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Why does the Dottie number $=\sqrt{1-\left(2\text I^{-1}_\frac12\left(\frac12,\frac32\right)\right)^2}$?

**Introduction:**

For some background information on the [**Dottie Number D**](https://en.m.wikipedia.org/wiki/Dottie_number), 

Cross posted from: 
>[Why does the Dottie Number equal a median of a Beta distribution?](https://math.stackexchange.com/questions/4389528/why-dottie-2-sqrti-1-frac12-frac-12-frac-32-i-1-frac12-frac-12-fr)

since the original question is unanswered.

Some definitions:

The “solution” to [Kepler’s  equation](https://en.m.wikipedia.org/wiki/Kepler%27s_equation) is [Kepler E](https://resources.wolframcloud.com/FunctionRepository/resources/KeplerE/):

$$M=E-e\sin(E)\iff x=y-a\sin(y)\implies y=\text E(a,x)$$

and [**Inverse Beta Regularized function $\text I^{-1}_z(a,b)$**](https://reference.wolfram.com/language/ref/InverseBetaRegularized.html), [Beta Regularized function $\text I_z(a,b)$](https://mathworld.wolfram.com/RegularizedBetaFunction.html), and [the Incomplete Beta function $\text B_z(a,b)$](https://mathworld.wolfram.com/IncompleteBetaFunction.html):


$$\text B_z(a,b)=\int_0^z t^{a-1} (1-t)^{b-1}dt, \text I_z(a,b)=\frac{\int_0^z t^{a-1} (1-t)^{b-1}dt}{\int_0^1 t^{a-1} (1-t)^{b-1}dt } =\frac{\text B_z(a,b)}{\text B(a,b)}$$


$$\text I_y(a,b)=z\implies y=\text I^{-1}_{0\le z\le 1}(a\ge 0,b\ge0)=\text I^{-1}_z(a,b)$$

where the restrictions on $z,a,b$ help find the [quantile](https://en.m.wikipedia.org/wiki/Quantile_function) of a [Beta type Distribution](https://en.m.wikipedia.org/wiki/Beta_distribution). The goal is to notice that:

$$\text B_z\left(\frac12,\frac32\right)=\int_0^z \sqrt{\frac 1x-1}dx=\sqrt{x-x^2}+\sin^{-1}\sqrt x\implies \text B_{\sin^2(z)}\left(\frac12,\frac32\right)\mathop=^{0\le z\le 1}z+\sin(z)\cos(z)=z+\frac12 \sin(2z)$$


and $$z+\frac12 \sin(2z)=y\mathop=^{x=2z} \frac x2+\frac 12 \sin(x)=y\implies x-(-\sin(x))=2y\implies x=\text E(-1,2y)\implies z=\frac {\text E(-1,2y)}2$$

while the dottie number:

$$x=\text D:x-\cos(x)=0\mathop\implies^{x=\frac \pi2-z}\frac \pi2-z-\cos\left(\frac\pi2-z\right)=\frac\pi2-z-\sin(z)=0\implies \frac \pi2=z-(-\sin(z))\implies z=\text E\left(-1,\frac\pi2\right)=\frac\pi 2-x\implies  \frac\pi2-\text E\left(-1,\frac\pi2\right) =x$$

However if $$\text B_{\sin^2(z)}\left(\frac12,\frac32\right)=z+\frac12 \sin(2z)=\text B\left(\frac12,\frac32\right) \text I_{\sin^2(z)}\left(\frac12,\frac32\right)=\frac \pi2 \text I_{\sin^2(z)}\left(\frac12,\frac32\right)=y \implies \sin^2(z)=\text I^{-1}_{\frac{2y}\pi}\left(\frac12,\frac32\right)\mathop\implies^{0\le z\le\frac \pi 2} \sin^{-1}\sqrt{\text I^{-1}_{\frac{2y}\pi}\left(\frac12,\frac32\right)}=z$$

Therefore:

>$$\text E(-1,x)=2\sin^{-1}\sqrt{\text I^{-1}_{\frac x\pi}\left(\frac12,\frac32\right)}=\text{hav}^{-1}\left(\text I^{-1}_{\frac x\pi}\left(\frac12,\frac32\right)\right)\,\implies \text I^{-1}_x\left(\frac12,\frac32\right)= \sin^2\left(\frac{\text E(-1,\pi x)}2\right)=\text {hav}(\text E(-1,x))$$

where appears the [Half Versed Sine](https://mathworld.wolfram.com/Haversine.html) and [Inverse Half Versed Sine](https://mathworld.wolfram.com/InverseHaversine.html)

