To solve this, we are asked to find the Fourier series for the function $f(x) = x$ on the interval $0 < x < 2\pi$.

General Formula for 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 coefficients $a_0$, $a_n$, and $b_n$ are defined as:

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

Step 1: Compute $a_0$

$a_0 = \frac{1}{\pi} \int_0^{2\pi} x \, dx$ $a_0 = \frac{1}{\pi} \left[ \frac{x^2}{2} \right]_0^{2\pi} = \frac{1}{\pi} \times \frac{(2\pi)^2}{2} = 2\pi$ Thus, $a_0 = 2\pi$.

Step 2: Compute $a_n$

$a_n = \frac{1}{\pi} \int_0^{2\pi} x \cos(nx) \, dx$ This requires integration by parts. Let $u = x$ and $dv = \cos(nx) \, dx$.

• $du = dx$
• $v = \frac{\sin(nx)}{n}$

Thus, applying integration by parts: $a_n = \frac{1}{\pi} \left[ \frac{x \sin(nx)}{n} \right]_0^{2\pi} - \frac{1}{\pi} \int_0^{2\pi} \frac{\sin(nx)}{n} \, dx$ Evaluating the boundary terms at 0 and $2\pi$, they vanish because $\sin(n \cdot 0) = \sin(n \cdot 2\pi) = 0$. The remaining integral is: $a_n = -\frac{1}{\pi n} \int_0^{2\pi} \sin(nx) \, dx = 0$ Thus, $a_n = 0$.

Step 3: Compute $b_n$

$b_n = \frac{1}{\pi} \int_0^{2\pi} x \sin(nx) \, dx$ Again, use integration by parts with $u = x$ and $dv = \sin(nx) \, dx$.

• $du = dx$
• $v = -\frac{\cos(nx)}{n}$

Applying integration by parts: $b_n = \frac{1}{\pi} \left[ -\frac{x \cos(nx)}{n} \right]_0^{2\pi} + \frac{1}{\pi} \int_0^{2\pi} \frac{\cos(nx)}{n} \, dx$ Evaluating the boundary terms at 0 and $2\pi$ again gives 0 because $\cos(n \cdot 0) = \cos(n \cdot 2\pi) = 1$. The remaining integral is: b_n = \frac{1}{n} \cdot 2 \quad \text{(since the integral evaluates to 2$for odd$n)}.

Final Fourier Series:

The Fourier series for $f(x) = x$ over $0 < x < 2\pi$ is: $f(x) = \pi - 2\sum_{n=1}^{\infty} \frac{(-1)^n}{n} \sin(nx)$

Would you like further clarification on any step or additional details? Here are five related questions to deepen your understanding:

1. How does the Fourier series differ when the function is defined on a symmetric interval, such as $[-\pi, \pi]$?
2. What happens to the Fourier series if we extend $f(x) = x$ periodically beyond $[0, 2\pi]$?
3. Could we represent non-periodic functions using Fourier series over a finite interval?
4. Why do the cosine terms vanish for this function in the Fourier series?
5. What are the practical applications of Fourier series in engineering and physics?

Tip: For odd functions, the Fourier series will only contain sine terms, while for even functions, it will only contain cosine terms.