How to check Fourier series solution (complex)

In summary, the complex Fourier series for f(t) is given by f(t)=t(1-t), 0<t<1, where t(1-t)=-\frac{1}{2n^2\pi^2} and c_n=-\frac{1}{2n^2\pi^2}. The coefficients, c_n, are found by integrating: c_n=\int_{0}^{1}t(1-t)e^{-i2\pi{n}t}dt.
  • #1
PhysicsMark
91
0

Homework Statement


Find the complex Fourier series for:

[tex]f(t)=t(1-t), 0<t<1[/tex]


Homework Equations



[tex]f(t)=\sum_{n=-\infty}^{\infty}c_n{e^{i\omega_n{t}}}[/tex]

[tex]c_n=\frac{1}{\tau}\int_{t_0}^{t_0+\tau}e^{-i\omega_n{t}}f(t)dt[/tex]

[tex]\omega_n=2\pi{n}\quad\tau=1[/tex]

The Attempt at a Solution



I solved for c_n. I want to check my answer. I can only think of checking it by graphing it out to a few (50 or so) terms. I tried to graph this in Maple with my value for c_n and it couldn't do it. After that, I tried to solve the entire problem in Maple and that also did not work.

I have a few more of these to do, and I'd like to make sure I am doing this correctly before I move on. Does anyone know how to check my value for the coefficient?
 
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  • #2
Do your coefficients depend on your choice of [itex]t_0[/itex]? Because the relevant equations you posted are not correct.
 
  • #3
I found the coefficients, c_n, by integrating:

[tex]c_n=\int_{0}^{1}t(1-t)e^{-i2\pi{n}t}dt[/tex]

Are you saying that this is not the correct method to find c_n?
 
  • #4
No, it isn't.

The Fourier coefficients are given by

[tex]c_n=\int_{-\frac{1}{2}}^{\frac{1}{2}}f(2\pi t)e^{-i\omega_n t}dt=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(t)e^{-int}dt[/tex]

which is not equivalent to what you've posted. Where did you find that incorrect equation for the coefficients?
 
  • #5
Thanks for pointing that out. I'll recalculate my coefficient value later today.

My tutorial lists the equation for c_n as:

[tex]c_n=\frac{1}{\lambda}\int_{x_0}^{x_0+\lambda}e^{-i{k_n}x}f(x)dx[/tex]

In a paragraph above this equation, it states that:

"...But most applications involve either functions of position or of time. In the former case, the period of the function, [tex]\lambda[/tex], is more conventionally called the wavelength, and [tex]k_n=\frac{2\pi{n}}{\lambda}[/tex] is the wave number for the n'th mode. If time is the variable, however, the period is called, indeed, the period, and is usually represented by [tex]\tau[/tex]. The n'th mode has the angular frequency, or often simply the frequency, [tex]\omega_n=\frac{2\pi{n}}{\tau}[/tex]."

It did not explicitly give the equation for c_n in this case. I must have made a mistake converting from a spatial to time variable.
 
  • #6
gabbagabbahey said:
No, it isn't.

The Fourier coefficients are given by

[tex]c_n=\int_{-\frac{1}{2}}^{\frac{1}{2}}f(2\pi t)e^{-i\omega_n t}dt=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(t)e^{-int}dt[/tex]

which is not equivalent to what you've posted. Where did you find that incorrect equation for the coefficients?
Those expressions are for a function with period 2π. If you rescale for a function with period τ and allow for a shift (since it really doesn't matter which particular cycle you integrate over), you get PhysicsMark's integral.
 
  • #7
vela said:
Those expressions are for a function with period 2π. If you rescale for a function with period τ and allow for a shift (since it really doesn't matter which particular cycle you integrate over), you get PhysicsMark's integral.

Thanks for the clarification, Vela. I spoke with my professor today, and he also said the original integral should be correct.

He went on to say that I should be able to plot it in Maple. So, I guess that means I need some more practice in Maple (That should be no surprise to Vela...https://www.physicsforums.com/showthread.php?t=391887).

Thanks to Vela and Gabbagabbahey for replying.
 
  • #8
In case you want to check your answer, here's what I found for cn for [itex]n\ne 0[/itex]:

[tex]c_n=-\frac{1}{2n^2\pi^2}[/tex]

When you plot the series, you should only need to use a handful of terms -- 50 would be overkill -- to see if it's summing to what you expect. Just remember to match the negative n's with positive n's so that the sum is real.
 
  • #9
Thanks a lot! That is also the answer I got.

I realized why maple wouldn't plot it, it was because I did not account for the "n" in the denominator. No wonder Maple was blabbering about a singularity.
 

FAQ: How to check Fourier series solution (complex)

What is a Fourier series solution?

A Fourier series solution is a mathematical representation of a periodic function using a sum of trigonometric functions. It is used to approximate a function by breaking it down into simpler components.

How do I check if a Fourier series solution is correct?

To check if a Fourier series solution is correct, you can compare it to the original function and see if they match for a given number of terms. You can also use mathematical techniques such as the Dirichlet conditions or the uniform convergence test.

Can a Fourier series solution be complex?

Yes, a Fourier series solution can be complex. In fact, complex Fourier series are often used to represent functions with complex-valued outputs.

What is the difference between a Fourier series and a Fourier transform?

A Fourier series is used to represent a periodic function, while a Fourier transform is used to represent a non-periodic function. Additionally, a Fourier series uses discrete frequencies, while a Fourier transform uses continuous frequencies.

Are there any tools or software for checking Fourier series solutions?

Yes, there are various tools and software available for checking Fourier series solutions. Some popular options include WolframAlpha, MATLAB, and Mathematica. These tools allow you to input a function and its Fourier series solution and compare them visually or using mathematical tests.

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