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

Compute its sine Fourier series:
f(x)∼∑n=1∞bnsin(ωnx)

We are tasked with finding the sine Fourier series for the function \( f(x) = e^x \) defined on the interval \( [0, 3] \) and extended as an odd function with period 6.

### Step 1: Setting up the Fourier sine series form

The sine Fourier series for an odd periodic function with period \( 2L \) is given by:

\[
f(x) \sim \sum_{n=1}^{\infty} b_n \sin \left( \frac{n \pi x}{L} \right)
\]

Here, the interval length \( L = 3 \), since the function has period 6. Thus, the Fourier sine series becomes:

\[
f(x) \sim \sum_{n=1}^{\infty} b_n \sin \left( \frac{n \pi x}{3} \right)
\]

### Step 2: Fourier sine coefficients

The coefficients \( b_n \) are computed using the formula:

\[
b_n = \frac{2}{L} \int_0^L f(x) \sin \left( \frac{n \pi x}{L} \right) dx
\]

Substituting \( L = 3 \) and \( f(x) = e^x \), we get:

\[
b_n = \frac{2}{3} \int_0^3 e^x \sin \left( \frac{n \pi x}{3} \right) dx
\]

### Step 3: Evaluating the integral for \( b_n \)

To compute \( b_n \), we use integration by parts or a standard result from Fourier series for the integral:

\[
I_n = \int_0^3 e^x \sin \left( \frac{n \pi x}{3} \right) dx
\]

We can solve this integral by applying integration by parts twice or using a table of integrals. The integral has the general form:

\[
\int e^{ax} \sin(bx) dx = \frac{e^{ax}(a \sin(bx) - b \cos(bx))}{a^2 + b^2}
\]

For our case, \( a = 1 \) and \( b = \frac{n \pi}{3} \). Thus, applying this formula over the limits \( 0 \) to \( 3 \), we get:

\[
b_n = \frac{2}{3} \left[ \frac{e^3 \left( \sin \left( n \pi \right) - \frac{n \pi}{3} \cos \left( n \pi \right) \right) }{1 + \left( \frac{n \pi}{3} \right)^2 } - \frac{\sin(0) - \frac{n \pi}{3} \cos(0)}{1 + \left( \frac{n \pi}{3} \right)^2} \right]
\]

Simplifying the terms with \( \sin(n \pi) = 0 \) and \( \cos(n \pi) = (-1)^n \), we obtain:

\[
b_n = \frac{2}{3} \cdot \frac{(-1)^{n+1} \frac{n \pi}{3} e^3 - \frac{n \pi}{3}}{1 + \left( \frac{n \pi}{3} \right)^2}
\]

This expression gives the sine Fourier coefficients \( b_n \).

### Final Sine Fourier Series

Thus, the sine Fourier series for the function \( f(x) = e^x \), extended as an odd function over \( [-3, 3] \) and period 6, is:

\[
f(x) \sim \sum_{n=1}^{\infty} \frac{2}{3} \cdot \frac{(-1)^{n+1} \frac{n \pi}{3} e^3 - \frac{n \pi}{3}}{1 + \left( \frac{n \pi}{3} \right)^2} \sin \left( \frac{n \pi x}{3} \right)
\]

Would you like further details or have any questions about this derivation?

### Related Questions:
1. How would this Fourier series change if the function were extended as an even function instead of an odd function?
2. What is the general process for calculating the Fourier series of a non-exponential function?
3. Can you explain the difference between the sine and cosine Fourier series for periodic functions?
4. How does extending a function as an odd function impact its Fourier series?
5. How would you approach the Fourier series expansion for a function that is not periodic?

### Tip:
When dealing with Fourier series, always check the symmetry of the function first. This helps determine whether the series will involve sine, cosine, or both.

Consider the function f(x)=e^x defined on the interval [0,3] and extend it as an odd function with period 6. Compute its sine Fourier series: f(x)∼∑n=1∞bnsin(ωnx).

Learn how to compute the sine Fourier series for the function f(x) = e^x defined on the interval [0,3] and extended as an odd function with period 6. This solution includes key steps such as setting up the sine series form, calculating the Fourier coefficients using integration, and applying properties of periodic functions.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation