The Wolfram Language’s Inverse Beta Regularized $\text I^{-1}_z(a,b)$ is a quantile function. This applicable yet obscure function appears in Excel as BETA.INV and a special case of it as InvT in AP Statistics. It’s special cases have not been well investigated yet, but they include elliptic, polynomial root, transcendental solution, and other functions.
The main definition of this function is:
$$\frac{\int_0^z t^{a-1} (1-t)^{b-1}dt}{\int_0^1 t^{a-1} (1-t)^{b-1}dt}=x \implies z=\text I^{-1}_x(a,b);0\le x\le 1 \text{ and }a,b>0$$
Note that there are many other “special cases” of inverse beta regularized which do not have closed forms, so they have no connections to other functions. Therefore, finding more special cases of the Incomplete Beta function will show it’s relationship to other (special) functions. Using properties of the Gauss Hypergeometric function and inverting also helps find special cases.
What are some other closed forms or named constants of special cases and limits of Inverse Beta Regularized not in the self answer?
Are there any identities or transformations for Inverse Beta Regularized?
If you found a closed form for an evaluation of inverse beta regularized mentioned in the self answer, but it has a different evaluation of $\text I^{-1}_x(a,b)$, then please answer too.
The self answer also has guiding questions which are part of the block-quoted question
Formula $1$:
$$x^p+cx+a=0\implies x=\frac{ap}{c(1-p)}\text I^{-1}_\frac{ a^{p-1}(p-1)}{c^p\left(\frac1p-1\right)^p}(-p,2)$$
with a tester of the formula.
and
$$x^r+ax+b=0\implies x=\frac{b r}{a(1-r)\text I^{-1}_\frac{b^{r-1}(r-1)}{a^r\left(\frac1r-1\right)^r}(r-1,2)}$$
If $0< r< 1$, simply substitute $x\to x^\frac 1r$. Test the formula here
In short, $$\text I^{-1}_z (n\in\Bbb N, b)$$ solves for some power function’s root which should have a closed form in terms of hypergeometric functions, the Bring Radical, Radicals, Multidimensional Theta functions, and Elliptic functions, but the relationships between these functions and Inverse Beta Regularized is complicated, so which ones are there? Also, there may be a way to use the result to find more than one solution too, but how?
Formula 2:
A quarter period of a trigonometric function:
$$\sqrt{\text I^{-1}_\frac{2x}\pi\left(\frac12,\frac12\right)}=\sin(x)\implies\text I^{-1}_x\left(\frac12,\frac12\right)=\sin^2\left(\frac\pi 2x\right)$$
Test the identity here
Formula 3:
A constant named the Dottie Number:
$$\text{Dottie Number}=\text D=\sin^{-1}\left(1-2\text I^{-1}_\frac12\left(\frac12,\frac32\right)\right)=\sqrt{1-\left(1-2 \text I^{-1}_\frac12\left(\frac12,\frac32\right)\right)^2}\implies\text I^{-1}_\frac12\left(\frac12,\frac32\right)=\frac{1-\sqrt{1-\text{D}^2}}{2}=\frac{1-\sin(\text D)}2$$
See the results here.
