User:Bdodo1992

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WOW! You managed to get the User page of Bdodo1992. This is a really bad page. Edit it to make it better. As you can see, I don't know how to use wikipedia language editting.

1 Profile
2 <Everything below this line is ficitional>
3 A Proof that π really is irrational
4 Userboxes

[edit] Profile

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(…so where is my $200?)

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So i screwed up my site... 68.83.127.31 from "User:Marco Polo" thx

[edit] <Everything below this line is ficitional>

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[edit] Me

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[edit] Esperanza

[edit] A Proof that π really is irrational

Theorem: $π$ 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

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

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

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

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

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 $π$ , 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.

[edit] Userboxes

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