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Q&A

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

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Introduction:

For some background information on the Dottie Number D,

Cross posted from:

Why does the Dottie Number equal a median of a Beta distribution?

since the original question is unanswered.

Some definitions:

The “solution” to Kepler’s equation is Kepler E:

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

and Inverse Beta Regularized function $\text I^{-1}_z(a,b)$, Beta Regularized function $\text I_z(a,b)$, and the Incomplete Beta function $\text B_z(a,b)$:

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

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

where the restrictions on $z,a,b$ help find the quantile of a Beta type 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 and Inverse Half Versed Sine

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 Half Covered Sine. Using $\text D=\cos^{-1}(\text D)$ we get a closed form with $0$ error and another closed form:

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

$$=\sqrt{1-\left(2\text I^{-1}_\frac12\left(\frac12,\frac32\right)-1\right)^2}$$

where $\sqrt{\text{quantile}(1-\text{quantile})}$ is not quite a statistical formula and where the arcsin formula works with an error of $10^{-179}$. Also, $\text I^{-1}_\frac12(a,b)$ finds the median of a Beta Distribution with $a,b$ as shape parameters.

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 $1-(2x-1)^2$ 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

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 constant

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.

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