Formula 4:
Notice the Jacobi Amplitude $\text{am}(x,k)$ with parameter $k$ which can derive other Jacobi Elliptic function identities where $\text L_1$ is the First Lemniscate Constant:
$$\sin^{-1}\sqrt[4]{\text I^{-1}_\frac{x}{\text L_1}\left(\frac14,\frac12\right)}=\text{am}(x,-1)$$
also note Jacobi SN, Jacobi NC, and the lemniscate case half period constant $\omega$
$$\text I^{-1}_x\left(\frac14,\frac12\right)=\text{sn}^4\left(\text L_1 x,-1\right)=\text{nc}^2\left(2\omega i(1-x),\frac12\right)$$
where there is a true identity. Next, with the Jacobi CN function:
$$\text I^{-1}_x\left(\frac12,\frac14\right)=4\text{sn}^2(\text L_1 x,2) \text{cn}^2(\text L_1 x,2)$$
and
$$\sec^{-1}\sqrt[4]{\text I^{-1}_{1-\frac x{2\omega }}\left(\frac14,\frac12\right)}=\text{am}\left(ix,\frac12\right)$$
which is also true and finally,
$$\frac12\sin^{-1}\sqrt{\text I^{-1}_\frac x{\text L_1}\left(\frac12,\frac14\right)}=\text{am}(x,2)$$
which is correct
For the derivative of the Jacobi amplitude, we have Jacobi DN:
$$\sqrt{\sqrt{\text I^{-1}_\frac x{\text L_1}\left(\frac14,\frac12\right)}+1}=\text{dn}(x,-1)\implies\text I^{-1}_x\left(\frac14,\frac12\right)=\left(\text{dn}^2(\text L_1x,-1)-1\right)^2$$
which is true.
$$\sqrt[4]{1-\text I^{-1}_\frac{x}{\text L_1}\left(\frac12,\frac14\right)}=\text{dn}(x,2)\implies \text I^{-1}_x\left(\frac12,\frac14\right)=1-\text{dn}^4\left(\text L_1x,2\right)$$
which works
similarly
*simplify?:
$$\frac{\sqrt{\frac1{\sqrt{\text I^{-1}_{1-\frac x{2\omega}}\left(\frac14,\frac12\right)}}+1}}{\sqrt2}=\text{dn}\left(ix,\frac12\right)\implies\text I^{-1}_x\left(\frac12,\frac14\right)=1-\frac1{\left(2\text{dn}^2\left(2\omega ix,\frac12\right)-1\right)^2}$$
which works. Finally, Jacobi Epsilon $\varepsilon(x,k)$ with the Incomplete Beta function $\text B_z(a,b)$
$$\frac14 \text B_{\text I^{-1}_\frac x{\text L_1}\left(\frac12,\frac14\right)}\left(\frac12,\frac34\right)=\varepsilon(x,2)\implies \text B_{\text I^{-1}_x\left(\frac12,\frac14\right)}\left(\frac12,\frac34\right) =4\varepsilon(\text L_1x,2)$$
which is true. Similarly,
$$x+\frac14 \text B_{\text I^{-1}_\frac x{\text L_1}\left(\frac14,\frac12\right)}\left(\frac34,\frac12\right)=\varepsilon(x,-1)\implies\text B_{\text I^{-1}_x\left(\frac14,\frac12\right)}\left(\frac34,\frac12\right)=4\varepsilon(\text L_1x,-1)-4\text L_1x $$
which is also true
finally, @Bertrand87’s answer and the ubiquitous constant=U:
*simplify:
$$\frac{\text L_1}{\sqrt2}+\frac{\text U}{2}-\frac{\text B_{\frac1{\text I^{-1}_{1-\frac{\sqrt2x}{\text L}}\left(\frac14,\frac12\right)}}\left(\frac12,\frac14\right)+ \text B_{\frac1{\text I^{-1}_{1-\frac{\sqrt2x}{\text L}}\left(\frac14,\frac12\right)}}\left(\frac12,\frac34\right)}{4\sqrt 2}=\varepsilon\left(i x,\frac12\right)\implies$$
Compare the value of this value with EllipticE(am(i,1/2),1/2)
