# Coefficients for an exponential Fourier Series

• ElPimiento
In summary, the conversation revolved around finding the Fourier series for a given function, and the use of Euler's Equation and vector projection in the solution. The final answer was found to be a messy expression, but upon further simplification, it was determined to be $$k_n = - \frac{2 \alpha ((-1)^n e^{- \pi \alpha} - 1)}{\pi (\alpha^2 + n^2)}.$$
ElPimiento
I'm kinda just hoping someone can look over my work and tell me if I'm solving the problem correctly. Since my final answer is very messy, I don't trust it.

1. Homework Statement

We're asked to find the Fourier series for the following function:
$$f(\theta)=e^{−\alpha \lvert \theta \rvert}}, \ \ \ \text{for} −\pi<\theta<\pi,$$
where ##f(\theta + \pi) = f(\theta)##. Except our Fourier series has the form:
$$f(\theta) = \sum_{-\infty}^{\infty} c_n e^{in\theta}$$

## Homework Equations

Variation on Euler:
$$\cos{\theta} = \frac{1}{2}(e^{i\theta} + e^{-i\theta})$$
Vector projection:
$$proj_\vec{v}(\vec{u}) = \frac{\langle\vec{v}, \vec{u}\rangle}{\langle\vec{v}, \vec{v}\rangle}$$

## The Attempt at a Solution

So, supposing ##n## is an Integer, I First realized that every positive ##n## has a negative partner, hence our special Fourier series formula can be rewritten as
\begin{align*} \sum_{-\infty}^{\infty} c_n e^{in\theta} & = c_0 + \sum_{-\infty}^{-1} (c_{n^{(-)}} e^{in\theta}) + \sum_{1}^{\infty} (c_{n^{(+)}} e^{in\theta}) \\ & = c_0 + \sum_{-\infty}^{\infty }k_n (e^{in\theta} + e^{-in\theta}) \\ & = \sum_{-\infty}^{\infty} k_n \cos(n\theta) \\ \end{align*}
Where I have used the variation on Euler's Equation (above) and set ##k_n = (c_{n^{(-)}} + c_{n^{(+)}})/2##.
Then using the inner product:
$$\langle f(x), g(x) \rangle = \int_{-\pi}^{\pi} dx f(x) g(x)$$
I found that the coefficients should be:
##
\begin{align*}
k_n & = \frac{\langle e^{-\alpha \lvert \theta \rvert}, \cos(n \theta) \rangle}{\langle \cos(n \theta), cos(n \theta)\rangle} \\
& = \frac{\int_{-\pi}^{\pi} e^{-\alpha \lvert \theta \rvert} cos(n \theta)} {\int_{-\pi}^{\pi} cos^2(n \theta)} \\
& = \frac{\int_{0}^{\pi} e^{-\alpha \theta} cos(n \theta)} {\int_{0}^{\pi} cos^2(n \theta)} \ \ \ \ \ (\because \text{all the functions are even}) \\
& = - \frac{4 n (\frac{\alpha \cos(\pi n) - n \sin(\pi n)}{(\alpha^2 + n^2)e^{(\pi \alpha)}} - \frac{\alpha}{\alpha^2 + n^2} )} {2 \pi n + \sin(2 \pi n)} \\
\end{align*}
##
Which is a mess.

I used cosines hoping it would simplify the problem (since they're even and the resulting series starts at 0 instead of ##-\infty##), but it doesn't look like it did much good. So is the method at least right?

Last edited:
ElPimiento said:
I'm kinda just hoping someone can look over my work and tell me if I'm solving the problem correctly. Since my final answer is very messy, I don't trust it.

1. Homework Statement

We're asked to find the Fourier series for the following function:
$$f(\theta)=e^{−\alpha \lvert \theta \rvert}}, \ \ \ \text{for} −\pi<\theta<\pi,$$
where ##f(\theta + \pi) = f(\theta)##. Except our Fourier series has the form:
$$f(\theta) = \sum_{-\infty}^{\infty} c_n e^{in\theta}$$

## Homework Equations

Variation on Euler:
$$\cos{\theta} = \frac{1}{2}(e^{i\theta} + e^{-i\theta})$$
Vector projection:
$$proj_\vec{v}(\vec{u}) = \frac{\langle\vec{v}, \vec{u}\rangle}{\langle\vec{v}, \vec{v}\rangle}$$

