integration proof (1 Viewer)

Constantspy977 · May 10, 2025

wtf is this

Constantspy977 · May 10, 2025

tywebb said:
wtf is what?

View attachment 47707

you should be able to see it now

fromthethickofit · May 10, 2025

tywebb · May 10, 2025

$\begin{aligned}\text{a. }I_n&=\textstyle\frac{q^{2n}}{n!}\int_{-\frac{\pi}{2}}^\frac{\pi}{2}\left(\frac{\pi^2}{4}-x^2\right)^n\frac{d}{dx}\sin x\ \!dx\\ &=\textstyle\frac{q^{2n}}{n!}\left[\left(\frac{\pi^2}{4}-x^2\right)^n\sin x\right]_{-\frac{\pi}{2}}^\frac{\pi}{2}-\frac{q^{2n}}{n!}\int_{-\frac{\pi}{2}}^\frac{\pi}{2}\sin x\frac{d}{dx}\left(\frac{\pi^2}{4}-x^2\right)^n\ \!dx\\ &=\textstyle\frac{2nq^{2n}}{n!}\int_{-\frac{\pi}{2}}^\frac{\pi}{2}x\left(\frac{\pi^2}{4}-x^2\right)^{n-1}\sin x\ \!dx\\ &=\textstyle\frac{2nq^{2n}}{n!}\int_{-\frac{\pi}{2}}^\frac{\pi}{2}x\left(\frac{\pi^2}{4}-x^2\right)^{n-1}\frac{d}{dx}(-\cos x)\ \!dx\\ &=\textstyle\frac{2nq^{2n}}{n!}\left[-x\left(\frac{\pi^2}{4}-x^2\right)^{n-1}\cos x\right]_{-\frac{\pi}{2}}^\frac{\pi}{2}-\frac{2nq^{2n}}{n!}\int_{-\frac{\pi}{2}}^\frac{\pi}{2}-\cos x\frac{d}{dx}\left(x\left(\frac{\pi^2}{4}-x^2\right)^{n-1}\right)\ \!dx\\ &=\textstyle\frac{2nq^{2n}}{n!}\int_{-\frac{\pi}{2}}^\frac{\pi}{2}\cos x\left(\left(\frac{\pi^2}{4}-x^2\right)^{n-1}-2(n-1)x^2\left(\frac{\pi^2}{4}-x^2\right)^{n-2}\right)\ \!dx\\ &=\textstyle\frac{2nq^{2n}}{n!}\int_{-\frac{\pi}{2}}^\frac{\pi}{2}\left(\frac{\pi^2}{4}-x^2\right)^{n-1}\cos x\ \!dx-\frac{4n(n-1)q^{2n}}{n!}\int_{-\frac{\pi}{2}}^\frac{\pi}{2}x^2\left(\frac{\pi^2}{4}-x^2\right)^{n-2}\cos x\ \!dx\\ &=\textstyle\frac{2q^{2n}}{(n-1)!}\int_{-\frac{\pi}{2}}^\frac{\pi}{2}\left(\frac{\pi^2}{4}-x^2\right)^{n-1}\cos x\ \!dx-\frac{4q^{2n}}{(n-2)!}\int_{-\frac{\pi}{2}}^\frac{\pi}{2}x^2\left(\frac{\pi^2}{4}-x^2\right)^{n-2}\cos x\ \!dx\\ \end{aligned}$

$\begin{aligned}\text{b. }I_n&=\textstyle\frac{2q^{2n}}{(n-1)!}\int_{-\frac{\pi}{2}}^\frac{\pi}{2}\left(\frac{\pi^2}{4}-x^2\right)^{n-1}\cos x\ \!dx-\frac{4q^{2n}}{(n-2)!}\int_{-\frac{\pi}{2}}^\frac{\pi}{2}\left(\frac{\pi^2}{4}-\left(\frac{\pi^2}{4}-x^2\right)\right)\left(\frac{\pi^2}{4}-x^2\right)^{n-2}\cos x\ \!dx\\ &=\textstyle\left(\frac{2q^{2n}}{(n-1)!}+\frac{4q^{2n}}{(n-2)!}\right)\int_{-\frac{\pi}{2}}^\frac{\pi}{2}\left(\frac{\pi^2}{4}-x^2\right)^{n-1}\cos x\ \!dx-\frac{q^{2n}\pi^2}{(n-2)!}\int_{-\frac{\pi}{2}}^\frac{\pi}{2}\left(\frac{\pi^2}{4}-x^2\right)^{n-2}\cos x\ \!dx\\ &=\textstyle\frac{2q^{2n}+4(n-1)q^{2n}}{(n-1)!}\int_{-\frac{\pi}{2}}^\frac{\pi}{2}\left(\frac{\pi^2}{4}-x^2\right)^{n-1}\cos x\ \!dx-\frac{q^{2(n-1)}p^2}{(n-2)!}\int_{-\frac{\pi}{2}}^\frac{\pi}{2}\left(\frac{\pi^2}{4}-x^2\right)^{n-2}\cos x\ \!dx\\ &=\textstyle\frac{(4n-2)q^2q^{2(n-1)}}{(n-1)!}\int_{-\frac{\pi}{2}}^\frac{\pi}{2}\left(\frac{\pi^2}{4}-x^2\right)^{n-1}\cos x\ \!dx-\frac{p^2q^2q^{2(n-2)}}{(n-2)!}\int_{-\frac{\pi}{2}}^\frac{\pi}{2}\left(\frac{\pi^2}{4}-x^2\right)^{n-2}\cos x\ \!dx\\ &=(4n-2)q^2I_{n-1}-p^2q^2I_{n-2} \end{aligned}$