for numerical proof
Formula 5:
Please note that sections for Weierstrass Zeta and Sigma, defined later, can also use these transformations
The Weierstrass $\wp(x;a,b)$ function has this first basic formula
$$\frac{\sqrt a}{2\sqrt{\text I^{-1}_{-\frac{\sqrt[4]a x}{2\omega}}\left(\frac14,\frac12\right)}}=\wp(x;{a,0})\implies\text I^{-1}_x\left(\frac14,\frac12\right)=\frac a{4\wp^2\left(\frac{2\omega x}{\sqrt[4]a};a,0\right)}$$
which works. There is also the Weierstrass $\wp’(x;a,b)$:
$$\frac{a^\frac34\left(1-2\text I^{-1}_{-\frac{\sqrt[4]a x}{2\omega}}\left(\frac14,\frac14\right)\right)}{\sqrt 2\text I^{-1}_{-\frac{\sqrt[4]a x}{2\omega}}\left(\frac14,\frac12\right)^\frac34}=\wp’(x,{a,0})\implies\frac{\left(\text I^{-1}_x\left(\frac14,\frac12\right)-1\right)^2}{\text I^{-1}_x\left(\frac14,\frac12\right)^3}=\frac4{a^3}\wp’^4\left(\frac{2\omega x}{\sqrt[4]a};a,0\right)$$
which is correct
using Weierstrass Zeta $\zeta(x;a,b)$, the Incomplete Elliptic Integral of the Second Kind $\text E(x,k)$, the Second Lemniscate Constant$=\text L_2$, and the Ubiquitous Constant=U:
$$-\frac{\sqrt[4]a}{\sqrt2}\left(\text E\left(\frac12\cos^{-1}\sqrt {\text I^{-1}_{-\frac{\sqrt[4]ax}{2\omega}}\left(\frac14,\frac12\right)},2\right) +\frac{2\text I^{-1}_{-\frac{\sqrt[4]ax}{4\omega}}\left(\frac14,\frac14\right)-1}{\sqrt[4]{\text I^{-1}_{-\frac{\sqrt[4]ax}{2\omega}}\left(\frac14,\frac12\right)}}\right)-\frac{\text U}2=\zeta(x;a,0)\implies \text E\left(\frac12\cos^{-1}\sqrt {\text I^{-1}_x\left(\frac14,\frac12\right)},2\right) +\frac{\sqrt{1-\text I^{-1}_x\left(\frac14,\frac12\right)}}{\sqrt[4]{\text I^{-1}_x\left(\frac14,\frac12\right)}}=\frac1{\sqrt[4]a}\left(\sqrt2\zeta\left(\frac{2\omega x}{\sqrt[4]a};a,0\right)- \text L_2 \right)$$
which works with this identity
with Weierstrass Sigma $\sigma(x;a,b)$ and The $\,_3\text F_2$ Hypergeometric function, Inverse Jacobi NS, and the complex conjugate:
$$\frac{\sqrt2}{\sqrt[4]a}\exp\left({\frac{\,_3\text F_2\left(1,1,\frac54;\frac74,2;\frac1{\text I^{-1}_\frac{\sqrt[4]ax}{2\omega}\left(\frac14,\frac12\right)}\right)}{12 \text I^{-1}_\frac{\sqrt[4]ax}{2\omega}\left(\frac14,\frac12\right)}-\text L_2\overline{\text{nc}^{-1}\left(\sqrt[4]{\text I^{-1}_\frac{\sqrt[4]ax}{2\omega}\left(\frac14,\frac12\right)},-1\right)}+\frac{\sqrt[4]a\text U}2x}\right)=\sigma(x;a,0)\implies $$
which comes from this correct result. Does $\,_3\text F_2\left(1,1,\frac54;\frac74,2;x\right)$ have a form in terms of simpler functions?
There is also a formula including $\text I^{-1}_x\left(\frac16,\frac12\right)$:
$$\implies$$
…
The last set of Weierstrass elliptic function formulas use $\text I^{-1}_x\left(\frac13,\frac12\right)$ and the second omega constant $\omega_2$:
$$\implies$$
…
Are there any formulas for the Weierstrass Utility functions?
