Expand the function $f (x)=\sin(\frac{3}{x})$ in Fourier series on the interval $[0,10]$

Question

Could you help me understand this a little better?

It asks me if the function $f (x)$ can be expanded in fourier series on the interval $[0,10]$

It confuses me a little more than anything to understand this part. I would like to understand more than anything what kind of conditions a function must fulfill so that it can be expanded in Fourier series. I know that there are expansions for a period $ [- \pi, \pi] $, $ [- L, L] $ and for an arbitrary period $ [a, b] $.

In this case it can be seen that it is an arbitrary period but it is not completely clear to me what must happen for me to say that a function can be expanded in Fouirer series in said interval. I would like to be able to solve this problem but more than anything I would like you to help me with a generalization about how I can know when a function can be expanded in Fourier series according to the interval that they give me.

I hope you can help me.

Answer 1

There are two problems here:

• $f(x)$ is not periodic.
• $f(x)$ is undefined at $0$. In fact, $f(x)$, though bounded, does not converge at $0$.

The first problem can be bypassed by choosing some period of length greater than $10$. Then the interval in question would fit within one period, and we don't care about its behavior outside that interval. We could choose to extend $f$ to a full period in any way that was convenient for calculation.

The second problem is more significant. Exactly as the question is stated now, it can be easily answered with "no". Because $f$ is not defined at $0$, we cannot express it as a fourier series on $[0,10]$, which includes $0$.

Even if one chooses not to be so pendantic and allow a representation that fails to match $f(x)$ at isolated points, this still doesn't work.

$f(x)$ is not integrable on $[0,10]$. The integral $\int_\epsilon^{10}$ exists for all $\epsilon > 0$, but the limit as $\epsilon \to 0$ doesn't converge. But Fourier series are always integrable, with the integral over a period being the constant term times the period.

Even interpreting the question to refer to formal Fourier series - that is, not requiring that the series converge at all, much less to $f(x)$ - still leaves the answer as "no". The connection of a formal Fourier series to its function is by the integral formula for its coefficients. But the coefficients of the cosine terms for this function are all undefined, since $f(x)\cos Ax$ is not integrable around $0$ for any constant $A$.