Communities

Writing
Writing
Codidact Meta
Codidact Meta
The Great Outdoors
The Great Outdoors
Photography & Video
Photography & Video
Scientific Speculation
Scientific Speculation
Cooking
Cooking
Electrical Engineering
Electrical Engineering
Judaism
Judaism
Languages & Linguistics
Languages & Linguistics
Software Development
Software Development
Mathematics
Mathematics
Christianity
Christianity
Code Golf
Code Golf
Music
Music
Physics
Physics
Linux Systems
Linux Systems
Power Users
Power Users
Tabletop RPGs
Tabletop RPGs
Notifications
Mark all as read
Q&A

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

+0
−0

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.

Why does this post require moderator attention?
You might want to add some details to your flag.
Why should this post be closed?

0 comment threads

0 answers

Sign up to answer this question »

This community is part of the Codidact network. We have other communities too — take a look!

You can also join us in chat!

Want to advertise this community? Use our templates!

Like what we're doing? Support us! Donate