Formula 6:
These identities do not give a special case of inverse beta regularized, but they generate many more special cases of it when applied to other formulas in this answer:
$$\text I^{-1}_x(a,b)=1-\text I^{-1}_{1-x}(b,a)\implies\text I^{-1}_x(b,a)=1-\text I^{-1}_{1-x}(a,b)$$
also, using this identity, one easily finds the following formulas where the $(a,a)$ hand side extends the domain of the $\left(a,\frac12\right)$ case:
$$\text I^{-1}_x(a,a)=\frac12-\frac12\sqrt{1-\text I^{-1}_{2x}\left(a,\frac12\right)}\implies\text I^{-1}_x\left(a,\frac12\right)=1-\left(2\text I^{-1}_\frac x2(a,a)-1\right)^2$$
therefore, we have some identities with InvT, from the question
Formula 7:
Remember to appropriately transform Inverse Beta Regularized if you want another period of a periodic function like in formulas $2,4,5$. Another way to extend the function is using these different formulas using
$$\implies$$
similarly,
$$\implies$$
which can be used to find this result among many others.
Formula 8:
The first limit of $\text I^{-1}_x(a,b)$ with a closed form is
$$1-\lim_{a\to0}\text I^{-1}_{-ax}(1,a),\lim_{a\to0}\frac1{\text I^{-1}_{1-ax}(a,1)}=e^x\implies\lim_{a\to0}\text I^{-1}_{ax}(1,a)=1-e^{-x}$$
which is true
similarly with the Lambert W function
$$-\lim_{a\to0}\text I^{-1}_{a\ln(-x)+a+1}(a,2)=\text W(x)\implies\lim_{a\to0}\text I^{-1}_{ax}(a,2)=\text W\left(-e^{-x-1}\right)$$
which work. Next,
$$2\lim_{a\to0}\text I^{-1}_{ax+\frac12}(a,a)-1=\tanh(x)\implies\lim_{a\to0}\text I^{-1}_{ax}\left(\frac12,a\right)=\tanh^2\left(\frac x2\right)$$
which are correct
Formula 9:
Inverse beta regularized generalizes Inverse Gamma Regularized $Q^{-1}(a,x)$ from this limit:
$$\lim_{b\to\infty}b\,\text I^{-1}_x(a,b)=Q^{-1}(a,1-x)\implies\lim_{b\to\infty}b\,\text I^{-1}_x(b,a)=1-Q^{-1}(a,x)$$
which is correct
Formula 10:
One special case of $Q^{-1}(a,x)$, and therefore a special case of $\text I^{-1}_x(a,b)$, is
$$\mp Q^{-1}\left(1,x^{\pm 1}\right)=\ln(x)$$
where one of each sign is taken which works. Next with the $-1$st branch of the Lambert W function,
$$-Q^{-1}(2,-ex)-1=\text W_{-1}(x)\implies Q^{-1}(2,x)=-\text W_{-1}\left(-\frac xe\right)-1$$
which is true. Also with the Inverse Error function,
$$\lim_{a\to\infty}\sqrt a\left(2\text I^{-1}_\frac{x+1}2(a,a)-1\right)=\text{erf}^{-1}(x)$$
which works and with inverse erfc,
$$Q^{-1}\left(\frac12,x\right)=\text{erfc}^{-1}(x)^2,0<x\le 1$$
which works and one of each sign is taken.
Formula 11:
While possibly not closed forms, maybe these unique limits of Inverse Gamma Regularized are named inverting the Exponential Integral function $\text{Ei}(x)$ and Logarithmic Integral function $\text{li}(x)$ for the specified branch:
$$y=-\lim_{a\to0}Q^{-1}(a,-ax)\implies \text{Ei}(y)=x,y<0$$
and
$$y=\lim_{a\to0}e^{-Q^{-1}(a,-ax)}\implies \text{li}(y)=x,0\le y<1 $$
therefore,
$$y=\lim_{a\to0}Q^{-1}(a,ax)\implies -\text{Ei}(-y)=x,y>0$$
which work. Are there any other closed form particular cases or identities for Inverse Gamma Regularized?