- #36

#### julian

Gold Member

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Problem 10.

A way of proving that ##\pi## is irrational is considering the integral

\begin{align*}

I_n = \frac{1}{n!} \int_0^\pi C^n [(\pi - x) x]^n \sin x dx .

\end{align*}

You show that ##0 < I_n < 1## for ##n## large enough. Then you show that for all ##n \geq 0## that ##I_n## is an integer if ##\pi = a/b## and if we choose ##C = b##, obtaining a contradiction.

It is trivial to use this method to prove that ##\pi^2## is irrational.

Let us quickly run through the details of this argument for proving ##\pi## to be irrational. We want to obtain a contradiction by putting

\begin{align*}

\pi = \frac{a}{b} .

\end{align*}

Write

\begin{align*}

I_n = \frac{1}{n!} \int_0^\pi C^n [(\pi - x) x]^n \sin x dx

\end{align*}

The polynomial ##C^n [(\pi - x) x]^n## is non-negative over ##[0, \pi]## and symmetric about the point ##\pi / 2## and has a maximum of ##(C \pi^2 /4)^n## there which implies ##0 < I_n < \pi \frac{(C \pi^2 / 4)^n}{n!}##, in which ##\pi \frac{(C \pi^2 / 4)^n}{n!}## can be made arbitrarily small by choosing ##n## large enough.

Integrating ##I_n## by parts twice gives

\begin{align*}

I_n = (4n - 2) C I_{n-1} - C^2 \pi^2 I_{n-2} \qquad (*)

\end{align*}

Also we have

\begin{align*}

I_0 = 2 \quad \text{and} \quad I_1 = 4C \qquad (**)

\end{align*}

By choosing ##C = b## we have, by our original assumption, that ##C^2 \pi^2 = b^2 \pi^2 = a^2 =##integer. We then have by ##(*)## and ##(**)## that all the ##I_n## are integers in contradiction with the fact that ##0 < I_n < 1## for ##n## large enough.

Now to prove that ##\pi^2## is irrational. We obtain a contradiction by putting

\begin{align*}

\pi^2 = \frac{p}{q} .

\end{align*}

By choosing ##C = q## we have that ##C^2 \pi^2 = q^2 \pi^2 = pq =##integer. We then have by ##(*)## and ##(**)## that all the ##I_n## are integers in contradiction with the fact that ##0 < I_n < 1## for ##n## large enough. As easy as pie.

A way of proving that ##\pi## is irrational is considering the integral

\begin{align*}

I_n = \frac{1}{n!} \int_0^\pi C^n [(\pi - x) x]^n \sin x dx .

\end{align*}

You show that ##0 < I_n < 1## for ##n## large enough. Then you show that for all ##n \geq 0## that ##I_n## is an integer if ##\pi = a/b## and if we choose ##C = b##, obtaining a contradiction.

It is trivial to use this method to prove that ##\pi^2## is irrational.

Let us quickly run through the details of this argument for proving ##\pi## to be irrational. We want to obtain a contradiction by putting

\begin{align*}

\pi = \frac{a}{b} .

\end{align*}

Write

\begin{align*}

I_n = \frac{1}{n!} \int_0^\pi C^n [(\pi - x) x]^n \sin x dx

\end{align*}

The polynomial ##C^n [(\pi - x) x]^n## is non-negative over ##[0, \pi]## and symmetric about the point ##\pi / 2## and has a maximum of ##(C \pi^2 /4)^n## there which implies ##0 < I_n < \pi \frac{(C \pi^2 / 4)^n}{n!}##, in which ##\pi \frac{(C \pi^2 / 4)^n}{n!}## can be made arbitrarily small by choosing ##n## large enough.

Integrating ##I_n## by parts twice gives

\begin{align*}

I_n = (4n - 2) C I_{n-1} - C^2 \pi^2 I_{n-2} \qquad (*)

\end{align*}

Also we have

\begin{align*}

I_0 = 2 \quad \text{and} \quad I_1 = 4C \qquad (**)

\end{align*}

By choosing ##C = b## we have, by our original assumption, that ##C^2 \pi^2 = b^2 \pi^2 = a^2 =##integer. We then have by ##(*)## and ##(**)## that all the ##I_n## are integers in contradiction with the fact that ##0 < I_n < 1## for ##n## large enough.

Now to prove that ##\pi^2## is irrational. We obtain a contradiction by putting

\begin{align*}

\pi^2 = \frac{p}{q} .

\end{align*}

By choosing ##C = q## we have that ##C^2 \pi^2 = q^2 \pi^2 = pq =##integer. We then have by ##(*)## and ##(**)## that all the ##I_n## are integers in contradiction with the fact that ##0 < I_n < 1## for ##n## large enough. As easy as pie.

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