View source for User:Marco Polo

==Friends==
[[Image:Wikistress3D 3 v3.jpg|thumb|200px|right|Stress Indicator]]
===Me===
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 [[Special:Contributions/Marco Polo|here]], but [[Special:Contributions/Di Gama|here]] for my recent contributions.
====Macro Polo====
*[[Main Page|Wikipedia]] - [[User:Marco Polo|Marco Polo]] ([[User talk:Marco Polo|talk]] | [[Special:Contributions/Marco Polo|contribs]])
*[[s:Main Page|Wikisource]] - [[s:User:Marco Polo|Marco Polo]] ([[s:User talk:Marco Polo|talk]] | [[s:Special:Contributions/Marco Polo|contribs]])
====Di Gama====
*[[Main Page|Wikipedia]] - [[User:Di Gama|Di Gama]] ([[User talk:Di Gama|talk]] | [[Special:Contributions/Di Gama|contribs]])
*[[wikt:Main Page|Wiktionary]] - [[wikt:User:Di Gama|Di Gama]] ([[wikt:User talk:Di Gama|talk]] | [[wikt:Special:Contributions/Di Gama|contribs]])
*[[s:Main Page|Wikisource]] - [[s:User:Di Gama|Di Gama]] ([[s:User talk:Di Gama|talk]] | [[s:Special:Contributions/Di Gama|contribs]])
*[[wikia:starwars:Main Page|Wookieepedia]] - [[wikia:starwars:User:Darth Gama|Darth Gama]] ([[wikia:starwars:User talk:Darth Gama|talk]] | [[wikia:starwars:Special:Contributions/Darth Gama|contribs]])
===Destroyer 45===
[[Main Page|Wikipedia]] - [[User:Destroyer 45|Destroyer 45]] ([[User talk:Destroyer 45|talk]] | [[Special:Contributions/Destroyer 45|contribs]])

[[wikia:starwars:Main Page|Wookieepedia]] - [[wikia:starwars:User:Star Destroyer 2500|Star Destroyer 2500]] ([[wikia:starwars:User talk:Star Destroyer 2500|talk]] | [[wikia:starwars:Special:Contributions/Star Destroyer 2500|contribs]])
===[[/Esperanza|Esperanza]]===
==A Proof that ''π'' really is irrational==

Theorem:  <math>\pi</math> is irrational

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

<math>f_n(x)=\frac{q^nx^n(\pi-x)^n}{n!}  =  \frac{x^n(p-qx)^n}{n!}</math>

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

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

<math>\operatorname{max}\ f_n(x) = f_n\left(\frac{\pi}{2}\right) = \frac{q^n\left(\frac{\pi}{2}\right)^{2n}}{n!}</math>

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