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 the Inverse T distribution function InvT(x=area,d=degrees of freedom) 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 an alternate form or simplification of an identity in the self answer then please answer too.
The self answer also has guiding questions which are part of the block-quoted question for other possible identities
Section $1$:
$$x^r+ax+b=0\implies x=\frac{br}{a(1-r)}\text I^{-1}_\frac{ b^{r-1}(r-1)}{a^r\left(\frac1r-1\right)^r}(-r,2)$$
with a tester of the formula.
and
$$x^r+ax+b=0\implies x=\frac{b r}{a(1-r)\text I^{-1}_\frac{b^{r-1}(r-1)}{a^r\left(\frac1r-1\right)^r}(r-1,2)}$$
Test the formula here. Also,
$$x^r+ax+b=0\implies x=\sqrt[r]\frac b{(r-1)\text I^{-1}_{-\frac{a\left(\frac{r-1}b\right)^{1-\frac1r}}r}\left(\frac1r-1,2\right)}$$
Test here. All $3$ formulas account for any real $r$. 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, 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 root too, but how?
Section 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
Section 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.
Section 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 proofSection 5: Please note that sections for Weierstrass Zeta and Sigma, defined later, can also use these transformations. There are multiple formulas for completion and understanding compositions of functions on Inverse Beta Regularized
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:
\begin{align}-\frac{\sqrt[4]a}{4\sqrt 2}\text B_{\text I^{-1}_{\frac{\sqrt[4]ax}{2\omega}}\left(\frac14,\frac12\right)}\left(-\frac14,\frac12\right)=-\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 B_{\text I^{-1}_x\left(\frac14,\frac12\right)}\left(-\frac14,\frac12\right)=-\frac{4\sqrt2}{\sqrt[4]a}\zeta\left(\frac{2\omega x}{\sqrt[4]a};a,0\right),\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)\end{align}
which works with this identity. Also with Weierstrass Sigma $\sigma(x;a,b)$, MeijerG, The $\,_3\text F_2$ Hypergeometric function, Inverse Jacobi NS, Beta Regularized $\text I_x(a,b)$, and the complex conjugate:
\begin{align}= \frac{\sqrt2}{\sqrt[4]a}\exp\left(\frac{\text G_{3,3}^{3,2}\left(^{0,\frac34,1}_{0,0,\frac14}\bigg|\text I^{-1}_\frac{\sqrt[4]ax}{2\omega}\left(\frac14,\frac12\right)\right)}{4\Gamma^2\left(\frac14\right)}+\frac{\sqrt[4]a\text U}2x\right) = \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)}-\frac\pi4\text I_\frac1{\text I^{-1}_\frac{\sqrt[4]ax}{2\omega}\left(\frac14,\frac12\right)}\left(\frac14,\frac12\right)+\frac{\sqrt[4]a\text U}2x}\right) =\frac{\sqrt2}{\sqrt[4]a}\exp\left(\lim_{b\to0}{\frac{\,_2\text F_1\left(b,\frac14;\frac34;\frac1{\text I^{-1}_\frac{\sqrt[4]ax}{2\omega}\left(\frac14,\frac12\right)}\right)-1}{4a}-\text L_2\overline{\text{ns}^{-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 \text G_{3,3}^{3,2}\left(^{0,\frac34,1}_{0,0,\frac14}\bigg|\text I^{-1}_x\left(\frac14,\frac12\right)\right)= 4\Gamma^2\left(\frac14\right)\left[\ln\left(\frac{\sqrt[4]a}{\sqrt2}\sigma\left(\frac{2\omega x}{\sqrt[4]a};a,0\right)\right)-\frac{\sqrt[4]a\text U}2\cdot \frac{2\omega x}{\sqrt[4]a}\right],,\end{align}
which is correct with this identity. Does $\lim_{b\to0}\frac{\,_2\text F_1\left(b,\frac14;\frac34;x\right)-1}{4a} $ have a form in terms of simpler functions?
