Difference between revisions of "User:John0512"

Revision as of 20:48, 1 October 2021

Orzorzorz Number

This is a number I coined. The exact value of this number is $\sum_{m=17}^{122}\sum_{k=17}^{122}\sum_{n=17}^{\min(m,k)} (2^{24})^{n-17}\cdot(2^{730})^{\min(m,k)-n}\cdot26^{12(m-17)+12(k-17)},$ and is approximately equal to $4.8\cdot10^{26639}$ . It's also approximately $2^{76650}\cdot26^{2520}$ to 210 significant figures. The formula used to find this number is an approximation of the number of ways mathimathz can happen, however I will not go into further detail about exactly how it was derived. Each one of the numbers in the formula has a good reason, however I am not revealing just yet. :)

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 This is a number I coined. The exact value of this number is <cmath>\sum_{m=17}^{122}\sum_{k=17}^{122}\sum_{n=17}^{\min(m,k)} (2^{24})^{n-17}\cdot(2^{730})^{\min(m,k)-n}\cdot26^{12(m-17)+12(k-17)},</cmath> and is approximately equal to <math>4.8\cdot10^{26639}</math>. It's also approximately <math>2^{76650}\cdot26^{2520}</math> to 210 significant figures. The formula used to find this number is an approximation of the number of ways mathimathz can happen, however I will not go into further detail about exactly how it was derived. Each one of the numbers in the formula has a good reason, however I am not revealing just yet. :)
-==Unnamed Theorem==
-I have something called the Unnamed Theorem (which I did not name as I have not confirmed that this theorem has not existed before).
-Claim: Given a set <math>S=\{1,2,3\cdots n\}</math> where <math>n</math> is a positive integer, the number of ways to choose a subset of <math>S</math> then permute said subset is <math>\lfloor n!\cdot e\rfloor.</math>
-Proof: The number of ways to choose a subset of size <math>i</math> and then permute it is <math>\binom{n}{i}\cdot i!</math>. Therefore, the number of ways to choose any subset of <math>S</math> is <cmath>\sum_{i=0}^n \binom{n}{i}\cdot i! = \sum_{i=0}^n \frac{n!}{(n-i)!}.</cmath> This is also equal to <math>\sum_{i=0}^n \frac{n!}{i!}</math> by symmetry across <math>i=\frac{n}{2}</math>. This is also <math>n! \cdot \sum_{i=0}^n \frac{1}{i!}.</math> Note that <math>e</math> is defined as <math>\sum_{i=0}^\infty \frac{1}{i!}</math>, so our expression becomes <cmath>n!(e-\sum_{i={n+1}}^\infty \frac{1}{i!}).</cmath> We claim that <math>\sum_{i={n+1}}^\infty \frac{1}{i!}<\frac{1}{n!}</math> for all positive integers <math>n</math>.
-Since the reciprocal of a factorial decreases faster than a geometric series, we have that <math>\sum_{i={n+1}}^\infty \frac{1}{i!}<\frac{1}{(n+1)!}+\frac{1}{(n+1)!(n+1)}+\frac{1}{(n+1)!(n+1)^2}\cdots</math>. The right side we can evaluate as <math>\frac{1}{n(n!)}</math>, which is always less than or equal to <math>\frac{1}{n!}</math>. This means that the terms being subtracted are always strictly less than <math>\frac{1}{n!}</math>, so we can simply write it as <cmath>\lfloor n!\cdot e\rfloor.</cmath>
-Example: How many ways are there 5 distinct clones of mathicorn to each either accept or reject me, then for me to go through the ones that accepted me in some order?
-Solution to example: This is equivalent to the Unnamed Theorem for <math>n=5</math>, so our answer is <math>\lfloor 120e \rfloor=\boxed{326}</math>.
-Solution 2: Since I am not orz, all 5 clones will reject me, so the answer is <math>\boxed{1}</math>. Note that this contradicts with the answer given by the Unnamed Theorem.

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Difference between revisions of "User:John0512"

Revision as of 20:48, 1 October 2021

Orzorzorz Number