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Convergence of Fourier series

Hi,

I have no idea what to say for this question, I know it is asking if the series diverges when differentiated but I’m not sure if it does!

Here is the question and here is the series ( I’ve checked the answer so the series is correct) solution says

The differential of this Fourier series converges everywhere except at the discontinuities ( at positions given by L/2 vt) where the series diverges, as the new series has coefficients that do not decrease as n increases. However the series should diverge here as the spatial and temporal derivatives of the original sawtooth function do. It exist at exactly these points, thus assuming that the (unphysical) initial displacement is true and we are allowed to trust that the (unphysical) initial displacement is true we are allowed to trust the (unphysical) gradients”

But this solution has no meaning to me

906A3A7B-FB78-4C6B-BF7A-E60C3850D741.jpg.jpeg
FEA84963-74A9-4EB6-8E14-D2B13479F01C.jpg.jpeg
(edited 1 year ago)
Original post by grhas98
Hi,

I have no idea what to say for this question, I know it is asking if the series diverges when differentiated but I’m not sure if it does!

Here is the question and here is the series ( I’ve checked the answer so the series is correct) solution says

The differential of this Fourier series converges everywhere except at the discontinuities ( at positions given by L/2 vt) where the series diverges, as the new series has coefficients that do not decrease as n increases. However the series should diverge here as the spatial and temporal derivatives of the original sawtooth function do. It exist at exactly these points, thus assuming that the (unphysical) initial displacement is true and we are allowed to trust that the (unphysical) initial displacement is true we are allowed to trust the (unphysical) gradients”

But this solution has no meaning to me

906A3A7B-FB78-4C6B-BF7A-E60C3850D741.jpg.jpeg
FEA84963-74A9-4EB6-8E14-D2B13479F01C.jpg.jpeg

It would help to see the full question / solution as some of it is a bit hard to read / some defs missing, but if thats the fourier series, then differentiating it the "n"s pop out of the trig terms and cancel with the 1/n in the coeffs. So the coeffs of the derivative will not decrease as n increases (high frequencies), hence the answer.
(edited 1 year ago)
Reply 2
Original post by mqb2766
It would help to see the full question / solution as some of it is a bit hard to read / some defs missing, but if thats the fourier series, then differentiating it the "n"s pop out of the trig terms and cancel with the 1/n in the coeffs. So the coeffs of the derivative will not decrease as n increases (high frequencies), hence the answer.


But the answer says it will converge except at the discontinuities?
Original post by grhas98
But the answer says it will converge except at the discontinuities?


It would help to see the full question / solution youre referring to, but a sawtooth typically has coeffs of (-1)^n /n and differentiating youd get something like the sum of (-1)^n equals 1 (derivative of a sawtooth) and the sum -1+1-1.... does not converge.
(edited 1 year ago)
Reply 4
Original post by mqb2766
It would help to see the full question / solution youre referring to, but a sawtooth typically has coeffs of 1/n and differentiating youd get something like the sum of (-1)^n equals 1 (derivative of a sawtooth) and the sum -1+1-1.... does not converge.


This is the full question,
Reply 5
Original post by mqb2766
It would help to see the full question / solution youre referring to, but a sawtooth typically has coeffs of 1/n and differentiating youd get something like the sum of (-1)^n equals 1 (derivative of a sawtooth) and the sum -1+1-1.... does not converge.


Here is the series typed up
E380EDC4-1309-4D62-B564-4DC9B757C30C.jpg.jpeg
Probably going to have to come back to it in the morning, but I cant see how you can justify convergence. The basic sawtooth spatial derivative seems to be as described above and cant see how multiplying (-1)^n by the sin/cos(2*pi*n*l/L) terms representing the translation ensures convergence.
For a simplified series, K=1, L=2pi and evaluating with l=1 so its not related to the period and chucking it into wolframalpha gave (with a bit of tweaking) gave as a partial sum to k for the derivative
https://www.desmos.com/calculator/wimiug4sop
Obviously things dont converge as k increases. It looks like the derivative is centered on -1 rather than 1 Ive probably Ive done a small typo somewhere but its what youd expect with the derivative being large and increasing wrt k at the jumps and not converging in the centre. So dont really understand the answer about it converging.

@DFranklin @ghostwalker any insight/something Ive missed?
(edited 1 year ago)
Original post by mqb2766
For a simplified series, K=1, L=2pi and evaluating with l=1 so its not related to the period and chucking it into wolframalpha gave (with a bit of tweaking) gave as a partial sum to k for the derivative
https://www.desmos.com/calculator/wimiug4sop
Obviously things dont converge as k increases. It looks like the derivative is centered on -1 rather than 1 Ive probably Ive done a small typo somewhere but its what youd expect with the derivative being large and increasing wrt k at the jumps and not converging in the centre. So dont really understand the answer about it converging.

@DFranklin @ghostwalker any insight/something Ive missed?


I mean, if you diff term by term, you get a series whose terms don't tend to 0 so it's obviously non-convergent.

Also, the correct Fourier series for the derivative in x is obviously just a constant (individual points of discontinuity can be ignored when calculating the series), so there's no way the series you get by diffing is "the Fourier series for the derivative".

If you instead calculated the Fourier series for pi/2-|x| on [-pi, pi] (so you have a continuous zig-zag), you'd get terms with n^2 in the denominator and it would work a lot closer to the description they give (although the series should still converge (to 0) at the points where the derivation is undefined). I wonder if they were thinking it was this case?

Also, if you're going very "maths methods", there's a sense in which sin(n pi x) converges to 0 (basically because its average value over any interval tends to 0). But that's a big stretch as an explanation here.
Original post by DFranklin
I mean, if you diff term by term, you get a series whose terms don't tend to 0 so it's obviously non-convergent.

Also, the correct Fourier series for the derivative in x is obviously just a constant (individual points of discontinuity can be ignored when calculating the series), so there's no way the series you get by diffing is "the Fourier series for the derivative".

If you instead calculated the Fourier series for pi/2-|x| on [-pi, pi] (so you have a continuous zig-zag), you'd get terms with n^2 in the denominator and it would work a lot closer to the description they give (although the series should still converge (to 0) at the points where the derivation is undefined). I wonder if they were thinking it was this case?

Also, if you're going very "maths methods", there's a sense in which sin(n pi x) converges to 0 (basically because its average value over any interval tends to 0). But that's a big stretch as an explanation here.


Thanks, agree with that. The derivative of the fourier series of the sawtooth doesnt converge, which is what the solution seems to be saying.

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