Consider the function f(x)=e^2x
 defined on the interval [0,3]
 and extend it as an odd function with period 6
.

Compute its sine Fourier series:

To find the sine Fourier series of the given function \( f(x) = e^{2x} \) defined on the interval \([0, 3]\) and extended as an odd periodic function with period 6, we proceed with the following steps:

### Step 1: Define the function over one period

Since \( f(x) \) is extended as an odd function with period 6, we have:
\[
f(x) = 
\begin{cases}
e^{2x}, & 0 \leq x \leq 3 \\
-e^{2x}, & -3 \leq x < 0
\end{cases}
\]

### Step 2: General form of the sine Fourier series

For a function defined on the interval \([-L, L]\), the sine Fourier series expansion is given by:
\[
f(x) \sim \sum_{n=1}^{\infty} b_n \sin\left( \frac{n \pi x}{L} \right),
\]
where
\[
b_n = \frac{2}{L} \int_0^L f(x) \sin\left( \frac{n \pi x}{L} \right) \, dx.
\]

Here, \( L = 3 \), so the Fourier coefficients become:
\[
b_n = \frac{2}{3} \int_0^3 e^{2x} \sin\left( \frac{n \pi x}{3} \right) \, dx.
\]

### Step 3: Compute \( b_n \)

We need to evaluate the integral:
\[
b_n = \frac{2}{3} \int_0^3 e^{2x} \sin\left( \frac{n \pi x}{3} \right) \, dx.
\]

Using integration by parts or a standard integral table, we find:
\[
\int_0^3 e^{2x} \sin\left( \frac{n \pi x}{3} \right) \, dx = \left[ \frac{e^{2x} \left( 2 \sin\left( \frac{n \pi x}{3} \right) - \frac{n \pi}{3} \cos\left( \frac{n \pi x}{3} \right) \right)}{4 + \left( \frac{n \pi}{3} \right)^2} \right]_0^3.
\]

Evaluating at the bounds and simplifying:
1. At \( x = 3 \), we get:
   \[
   e^{6} \left( \frac{2 \sin(n \pi) - \frac{n \pi}{3} \cos(n \pi)}{4 + \left( \frac{n \pi}{3} \right)^2} \right) = e^6 \left( \frac{- \frac{n \pi}{3} (-1)^n}{4 + \left( \frac{n \pi}{3} \right)^2} \right) = \frac{n \pi e^6 (-1)^n}{3 \left( 4 + \frac{n^2 \pi^2}{9} \right)}.
   \]

2. At \( x = 0 \), we get:
   \[
   \frac{2 \sin(0) - \frac{n \pi}{3} \cos(0)}{4 + \left( \frac{n \pi}{3} \right)^2} = \frac{- \frac{n \pi}{3}}{4 + \left( \frac{n \pi}{3} \right)^2}.
   \]

Combining these, we obtain:
\[
b_n = \frac{2}{3} \left( \frac{n \pi e^6 (-1)^n}{3 \left( 4 + \frac{n^2 \pi^2}{9} \right)} - \frac{-n \pi}{3 \left( 4 + \frac{n^2 \pi^2}{9} \right)} \right).
\]

Simplifying further:
\[
b_n = \frac{2 n \pi \left( e^6 (-1)^n + 1 \right)}{9 \left( 4 + \frac{n^2 \pi^2}{9} \right)}.
\]

### Step 4: Write the Fourier series

The sine Fourier series for \( f(x) \) is:
\[
f(x) \sim \sum_{n=1}^{\infty} \frac{2 n \pi \left( e^6 (-1)^n + 1 \right)}{9 \left( 4 + \frac{n^2 \pi^2}{9} \right)} \sin\left( \frac{n \pi x}{3} \right).
\]

Let me know if you need further details or have any questions.

Here are some additional questions for further practice:
1. What would be the cosine Fourier series of the function \( f(x) = e^{2x} \)?
2. How would the Fourier series change if the function was extended as an even function?
3. Can you find the Fourier coefficients if the function was defined over a different interval, say \( [0, 2] \)?
4. What is the general approach for computing the Fourier series of a piecewise function?
5. How do the properties of even and odd functions simplify the computation of Fourier coefficients?

**Tip:** When working with Fourier series, always consider the function's symmetry, as it can significantly simplify the calculations.

Consider the function f(x)=e^2x defined on the interval [0,3] and extend it as an odd function with period 6. Compute its sine Fourier series.

Compute the sine Fourier series for the function f(x)=e^2x, defined on [0,3] and extended as an odd periodic function with period 6. The solution covers key concepts in Fourier analysis, including periodic functions, integration by parts, and the properties of odd functions.

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