# Find Fourier coefficients - M. Chester text

• GreyNoise
In summary: L}}\\ {} & \underset{k = 1}{\textstyle \frac{1}{L}}\\ {} & \underset{k = L}{\textstyle \frac{1}{L}}\\ {} & \underset{k = \infty}{\textstyle \frac{1}{L}}\\ {} & \underset{x \in [0,L]}{\textstyle f(x)}\\ \end{pmatrix}}$$we can write this as$$f(x) = \sum
GreyNoise
Gold Member

## Homework Statement

I am self studying an introductory quantum physics text by Marvin Chester Primer of Quantum Mechanics. I am stumped at a problem (1.10) on page 11. We are given

$f(x) = \sqrt{ \frac{8}{3L} } cos^2 \left ( \frac {\pi}{L} x \right )$

and asked to find its Fourier coefficients $C_0$ and $C_1$ from

$C_k = \frac{1}{L} \displaystyle {\int_0^L} e^{-ikx} f( x ) dx$

I found $C_0 = \sqrt{ \frac{2}{3L} }$; it was easy as k=0 greatly simplified the integral. But I am completely baffled at an answer for the $C_1$ coefficient. The text has the answer as $\frac{1}{ \sqrt{6L} }$, and I do not get anything remotely close that; I keep getting answers with nonzero imaginary parts and $\pi$ in the terms.

## Homework Equations

$f(x) = \sqrt{ \frac{8}{3L} } cos^2 \left (\frac{\pi}{L} x \right)$

$C_k = \frac{1}{L} \displaystyle {\int_0^L} e^{-ikx} f( x ) dx$

$C_1 = \frac{1}{ \sqrt{6L} }$

## The Attempt at a Solution

[/B]
To begin, I substituted f(x) into the equation for $C_k$ for k = 1

$C_1 = \frac{1}{L} \displaystyle {\int_0^L e^{-ix} \sqrt{ \frac {8}{3L} }} cos^2 \left ( \frac {\pi}{L} x \right ) dx$

Then I factored the constants out of the integral

$C_1 = \frac{1}{L} \sqrt{ \frac {8}{3L} } \displaystyle {\int_0^L e^{-ix}} cos^2 \left ( \frac {\pi}{L} x \right ) dx$

Then I attempted to solve the integral

$\displaystyle {\int_0^L e^{-ix}} cos^2 \left ( \frac {\pi}{L} x \right ) dx$

I tried doing it by parts, then I tried by substituting $e^{-ix} = cos (x) - i sin (x)$ and distributing $cos^2 \left ( \frac {\pi}{L} x \right )$ into that and integrating the result, but I only generated 4 pages of blundering chicken scratch that gave me things like $\textstyle {\frac {2i}{3L} \sqrt{ \frac{8}{3L}} }$; another attempt led to $\textstyle {\frac {1}{\pi} \frac{2i}{3} }$. I even went to my TI-89 and the Wolfram website and got an analytical solution to the integral, but the definite integrals they returned gave complex numbers that did NOT look anything like $\frac{1}{ \sqrt{6L} }$. Can anyone tell me; am I even setting up the integral correctly?

GreyNoise said:
Can anyone tell me; am I even setting up the integral correctly?

Double check your form defining the Fourier coefficents. The ##e^{-ikx}## doesn't have period ##L## for ##k=1##. Shouldn't it be something more like ##e^{-i\frac{2k\pi}{L}x}##? ##e^{-ikx}## is used in the transform over the whole line.

Last edited:
Dick said:
Double check your form defining the Fourier coefficents. The ##e^{-ikx}## doesn't have period ##L## for ##k=1##. Shouldn't it be something more like ##e^{-i\frac{2k\pi}{L}x}##? ##e^{-ikx}## is used in the transform over the whole line.

Thnx much for that Dick! That was the fix I needed. Now it brings to my mind a question about the form of the exponential argument. The author
wrote
$$C_k = \frac{1}{L} \int_0^L e^{-ikx} f(x) dx$$
and I used the form literally, but now I am concerned that I missed a basic Fourier series concept. The author had written (before the above equation) that
$$kL = 2 \pi n$$
So the form I should have used would appear as
$$C_n = \frac{1}{L} \int_0^L e^{-i \frac{2 \pi n}{L} x} f(x) dx$$
where I substituted ## k = \textstyle \frac{2 \pi n}{L} ## in the exponent's argument. Am I correct on this point?