There is also a formula including $\text I^{-1}_x\left(\frac16,\frac12\right)$ and the second omega constant $\omega_2$:
$$\frac{\sqrt[3]b}{2^\frac23 \sqrt[3]{\text I^{-1}_\frac{\sqrt[6]ax}{\omega_2}\left(\frac16,\frac12\right)}}=\wp(x;0,b)\implies\text I^{-1}_x\left(\frac16,\frac12\right)=\frac b{4\wp^3\left(\frac{\omega_2x}{\sqrt[6]b};0,b\right)}$$ which is true
and
$$-\sqrt{\frac b{{\text I^{-1}_\frac{\sqrt[6]bx}{\omega_2}\left(\frac16,\frac12\right)}}-1}=\wp’(x;0,b)\implies \text I^{-1}_x\left(\frac16,\frac12\right)=\frac b{\wp’^2\left(\frac{\omega_2x}{\sqrt[6]b};0,b\right)+b}$$
which works
these formulas should have a form with Inverse Weierstrass P $\wp^{-1}(x;a,b)$:
$$-\frac{\sqrt[6]b\text B_{\text I^{-1}_\frac{\sqrt[6]bx}{\omega_2}\left(\frac16,\frac12\right)}\left(-\frac16,\frac12\right)}{6\sqrt[3]2}=\zeta(z;0,b)\implies \text B_{\text I^{-1}_x\left(\frac16,\frac12\right)}\left(-\frac16,\frac12\right)=-\frac{6\sqrt[3]2}{\sqrt[6]b}\zeta\left(\frac{\omega_2 x}{\sqrt[6]b};0,b\right)$$
which is true
finally with the first omega constant$=\omega_1$ * $$\frac1{\sqrt[6]b}\exp\left(\frac{\pi \sqrt[6]bx}{4\text{Im}(\omega_1)}-\frac{\text G_{3,3}^{3,2}\left(^{0,\frac23,1}_{0,0,\frac16}\bigg|\text I^{-1}_\frac{\sqrt[6]bx}{\omega_2}\left(\frac16,\frac12\right)\right)}{24 \sqrt\pi2^\frac23 \text{Im}(\omega_1) }-\frac\pi{6\sqrt3}\right) =\sigma(x;0,b)\implies ,,$$
which works
The last set of Weierstrass elliptic function formulas use $\text I^{-1}_x\left(\frac13,\frac12\right)$. To avoid complex numbers, $-B$ is used in a weierstrass function’s parameter:
$$-\frac{\sqrt[3]{B\text I^{-1}_{\frac{3\sqrt[6]Bx}{2\text{Im}(\omega_1)}-2}\left(\frac13,\frac12\right)}}{2^\frac23}=\wp(x;0,-B)\implies\text I^{-1}_x\left(\frac13,\frac12\right)=-\frac4B\wp^3\left(\frac{2\text{Im}(\omega_1)(x+2)}{3\sqrt[6]B};0,-B\right)$$
which is true
also,
$$\sqrt B\left(2\text I^{-1}_{\frac{3\sqrt[6]Bx}{4\text{Im}(\omega_1)}-1}\left(\frac13,\frac13\right)-1\right)=\wp’(x;0,-B)\implies=\text I^{-1}_x\left(\frac13,\frac13\right)=\frac{\wp’\left(\frac{4\text{Im}(\omega_1)(x+1)}{3\sqrt[6]B};0,-B\right)}{2\sqrt B}+\frac12$$
which works
with Baxter’s Four-Coloring Constant=B,
$$\sqrt[6]B\left(\frac{\text B_{\text I^{-1}_{\frac{3\sqrt[6]Bx}{2\text{Im}(\omega_1)}-2}\left(\frac13,\frac12\right)}\left(\frac23,\frac12\right)}{6\sqrt[3]2}+\frac1{\text B}\right)=\zeta(x;0,-B)\implies \text B_{\text I^{-1}_x\left(\frac13,\frac12\right)}\left(\frac23,\frac12\right)=6\sqrt[3]2\left(\frac{\zeta\left(\frac{2\text{Im}(\omega_1)(x+2)}{3\sqrt[6]B};0,-B\right)}{\sqrt[6]B}-\frac1{\text B}\right)$$
which is correct
\begin{align}\implies\end{align}
Are there any formulas for the Weierstrass utility functions or modular forms in terms of Inverse Beta Regularized?
Section 6:
Although possibly not a special case of $\text I^{-1}_x (a,b)$, its definition shows that it gives a symmetric part of a period of a hyperelliptic function. What is inverse beta regularized in terms of Riemann Theta function? For more information, please see this question.
Section 7:
This identity does not give a special case of inverse beta regularized, but it generates many more special cases when applied to other formulas in this answer:
$$\text I^{-1}_x(a,b)=1-\text I^{-1}_{1-x}(b,a)$$
InvT, from the question, represents most special cases of inverse beta regularized in this answer with the sign function
$$\sqrt d\text{sgn}\left(x-\frac12\right)\sqrt{\frac1{1-\left(2\text I^{-1}_x\left(\frac d2,\frac d2\right)-1\right)^2}-1}=\text{InvT}(x,d)$$
which works . Therefore, with this identity, we have these conversion formulas and identities applicable to other formulas in this answer
$$\text I^{-1}_x(a,a)=\frac12-\frac12\sqrt{1-\text I^{-1}_{2x}\left(a,\frac12\right)}=\frac12\left(\frac{\text{sgn}\left(x-\frac12\right)}{\sqrt{\frac{2a}{\text{InvT}(x,2a)}+1}}+1\right)\implies\text I^{-1}_x\left(a,\frac12\right)=1-\left(2\text I^{-1}_\frac x2(a,a)-1\right)^2=1-\frac1{\frac{2a}{\text{InvT}^2\left(\frac x2,2a\right)}+1}$$
which is true
Section 8:
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
$$$$
similarly,
$$$$
which can be used to find this result among many others.
$$$$
and
$$$$ Section 9:
The first limit of $\text I^{-1}_x(a,b)$ with a closed form is
$$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)$$
which are correct
Section 10:
Inverse beta regularized generalizes Inverse Gamma Regularized $Q^{-1}(a,x)$ from this limit:
$$\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)$$
which is correct
Section 11:
One special case of $Q^{-1}(a,x)$, and therefore a special case of $\text I^{-1}_x(a,b)$, is
$$\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$$
which works and one of each sign is taken.
Section 12:
While possibly not closed forms, maybe these unique limits of Inverse Gamma Regularized are named inverting the Exponential Integral function $\text{Ei}(x)$ and Logarithmic Integral function $\text{li}(x)$ for the specified branch:
$$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?