Invertibility of specific function

analysiscalculusinverse function

This is my first post. I'm not a mathematician, just an electronics engineer who loves mathematics. In one of my projects, I arrived at the following function:

$$V\left(\varphi\right)=\frac{A\sqrt{\pi-\varphi+\sin{\varphi\cos{\varphi}}}}{\sqrt{2\pi}}$$

The project requires $V\left(\varphi\right)$ to be inverted, to obtain angle $\varphi$ (unknown), from a voltage $V$ (known). $V\left(\varphi\right)$ is continuous and strictly monotonic (descending), so an inverse mapping should exist. I tried to invert it symbolically, but couldn't arrive at a closed-form solution for $\varphi\left(V\right)$. By "closed-form solution" I mean a formula that takes me directly (algebraically) from $V$ to $\varphi$. I ended up using MATLAB to compute it numerically, and the project was successfully completed. That was 11 years ago (in 2009).

Recently, that inversion problem resurfaced. Out of pure curiosity, I asked my cousin (a mathematician) to attempt to symbolically invert the above function, but he also couldn't do it, and couldn't even give me an answer as to the existence of such solution. So, my questions are the following:

Does a closed-form expression for $\varphi\left(V\right)$ exist?
If the answer to (1) is YES, can someone provide that function, or point me to a method for deriving it?
If the answer to (1) is NO, what is the formal reason for it? Is there a way to show/prove that such solution does not exist?

This is not a homework question, and the associated practical problem has already been solved numerically. This post was made out of pure curiosity about the invertibility of functions of the form of $V\left(\varphi\right)$. Many thanks to all for your replies.

Adding some graphics, in order to better illustrate the problem.

The following figure shows how the function we're looking for, $\varphi\left(V\right)$, looks like. Notice that there appears to be no symmetry in this function. The values of $\varphi$ lie in the interval $[0,\pi]$, while the values of $V\left(\varphi\right)$ lie in the interval $[0,\frac {A} {\sqrt2}]$.

Based on the comment by @user2661923, the problem can be reduced to the inversion of function $k=x-\sin(x)$. In this case, the values of both $x$ and $k$ lie in the interval $[0,2\pi]$. The following figure shows a plot of the inverse of $k=x-\sin(x)$, together with a plot of the function itself (dashed line). Plotting them both on the same graph is useful, since they both have the same range for their independent and dependent variables. It can be observed that now there is a clear symmetry of this function at the point $(\pi,\pi)$, thanks to the removal of the square root term. This means that we only need to deal with the interval $[0,\pi]$, and use symmetry on that result to obtain the other half ($[\pi,2\pi]$).

Best Answer

Using @user2661923's comments, we end with $$k=x-\sin(x)$$

The first thing we could do is a Taylor expansion around $x=0$ ( to get $$k=\frac{x^3}{6}-\frac{x^5}{120}+\frac{x^7}{5040}-\frac{x^9}{362880}+\frac{x^{11}}{39 916800}-\frac{x^{13}}{6227020800}+O\left(x^{15}\right)$$ which is very accurate.

Then a series reversion leads to
$$x=t+\frac{t^3}{60}+\frac{t^5}{1400}+\frac{t^7}{25200}+\frac{43 t^9}{17248000}+O\left(t^{11}\right)$$ where $t=\sqrt[3]{6k}$.

Edit

Doing the same around $x=\frac \pi 2$ , we should get $$x=\frac{\pi }{2}+t-\frac{t^2}{2}+\frac{t^3}{2}-\frac{7 t^4}{12}+\frac{3 t^5}{4}-\frac{46 t^6}{45}+\frac{29 t^7}{20}-\frac{21341 t^8}{10080}+O\left(t^{9}\right)$$ where $t=k+1-\frac \pi 2$.

Doing the same around $x=\pi $ , we should get $$x=\pi +t+\frac{t^3}{12}+\frac{t^5}{60}+\frac{43 t^7}{10080}+\frac{223 t^9}{181440}+\frac{60623 t^{11}}{159667200}+\frac{764783 t^{13}}{6227020800}+O\left(t^{15}\right)$$ where $t=\frac{k-\pi}2$.

Related Solutions

Inverse of $y = {x}/{(1+x)^{1+ \omega}}$ with $x, \omega >0$

Let $$z=f(x)=\frac{x}{(1+x)^{1+\omega}}$$ and $$x=g(z)\equiv f^{-1}(z)=\sum^\infty_{n=1}g_n\frac{z^n}{n!}$$

Easily, by Lagrange inversion theorem about $z=f(a=0)$, $$g_n=(n+n\omega)^{\underline{n-1}}$$

Therefore, $$f^{-1}(z)=\sum^\infty_{n=1}\frac{(n+n\omega)^{\underline{n-1}} }{(n-1)!}\frac{z^n}n=\sum^\infty_{n=1}\binom{n+n\omega}{n-1}\frac{z^n}n$$

I doubt if you can get anything neater than this.

