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I got the Fourier series as $(e-e^{-1})(\frac 1 2+\sum \limits _{n=1} ^\infty \frac {(-1)^n(\cos(n\pi x)-n\pi \sin(n\pi x)} {1+n^2\pi^2})$.

Although I've seen the answer online as being $\sum \limits _{n=0} ^\infty \frac {(-1)^n(e-e^-1)(\cos(n\pi x)-n\pi \sin(n\pi x))} {1+n^2\pi^2}$ but I don't see why there's no $\frac 1 2$ coefficient on $a_0$ in the answer I found on the internet.

Am I right or have I made a mistake?

Alex M.
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Goods
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  • There is a coefficient $a_0$ in the answer found on the Internet : it is twice you've found. – Nicolas May 16 '15 at 13:31
  • i'm sorry I wasn't clear. The integral of $e^x$ over $x=1$ to $x=-1$ is $e-e^-1$ which is a0. The first term is a0/2 in a fourier series. The answer I found appears to admit this factor of 1/2 but I don't see why – Goods May 16 '15 at 13:38
  • As a side-note: if your exponents have more than one character (like $-1$), you have to enclose them between matching { and }. Fractions are written as "\frac {numerator} {denominator}"; if any of them has only one character, you may omit the curly brackets (and this is a general rule, not just for fractions). – Alex M. May 16 '15 at 14:03

1 Answers1

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The correct coefficients are $a_0 = \frac {\mathbb e - \mathbb e ^{-1}} 2, \space a_n = (\mathbb e - \mathbb e ^{-1}) \frac {(-1)^n} {1+n^2 \pi^2}, \space b_n = -(\mathbb e - \mathbb e ^{-1})n\pi \frac {(-1)^n} {1+n^2 \pi^2}$.

Alex M.
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  • I don't have mathematica so I can't check but the online series is $\sum \limits _{n=0} ^\infty \frac {(-1)^n e-e^-1} {(1 + n^2 \pi^2}$ for $x=0$ also just using plain old simple integration a0 is surely $e-e^-1$? I miss typed them originally so I probably made the correction whilst you were checking sorry! – Goods May 16 '15 at 13:49
  • @goods: I apologize, there were some mistakes in my answer that I have corrected. – Alex M. May 16 '15 at 13:54
  • great! phew your new answers agree with what I got! so the online answer must have been wrong – Goods May 16 '15 at 13:56