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The question given to me was: Find the Fourier series for $f(x) = e^x$ over the range $-1\lt x\lt 1$ and find what value the expansion will have when $x = 2$?

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

My attempt to compute these values: $$f(2)=\sinh1\left(1+2\sum_{n=1}^{\infty}\frac{(-1)^{n}}{(n\pi)^2+1}\right)$$

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

The answer given was: The series will converge to the same value as it does at $x = 0$, i.e. $f(0) = 1$.

Could someone please explain to me the difference between:

What value will the expansion take at a given $x$ and the value the series converges to for a given $x$?

Many Thanks

BLAZE
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2 Answers2

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That answer is wrong.

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 has a jump-type discontinuity at $x = 1$, the Fourier series at this point will converge to $\frac{1}{2}(e + e^{-1})$.

Dominik
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  • Thanks for your reply, so i guess you're saying at $x=1$ the series converges to $\sinh1$ is that it? – BLAZE Sep 20 '15 at 06:58
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    It's actually $\cosh(1)$. – Dominik Sep 20 '15 at 07:02
  • which answer is wrong? Can you elaborate on your answer please as if what you are saying is true (and I'm not suggesting it isn't) then the book I'm reading is wrong – BLAZE Sep 20 '15 at 07:03
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    The value at $x = 1$ isn't the same as the value at $x = 0$. – Dominik Sep 20 '15 at 07:04
  • that makes more sense, thanks Dominik – BLAZE Sep 20 '15 at 07:06
  • I don't know any way to calculate the value aside from the Dirichlet condition. – Dominik Sep 20 '15 at 12:38
  • Please elaborate further. I need more information as to how you calculated that the series converges to $\cosh1$ via Dirichlet condition? – BLAZE Sep 20 '15 at 21:15
  • Did you take a look at the article? It describes exactly how the value of the Fourier series is connected to the value of the function. I just plugged in the corresponding values. – Dominik Sep 20 '15 at 21:28
  • yes I read the link you posted in your answer, but could you please show me in your answer those values being plugged into the relevant formula? Sorry, I know this seems futile and obvious to you, but this is the first time I have come across these Dirichlet conditions and it's not obvious to me. I just need to see your working on how you arrived at $\cosh1$, instead of an explanation for it. Thanks again. – BLAZE Sep 21 '15 at 01:54
  • The right-hand limit is $e^{-1}$ and the left-hand limit is $e$. Just take a look at the graph of the function and it will be pretty obvious. I really can't explain it in any more detail. – Dominik Sep 21 '15 at 07:46
  • Okay thanks, one more question: How did you get ${e}^{-1}$ for the right-hand limit and ${e}$ for the left-hand limit? – BLAZE Sep 21 '15 at 08:03
  • You seem to have misread the question. It asked (from the initial version on) for the behaviour of the Fourier series at $x = 2$. And since the Fourier series has period $2$, it clearly behaves at $2$ exactly as it does at $0$. You discuss the behaviour at $1$. – Daniel Fischer Sep 02 '16 at 13:25
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Hint and suggestion.

  1. This is wrong: $${f(2)=e^2=\sinh1\left(1+2\sum_{n=1}^{\infty}\frac{(-1)^{2n}}{(n\pi)^2+1}\right)}$$ Erase it

  2. Grab your pen and recalculate $f(2)$. If you do this with care, you will soon end up with $f(0)$.

  3. Scroll down for a solution.

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Solution:

Put $f(x) = e^x$, and let $\widetilde{f}$ denote the the Fourier series expansion of $f$ over the interval [-1,1].

Next, according to the theory of Fourier series (for example we can use to the Dirichlet theorem) tells us that we will have

$$\widetilde{f}(x) = f(x), \qquad \text{for all $x$ with $-1<x<1$}\tag1$$ $$\widetilde{f}(1) = \frac{f(1)+f(-1)}{2}, \qquad \text{for all $x$ with $-1<x<1$}\tag2$$ and $$\widetilde{f}(2n + x) = \widetilde{f}(x), \qquad \text{for all real $x$ and integers $n$}\tag3$$

(In our case we do not need any limits in (2)).

Now, regarding the value at $x=2$. The value of $f(2)$ is $e^2$, and does not depend on the Fourier series at all, while $$\widetilde{f}(2) = \widetilde{f}(0) = f(0) = 1$$ due to (3) and (1), respectively. On the one hand, according to the expansion of yours, $$\widetilde{f}(0) = \sinh1\left(1+2\sum_{n=1}^{\infty}\frac{(-1)^{n}}{(n\pi)^2+1}\right)$$

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