I have to find the Fourier-coefficients of $f(x)=\text{exp}(x)$ for $-1 < x <1$ and evaluate the value of this series at $x=2$.
(EDIT: with period 2)
I have calculated the coefficients to
$$a_n = \frac{(e^2-1)}{e\pi^2 n^2 +e} (-1)^n$$ for the cosine-termes and
$$b_n = \frac{(1-e^2)\pi n}{e\pi^2 n^2 +e}(-1)^n$$
for the sine-termes.
My first question would be, if these are correct?
Assuming they are correct, I get the following series
$$f(x)=\frac{a_0}{2}+\sum_{n\geq1}\frac{(-1)^n}{e\pi^2 n^2 +e}
\left((e^2-1)\cos(\pi n x)+(1-e^2)\pi n \sin(\pi n x)\right)$$
Now, how can I calculate its value for $x=2$? In this case the series gets a little bit easier but still, I dont know how to calculate its value.
And last but not least, when I integrate the series term by term, I dont really get the old series again.
I am very grateful for any kind of help!
Thank you!
Best Answer
Your formulas for $a_n$ and $b_n$ are correct. To calculate $f(2)= \frac{e^2-1}{2e} + \frac{e^2-1}{e} \sum_{n=1}^\infty \frac{(-1)^n}{1+ \pi^2 n^2}$ you just notice that it is the same sum as for $f(0)=1$. It may be possible to calculate this sum independently, but I doubt you're supposed to do that.
The reason why you're not obtaining the previous series because when you integrate term $a_0$ you get a non-constant function, which causes the difference. For $|x|<1$ we have: \begin{align} e^x &= 1 + \int_0^x f(y)dy = 1 + \int_0^x \Big(\frac{e^2-1}{2e} + \frac{e^2-1}{e} \sum_{n=1}^\infty \frac{(-1)^n\big(\cos (\pi n y) - \pi n \sin(\pi n y)\big)}{1+\pi^2n^2}\Big) dy = \\ &= 1 + \frac{e^2-1}{2e} x + \frac{e^2-1}{e}\sum_{n=1}^\infty \frac{(-1)^n}{1+\pi^2n^2}\big(\frac{\sin (\pi n x) + \pi n (\cos(\pi n x) - 1)}{\pi n}\big) = \\ &= 1 - \frac{e^2-1}{e}\sum_{n=1}^\infty\frac{(-1)^n}{1+\pi^2 n^2} + \frac{e^2-1}{2e} x + \frac{e^2-1}{e}\sum_{n=1}^\infty \frac{(-1)^n}{1+\pi^2n^2}\big(\cos(\pi n x) + \frac{1}{\pi n}\sin (\pi n x)\big) = \\ &= \frac{e^2-1}{2e} + \frac{e^2-1}{2e} x + \frac{e^2-1}{e}\sum_{n=1}^\infty \frac{(-1)^n}{1+\pi^2n^2}\big(\cos(\pi n x) + \frac{1}{\pi n}\sin (\pi n x)\big)\end{align} It is equivalent with the original formula $$e^x = \frac{e^2-1}{2e} + \frac{e^2-1}{e} \sum_{n=1}^\infty \frac{(-1)^n}{1+\pi^2n^2}\big(\cos (\pi n x) - \pi n \sin(\pi n x)\big) $$ because (for $|x|<1$): $$ x = 2\sum_{n=1}^\infty \frac{(-1)^{n+1}}{\pi n}\sin(\pi n x) = 2\sum_{n=1}^\infty \frac{(-1)^n}{1+\pi^2 n^2} \big(-\pi n - \frac{1}{\pi n}\big)\sin(\pi n x)$$