# Fourier series of e^x

by twoflower
Tags: fourier, series
 P: 368 Hi all, I've been having little problems getting Fourier series of $e^x$. I have given $$f(x) = e^{x}, x \in [-\pi, \pi)$$ Then $$a_0 = \frac{1}{\pi}\int_{-\pi}^{\pi} e^{x}\ dx = \frac{2\sinh \pi}{\pi}$$ $$a_{n} = \frac{1}{\pi}\int_{-\pi}^{\pi} e^{x}\cos (nx)\ dx = \frac{1}{\pi}\left\{\left[ \frac{1}{n}\ e^{x} \sin (nx)\right]_{-\pi}^{\pi} - \frac{1}{n}\int_{-\pi}^{\pi} e^{x}\sin (nx)\ dx\right\} = \frac{(-1)^{n+1}\left(e^{-\pi} - e^{\pi}\right)}{\pi(1+n^2)}$$ $$b_{n} = \frac{1}{\pi}\int_{-\pi}^{\pi}e^{x}\sin (nx)\ dx = \frac{1}{\pi}\left\{ \left[ -\frac{1}{n}\ e^{x}\cos (nx)\right]_{-\pi}^{\pi} + \frac{1}{n}\int_{-\pi}^{\pi} e^{x}\cos(nx)\ dx\right\} = \frac{n(-1)^{n}\left(e^{-\pi} - e^{\pi}\right)}{\pi(1+n^2)}$$ So the Fourier series looks like this: $$S(f,x) = \frac{\sinh \pi}{\pi}\ +\ \frac{2\sinh \pi}{\pi}\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{1+n^2}\left(\cos(nx) - n\sin(nx))$$ Anyway, our professor gave us another right (I hope so) result: $$S(f,x) = \frac{\sinh \pi}{\pi}\ +\ \frac{2\sinh \pi}{\pi}\sum_{n=1}^{\infty} \frac{(-1)^{n}}{1+n^2}\left(\cos(nx) - n\sin(nx))$$ Obviously my series is just of opposite sign than it should be, but I can't find the mistake, could you help me please?
 P: 618 twoflower, Am I missing something? The two solutions look the same to me!
 P: 695 jdavel, twoflower's solution has (-1)^(n + 1) inside the sum, while the professor's solution has (-1)^n.
P: 618
Fourier series of e^x

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

Doh!
 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).
 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 $$\sum1/(1 + n^2)$$ = 1/2(picoth(pi) -1), however when i calculate it to be 1/2(picoth(pi) + pi - 1), can anyone help?
 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 !!
 P: 42 Sinh(x)= 1/2 [e^x - e^(-x)]

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