Properties of the Fourier series & integral (Taylor, Modern Physics, 2 ed, Prob 6.33)

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Homework Statement


For an even function, the Fourier series takes the form
[itex]^{\infty}_{n=0}\Sigma A_n cos(\frac{2\pi n x}{\lambda})[/itex]
where [itex]\lambda[/itex] is the wavelength of the function. In this problem you will see how to find the Fourier coefficients [itex]A_n[/itex].

a.) Prove that
[itex]A_0 = \frac{1}{\lambda} \int^{\lambda}_0 F(x)dx[/itex]
(Hint: Integrate F(x) from x = 0 to x = [itex]\lambda[/itex].)

b.) Prove that, for m > 0,
[itex]A_m = \frac{2}{\lambda} \int^{\lambda}_0 F(x) cos(\frac{2\pi m x}{\lambda}) dx[/itex]
where Am (rather than An) is the Fourier coefficient for reasons that will become apparent in your proof. [Hint: Multiply both sides of equation (6.57) by [itex]cos (\frac{2\pi m x}{\lambda})[/itex], and integrate from 0 to λ.]


Homework Equations



Equation (6.57): [itex]F(x) = ^\infty _{n=0} \Sigma A_n cos(\frac{2\pi n x}{\lambda})[/itex]
Final result (a) should be: [itex]A_0 = \frac{1}{\lambda} \int^{\lambda}_0 F(x) dx[/itex]
Final result (b) should be: [itex]A_m = \frac{2}{\lambda} \int^{\lambda}_0 F(x) cos(\frac{2\pi m x}{\lambda}) dx[/itex]
For a square wave: [itex]A_n = \frac{2}{\pi n} cos(\frac{\pi a n}{\lambda})[/itex] (Eq. 6.26)

The Attempt at a Solution



For part (a) I tried integrating as suggested. I ended up with

[itex]\int^{\lambda}_0 A_n cos(\frac{2\pi n x}{\lambda})dx = \frac{\lambda A_{n} sin(2\pi n)}{2\pi n}[/itex]
...which doesn't do anything for me. Substituting 0 for n sets the whole sine argument equal to 0 (which is problematic in its own right, given the known expected outcome) but even worse puts a 0 in the denominator. If I substitute the function for An for an even square wave, I get

[itex]\frac{2}{\pi n} * cos(\frac{\pi a n}{\lambda}) * \frac{sin({2\pi n})}{2\pi n}[/itex]

...which still doesn't do a thing for me, for the same reasons. I'm pretty badly lost here, and I'm not sure where to take the problem.
 

Answers and Replies

  • #2
I like Serena
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What if you pick a function for F(x) such that only one of the An coefficients it not zero?
Say F(x)=L.
 
  • #3
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I'm afraid I don't quite understand what you mean. I should replace F with an arbitrary function L?
 
  • #4
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Not a function L, but an arbitrary constant L.

Since F(x) is supposed to be written as the sum of cosines, for a constant function each cosine term must be zero, except for the constant term A0.

So if you calculate A0 with the integral expression, you should find the constant L again.
 
  • #5
ehild
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For part (a) I tried integrating as suggested. I ended up with

[itex]\int^{\lambda}_0 A_n cos(\frac{2\pi n x}{\lambda})dx = \frac{\lambda A_{n} sin(2\pi n)}{2\pi n}[/itex]
What is sin(2πn), (do not forget that n is integer)?

ehild
 
  • #6
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Not a function L, but an arbitrary constant L.

Since F(x) is supposed to be written as the sum of cosines, for a constant function each cosine term must be zero, except for the constant term A0.

So if you calculate A0 with the integral expression, you should find the constant L again.
How would the cosine terms be zero if they're cos(2πnx)? Wouldn't that require that x is always equal to 1/4 + some integer m, or something along those lines? I don't know how that would work unless there was some phase angle introduced to the argument, for n = 0.

