User:Marco Polo

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If Wikimedia had invented its "multi-login" setup before I had created this account, I would not be in this predicament, but alas, it was not to be. I am known under two (2) usernames. At the risk of being accused as a sock-puppeteer, I make this known here. The reason is that this username is taken on a few sister sites, so I use an alternate name there, but once the multi-login came about, My alternate obtained similarly-named accounts here and elsewhere. As a result, I now contribute using both users, as whim strikes me, but mostly my new one, since it does not have any name clashes. So look not here, but here for my recent contributions.

Macro Polo

• Wikipedia - Marco Polo (talk | contribs)
• Wikisource - Marco Polo (talk | contribs)

Di Gama

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• Wiktionary - Di Gama (talk | contribs)
• Wikisource - Di Gama (talk | contribs)
• Wookieepedia - Darth Gama (talk | contribs)

Destroyer 45

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Wookieepedia - Star Destroyer 2500 (talk | contribs)

A Proof that π really is irrational

Theorem: $\pi$ is irrational

Proof: Suppose $\pi =\frac{p}{q}$, where $p$ and $q$ are integers. Consider the functions $f_n(x)$ defined on $[0,\pi ]$ by

$f_n(x)=\frac{q^n{x}^n(\pi -x{)}^n}{n!}=\frac{x^n(p-qx{)}^n}{n!}$

Clearly $f_n(0)=f_n(\pi )=0$ for all $n$. Let $f_n^{(m)}(x)$ denote the $m$-th derivative of $f_n(x)$. Note that

$f_n^{(m)}(0)=-f_n^{(m)}(\pi )=0\text{ for }m<=n\text{ or for }m>2n$; otherwise some integer

$\max f_n(x)=f_n\left(\frac{\pi }{2}\right)=\frac{q^n{\left(\frac{\pi }{2}\right)}^{2n}}{n!}$

By repeatedly applying integration by parts, the definite integrals of the functions $f_n(x)\sin x$ can be seen to have integer values. But $f_n(x)\sin x$ are strictly positive, except for the two points 0 and $\pi$, and these functions are bounded above by $\frac{1}{\pi }$ for all sufficiently large n. Thus for a large value of n, the definite integral of $f_n(x)\sin x$ is some value strictly between 0 and 1, a contradiction.

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