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This is a follow up topic from this previous post.

The Fourier series for $f(x)=e^x$ is $$f(x)=e^x=\sinh1\left(1+2\sum_{n=1}^{\infty}\frac{(-1)^n}{(n\pi)^2+1}\left(\cos(n\pi x)-n\pi\sin(n\pi x)\right)\right)$$

The value of the Fourier series $f(x)=e^x$ at $x=2$ and $x=1$ are the same: $$f(2)=\sinh1\left(1+2\sum_{n=1}^{\infty}\frac{(-1)^{n}}{(n\pi)^2+1}\right)$$

$$f(0)=e^0=\sinh1\left(1+2\sum_{n=1}^{\infty}\frac{(-1)^{n}}{(n\pi)^2+1}\right)$$

Could someone please prove that $$\sinh1\Bigl(1+2{\sum}_{n=1}^{\infty}\frac{(-1)^{n}}{(n\pi)^2+1}\Bigr)=1?$$ The interval is $-1\lt x \lt 1$

Thank you.

Community
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BLAZE
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1 Answers1

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This answer is, on purpose, a word-by-word copy of the answer by @Dominik (to show OP that this is essentially the same question as discussed in the comments), with the small necessary changes of course. I hope Dominik does not care.

Define the function $f$ to be $\exp(x)$ on $[-1, 1)$ and continue it periodically. Then by the Dirichlet condition the Fourier series of $f$ converges for all $x$ to $\frac{1}{2}(f(x+) + f(x-))$. Since the function is continuous at $x = 0$, the Fourier series at this point will converge to $e^0=1$.

Hence, $$ \sinh1\Bigl(1+2{\sum}_{n=1}^{\infty}\frac{(-1)^{n}}{(n\pi)^2+1}\Bigr)=1. $$

Edit Upon request. $f(x+)$ denotes the right limit $\lim_{t\to x+}f(t)$ and $f(x-)$ denotes the left limit $\lim_{t\to x-}f(t)$. In the case $f$ is continuous at $x$, one has $f(x+)=f(x-)=f(x)$.

mickep
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  • Great, now I need you to do one more thing show in that answer the substitution of the values into $\frac{1}{2}(f(x+) + f(x-))$. This is what I asked of Dominik but no such luck. The reason I ask is because I don't know what $f(x+)$ and $f(x-)$ actually mean. If you do this I will consider this question concluded. – BLAZE Sep 22 '15 at 12:22
  • Did you read the link Dominik gave: https://en.wikipedia.org/wiki/Dirichlet_conditions? The definitions are in there. – mrf Sep 22 '15 at 12:24
  • @mrf Yes I did it was in Dominiks post – BLAZE Sep 22 '15 at 12:24
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    You could have mentioned in your question that the notations $f(x+)$ and $f(x-)$ were the problem. Nevertheless, not it is updated with that information included. – mickep Sep 22 '15 at 12:25
  • @mickep Thanks, that was all I wanted to see. – BLAZE Sep 22 '15 at 12:27