GreyNoise said:
Thnx much for that Dick! That was the fix I needed. Now it brings to my mind a question about the form of the exponential argument. The author
wrote
$$C_k = \frac{1}{L} \int_0^L e^{-ikx} f(x) dx$$
and I used the form literally, but now I am concerned that I missed a basic Fourier series concept. The author had written (before the above equation) that
$$kL = 2 \pi n$$
So the form I should have used would appear as
$$C_n = \frac{1}{L} \int_0^L e^{-i \frac{2 \pi n}{L} x} f(x) dx$$
where I substituted ## k = \textstyle \frac{2 \pi n}{L} ## in the exponent's argument. Am I correct on this point?

Not quite, but it depends in part on how you plan to use the ##C_n##.

Basically, it is always easiest to start with an orthonormal set of functions ##\{ u_k(x) \}##, which have the inner-product properties that
## \langle u_j, u_k \rangle = 0 ## if ##k \neq j## and ##\langle u_k,u_k \rangle = 1## for all ##k##. Here, ##\langle \cdot , \cdot \rangle## denotes the inner product for functions on ##[0,L]##:
$$\langle f,g \rangle = \int_0^L f^*(x) g(x) \, dx,$$
where ##f^*## is the complex congugate of ##f##.

You can easily verify that the functions ##u_k(x) =(1/\sqrt{L}) \: e^{i 2 \pi k x/L}, \; k = 0, \pm 1, \pm 2, \ldots ## constitute an orthonormal set. So, if we expand a function ##f(x)## as
$$f(x) = \sum_{k=-\infty}^{\infty} c_k u_k(x),$$
we get (with almost no work) that the coefficients are
$$c_k = \langle u_k , f \rangle = \int_0^L u^*(x) f(x) \, dx = \frac{1}{\sqrt{L}} \int_0^L e^{-i 2 \pi k x/L} f(x) \; dx.$$

In other words, we can either use the expansion ##\sum c_k u_k## or use the expansion ##\sum_k \sqrt{L} C_k e^{i 2 \pi k x/L}## with your ##C_k## (because ##\text{my} \: c_k = \sqrt{L} \times \text{your} \: C_k##). The factor ##\sqrt{L}## needs to go somewhere, and IMHO it is easier to put it into the definition of the basis functions ##\{ u_k \}##.

Note added in edit: it turns out that what YOU wrote is also correct! You are using expansion functions ##f_k(x) = e^{i 2 \pi x k/L}, \; k = 0, \pm 1, \pm 2, \ldots## and coefficients ##C_k = (1/L) \int_0^L f^*_k(x) f(x) \, dx,##, so your ##k##th term of the expansion has the form
$$\text{term}\; k = \frac{1}{L} \int_0^L f^*_k(t) f(t) \, dt \times f_k(x),$$
and this is the same as the ##k##th term ##c_k u_k(x)## given above.

Last edited:
Sorry for delayed response; thnx are in order; I was away on vacation. Thank you Dick and Ray. The posts got me back on top of the curve (my Fourier forays are ten years ago now), so I committed them to my notes. I might be back for more regarding the M. Chester text I am reading. Thnx again.

## 1. What are Fourier coefficients and why are they important?

Fourier coefficients are the coefficients used in a Fourier series to represent a periodic function as a sum of sine and cosine waves. They are important because they allow us to understand and analyze complex periodic functions, such as sound waves and signals, in terms of simpler components.

## 2. How do you find Fourier coefficients?

To find Fourier coefficients, you can use the Fourier series formula, which involves integrating the function over one period and multiplying it by sine and cosine functions at different frequencies. Alternatively, you can use Fourier transform techniques, which use complex numbers and the frequency domain to find the coefficients.

## 3. What is the M. Chester text used for in finding Fourier coefficients?

The M. Chester text is a specific function that is often used as an example in finding Fourier coefficients. It is a periodic function that is relatively simple, making it a good starting point for understanding and practicing the process of finding Fourier coefficients.

## 4. Can you use Fourier coefficients to approximate any function?

Yes, the Fourier series can be used to approximate any periodic function with a finite number of terms. However, the accuracy of the approximation depends on the smoothness of the function and the number of terms used in the series.

## 5. How are Fourier coefficients used in real-world applications?

Fourier coefficients have various real-world applications, including signal processing, image compression, and data analysis. They are also used in fields such as physics, engineering, and economics to analyze and model periodic phenomena and systems.

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