f(x) = x^2$ over the interval $(0, 2\pi)$.

Steps to find the Fourier series:

1. Fourier Series Representation: The Fourier series of a function $f(x)$ 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 coefficients $a_0$, $a_n$, and $b_n$ are calculated as follows:

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

2. Given Function: $f(x) = x^2$ on $(0, 2\pi)$.

3. Compute $a_0$: $a_0 = \frac{1}{\pi} \int_0^{2\pi} x^2 \, dx$ Solve this integral: $a_0 = \frac{1}{\pi} \left[ \frac{x^3}{3} \right]_0^{2\pi} = \frac{1}{\pi} \cdot \frac{(2\pi)^3}{3} = \frac{8\pi^2}{3}$ So, $a_0 = \frac{8\pi^2}{3}$.

4. Compute $a_n$: $a_n = \frac{1}{\pi} \int_0^{2\pi} x^2 \cos(nx) \, dx$ This integral involves integrating by parts. The result of this computation is: $a_n = \frac{4(-1)^n}{n^2}$

5. Compute $b_n$: $b_n = \frac{1}{\pi} \int_0^{2\pi} x^2 \sin(nx) \, dx$ By symmetry, since $x^2$ is an even function and $\sin(nx)$ is an odd function, the result of this integral is $b_n = 0$.

Final Fourier Series:

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

Would you like more detailed steps for the integration, or do you have any other questions?

Related Questions:

1. How do you integrate by parts for functions like $x^2 \cos(nx)$?
2. What are some practical applications of Fourier series?
3. How does the Fourier series change for different intervals?
4. How can we use symmetry to simplify the Fourier series computations?
5. What happens to the Fourier series of non-periodic functions?

Tip:

When solving Fourier series, always check if the function is even or odd, as this can simplify the calculation by making some terms zero.