# Derivatives and integrals of Fourier series Watch

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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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#2

(Original post by

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...

**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...

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(Original post by

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 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.

**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 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.

_{n}and b

_{n }of the original series?

can i quote as a standard result then as regards the term wise derivative?

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#4

(Original post by

are those a

can i quote as a standard result then as regards the term wise derivative?

**Goods**)are those a

_{n}and b_{n }of the original series?can i quote as a standard result then as regards the term wise derivative?

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#5

**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...

|x| differentiates to signum(x)

|x| integrates to

^{1}/

_{2}x

^{2}signum(x)

both can be written as piecewise continuous functions.

the only thing you have to be careful is creating an a

_{0}term when performing these operations, which does not apply to this example as |x| is even

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#6

(Original post by

More explicitly, you can use the Weierstrass M-test.

**Smaug123**)More explicitly, you can use the Weierstrass M-test.

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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#7

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Not sure how the M-test works here, the obvious choice for M_n is O(1/n) and then 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...

**DFranklin**)Not sure how the M-test works here, the obvious choice for M_n is O(1/n) and then 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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#8

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

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.

**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.

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#10

(Original post by

is |x| continuous and piecewise-smooth? and what are the criteria for it to be so?

**Goods**)is |x| continuous and piecewise-smooth? and what are the criteria for it to be so?

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