User:GFauxPas/Sandbox

Welcome to my sandbox, you are free to play here as long as you don't track sand onto the main wiki. --GFauxPas 09:28, 7 November 2011 (CST)

User:GFauxPas/Sandbox/Zeta2/lnxln1-x/existence

User:GFauxPas/Sandbox/Zeta2/lnxln1-x/integrand

User:GFauxPas/Sandbox/Zeta2/lnxln1-x/evaluation

User:GFauxPas/Sandbox/Zeta2/FourierSeries/

User:GFauxPas/Sandbox/Zeta2/Informal Proof

Theorem
Let $m, n \in \N_{>0}$ be natural numbers

Let $m^{\overline n}$ be $m$ to the power of $n$ rising.

Then:


 * $m^{\overline n} \equiv 0 \bmod n!$

That is, the factorial of $n$ divides the product of $n$ successive numbers.

Proof
The proof proceeds by induction on $m$.

Basis for the Induction
The case $n = 1$ is verified as follows:

This is true, as $m^{\overline{1}} = m \times 1$.

This is the basis for the induction.

Induction Hypothesis
Fix $n \in \N$ with $n \ge 1$.

Assume: $m^{\overline n} \equiv 0 \bmod n!$

This is our induction hypothesis.

Induction Step
This is our induction step:

Now, We conclude to-be-proved property holds for $n+1$.

The result follows by the Principle of Mathematical Induction.

I don't know why, but something about this proof strikes me as wrong. Is it correct? --GFauxPas (talk) 13:25, 9 November 2014 (UTC)


 * You ought to prove $m^{\overline{n+1}} \equiv 0 \bmod (n+1)!$. &mdash; Lord_Farin (talk) 15:20, 9 November 2014 (UTC)


 * D'oh! --GFauxPas (talk) 15:36, 9 November 2014 (UTC)