Inverse Arc Length of a Parabola

Question

Is the inverse for the arc length of a parabola (say, $f(x)=\dfrac{x^2}{2}$) not discovered, or not possible to express given elementary functions and product log ($W(x)$)? If the latter is so, is there a proof?

I ask because Wolfram Alpha offers not an expression, even using the product log, of the inverse arc length $L$ of a parabola of the simplest form: the inverse of $$ L=\frac{x\sqrt{1+x^2}}2+\frac{\ln(x+\sqrt{1+x^2})}2.$$

I've also seen other questions of a similar nature, but none address why there is no inverse; is it a lack of an ingenious way to find it or a lack of possibility (or a lack of an ingenious proof of the latter)

Answer 1

The original equation satisfies if $L>0$ and $x$ is a solution then for $L<0$, $-x$ is the solution. Therefore we will only consider $L>0$.

First, as

\begin{equation} \log\left(x^{2}+\sqrt{1+x^{2}}\right)=\textrm{arcsinh}\left(x\right) \end{equation}

the equation is equivalent to

\begin{equation} 2L=x\sqrt{1+x^{2}}+\textrm{arcsinh}\left(x\right) \end{equation}

Transforming the equation with the change of variable $x\rightarrow\sqrt{1-1/z}$ and multiplying by z we obtain the equation

\begin{equation} -2Lz+z\sqrt{1-\frac{1}{z}}\sqrt{2-\frac{1}{z}}+z\cdot\textrm{arcsinh}\left(\sqrt{1-\frac{1}{z}}\right)=0 \end{equation}

Noting that the solution of this equation is positive we can simplify this to expression

\begin{equation} -2Lz+\sqrt{z-1}\sqrt{2z-1}+z\cdot\textrm{arcsinh}\left(\sqrt{1-\frac{1}{z}}\right)=0 \end{equation}

Defining the function $f(z)=-2Lz+\sqrt{z-1}\sqrt{2z-1}+z\cdot\textrm{arcsinh}\left(\sqrt{1-\frac{1}{z}}\right)$ and applying the Burniston-Siewert method to solve transcendental equations to this equation we obtain

\begin{equation} x=\sqrt{1-\frac{1}{z_{0}}},\quad z_{0}=\frac{\sqrt{2}}{\sqrt{2}+\textrm{arcsinh}\left(1\right)-2L}+m \end{equation}

where

\begin{equation} m={\displaystyle \frac{1}{2\pi}\int\limits _{0}^{1}\theta_{1}(t)+\theta_{2}(t)\,dt} \end{equation}

with

\begin{equation} \theta_{1}(t)=\arctan\left(\frac{\pi t}{4\sqrt{\frac{2-t}{2}}\sqrt{1-t}+2t\left(2L-\textrm{arcsinh}\left(\sqrt{\frac{2-2t}{t}}\right)\right)}\right) \end{equation}

\begin{equation} \theta_{2}(t)=\arctan\left(\frac{2\sqrt{t}\sqrt{\frac{1-t}{2}}+\left(t+1\right)\textrm{arcsin}\left(\sqrt{\frac{1-t}{1+t}}\right)}{2L(t+1)}\right) \end{equation}

Putting everything into a single expression, the inverse for the arc length of a parabola (let's say $L(x)$) is

\begin{equation} L^{-1}(x)=\left[1-\left[\frac{\sqrt{2}}{\sqrt{2}+\textrm{arcsinh}\left(1\right)-2x}+\frac{1}{2\pi}\int\limits _{0}^{1}\theta_{1}(t,x)+\theta_{2}(t,x)\,dt\right]^{-1}\right]^{1/2} \end{equation}

with

\begin{equation} \theta_{1}(t,x)=\arctan\left(\frac{\pi t}{4\sqrt{\frac{2-t}{2}}\sqrt{1-t}+2t\left(2x-\textrm{arcsinh}\left(\sqrt{\frac{2-2t}{t}}\right)\right)}\right) \end{equation}

\begin{equation} \theta_{2}(t,x)=\arctan\left(\frac{2\sqrt{t}\sqrt{\frac{1-t}{2}}+\left(t+1\right)\textrm{arcsin}\left(\sqrt{\frac{1-t}{1+t}}\right)}{2x(t+1)}\right) \end{equation}

or, completely

an inverse graph

Again, this inverse is for $x>0$, for $x<0$ the inverse will be $-L^{-1}(x)$

Ref:

-E. E. Burniston, C.E. Siewert. The use of Riemann problems in solving a class of transcendental equations. Mathematical Proceedings of the Cambridge Philosophical Society. 73. 111 - 118. (1973).

-Henrici P., Applied and computational complex analysis, Volume III. Wiley, New York. 183-191 (1986).

Answer 2

You want to solve for $x$ the equation $$2L={x\sqrt{1+x^2}}+{\log(x+\sqrt{1+x^2})}$$ If $L$ is large, so $x$ and we have the approximation $$2L=x^2+\frac{1}{2} ( \log (x^2)+1+2 \log(2))+O\left(\frac{1}{x^2}\right)$$ which gives the estimate $$x_0=\sqrt{\frac 12 W\left(\frac{1}{2} e^{4 L-1}\right) }$$ where $W(.)$ is the pricipal branch of Lambert function.

Consider now the function $$f(x)={x\sqrt{1+x^2}}+{\log(x+\sqrt{1+x^2})}-2L$$ and its derivatives.

$$f'(x)=2 \sqrt{x^2+1} \qquad \text{and} \qquad f''(x)=\frac{2 x}{\sqrt{x^2+1}}$$

We have $$f(x_0) >0 \qquad \text{and} \qquad f''(x_0) >0$$

So, by Darboux theorem, we know that $x_0$ is an overestimate of the solution which will be reached without any overshoot if we invoke Newton method.

Now, using the first estimate of a Newton-like method of order $n$, we have an explicit formula for $x_1^{(n)}$.

More than likely $$x_1^{(2)}=x_0-\frac {f(x_0)}{f'(x_0)}$$ would be sufficient for pratical problems.

Answer 3

$$L=\frac{x\sqrt{1+x^2}}{2}+\frac{\ln(x+\sqrt{1+x^2})}{2}$$

We see, this equation is a polynomial equation of more than one algebraically independent monomials ($x\sqrt{1+x^2},\ln(x+\sqrt{1+x^2})$) and with no univariate factor. We therefore don't know how to solve the equation for $x$ by rearranging by applying only finite numbers of elementary functions (elementary operations) we can read from the equation.

For applying Lambert W, we would have to transform the equation into the form $$f(x)e^{f(x)}=\text{constant},\tag{1}$$ but we get

$x\to \frac{1}{2}\frac{-1+e^t-1}{e^\frac{t}{2}}$: $$\left(e^t\right)^4+(-64L^2+64Lt-16t^2-2)\left(e^t\right)^2=-1.$$ $$\left(\left(e^t\right)^2+(8L-4t)e^t-1\right)\left(-\left(e^t\right)^2+(8L-4t)e^t+1\right)=0$$ $$\left(e^t\right)^2+(8L-4t)e^t-1=0\ \ \ \ \ \ \ \ \ \ \ \ \left(e^t\right)^2-(8L-4t)e^t-1=0\tag{2}$$

We see, for algebraic $L$, both equations are irreducible algebraic equations of both $t$ and $e^t$. Lin 1983 states that such equations don't have solutions that are an elementary number.

We don't see how we could transform the equations in (2) into the form of equation (1). We cannot apply Lambert W therefore.
To get a prove, we would have to prove that no corresponding transformation exists.

It seems we also cannot apply generalised Lambert W.

To solve the equation in closed form, we therefore have no choice but to try other Special functions.

Maybe the equation can be solved in terms of Lambert-Tsallis function.