I was plotting random functions on Desmos and then I noticed an interesting phenomenon with the function $x^2 \sin(1/x).$ When I zoomed in at the number $1/k$ by a factor of $k^2$, where $-k$ is a large number, I noticed that the graph looked like a sine wave with amplitude $1$ and period $2\pi.$ Is there any way to formalize the idea of "looks like a sine wave?" If so, does it really look like the sine wave that I described?
I thought of taking a limit as $-k \rightarrow \infty,$ but phase differences wreck this plan. I don't see how I can normalize these phase differences.
Best Answer
Let's see.
Of course, $x\mapsto x^2\cdot\sin(1/x)$ will never be exactly 'like' a sine wave. However, as $x\to 0$, there are some asymptotic relations which hold.
First of all, when looking at the graph, you notice the amplitude is essentially constant. Let's fix a small number $a$ (for you, $a=1/k$). Near $a$, the difference between $x^2$ and $a^2$, if $x=a+h$, is $2ah+h^2$ which, for $a$ and $h$ small, is essentially negligible. Especially since this is being multiplied by a number bounded in $[-1,1]$, so the error can't blow up at all.
Therefore, near $a$, I approximate $x^2\sin(1/x)\approx a^2\sin(1/x)$ with reasonably good accuracy. What do we do about the $1/x$ term?
Again, it is good to examine how this changes. For $x=a+h$, we have: $$\frac{1}{a+h}-\frac{1}{a}=-\frac{h}{a(a+h)}=-\frac{h}{a^2}\cdot\frac{1}{1+h/a}$$
So, when $|h|\ll a$, this difference is essentially $-h/a^2$. So I notice: $$\sin(1/x)\approx\sin\left(\frac{1}{a}-\frac{h}{a^2}\right)$$Now this is looking linear in $h$. Consider the function: $$x\mapsto\sin\left(\frac{1}{a}-\frac{x-a}{a^2}\right)=\sin\left(\frac{2}{a}-\frac{x}{a^2}\right)$$Near $x=a$, this will roughly equal $\sin(1/x)$.
We then approximate the curve with: $$x\mapsto a^2\cdot\sin\left(\frac{2}{a}-\frac{x}{a^2}\right)$$
If you let $a=-0.001$ in your graph, you will notice very good results. It will only get better as $a\to0$.
See the results for $a=-10^{-4}$.