Fourier Series Expansion of f(x) = x/k in interval (0, 2 pi)

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.

This page provides a detailed step-by-step solution to the Fourier Series expansion of the function f(x) = x/k over the interval (0, 2π). Learn key mathematical concepts like integration by parts and trigonometric function expansions, suitable for undergraduate mathematics students.

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