How is this possible? (simple fourier series)

  • #1
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Main Question or Discussion Point

Hello,

Given a well-behaved function [tex]f:[0,1] \to \mathbb C[/tex] with f(0)=0, is it then totally equivalent to write it as a sum of [tex]e^{2 \pi i n x}[/tex] ([tex]n \in \mathbb Z[/tex]) or as a sum of [tex]\sin{\pi n x}[/tex] (n = 1,2,3,...) -- this last one by defining [tex]f:[-1,1][/tex] as an uneven function and then applying the first method?

The reason I ask is because when solving
[tex]-\frac{\partial^2 u(x)}{\partial x^2} = q(x)[/tex]
with the first method (to u(x), because q(x) is given), I gained no information about the zeroth fourier coëfficient [tex]u_0[/tex] (because the relation between fourier coëfficients is [tex]4 \pi n^2 u_n = q_n[/tex]),
while with the second method I had nothing of u left undetermined.

Can this be?

Thank you.
 

Answers and Replies

  • #2
AlephZero
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If u(x) is a solution of your example PDE, then u(x) + C is also a solution for any constant C.

So it makes sense that your Fourier solution method did not give you any information about the value of C (which is the zero fourier coefficient of course). You need to use the boundary conditions to fix the value of C.

Your first statement is not quite true, because a function and the sum of its Fourier series can be different at a set of discrete points. So they are not "totally equivalent", but they are equivalent in the sense that you were probably thinking about, if you are an engineer or physicist (but not a pure mathematician).
 
  • #3
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Hm, two questions regarding your response:

- Then shouldn't I also get the free parameter in the second method?

- Shouldn't I get two parameters in both methods? Because not only u(x) + C is a solution, but also u(x) + C + Dx.

NOTE: hm, you mean f(x) is not equivalent to its fourier decomposition? I understand that, but I was asking if the exp(inx) decomposition is equivalent to the sine-decomposition.
 
  • #4
AlephZero
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Hm, two questions regarding your response:

- Then shouldn't I also get the free parameter in the second method?

- Shouldn't I get two parameters in both methods? Because not only u(x) + C is a solution, but also u(x) + C + Dx.

NOTE: hm, you mean f(x) is not equivalent to its fourier decomposition? I understand that, but I was asking if the exp(inx) decomposition is equivalent to the sine-decomposition.
OK, I see what you were really asking now.

Yes the sine-series decomposition is equivalent to a full (sine and cosine) decomposition of an odd function over twice the interval, because the cosine terms are all 0 for an odd function.

But when you do this, you will only find solutions that are periodic odd functions. That is OK for a particular solution, because you are also extending q(x) to be a periodic odd function. But you lose the arbitrary general solution C + Dx, because the only values of C + Dx that satisfy the conditions you assumed are C = D = 0.
If C is non-zero, C is periodic, but not an odd function.
If D is non-zero, Dx is an odd function, but not periodic.

The reason you got the equation 0.u_0 = q_0 = 0, is because there is no "zero sine term" in a Fourier series. Sin 0x is identically zero, so the value of the coeffient is irrelevant.

If you have a general complex fourier series, the e^(0iz) term is always equal to 1, but the coefficient of the "1" can be any complex number, so you don't lose information in that case.
 
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