Fourier Series Help

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:
49C1B86C-0A62-4D7C-B118-8A204AAD3BB0.jpg.jpeg

49C1B86C-0A62-4D7C-B118-8A204AAD3BB0.jpg.jpeg

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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Reply 1

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

(edited 3 years ago)

Reply 2

BrandonS15

Original post by mqb2766

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

(edited 3 years ago)

Reply 3

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

Reply 4

mqb2766

Original post by 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

What is the periodic/repetition interval?
-pi to pi
or
0 to pi
Or ...

(edited 3 years ago)

Reply 5

BrandonS15

Original post by mqb2766

What is the periodic/repetition interval?
-pi to pi
or
0 to pi
Or ...

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

Reply 6

mqb2766

Original post by BrandonS15

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

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

(edited 3 years ago)

Reply 7

BrandonS15

Original post by mqb2766

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

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

Reply 8

mqb2766

Original post by BrandonS15

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

See the link in #2. It does not converge at x=(2k+1)pi for this problem (the discontinuities) where k is an integer.

(edited 3 years ago)

Reply 9

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.

(edited 3 years ago)

Reply 10

davros

Original post by BrandonS15

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

Reply 11

BrandonS15

Original post by davros

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

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

Reply 12

BrandonS15

Original post by 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.

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

(edited 3 years ago)

Reply 13

mqb2766

Original post by BrandonS15

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.

(edited 3 years ago)

Reply 14

DFranklin

Original post by BrandonS15

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

I was meaning the periodic Fourier expansion of the sum will be continuous (because if g(x) = f(x)+f(-x) then g(pi)=g(-pi)).

Reply 15

BrandonS15

Original post by mqb2766

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

Reply 16

mqb2766

Original post by BrandonS15

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

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

(edited 3 years ago)

Reply 17

BrandonS15

Original post by 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.

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

Reply 18

DFranklin

Original post by 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?
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?

Reply 19

mqb2766

Original post by BrandonS15

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

The (repeated/periodic) function
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.