# Fourier Series Help

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Hello,

I’ve found the Fourier series for e^x and now I realise I can form two closed-form values for two infinite summations if I let x=0 and x=pi (sin(xn) = 0 for all n) For the case x = pi I’m getting the following result:

I checked the summation from 1 to inf of 1/(1+n^2) on Wolfram and it gives me 1.07 something while my value is 0.5 something,

I’m not sure where I went wrong, I feel like there is a restriction to what values I can put for x (due to periodic) but I’m not sure.

Any help is appreciated, many thanks

I’ve found the Fourier series for e^x and now I realise I can form two closed-form values for two infinite summations if I let x=0 and x=pi (sin(xn) = 0 for all n) For the case x = pi I’m getting the following result:

I checked the summation from 1 to inf of 1/(1+n^2) on Wolfram and it gives me 1.07 something while my value is 0.5 something,

I’m not sure where I went wrong, I feel like there is a restriction to what values I can put for x (due to periodic) but I’m not sure.

Any help is appreciated, many thanks

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#2

What is the actual function you're finding the Fourier series of?

You realise that the Fourier series does not converge pointwise at the jumps (discontinuities), which is what you may be asking about?

https://en.m.wikipedia.org/wiki/Gibbs_phenomenon

You realise that the Fourier series does not converge pointwise at the jumps (discontinuities), which is what you may be asking about?

https://en.m.wikipedia.org/wiki/Gibbs_phenomenon

Last edited by mqb2766; 3 weeks ago

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(Original post by

What is the actual function you're finding the Fourier series of?

You realise that the Fourier series does not converge pointwise at the jumps (discontinuities), which is what you may be asking about?

https://en.m.wikipedia.org/wiki/Gibbs_phenomenon

**mqb2766**)What is the actual function you're finding the Fourier series of?

You realise that the Fourier series does not converge pointwise at the jumps (discontinuities), which is what you may be asking about?

https://en.m.wikipedia.org/wiki/Gibbs_phenomenon

Last edited by BrandonS15; 3 weeks ago

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I’m evaluating at x = pi because (like x=0) it removed the other infinite summation with sines so just like x=0 I’m trying to use x=pi to form the infinite summation of (1/(1+n^2)) as shown in the photo

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#5

(Original post by

Yes sorry, it’s of e^x and I’m evaluating at x=pi because like x=0 It gets rid of the second infinite sum of sines

**BrandonS15**)Yes sorry, it’s of e^x and I’m evaluating at x=pi because like x=0 It gets rid of the second infinite sum of sines

-pi to pi

or

0 to pi

Or ...

Last edited by mqb2766; 3 weeks ago

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#7

(Original post by

-pi to pi as I did in the integrals for computing the Fourier coefficients

**BrandonS15**)-pi to pi as I did in the integrals for computing the Fourier coefficients

Last edited by mqb2766; 3 weeks ago

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(Original post by

So at x=pi you're evaluating it at a point where the series does not converge to e^x?

**mqb2766**)So at x=pi you're evaluating it at a point where the series does not converge to e^x?

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#9

(Original post by

How do I know the series doesn’t converge to e^x at x = pi but does at x = 0?

**BrandonS15**)How do I know the series doesn’t converge to e^x at x = pi but does at x = 0?

Last edited by mqb2766; 3 weeks ago

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#10

It *will* converge to the average of the values at -pi, pi; whether you're allowed to quote that in your course I don't know.

Edit: although f(x)+f(-x) will be continuous at pi and so that series will converge, so it's not hard to prove if you need to.

Edit: although f(x)+f(-x) will be continuous at pi and so that series will converge, so it's not hard to prove if you need to.

Last edited by DFranklin; 3 weeks ago

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#11

(Original post by

I’m evaluating at x = pi because (like x=0) it removed the other infinite summation with sines so just like x=0 I’m trying to use x=pi to form the infinite summation of (1/(1+n^2)) as shown in the photo

**BrandonS15**)I’m evaluating at x = pi because (like x=0) it removed the other infinite summation with sines so just like x=0 I’m trying to use x=pi to form the infinite summation of (1/(1+n^2)) as shown in the photo

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(Original post by

What's the actual problem you've been asked to solve here? Can you upload a picture of the question?