## The Attempt at a Solution

So, supposing ##n## is an Integer, I First realized that every positive ##n## has a negative partner, hence our special Fourier series formula can be rewritten as
\begin{align*} \sum_{-\infty}^{\infty} c_n e^{in\theta} & = c_0 + \sum_{-\infty}^{-1} (c_{n^{(-)}} e^{in\theta}) + \sum_{1}^{\infty} (c_{n^{(+)}} e^{in\theta}) \\ & = c_0 + \sum_{-\infty}^{\infty }k_n (e^{in\theta} + e^{-in\theta}) \\ & = \sum_{-\infty}^{\infty} k_n \cos(n\theta) \\ \end{align*}
Where I have used the variation on Euler's Equation (above) and set ##k_n = (c_{n^{(-)}} + c_{n^{(+)}})/2##.
Then using the inner product:
$$\langle f(x), g(x) \rangle = \int_{-\pi}^{\pi} dx f(x) g(x)$$
I found that the coefficients should be:
##
\begin{align*}
k_n & = \frac{\langle e^{-\alpha \lvert \theta \rvert}, \cos(n \theta) \rangle}{\langle \cos(n \theta), cos(n \theta)\rangle} \\
& = \frac{\int_{-\pi}^{\pi} e^{-\alpha \lvert \theta \rvert} cos(n \theta)} {\int_{-\pi}^{\pi} cos^2(n \theta)} \\
& = \frac{\int_{0}^{\pi} e^{-\alpha \theta} cos(n \theta)} {\int_{0}^{\pi} cos^2(n \theta)} \ \ \ \ \ (\because \text{all the functions are even}) \\
& = - \frac{4 n (\frac{\alpha \cos(\pi n) - n \sin(\pi n)}{(\alpha^2 + n^2)e^{(\pi \alpha)}} - \frac{\alpha}{\alpha^2 + n^2} )} {2 \pi n + \sin(2 \pi n)} \\
\end{align*}
##
Which is a mess.

I used cosines hoping it would simplify the problem (since they're even and the resulting series starts at 0 instead of ##-\infty##), but it doesn't look like it did much good. So is the method at least right?
Why don't you simplify ##\cos(\pi n), \sin(\pi n)## and ##\sin (2 \pi n)##? Remember that ##n## is an integer.

When you do that you get a result that agrees with that obtained using Maple.

ElPimiento
Oh duh, I should've realized that >~<.
Thanks though! That cleaned it up nicely:
$$k_n = -\frac{2 \alpha (e^{\pi \alpha} - 1)}{\pi (\alpha^2 + n^2)}$$

ElPimiento said:
Oh duh, I should've realized that >~<.
Thanks though! That cleaned it up nicely:
$$k_n = -\frac{2 \alpha (e^{\pi \alpha} - 1)}{\pi (\alpha^2 + n^2)}$$
Woops! got a little too excited with the cancelation:
\begin{align*} k_n & = -\frac{2 \, {\left(\frac{\left(-1\right)^{n} \alpha}{\alpha^{2} e^{\left(\pi \alpha\right)} + n^{2} e^{\left(\pi \alpha\right)}} - \frac{\alpha}{\alpha^{2} + n^{2}}\right)}}{\pi} \\ & = - \frac{2 \alpha ((-1)^n e^{- \pi \alpha} - 1)}{\pi (\alpha^2 + n^2)} \\ \end{align*}

Last edited:

## 1. What is an exponential Fourier series?

An exponential Fourier series is a mathematical tool used to represent a periodic function as a sum of exponential functions. It is a special case of the Fourier series and is commonly used in signal processing and physics.

## 2. What are coefficients in an exponential Fourier series?

Coefficients in an exponential Fourier series refer to the constant values that are multiplied by each exponential function in the series. These coefficients determine the amplitude and phase of each exponential function and ultimately, the shape of the periodic function being represented.

## 3. How do you calculate the coefficients for an exponential Fourier series?

The coefficients for an exponential Fourier series can be calculated using a formula that involves integrating the periodic function being represented with each exponential function in the series. The resulting integral values are then used to determine the coefficients for each exponential function.

## 4. What is the significance of the coefficients in an exponential Fourier series?

The coefficients in an exponential Fourier series help determine the shape and characteristics of the periodic function being represented. They also allow us to analyze the frequency content of the function and make predictions about its behavior.

## 5. Can an exponential Fourier series accurately represent any periodic function?

Yes, an exponential Fourier series can accurately represent any periodic function as long as the function satisfies certain conditions, such as being piecewise continuous and having a finite number of maxima and minima within a given interval.

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