# Fourier Series Help

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#1
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
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3 weeks ago
#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
Last edited by mqb2766; 3 weeks ago
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#3
(Original post by 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
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
Last edited by BrandonS15; 3 weeks ago
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#4
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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3 weeks ago
#5
(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 ...
Last edited by mqb2766; 3 weeks ago
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#6
(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
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3 weeks ago
#7
(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?
Last edited by mqb2766; 3 weeks ago
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#8
(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?
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3 weeks ago
#9
(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.
Last edited by mqb2766; 3 weeks ago
0
3 weeks ago
#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.
Last edited by DFranklin; 3 weeks ago
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3 weeks ago
#11
(Original post by 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
What's the actual problem you've been asked to solve here? Can you upload a picture of the question?
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#12
(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
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#13
(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
Last edited by BrandonS15; 3 weeks ago
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3 weeks ago
#14
(Original post by 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
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.
Last edited by mqb2766; 3 weeks ago
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3 weeks ago
#15
(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)).
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#16
(Original post by 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.
Does that mean I can only evaluate it at x = 0?
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3 weeks ago
#17
(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.
Last edited by mqb2766; 3 weeks ago
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#18
(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
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3 weeks ago
#19
(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?
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3 weeks ago
#20
(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.
Last edited by mqb2766; 3 weeks ago
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