To find the Fourier series of the function $f(x) = e^{-x}$ defined on the interval $[-2\pi, 2\pi]$, we can follow the general approach for finding the Fourier series of a periodic function.

Fourier Series Formula:

The Fourier series of a function $f(x)$ on an interval $[-L, L]$ is given by:

$f(x) = a_0 + \sum_{n=1}^{\infty} \left( a_n \cos \left( \frac{n\pi x}{L} \right) + b_n \sin \left( \frac{n\pi x}{L} \right) \right)$ Where the Fourier coefficients $a_n$ and $b_n$ are given by:

$a_0 = \frac{1}{2L} \int_{-L}^{L} f(x) \, dx$ $a_n = \frac{1}{L} \int_{-L}^{L} f(x) \cos \left( \frac{n\pi x}{L} \right) \, dx$ $b_n = \frac{1}{L} \int_{-L}^{L} f(x) \sin \left( \frac{n\pi x}{L} \right) \, dx$

In this case, we have $L = 2\pi$, so the Fourier series will take the form:

$f(x) = a_0 + \sum_{n=1}^{\infty} \left( a_n \cos(nx) + b_n \sin(nx) \right)$

Step 1: Compute $a_0$

We start by calculating the coefficient $a_0$:

$a_0 = \frac{1}{2\pi} \int_{-2\pi}^{2\pi} e^{-x} \, dx$

The integral of $e^{-x}$ is straightforward:

$\int e^{-x} \, dx = -e^{-x}$

Evaluating the definite integral:

= \frac{1}{2\pi} \left( -e^{-2\pi} + e^{2\pi} \right) = \frac{1}{2\pi} \left( e^{2\pi} - e^{-2\pi} \right)$$ Using the fact that $$e^{2\pi} - e^{-2\pi} = 2\sinh(2\pi)$$, we have: $$a_0 = \frac{1}{2\pi} \cdot 2\sinh(2\pi) = \frac{\sinh(2\pi)}{\pi}$$ ### Step 2: Compute $$a_n$$ Now we calculate the coefficient $$a_n$$ for $$n \geq 1$$: $$a_n = \frac{1}{2\pi} \int_{-2\pi}^{2\pi} e^{-x} \cos(nx) \, dx$$ This is a standard integral. Using the integration by parts formula or looking up standard results for the Fourier cosine transform of $$e^{-x}$$, we find that: $$a_n = \frac{1}{2\pi} \cdot \frac{2e^{-2\pi} + 2e^{2\pi}}{n^2 + 1}$$ This simplifies to: $$a_n = \frac{2\sinh(2\pi)}{\pi(n^2 + 1)}$$ ### Step 3: Compute $$b_n$$ Finally, we compute the coefficient $$b_n$$: $$b_n = \frac{1}{2\pi} \int_{-2\pi}^{2\pi} e^{-x} \sin(nx) \, dx$$ Again, using standard integrals or integration by parts, we find that: $$b_n = 0$$ ### Final Fourier Series: Combining all of the terms, the Fourier series of $$f(x) = e^{-x}$$ on $$[-2\pi, 2\pi]$$ is: $$f(x) = \frac{\sinh(2\pi)}{\pi} + \sum_{n=1}^{\infty} \frac{2\sinh(2\pi)}{\pi(n^2 + 1)} \cos(nx)$$ --- Would you like any further details on this process or a deeper explanation of any specific step? ### Related Questions: 1. How do you derive the Fourier series for functions defined on intervals other than $$[-2\pi, 2\pi]$$? 2. How can we compute Fourier coefficients for piecewise functions? 3. What is the importance of the Fourier transform in signal processing? 4. How does the convergence of Fourier series behave for different types of functions? 5. Can you explain the difference between the Fourier series and Fourier transform in terms of periodicity? **Tip:** When working with Fourier series, always check for symmetries (even or odd functions) to simplify the calculation of coefficients.