User:Renerpho/sandbox

Integral

$\int _{0}^{\infty }{\frac {\sin {t}}{t}}{\text{ dt}}={\frac {\pi }{2}}$

$\int _{0}^{\infty }{{\frac {\sin {t}}{t}}\cdot {\frac {\sin {\left({\frac {t}{101}}\right)}}{\left({\frac {t}{101}}\right)}}}{\text{ dt}}={\frac {\pi }{2}}$

$\int _{0}^{\infty }{{\frac {\sin {t}}{t}}\cdot {\frac {\sin {\left({\frac {t}{101}}\right)}}{\left({\frac {t}{101}}\right)}}\cdot {\frac {\sin {\left({\frac {t}{201}}\right)}}{\left({\frac {t}{201}}\right)}}}{\text{ dt}}={\frac {\pi }{2}}$

$\vdots \!\,$

$\int _{0}^{\infty }{{\frac {\sin {t}}{t}}\cdot {\frac {\sin {\left({\frac {t}{101}}\right)}}{\left({\frac {t}{101}}\right)}}\cdot {\frac {\sin {\left({\frac {t}{201}}\right)}}{\left({\frac {t}{201}}\right)}}\cdots {\frac {\sin {\left({\frac {t}{100n+1}}\right)}}{\left({\frac {t}{100n+1}}\right)}}}{\text{ dt}}={\frac {\pi }{2}}$

True for every $n<15341178777673149429167740440969249338310889,$

but false for all $n$ larger than that.

$\int _{0}^{\infty }{\frac {\sin {t}}{t}}{\text{ dt}}={\frac {\pi }{2}}$

$\int _{0}^{\infty }{{\frac {\sin {t}}{t}}\cdot {\frac {\sin {\left({\frac {t}{2}}\right)}}{\left({\frac {t}{2}}\right)}}}{\text{ dt}}={\frac {\pi }{2}}$

$\int _{0}^{\infty }{{\frac {\sin {t}}{t}}\cdot {\frac {\sin {\left({\frac {t}{2}}\right)}}{\left({\frac {t}{2}}\right)}}\cdot {\frac {\sin {\left({\frac {t}{4}}\right)}}{\left({\frac {t}{4}}\right)}}}{\text{ dt}}={\frac {\pi }{2}}$

$\vdots \!\,$

$\int _{0}^{\infty }{{\frac {\sin {t}}{t}}\cdot {\frac {\sin {\left({\frac {t}{2}}\right)}}{\left({\frac {t}{2}}\right)}}\cdot {\frac {\sin {\left({\frac {t}{4}}\right)}}{\left({\frac {t}{4}}\right)}}\cdots {\frac {\sin {\left({\frac {t}{2^{n}}}\right)}}{\left({\frac {t}{2^{n}}}\right)}}}{\text{ dt}}={\frac {\pi }{2}}$

True for all $n$

New stuff

Behind this "magic" is the following: Let $\left(a_{n}\right)$ be a sequence of positive real numbers, $a_{0}=1.$ Then $\int _{0}^{\infty }{\prod _{n=0}^{N}{\frac {\sin {a_{n}t}}{a_{n}t}}}{\text{ dt}}={\frac {\pi }{2}}$ if and only if $\sum _{1}^{N}a_{n}\leq 1.$

$\sum _{n=1}^{N}{\frac {1}{100n+1}}$ diverges, but very slowly (it grows logarithmically). The sum is smaller than $1$ until $N$ gets to approximately $10^{43}.$

In contrast, $\sum _{n=1}^{N}{\frac {1}{2^{n}}}$ converges, and $\sum _{n=1}^{N}{\frac {1}{2^{n}}}=1-{\frac {1}{2^{N}}}<1$ for all $N.$

An example that diverges even slower is $a_{n}={\frac {1}{\left(100n+1\right)\cdot \ln {\left(100n+1\right)}}}.$ This will not fail until $N\approx 10^{10^{43}}.$

Stuff, continued

1±1

5±1

18±1

52±1

149±3

411+45
−41

1100+270
−220

2850+1200
−870

7210+5000
−3000

17900+19000
−9200

43500+66000
−26000

104000+230000
−71000

Haumea resonance

The resonant angle $\phi$ in this case is

$\phi ={\rm {12\cdot \lambda -{\rm {7\cdot \lambda _{\rm {N}}-{\rm {5\cdot \varpi -{\rm {1\cdot \Omega }}}}}}}}$

This angle is librating intermittently, hence why Buie does not classify Haumea as a resonant object.[1] Haumea's ascending node $\Omega$ precesses with a period of about 4.4 million years. It seems like Haumea's 7:12 resonance is broken twice per cycle, once every 2.2 million years, and is reestablished again a few hundred thousand years later. The resonance will next be broken about 250,000 years from now. See here for the results of my orbit simulation.

Haumea and the other objects in the Haumea family occupy a region of the Kuiper belt where multiple resonances (including the 3:5, 4:7, 7:12, 10:17 and 11:19 mean motion resonances) interact, leading to the orbital diffusion of that collision family.[2] While Haumea is in a weak 7:12 resonance, other objects in the Haumea family are known to temporarily occupy some of the other resonances. For instance, (19308) 1996 TO66, the first member of the Haumea family to be discovered, is in an intermittent 11:19 resonance.

Pi day 2021

Mathematicians: Let's calculate 50 trillion digits of $\pi$.
Astronomers: Let's assume $\pi \approx 1$, but also $\pi ^{2}\approx 10$.