Derivatives and integrals of Fourier series Watch

Goods
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I've got the series and the derivatives and integrals. How can I show what they sum to?

Or can I simply say that the Integral sums to the integral of |x| for |x|< l because the terms decrease more quickly and so it converges.

If the derivative converges can I say likewise for x=-l,l,0 its undefined for 0<x<l it's 1 and -l<x<0 it's -1. Also how can I show it converges? as the terms decrease more slowly than for the original fourier series and so its not necessarily going to converge...
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Smaug123
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(Original post by Goods)


I've got the series and the derivatives and integrals. How can I show what they sum to?

Or can I simply say that the Integral sums to the integral of |x| for |x|< l because the terms decrease more quickly and so it converges.

If the derivative converges can I say likewise for x=-l,l,0 its undefined for 0<x<l it's 1 and -l<x<0 it's -1. Also how can I show it converges? as the terms decrease more slowly than for the original fourier series and so its not necessarily going to converge...
It is a fact that the Fourier series of the integral is the termwise integral of the Fourier series, and the Fourier series of the derivative is the termwise derivative of the Fourier series, assuming \sum_{n=0}^{\infty} n (a_n + b_n) converges. All you need is uniform convergence of the series, and that condition turns out to be sufficient. More explicitly, you can use the Weierstrass M-test.
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Goods
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(Original post by Smaug123)
It is a fact that the Fourier series of the integral is the termwise integral of the Fourier series, and the Fourier series of the derivative is the termwise derivative of the Fourier series, assuming \sum_{n=0}^{\infty} n (a_n + b_n) converges. All you need is uniform convergence of the series, and that condition turns out to be sufficient. More explicitly, you can use the Weierstrass M-test.
are those an and bn of the original series?

can i quote \sum_{n=0}^{\infty} n (a_n + b_n) as a standard result then as regards the term wise derivative?
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Smaug123
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(Original post by Goods)
are those an and bn of the original series?

can i quote \sum_{n=0}^{\infty} n (a_n + b_n) as a standard result then as regards the term wise derivative?
It is, yes. You probably want to say "by the Weierstrass M-test" or something just to dress it up a bit.
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TeeEm
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(Original post by Goods)


I've got the series and the derivatives and integrals. How can I show what they sum to?

Or can I simply say that the Integral sums to the integral of |x| for |x|< l because the terms decrease more quickly and so it converges.

If the derivative converges can I say likewise for x=-l,l,0 its undefined for 0<x<l it's 1 and -l<x<0 it's -1. Also how can I show it converges? as the terms decrease more slowly than for the original fourier series and so its not necessarily going to converge...
I have not done highly analytical stuff on fourier but I can tell you that

|x| differentiates to signum(x)
|x| integrates to 1/2 x2 signum(x)


both can be written as piecewise continuous functions.


the only thing you have to be careful is creating an a0 term when performing these operations, which does not apply to this example as |x| is even
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DFranklin
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(Original post by Smaug123)
More explicitly, you can use the Weierstrass M-test.
Not sure how the M-test works here, the obvious choice for M_n is O(1/n) and then \sum M_n diverges.

Edit: in fact, I don't think the M-test *can* work here. If it does, then the sum converges uniformly, and so the result is continuous. But the derivative of |x| is not continuous...
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Smaug123
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(Original post by DFranklin)
Not sure how the M-test works here, the obvious choice for M_n is O(1/n) and then \sum M_n diverges.

Edit: in fact, I don't think the M-test *can* work here. If it does, then the sum converges uniformly, and so the result is continuous. But the derivative of |x| is not continuous...
Ah. Curses.
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DFranklin
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FWIW, from wording of the question I don't think proof is required. There is actually a relevant theorem: given a continuous and piecewise-smooth function f, differentiating the Fourier series for f gives the correct Fourier series for f', but I think the proof is quite technical.
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(Original post by DFranklin)
FWIW, from wording of the question I don't think proof is required. There is actually a relevant theorem: given a continuous and piecewise-smooth function f, differentiating the Fourier series for f gives the correct Fourier series for f', but I think the proof is quite technical.
is |x| continuous and piecewise-smooth? and what are the criteria for it to be so?
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DFranklin
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(Original post by Goods)
is |x| continuous and piecewise-smooth? and what are the criteria for it to be so?
Google is your friend.
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