Inverse of $y=\frac{\sin^2(x)}{x^2}$?

Question

I'm having trouble finding the inverse of $y=\dfrac{\sin^2(x)}{x^2}$.

So far what I have is $$\begin{align} \frac{\sin^2(y)}{y^2} &= x \\[4pt] \sin^2(y) &= xy^2 \\[4pt] \sin(y) &= \sqrt{xy^2} = \sqrt{x}y \\[4pt] y &= \sin^{-1}(\sqrt{x}y) \end{align}$$

But I don't know how to take it any further than that. Does an inverse exist?

Answer 1

Technically, an inverse does not exist because this function is not one-to-one. Can you see that this function hits zero infinitely many times, for example? Therefore, there is no reasonable value to assign to $ f^{-1}(0) $. As a commenter, Andrei, pointed out, you won't get any closed-form inverse relations either.

Answer 2

The inverse function cannot be explicitly obtained, however it can be graphed as a reflection about the line $x=y.$

One can name them perhaps as SqSinc(x), SqSinc$ ^{-1} $x and so on..depending on application, utility, frequency of use ..

The inverse function can obtained by swapping $x$ and $y\,$ as:

$x=\dfrac{\sin^2(y)}{y^2}\,$.

This can be effected in the power series expansions as well ..

Related functions are labelled in the following graph:

Answer 3

Well inverse exist if this becomes one to one function

For that $x>0$

Now it becomes one to one function

To take the inverse there are two things you should go through

One is making a variable a subject and second is interchanging variables

First part only works when you can isolate a variable and in this case you can't isolate x

So just interchange x and y

You get $(\frac{\sin y}{y})^2=x \ \ , \ y>0$

Answer 4

In order to define an inverse function you need to specify a sub-interval of $\mathbb{R}$ over which $f(x)=\left(\frac{\sin x}{x}\right)^2$ is injective: $[0,\pi]$ is a good choice since $f'(x)\leq 0$ over there. So I am going to assume that you want a manageable expression for the inverse function of $\left(\frac{\sin x}{x}\right)^2$, intended as a map from $[0,\pi]$ to $[0,1]$. We have $$ \sqrt{y} = \frac{\sin x}{x} = \sum_{k\geq 0}\frac{(-1)^k x^{2k}}{(2k+1)!}. $$ Let $\sqrt{y}=1-\frac{w}{6}$ and $x^2=z$. The problem becomes to express $z$ in terms of $w$, given $$ w = \sum_{k\geq 1}\frac{6(-1)^{k+1} z^k}{(2k+1)!} = z+o(z).$$ Here we may apply the Lagrange inversion theorem, leading to $$ z = \sum_{n\geq 1}\frac{w^n}{n6^n}\cdot [w^{n-1}]\left(\frac{w}{1-\frac{\sin\sqrt{w}}{\sqrt{w}}}\right)^n=\sum_{n\geq 1}\frac{w^n}{n 6^n}\cdot\operatorname*{Res}_{w=0}\left(\frac{\sqrt{w}}{\sqrt{w}-\sin\sqrt{w}}\right)^n.$$ The computation of the first residues leads to $$ z = w+\frac{1}{20}w^2 +\frac{2}{525}w^3+\frac{13}{37800}w^4+\ldots$$ then to $$\boxed{ x=\sqrt{6(1-\sqrt{y})+\frac{9}{5}(1-\sqrt{y})^2+\frac{144}{175}(1-\sqrt{y})^3+\frac{78}{175}(1-\sqrt{y})^4+\ldots} }$$ for $y\in(0,1)$. The approximation given by the truncation at $(1-\sqrt{y})^4$ is fairly good:

In the diagram above, the yellow line represents the actual inverse and the blue line the fourth-order approximation obtained through residues.