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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 $\pi$?

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  • $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ 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
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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.)

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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,

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

and

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

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

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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 $$y\approx -5.99\times 10^{-(n+4)}$$

Edit

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.

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