Therefore:

$$\text D=  \frac\pi2-\text E\left(-1,\frac\pi2\right) =\frac \pi2-2\sin^{-1}\sqrt{\text I^{-1}_\frac12\left(\frac 12,\frac 32\right)}=\sin^{-1}\left(1-2 \text I^{-1}_\frac12\left(\frac 12,\frac 32\right)\right) $$

which is the [Inverse](https://www.wolframalpha.com/input?i=inverse+hacoversin%28x%29) [Half Covered Sine](https://mathworld.wolfram.com/Hacoversine.html). Using $\text D=\cos^{-1}(\text D)$ we get [**a perfect approximation with this computation**](https://www.wolframalpha.com/input?i=2+sqrt%28inversebetaregularized%281%2F2%2C1%2F2%2C3%2F2%29-inversebetaregularized%281%2F2%2C1%2F2%2C3%2F2%29%5E2%29-dottie+number) and another closed form:

**The Main Result:**
$$\text{Dottie Number}=\text D=\text{hacoversin}^{-1}\,\text I^{-1}_\frac12\left(\frac 12,\frac 32\right)=2\sqrt{\text I^{-1}_\frac12\left(\frac 12,\frac 32\right)-\text I^{-1}_\frac12\left(\frac 12,\frac 32\right)^2}$$

where $\sqrt{\text{quantile}(1-\text{quantile})}$ is not quite a statistical formula and where the arcsin formula works with an [**error of $10^{-179}$**](https://www.wolframalpha.com/input?i=sin%5E%28-1%29%281+-+2+inversebetaregularized%281%2F2%2C1%2F2%2C3%2F2%29%29-dottie+number). Also, $\text I^{-1}_\frac12(a,b)$ finds the [median of a Beta Distribution](https://en.m.wikipedia.org/wiki/Beta_distribution#/editor/14) with $a,b$ as [shape parameters](https://en.m.wikipedia.org/wiki/Beta_distribution#/editor/10). 

**The Question:**

Even though the calculations are an explanation, 

what is the intuition behind the Dottie number being the inverse half covered sine of the median of a Beta distribution with half integer shape parameters?

Phrased differently, why is the dottie number the positive square root of $x-x^2$ where $x$ is the median of a Beta distribution with half integer shape parameters?

Please correct me and give me feedback!

**Side Note $1$:**

Therefore we have a closed form for a Inverse Beta Regularized function special case:

Median of $\frac 2\pi \sqrt{\frac 1x-1}$:

[![enter image description here][1]][1]

where the median is the point such that half of the area of a curve is the red region and the other half is the blue region. The $\color{red}{\text {red}}$ horizontal line is the median:

$$\text{median}=\text I^{-1}_\frac12\left(\frac12,\frac32\right)=\frac {\text D_\text{DHA}^2}4=\text{hacoversin}(\text D)= \frac{1-\sqrt{1-\text D^2}}2=0.16319398540839259232…\implies \int_0^{\frac {\text D_\text{DHA}^2}4} \sqrt{\frac1x-1}\,dx=\int_0^{\text I^{-1}_\frac12\left(\frac12,\frac32\right)} \sqrt{\frac1x-1}\,dx=\frac\pi4 $$

where $\text D_\text{DHA}= \sqrt{1+\text D} - \sqrt{1-\text D} = \sqrt 2\sqrt{1-\sqrt{1-\text D^2}} $ is the [offset at which $2$ unit disks overlap by half of each’s area](https://mathworld.wolfram.com/Circle-CircleIntersection.html) [constant](https://m.youtube.com/watch?v=LV9LAYl0kYM)

Notice how the dottie number appears in the offset when a unit disk is *half* over another one and the median, where the areas from $0$ to the median is *half* of the area under the curve, of a beta distribution.

  [1]: https://i.stack.imgur.com/1Hewl.jpg
**Please Note:**

Sorry if the MathJax does not show up; it is my first time on codidact, so please see the “cross posted link” at the top for the original source.