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find the fourier series of the given function in the given range (negative pi to positive pi)

The given function \( f(x) \) is defined piecewise as:

\[
f(x) = \begin{cases} 
-K & \text{for } -\pi \leq x < 0, \\
K & \text{for } 0 \leq x < \pi,
\end{cases}
\]
with periodicity \( f(x + 2\pi) = f(x) \).

This is a periodic function with period \( 2\pi \), and we are asked to find its Fourier series in the interval \( [-\pi, \pi] \).

### Step 1: Fourier Series Expansion
The Fourier series for a function with period \( 2\pi \) 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 Fourier coefficients \( a_0 \), \( a_n \), and \( b_n \) are calculated as:

\[
a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \, dx,
\]

\[
a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos(nx) \, dx,
\]

\[
b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin(nx) \, dx.
\]

### Step 2: Calculate \( a_0 \)

The coefficient \( a_0 \) is the average value of the function over one period:

\[
a_0 = \frac{1}{\pi} \left( \int_{-\pi}^{0} (-K) \, dx + \int_{0}^{\pi} K \, dx \right).
\]

Let me calculate this.The coefficient \( a_0 \), which represents the average value of the function over the interval \( [-\pi, \pi] \), is:

\[
a_0 = 0
\]

This result is expected because the function is symmetric around the origin, with equal positive and negative contributions that cancel each other out.

### Step 3: Calculate \( a_n \)

Now, let's compute \( a_n \):

\[
a_n = \frac{1}{\pi} \left( \int_{-\pi}^{0} (-K) \cos(nx) \, dx + \int_{0}^{\pi} K \cos(nx) \, dx \right).
\]

Let me calculate this next.The coefficient \( a_n \) also evaluates to:

\[
a_n = 0
\]

This means there are no cosine terms in the Fourier series, indicating that the function is odd.

### Step 4: Calculate \( b_n \)

Now, let's compute \( b_n \):

\[
b_n = \frac{1}{\pi} \left( \int_{-\pi}^{0} (-K) \sin(nx) \, dx + \int_{0}^{\pi} K \sin(nx) \, dx \right).
\]

Let me calculate this next.The coefficient \( b_n \) is:

\[
b_n = \begin{cases}
\frac{2K(1 - (-1)^n)}{\pi n} & \text{if } n \neq 0, \\
0 & \text{if } n = 0.
\end{cases}
\]

### Step 5: Fourier Series

Since \( a_0 = 0 \), \( a_n = 0 \), and we have the expression for \( b_n \), the Fourier series for the given function is:

\[
f(x) = \sum_{n=1}^{\infty} \frac{2K(1 - (-1)^n)}{\pi n} \sin(nx).
\]

This is the Fourier sine series representation of the function over the interval \( [-\pi, \pi] \).

Let me know if you need further details or explanations.

---

Here are 5 related questions you might find interesting:

1. How would the Fourier series change if \( f(x) \) was not odd?
2. What happens to the Fourier series if \( K \) is negative?
3. Can you compute the Fourier series for other periodic functions, like a sawtooth wave?
4. How do the Fourier sine and cosine series differ for non-symmetric functions?
5. Can this Fourier series be used to approximate the function for values of \( x \) outside the interval \( [-\pi, \pi] \)?

**Tip**: When working with Fourier series, always check the symmetry of the function to simplify your calculations (e.g., even functions only have cosine terms, and odd functions only have sine terms).

Find the Fourier series of the given function in the range (-π, π).

This page provides a detailed solution for finding the Fourier series of a piecewise function over the interval (-π, π). It includes step-by-step calculations for the Fourier coefficients a0, an, and bn, demonstrating the use of trigonometric integrals.

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