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

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?

• If you take the square root, the problem reduces to finding the inverse of the $\textbf{sinc}$ function. Which there is no simple expression. See stackoverflow.com/questions/30194652/… – Andrei Oct 17 at 4:48
• @Andrei: not simple, but not unbearable either (please see below). – Jack D'Aurizio Oct 17 at 19:26

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.

• Impressive. OP should accept this answer, it is excellent – Jake Mirra Oct 18 at 14:26

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.

• Technically, if you intend $\frac{\sin^2 x}{x^2}$ as a map from $[0,\pi]$ to $[0,1]$, there is an inverse function, and it also has a reasonable algebraic expression. – Jack D'Aurizio Oct 17 at 19:36

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

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 \end{align}

When you took the square root of both sides in third step, you forgot to add on a plus and minus sign

$$\sin(y) = \pm\sqrt{xy^2}$$

Although, simplifying this expression further will not produce an explicit form of $$y^{-1}$$. It is best to graph the inverse function with a computer program such as Desmos.

Then observe from the graph above that $$f^{-1}(0)$$ will have more than one function value. Therefore, $$f^{-1}(x)$$ isn't one-to-one. So, an inverse function does not exist.

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: