To find the Fourier Series expansion of the function $f(x) = \frac{x}{k}$ in the interval $(0, 2\pi)$, we need to express the function as a sum of sine and cosine terms (for real Fourier series) or exponentials (for complex Fourier series). Let's start with the real Fourier series expansion.

Real Fourier Series Form

The Fourier series of a function $f(x)$ defined on $(0, 2\pi)$ 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 determined by the following integrals:

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

Step 1: Compute $a_0$

We start by computing $a_0$:

$a_0 = \frac{1}{\pi} \int_0^{2\pi} \frac{x}{k} \, dx$

$a_0 = \frac{1}{k\pi} \int_0^{2\pi} x \, dx = \frac{1}{k\pi} \left[ \frac{x^2}{2} \right]_0^{2\pi}$

$a_0 = \frac{1}{k\pi} \times \frac{(2\pi)^2}{2} = \frac{2\pi}{k}$

Thus:

$a_0 = \frac{2\pi}{k}$

Step 2: Compute $a_n$

Next, we compute $a_n$:

$a_n = \frac{1}{\pi} \int_0^{2\pi} \frac{x}{k} \cos(nx) \, dx$

We can solve this integral using integration by parts. After computation, this integral evaluates to zero because it involves $x \cos(nx)$, which oscillates and cancels out over the interval $(0, 2\pi)$.

Thus:

$a_n = 0 \quad \text{for all} \, n \geq 1$

Step 3: Compute $b_n$

Finally, we compute $b_n$:

$b_n = \frac{1}{\pi} \int_0^{2\pi} \frac{x}{k} \sin(nx) \, dx$

Using integration by parts, we have:

$b_n = \frac{1}{k\pi} \left[ -\frac{x \cos(nx)}{n} \Bigg|_0^{2\pi} + \frac{1}{n} \int_0^{2\pi} \cos(nx) \, dx \right]$

The first term evaluates to zero because $\cos(nx)$ at the limits $0$ and $2\pi$ vanishes. The second integral also vanishes because it is the integral of a cosine over a full period $(0, 2\pi)$.

Thus:

$b_n = \frac{2}{kn}$

Final Fourier Series Expansion

Therefore, the Fourier series expansion of $f(x) = \frac{x}{k}$ in the interval $(0, 2\pi)$ is:

$f(x) = \frac{\pi}{k} - \frac{2}{k} \sum_{n=1}^{\infty} \frac{(-1)^n}{n} \sin(nx)$

This is the Fourier series expansion of $f(x)$ in the given interval.

Would you like more details on any step?

Related Questions:

1. What is the general formula for Fourier series expansion of any periodic function?
2. How does changing the interval affect the Fourier series coefficients?
3. How is the convergence of a Fourier series determined for a given function?
4. What role do symmetry properties (odd/even) play in simplifying Fourier series?
5. How can Fourier series be applied in solving partial differential equations?

Tip: Fourier series can help approximate complex waveforms by decomposing them into simpler trigonometric functions.