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Fourier series of e^x

by twoflower
Tags: fourier, series
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twoflower
#1
May11-05, 07:03 AM
P: 368
Hi all,

I've been having little problems getting Fourier series of [itex]e^x[/itex].

I have given

[tex]
f(x) = e^{x}, x \in [-\pi, \pi)
[/tex]

Then

[tex]
a_0 = \frac{1}{\pi}\int_{-\pi}^{\pi} e^{x}\ dx = \frac{2\sinh \pi}{\pi}
[/tex]

[tex]
a_{n} = \frac{1}{\pi}\int_{-\pi}^{\pi} e^{x}\cos (nx)\ dx =

\frac{1}{\pi}\left\{\left[ \frac{1}{n}\ e^{x} \sin (nx)\right]_{-\pi}^{\pi} -

\frac{1}{n}\int_{-\pi}^{\pi} e^{x}\sin (nx)\ dx\right\} =

\frac{(-1)^{n+1}\left(e^{-\pi} - e^{\pi}\right)}{\pi(1+n^2)}
[/tex]

[tex]
b_{n} = \frac{1}{\pi}\int_{-\pi}^{\pi}e^{x}\sin (nx)\ dx = \frac{1}{\pi}\left\{

\left[ -\frac{1}{n}\ e^{x}\cos (nx)\right]_{-\pi}^{\pi} +

\frac{1}{n}\int_{-\pi}^{\pi} e^{x}\cos(nx)\ dx\right\} =

\frac{n(-1)^{n}\left(e^{-\pi} - e^{\pi}\right)}{\pi(1+n^2)}
[/tex]

So the Fourier series looks like this:

[tex]
S(f,x) = \frac{\sinh \pi}{\pi}\ +\ \frac{2\sinh \pi}{\pi}\sum_{n=1}^{\infty}

\frac{(-1)^{n+1}}{1+n^2}\left(\cos(nx) - n\sin(nx))
[/tex]

Anyway, our professor gave us another right (I hope so) result:

[tex]
S(f,x) = \frac{\sinh \pi}{\pi}\ +\ \frac{2\sinh \pi}{\pi}\sum_{n=1}^{\infty}

\frac{(-1)^{n}}{1+n^2}\left(\cos(nx) - n\sin(nx))
[/tex]

Obviously my series is just of opposite sign than it should be, but I can't find
the mistake, could you help me please?
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jdavel
#2
May11-05, 08:55 AM
P: 618
twoflower,

Am I missing something? The two solutions look the same to me!
Muzza
#3
May11-05, 08:57 AM
P: 695
jdavel, twoflower's solution has (-1)^(n + 1) inside the sum, while the professor's solution has (-1)^n.

jdavel
#4
May11-05, 10:01 AM
P: 618
Fourier series of e^x

Quote Quote by Muzza
jdavel, twoflower's solution has (-1)^(n + 1) inside the sum, while the professor's solution has (-1)^n.
Muzza,

Doh!
StatusX
#5
May11-05, 10:18 AM
HW Helper
P: 2,567
You must have made a mistake in the integrals. It's hard to tell because you've skipped 3 or 4 steps in your post, but it looks like at least your bottom integral is wrong (wrong sign) for the case where n is odd (the only case I felt like doing in my head).
amda
#6
Jan24-10, 12:17 PM
P: 1
as far as i can tell, when you have taken (e^-pi - e^pi) out from the summation you have left it as sinh(pi) whereas it should be -sinh(pi), this should then match with the professors solution.
I have a similar question, using this series i must show that [tex]\sum1/(1 + n^2)[/tex] = 1/2(picoth(pi) -1), however when i calculate it to be 1/2(picoth(pi) + pi - 1), can anyone help?
SDmathlover
#7
Apr7-11, 12:14 AM
P: 1
Why is integration of e^x=2sinh (pie)/pie ?
i am not able to understand this concept can some body enlighten this for me !!
TheFurryGoat
#8
Aug11-11, 03:01 AM
P: 42
Sinh(x)= 1/2 [e^x - e^(-x)]


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