c. $I_0=\textstyle\int_{-\frac{\pi}{2}}^\frac{\pi}{2}\cos x\ \!dx=[\sin x]_{-\frac{\pi}{2}}^\frac{\pi}{2}=1- -1=2\in\Bbb Z$ and

$\begin{aligned}I_1&=\textstyle q^2\int_{-\frac{\pi}{2}}^\frac{\pi}{2}\left(\frac{\pi^2}{4}-x^2\right)\cos x\ \!dx\\ &=\textstyle\frac{p^2}{4}[\sin x]_{-\frac{\pi}{2}}^\frac{\pi}{2}-q^2\int_{-\frac{\pi}{2}}^\frac{\pi}{2}x^2\cos x\ \!dx\\ &=\textstyle\frac{p^2}{4}(1- -1)-q^2\cdot\frac{\pi^2-8}{2}\\ &=\textstyle\frac{p^2}{2}-\frac{p^2}{2}+4q^2\\ &=4q^2\in\Bbb Z \end{aligned}$

If $I_{k-2},I_{k-1}\in\Bbb Z$ then $I_k=(4k-2)q^2I_{k-1}-p^2q^2I_{k-2}\in\Bbb Z$ since $(4k-2)q^2,p^2q^2\in\Bbb Z$

Now since $I_0,I_1\in\Bbb Z$ then by induction then for all integers $n\ge0,I_n$ is an integer.

d. For $\textstyle-\frac{\pi}{2}<x<\frac{\pi}{2}$ and $\textstyle x\ne0,0<\left(\frac{\pi^2}{4}-x^2\right)^n<\left(\frac{\pi^2}{4}\right)^n$ and $\textstyle0<\cos x<1$

by way of example with n=3 you can see the area under curve is less than area under line

so

$\textstyle0<I_n<\frac{q^{2n}}{n!}\int_{-\frac{\pi}{2}}^\frac{\pi}{2}\left(\frac{\pi^2}{4}\right)^n\ \!dx=\frac{(q\pi)^{2n}}{n!4^n}[x]_{-\frac{\pi}{2}}^\frac{\pi}{2}=\frac{p^{2n}}{n!4^n}(\frac{\pi}{2}- -\frac{\pi}{2})=\frac{p}{q}\left(\frac{p}{2}\right)^{2n}\frac{1}{n!}$

e. Since for sufficiently large $\textstyle n,\frac{p}{q}\left(\frac{p}{2}\right)^{2n}\frac{1}{n!}<1$ then $0<I_n<1$ but with the assumption that $\pi$ is rational we proved $I_n$ is an integer but there is no integer between 0 and 1 - contradiction.

Hence $\pi$ is irrational.

tywebb · May 10, 2025

or something like that

i haven't checked for typos but that at least gives the general gist of how things will go with this

WeiWeiMan · May 10, 2025

Constantspy977 said:
View attachment 47708
wtf is this

follow the steps

barnyard · May 10, 2025

Constantspy977 said:
View attachment 47708
wtf is this

lowk for a, b, and e it all tells you exactly what to do
c is just solve I_0 and I_1 and then say that operations with integers make integers when there is no division lol
i only struggled a little with d because i was trying to use b but got there eventually by realising they wanted me to use the original definition of I_n again due to the 1/n! and ended up with the same thing as @tywebb (btw i couldnt see any typos)
pretty fun question i rate it

ivanradoszyce · May 13, 2025

Great question.

Can you please tell me where this question came from. If it comes from a text book, I would interested in purchasing it.

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integration proof (1 Viewer)

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