Given a Fourier series $f(x)$: What's the difference between the value the expansion takes for given $x$ and the value it converges to for given $x$?

Question

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

Answer 1

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 $\frac{1}{2}(e + e^{-1})$.

Answer 2

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