What is sin(2πn), (do not forget that n is integer)?

ehild
Right, sin(2πn) is 0 for all integer values of n. That's what was so problematic for me. Is the point supposed to be that A0 is always equal to 0?
 
  • #7
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The Fourier theorem states that any even function F(x) with wavelength λ can be written as:
F(x)=A0 + A1 cos(2pi x/λ) + A2 cos(2pi 2x/λ) + A3 cos(2pi 3x/λ) + ...

The coefficients A0, A1, A2, A3, ... are unique.

If F(x) is the constant function, it can be written as F(x)=A0.
This means that A1, A2, A3, ... will have to be zero, making all the cosine terms zero.

Can you calculate A0 with the integral formula for A0?
 
  • #8
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So that would mean I have

[itex]F(x) = A_0 cos(\frac{2\pi n x}{\lambda}) = A_0 cos(0) = A_0[/itex]

which in turn means that

[itex]\int^{\lambda}{0} F(x)dx = \int^{\lambda}_0 A_0 dx = \lambda A_0[/itex]

which then gives

[itex]\lambda A_0 = \int^{\lambda}_0 F(x)dx \Rightarrow A_0 = \frac{1}{\lambda} \int^{\lambda}_0 F(x)dx[/itex]
?

Hey, that's the answer I'm told to expect. Thanks. I feel kind of like a moron right about now. I'm in an intro class for PDEs and somehow I forgot the basics of freshman-level calc. :|
Thanks for the help, lol. I'll put up an attempt at part b.) in a while.
 
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  • #9
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Congrats! ;)
 
  • #10
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Okay, so part b.) was actually pretty easy and straightforward. I did it in like 3 minutes.

I multiplied both sides by [itex]cos(\frac{2\pi m x}{\lambda}[/itex] to get

[itex]F(x) cos(\frac{2\pi m x}{\lambda}) = \Sigma_{n=0}^{\infty} A_m cos(\frac{2\pi m x}{\lambda}) cos(\frac{2\pi n x}{\lambda})[/itex]

and from there just proved that since

[itex]cos(\frac{2\pi m x}{\lambda}) cos(\frac{2\pi n x}{\lambda}) = \frac{1}{2} cos(\frac{2\pi x}{\lambda} (m-n)) + cos(\frac{2\pi x}{\lambda} (m+n))[/itex]

and m & n are defined as integers,

[itex]m \neq n \Rightarrow \int^{\lambda}{0} cos(\frac{2\pi m x}{\lambda}) cos(\frac{2\pi n x}{\lambda})dx \equiv \frac{1}{2} sin(2\pi) + sin(2\pi) = 0[/itex]

and that

[itex]m = n \Rightarrow \int^{\lambda}_0 cos(\frac{2\pi m x}{\lambda}) cos(\frac{2\pi n x}{\lambda})dx = \int^{\lambda}_0 cos^{2}(\frac{2\pi m x}{\lambda})dx = \int^{\lambda}_0 \frac{1}{2} (1 + cos(\frac{4\pi m x}{\lambda}) ) = \frac{\lambda}{2} + \frac{\lambda}{4\pi m} sin(4\pi m) = \frac{\lambda}{2} + k sin(0) = \frac{\lambda}{2}[/itex].

Since by the definitions implied in the proof, m = n, you multiply the integrand for that case by Am and get

[itex]A_m \frac{\lambda}{2} = \int^{\lambda}_0 F(x) cos(\frac{2\pi m x}{\lambda})dx \Rightarrow A_m = \frac{2}{\lambda} \int^{\lambda}_0 F(x) cos(\frac{2\pi m x}{\lambda})dx[/itex]...

...quod erat demonstrandum.

Just let me know if I fouled it up somewhere or got the right answer but got there by doing something naughty or non-rigorous. ;)
 
  • #11
I like Serena
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Well, you've got a typo at sin(4πm), which should be sin(4πm/lambda).
But otherwise... looks good! ;)
 
Last edited:
  • #12
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That wouldn't be a typo. That's the result of evaluating the integral at x = lambda. :)
 

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