**davros**)What's the actual problem you've been asked to solve here? Can you upload a picture of the question?

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(Original post by

It *will* converge to the average of the values at -pi, pi; whether you're allowed to quote that in your course I don't know.

Edit: although f(x) f(-x) will be continuous at pi and so that series will converge, so it's not hard to prove if you need to.

**DFranklin**)It *will* converge to the average of the values at -pi, pi; whether you're allowed to quote that in your course I don't know.

Edit: although f(x) f(-x) will be continuous at pi and so that series will converge, so it's not hard to prove if you need to.

Last edited by BrandonS15; 3 weeks ago

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#14

(Original post by

No problem, I’m just experimenting and I thought I could plug pi into the Fourier series of e^x to obtain and infinite series in closed form but I don’t fully understand why it’s discontinuous at x = pi on the -pi to pi interval

**BrandonS15**)No problem, I’m just experimenting and I thought I could plug pi into the Fourier series of e^x to obtain and infinite series in closed form but I don’t fully understand why it’s discontinuous at x = pi on the -pi to pi interval

e^pi

And

e^(-pi)

At the point x=pi, by considering the boundaries of the neighbouring intervals [-pi,pi] and [pi,3pi]. It can't be multivalued, so converges to the midpoint at the interval boundaries.

Last edited by mqb2766; 3 weeks ago

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#15

(Original post by

How do you know it’s continuous or not? e^x is continuous for all x

**BrandonS15**)How do you know it’s continuous or not? e^x is continuous for all x

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(Original post by

You'd want the Fourier series to be equal to both

e^pi

And

e^(-pi)

At the point x=pi, by considering the boundaries of the neighbouring intervals [-pi,pi] and [pi,3pi]. It can't be multivalued, so converges to the midpoint at the interval boundaries.

**mqb2766**)You'd want the Fourier series to be equal to both

e^pi

And

e^(-pi)

At the point x=pi, by considering the boundaries of the neighbouring intervals [-pi,pi] and [pi,3pi]. It can't be multivalued, so converges to the midpoint at the interval boundaries.

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#17

(Original post by

Does that mean I can only evaluate it at x = 0?

**BrandonS15**)Does that mean I can only evaluate it at x = 0?

x = (2k+1)pi

Where is an integer.

At all other values (including 0) the Fourier series converges to the periodic version of e^x.

Last edited by mqb2766; 3 weeks ago

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(Original post by

Of course not. The only "difficult" points are where the periodic function is discontinuous (or multivalued) at

x = (2k+1)pi

At all other values (including 0) the Fourier series converges to e^x.

**mqb2766**)Of course not. The only "difficult" points are where the periodic function is discontinuous (or multivalued) at

x = (2k+1)pi

At all other values (including 0) the Fourier series converges to e^x.

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#19

(Original post by

The Fourier series repeats every -pi to pi right? So I’m not understanding why it’s dis continuous at those points

**BrandonS15**)The Fourier series repeats every -pi to pi right? So I’m not understanding why it’s dis continuous at those points

What is e^x when x=pi?

Since the Fourier series (call it F(x)) representation has period 2pi, F(-pi) must equal F(pi).

Do you see the problem here?

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#20

**BrandonS15**)

The Fourier series repeats every -pi to pi right? So I’m not understanding why it’s dis continuous at those points

e^x

Is not continuous at those points. Have you sketched/plotted it for few period intervals? You can see the discontinuities at those points?

The Fourier series is also periodic. It converges to e^x in the interior of each interval. It must jump at those points to represent the original function. That's what causes the Gibbs/ringing effect when you have a truncated Fourier seies. In the limit, the oscillations die down, but it still has to jump at the interval boundaries.

Last edited by mqb2766; 3 weeks ago

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