Based on:
and
Use:
$$\text{ArcLength}(f(x))=\int\sqrt{f’(x)^2+1}dx\\\implies \text{InverseArcLength}(f(x))=\text{inverse function of }\text{ArcLength}(f(x))\\=f^{-1}\left(\pm\sqrt{\left(\frac{d\text{ArcLength}(f(x))}{dx}\right)^2-1}\right)$$ Inverse of the Arc Length of $ax^n$:
From the bottom reference using the Incomplete Beta Function and the quantile Inverse Beta Regularized $\text I^{-1}_z(a,b)$ functions:
$$\text{ArcLength}(ax^n)=\frac{x\sqrt[2n-2]{-a^2n^2x^{2n-2}}\text B_{-a^2n^2x^{2n-2}}\left(\frac1{2n-2},\frac32\right)}{2n-2}\mathop=^{a,x,n>0}\frac{\sqrt[2-2n]{-1}\sqrt[1-n]{an} \text B_{-a^2n^2x^{2n-2}}\left(\frac1{2n-2},\frac32\right)}{2n-2} \implies \boxed{\text{InverseArcLength}(ax^n)\mathop=^{n>1} \frac{\sqrt[2n-2]{\text I^{-1}_{\frac{\sqrt[2n-2]{-1}\sqrt[n-1]{an}(2n-2) x}{\text B\left(\frac1{2n-2},\frac32\right)}}\left(\frac1{2n-2},\frac32\right)}}{\sqrt[2n-2]{-1}\sqrt[n-1]{an}}}$$
There are a few ways to get around the domain of the function: If $n$ is a positive irreducible rational number with an even denominator and odd numerator, then, there are no complex number problems. Also, the domain for $n$ can be extended by finding the arc length of $\sqrt[n]x$ by inverting it in terms of $x$.
We cannot find the inverse of the arc length of $x^2$, unless solving for when the arc length equal a complex value, because of the limited domain of $\text I^{-1}_z(a,b)$, but the arc length of $i\,f(x);x,f(x)\in\Bbb R$ should have a similar arc length to $f(x)$. Solve:
$$\text{ArcLength}(ix^2)=\int \sqrt{1-4x^2}dx=\frac12 \sqrt{1-4x^2}+\frac14\sin^{-1}(2x)=-\frac\pi{16}$$
Plug into the boxed formula and use
Why does the Dottie Number D$=\sqrt{1-\left(2\text I^{-1}_\frac12(\frac 12,\frac 32)-1\right)^2}$?
$$x=-\frac12\sqrt{\text I^{-1}_\frac12\left(\frac12,\frac32\right)}=-\frac{\sqrt{1-\sqrt{1-\text D^2}}}{2\sqrt 2}$$
to get proof of the formula:
$$\text{ArcLength}(ix^2)\text{ from }0\text{ to } \frac{\sqrt{1-\sqrt{1-\text D^2}}}{2\sqrt 2} =\int_0^ {\frac{\sqrt{1-\sqrt{1-\text D^2}}}{2\sqrt 2}} \sqrt{1-4x^2}=-\frac\pi{16}$$
Inverse of the Arc Length of $a^x$:
Remember that Wolfram Alpha has BesselJZero $\text j_{n,k}$. Both of the following formulas have less restrictions than the above boxed formula. This formula also finds the inverse of the arc length of a logarithmic function :
$$\text{ArcLength}(a^x)=\int \sqrt{\ln^2(a)a^{2x}+1}dx=\frac{\sqrt{\ln^2(a)a^{2x}+1}-\tanh^{-1}\left(\sqrt{\ln^2(a)a^{2x}+1}\right)}{\ln(a)}\implies \boxed{\text{InverseArcLength}(a^x)=\frac{\ln\left(-\frac{\text j^2_{\frac32,\frac{i\ln(a)x}{\pi}}+1}{\ln^2(a)}\right)}{2\ln(a)}}$$
Let $a=-1,x=-2$:
$$\text{ArcLength}((-1)^x)\text{ from }0\text{ to }-\frac{i\ln\left(\frac{1+\text j^2_{\frac32,2}}{\pi^2}\right)}{2\pi}=\int_0^{-0.2890486…i}\sqrt{\pi^2(-1)^{2x+1}+1}dx=-2$$
where $\text j_{\frac32,2}$ is the second smallest solution\fixed point of $\tan(x)=x$
Using the Inverse Beta Regularized formula:
$$\boxed{\text{InverseArcLength}(a^x)=\lim_{b\to0}\frac{\ln\left(-\frac{\sqrt{\text I^{-1}_{-b\ln(a)x}\left(\frac32,\frac b2\right)-1}}{\ln(a)}\right)}{\ln(a)}}$$
Let $a=-1,x=2i:$
$$\text{ArcLength}((-1)^x)\text{ from }0\text{ to }-\lim_{b\to0}\frac{i\ln\left(\frac{i\sqrt{\text I^{-1}_{2\pi b}\left(\frac32,\frac b2\right)-1}}{\pi}\right)}{\pi}=2i$$
After simplification, we have proof for the formula.
Both formulas respectively solve $\text{ArcLength}(a^x)=iy$ or $\text{ArcLength}(a^x)=y$ with $y\in\Bbb R$
The simple question:
Are there any applications for the inverse of the arc length of a power and exponential function?
Some ideas that come to mind are:
- Arc Length Parametrization
- “Straightening” a Curve
- Having “trigonometric function” versions for exponential and power functions like the Jacobi Elliptic functions
- Anything else?
