Difference between revisions of "User:John0512"

Revision as of 22:39, 14 September 2021

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 $S=\{1,2,3\cdots n\}$ where $n$ is a positive integer, the number of ways to choose a subset of $S$ then permute said subset is $\lfloor n!\cdot e\rfloor.$

Proof: The number of ways to choose a subset of size $i$ and then permute it is $\binom{n}{i}\cdot i!$ . Therefore, the number of ways to choose any subset of $S$ is $\sum_{i=0}^n \binom{n}{i}\cdot i! = \sum_{i=0}^n \frac{n!}{(n-i)!}.$ This is also equal to $\sum_{i=0}^n \frac{n!}{i!}$ by symmetry across $i=\frac{n}{2}$ . This is also $n! \cdot \sum_{i=0}^n \frac{1}{i!}.$ Note that $e$ is defined as $\sum_{i=0}^\infty \frac{1}{i!}$ , so our expression becomes $n!(e-\sum_{i={n+1}}^\infty \frac{1}{i!}).$ We claim that $\sum_{i={n+1}}^\infty \frac{1}{i!}<\frac{1}{n!}$ for all positive integers $n$ .

Since the reciprocal of a factorial decreases faster than a geometric series, we have that $\sum_{i={n+1}}^\infty \frac{1}{i!}<\frac{1}{(n+1)!}+\frac{1}{(n+1)!(n+1)}+\frac{1}{(n+1)!(n+1)^2}\cdots$ . The right side we can evaluate as $\frac{1}{n(n!)}$ , which is always less than or equal to $\frac{1}{n!}$ . This means that the terms being subtracted are always strictly less than $\frac{1}{n!}$ , so we can simply write it as $\lfloor n!\cdot e\rfloor.$

Example: How many ways are there 5 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 $n=5$ , so our answer is $\lfloor 120e \rfloor=\boxed{326}$ .

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-i am bad at math
+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 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>.

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Difference between revisions of "User:John0512"

Revision as of 22:39, 14 September 2021