
#1
May1105, 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? 



#2
May1105, 08:55 AM

P: 618

twoflower,
Am I missing something? The two solutions look the same to me! 



#3
May1105, 08:57 AM

P: 696

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




#4
May1105, 10:01 AM

P: 618

Fourier series of e^xDoh! 



#5
May1105, 10:18 AM

HW Helper
P: 2,566

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).




#6
Jan2410, 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? 



#7
Apr711, 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 !! 



#8
Aug1111, 03:01 AM

P: 42

Sinh(x)= 1/2 [e^x  e^(x)]



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