Solve functional equation $ h(y)+h^{-1}(y)=2y+y^2 $

Unfortunately, your proposed functional equation: $$ h(y)+h^{-1}(y)=2y+y^2 \tag{1} $$ has no differentiable solutions near $\,y=0.\,$ Suppose that $$ h(y) = 0 + a_1 y + O(y^2). \tag{2} $$ Then its inverse function is $$ h^{-1}(y) = 0 + \frac1{a_1} y + O(y^2). \tag{3} $$ Substituting equations $(2)$ and $(3)$ into $(1)$ gives $$ h(y)\!+\!h^{-1}(y) = \frac{a_1^2+1}{a_1}y \!+\! O(y^2) = 2y \!+\! O(y^2). \tag{4}$$ The only solutions for $\,a_1\,$ are $\,a_1=1\,$ and $\,a_1=-1.\,$ However, we are given that $$ 0<h(x)<x \tag{5}$$ and thus the only possible value is $\,a_1=1.\,$ Suppose we want more terms in the power series $$ h(y) = 0 + y + a_2y^2 + O(y^3). \tag{6}$$ The inverse function is now $$ h^{-1}(y) = 0 + y - a_2y^2 + O(y^3). \tag{7}$$ Adding these two equations together is the equation $$ h(y) + h^{-1}(y) = 0 + 2y + 0y^2 + O(y^3). \tag{8}$$ This implies that equation $(1)$ can not be satisfied.

There is a possibility that there exists an exponent $\,e\,$ not an integer such that $\, h(y) = 0 + y + c y^e + \cdots \,$ which perhaps should be studied according to comments.

In fact, define $$ g(x) : = \sqrt{h(x^2)}. \tag{9}$$ Then solving for a power series expansion gives $$ g(x) = x - \frac{x^2}{\sqrt{6}} + \frac{x^3}6 + O(x^4) \tag{10}$$ which implies that $$ h(x^2) = (x^2+x^4/2)+\frac1{\sqrt{6}}f(x) \tag{11} $$ where $$ f(x) \!=\! -2 x^3 \!-\!\frac{11x^5}{24} \!+\! \frac{117x^7}{1280} \!-\!\frac{5491x^9}{110592} \!+\! \frac{156538363x^{11}}{3715891200} \!+\! O(x)^{13}. \tag{12}$$

A Wolfram Language code to calculate $\,g(x)\,$ is

ClearAll[x, g, gx];
gx[3] = x - 1/Sqrt[6]*x^2 + O[x]^3;
Do[g = Normal[gx[n]] + O[x]^(2+n); gx[n+1] = Simplify[
    g + (x^2 + (Normal[g]/.x -> g)^2 - g^4 - 2*g^2) * 3 /
    ((4+n)*x^2*Sqrt[6])], 
  {n, 3, 6}]

As some comments suggest, it seems that the power series for $\,f(x)\,$ has zero radius of convergence. This makes finding properties of it difficult. Perhaps we need to let $\,x\,$ approach $\,\infty.$ In this case we find that $\,h(x) \approx \sqrt{x}\,$ with infinite number of other terms. Using equation $(1)$ we can find the expansion $$ h(x) = x^{1/2} - 1 + \frac12 x^{-1/4} + \frac1x s(x) \tag{13} $$ where $$ s(x) := \sum_{k=2}^\infty 2^{-k}(x^{-\frac32 2^{-k}} - x^{-2^{-k}} ). \tag{14} $$ This gives moderately good approximations as $\,x\,$ gets large and down to $1.$

One method that leads to equations $(13)$ and $(14)$ is as follows. From equation $(1)$ we immediately get $$ h^{-1}(y) = y^2 + 2y - h(y) \tag{15} $$ and if $\,y=h(x),\,$ then $$ x = h(x)^2 + 2h(x) - h(h(x)). \tag{16} $$ We start with an approximation and try to find what additional term will satisfy equation $(16)$. So we guess $$ h(x) = x^{1/2} + cx^e + \cdots \tag{17} $$ where $\,\cdots\,$ denotes terms with smaller exponents. Substitute equation $(17)$ in equation $(16)$ to get $$ x \!=\! (x \!+\! 2cx^{1/2+e} \!+\! \cdots) \!+\! 2x^{1/2} \!+\! \cdots \!=\! x \!+\! 2x^{1/2}(cx^e \!+\! 1) + \cdots. \tag{18} $$ This implies $\,0 = cx^e + 1.\,$ Solving this we get our next guess as $$ \,h(x) = x^{1/2} - 1 + cx^e + \cdots. \tag{19} $$ Repeating this process leads to equations $(13)$ and $(14)$.

The series in $(14)$ appears to converge but I have no proof, only numerical evidence. I also have no proof that $(13)$ satisfies the functional equation $(1)$.

An answer to this question by 'Semiclasical' contains sequence recursions $$ u_n = u_{n-1} - j_{n-1} \quad \text{ and } \quad j_n = j_{n-1} - u_n^2. \tag{20} $$ with the property that $\, u_n = h(u_{n-1})\,$ and that $\,u_n \to 0^+.\,$ I found that for suitable starting values of $\,u_1\,$ and $\,j_1\,$ we have $$ u_n = 6n^{-2} - \frac{15}2n^{-4} + \frac{663}{40} n^{-6} - \frac{43647}{800}n^{-8} + \cdots. \tag{21} $$

An answer to this question by 'Cesareo' uses recursion to construct a sequence of functions $\,h_n(x)\,$ which seems to converge to a global solution. It would be nice to give a proof of this.

Best Answer

Related Solutions

Inverse of $y = {x}/{(1+x)^{1+ \omega}}$ with $x, \omega >0$

Solve functional equation $ h(y)+h^{-1}(y)=2y+y^2 $

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