Is this even possible? Question about Fourier Series...

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Today I had a maths exam with a question which was worded something like:

Write ##sin(3x-x_0)## as its Fourier representation. By doing a suitable integral or otherwise, find the possible values of its fourier coefficients. You may find the following useful:

##sin(\alpha-\beta) = sin(\alpha)cos(\beta) - cos(\alpha)sin(\beta)##.

5 marks.

I simply do not understand what was expected of me here. How is it even possible, or at least necessary, to write a basic sin function as a Fourier series? I thought the idea of the Fourier series was to write functions as a series of trig terms, so it makes no sense applying the concept to a function which is already trigonometric.

Does anyone have an alternative view on this?

Thanks.
 

Answers and Replies

  • #2
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Today I had a maths exam with a question which was worded something like:

Write ##sin(3x-x_0)## as its Fourier representation. By doing a suitable integral or otherwise, find the possible values of its fourier coefficients. You may find the following useful:

##sin(\alpha-\beta) = sin(\alpha)cos(\beta) - cos(\alpha)sin(\beta)##.

5 marks.

I simply do not understand what was expected of me here. How is it even possible, or at least necessary, to write a basic sin function as a Fourier series? I thought the idea of the Fourier series was to write functions as a series of trig terms, so it makes no sense applying the concept to a function which is already trigonometric.

Does anyone have an alternative view on this?

Thanks.
The Fourier series for sin(x) is, as you recognize, very simple, since the series would consist of one term. But your problem is not as simple, with the function being ##\sin(3x - x_0)##. Using the hint, this function can be written as ##\sin(3x)\cos(x_0) - \sin(x_0)\cos(3x)##, where ##\cos(x_0)## and ##\sin(x_0)## are constants. This expanded form makes it much simpler to find the coefficients of the terms in the Fourier series.
 
  • #3
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The Fourier series for sin(x) is, as you recognize, very simple, since the series would consist of one term. But your problem is not as simple, with the function being ##\sin(3x - x_0)##. Using the hint, this function can be written as ##\sin(3x)\cos(x_0) - \sin(x_0)\cos(3x)##, where ##\cos(x_0)## and ##\sin(x_0)## are constants. This expanded form makes it much simpler to find the coefficients of the terms in the Fourier series.
Thanks for the response.

I can see that ##\cos(x_0)## and ##\sin(x_0)## are constants but I can't see how that would make it any easier to solve.

Surely the integrals for finding the coefficients are still going to be incredibly ugly.

For example, in finding the coefficient ##a_n##, where I've used the standard definition taught in my classes:

##a_n = \frac{2}{L} \! \int_{-\frac{-L}{2}}^{\frac{L}{2}} \! f(x)\cos(\frac{2n\pi x}{L}) \, \mathrm{d}x##

##a_n = \frac{2}{L} \! \int_{-\frac{-L}{2}}^{\frac{L}{2}} \! \Big(\sin(3x)\cos(x_0) - \sin(x_0)\cos(3x)\Big)\cos(\frac{2n\pi x}{L}) \, \mathrm{d}x##

where ##L = \frac{2\pi}{3}## for ##\sin(3x-x_0)##

I've punched variations of this into Mathematica and none of it seems to show anything neat or easy.
 
  • #4
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You're making it way too difficult. What is a Fourier series? It's a sum of trigonometric functions times coefficients. What are coefficients? Coefficients are constants.

This type of problem is the Fourier equivalent of problems like asking what the Maclaurin series of [itex](3x - a)^3[/itex] is.

Granted, with Fourier series the answer changes if the problem states an L or period different from the period of the expression in question, but the way you stated it, that subtlety doesn't apply in this case.
 
  • #5
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4,975
Thanks for the response.

I can see that ##\cos(x_0)## and ##\sin(x_0)## are constants but I can't see how that would make it any easier to solve.

Surely the integrals for finding the coefficients are still going to be incredibly ugly.

For example, in finding the coefficient ##a_n##, where I've used the standard definition taught in my classes:

##a_n = \frac{2}{L} \! \int_{-\frac{-L}{2}}^{\frac{L}{2}} \! f(x)\cos(\frac{2n\pi x}{L}) \, \mathrm{d}x##

##a_n = \frac{2}{L} \! \int_{-\frac{-L}{2}}^{\frac{L}{2}} \! \Big(\sin(3x)\cos(x_0) - \sin(x_0)\cos(3x)\Big)\cos(\frac{2n\pi x}{L}) \, \mathrm{d}x##

where ##L = \frac{2\pi}{3}## for ##\sin(3x-x_0)##
The integral above can be simplified (and should be) to
##a_n = \frac{2}{L} [ \int_{-\frac{-L}{2}}^{\frac{L}{2}} (\sin(3x) \cos(\frac{2n\pi x}{L}) dx - \sin(x_0) \int_{-\frac{-L}{2}} ^ {\frac{-L}{2}}\cos(3x) \cos(\frac{2n\pi x}{L}) \, dx]##
I think you'll find that a lot of the coefficients are zero.
sa1988 said:
I've punched variations of this into Mathematica and none of it seems to show anything neat or easy.
 
  • #6
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The Fourier series (without further conditions) is expected to be a series whose summands are each a constant, or a constant times sin(nx) for some integer n ≥ 1, or a contant times cos(nx) for some integer n ≥ 1.

IMPORTANT: Note that sin(3x - x0) is not in this form!

You need to do something to either calculate the coefficients of the Fourier series of sin(3x - x0) by the usual integral formula, or find a simpler way to express sin(3x - x0) in terms of what is mentioned in the first paragraph.

Do not forget that x0 is a constant! Which means that any function of this constant is also a constant, and therefore can be easily manipulated like a constant in any calculations — if need be!
 

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