I was graphing random functions and came across this one

$$y=\frac{\log \left(3x^2\right)}{\log \left(\left(\sin ^{-1}\left(\sin \left(x\right)\right)\right)^{12^x}\right)}$$

When graphing it:enter image description here

I found that a root is $\pi$ however when plugging this into our function the demoninator becomes $\log(0)$ which is undefined. How can the root be $\pi$?

  • $\begingroup$ Just as you say: the function is not defined at $\pi$, so $\pi$ is not a root. $\endgroup$ – Matthew Conroy Nov 19 '16 at 6:26
  • $\begingroup$ @MatthewConroy Uhhh I thought a root is something which makes a function = zero? $\endgroup$ – bigfocalchord Nov 19 '16 at 6:33
  • 1
    $\begingroup$ The function is not equal to zero when $x=\pi$ because the function is undefined at $x=\pi$. We conclude from this that $\pi$ is not a root. $\endgroup$ – Matthew Conroy Nov 19 '16 at 6:35
  • $\begingroup$ If you don't mind, I shall reuse this function which is quite interesting. Cheers :-) $\endgroup$ – Claude Leibovici Nov 20 '16 at 2:58
  • 1
    $\begingroup$ The variations are so slow close to $\pi$ ! The asymptotics is quite nice. As you see, we can deduce everything from the last edit of my answer basically with no computation.. $\endgroup$ – Claude Leibovici Nov 20 '16 at 3:02

If $x=\pi$, $\sin(\pi)=0$, so $\sin^{-1}\left(\sin(\pi)\right)=0$.

And then $\left(\sin^{-1}\left(\sin(\pi)\right)\right)^{12^\pi}=0$

And then if you try taking $\log$ of this, you can't. But if $x$ had been just shy of $\pi$, then you'd be taking $\log$ of a number barely larger than $0$. So the $\log$ would be a huge negative number.

Then $\frac{\log(3x^2)}{\log(\cdots)}$ is like $\frac{c}{-\infty}$. (This is very informal.)


When $x$ is near $\pi$, the numerator is approximately $\ln 3\pi^2\approx 3.388$. Suppose that $x=\pi-\epsilon$, where $\epsilon$ is a small positive number. Then $\sin^{-1}\sin x=\sin\epsilon\approx\epsilon$, and $\epsilon^{12^x}$ is a very small positive number. Thus,

$$\lim_{x\to\pi^-}\ln\left(\left(\sin^{-1}\sin x\right)^{12^x}\right)=-\infty\;,$$


$$\lim_{x\to\pi^-}\frac{\ln\left(3x^2\right)}{\ln\left(\left(\sin^{-1}\sin x\right)^{12^x}\right)}=0\;,$$

approached from below, just as the graph indicates. The function is undefined for $x\ge\pi$, however.


As said in answers, let $$x=\pi-10^{-k}$$ and let us compute the value of the function $y$ for different values of $k$. We should get $$\left( \begin{array}{cc} k & y \\ 5 & -0.0001197920 \\ 10 & -0.0000598948 \\ 15 & -0.0000399299 \\ 20 & -0.0000299474 \\ 25 & -0.0000239579 \\ 30 & -0.0000199649 \\ 35 & -0.0000171128 \\ 40 & -0.0000149737 \\ 45 & -0.0000133100 \\ 50 & -0.0000119790 \end{array} \right)$$ which goes very, very slowly to $0$ but will never reach it since $y$ is not defined for $x=\pi$. For $k=1000$, we should get something like $y\approx -5.99\times 10^{-7}$.

Going deeper and setting $k=10^n$, it seems that the value of $y$ is given by $$y\approx -5.99\times 10^{-(n+4)}$$


Making things more formal, let us consider

$$y=\frac{\log \left(3x^2\right)}{\log \left(\left(\sin ^{-1}\left(\sin \left(x\right)\right)\right)^{12^x}\right)}=\frac{\log \left(3x^2\right)}{{12^x}\log \left(\sin ^{-1}\left(\sin \left(x\right)\right)\right)}$$ and use Taylor series (an compositions of them) around $\epsilon=0$ with $x=\pi-\epsilon$ $$\sin(x)=\sin(\pi-\epsilon)=\sin(\epsilon)=\epsilon +O\left(\epsilon ^3\right)$$ $$\sin^{-1}(\sin(x))=\epsilon +O\left(\epsilon ^3\right)$$ $$\log \left(\sin ^{-1}\left(\sin \left(x\right)\right)\right)=\log (\epsilon )+O\left(\epsilon ^2\right)$$ $$12^x=12^\pi\times 12^{-\epsilon}=12^\pi\left(1-\epsilon \log (12)+O\left(\epsilon ^2\right)\right)$$ All of the above makes the denominnator to be $$\log \left(\left(\sin ^{-1}\left(\sin \left(x\right)\right)\right)^{12^x}\right)=12^\pi\left(\log (\epsilon )-\epsilon \log (12) \log (\epsilon )+O\left(\epsilon ^2\right)\right)$$ For the numerator $$\log(3x^2)=\log \left(3 \pi ^2\right)-\frac{2 \epsilon }{\pi }+O\left(\epsilon ^2\right)$$ All of that finally makes $$y=\frac{12^{-\pi } \log \left(3 \pi ^2\right)}{\log (\epsilon )}+O\left(\epsilon \right)$$ Setting $\epsilon=10^{-k}$ then leads to $$y\approx -\frac{12^{-\pi } \log \left(3 \pi ^2\right)}{\log (10 )}\times \frac 1 k\approx -\frac {5.98948 \times 10^{-4}} k$$ from which all above results.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.