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For each non-negative integer $n$ let $I_n(\theta)=\int^1_{-1}(1-x^2)^ncos(\theta x) dx$. Prove that ${\theta}^2I_n=2n(2n − 1)I_{n-1}−4n(n−1)I_{n-2}$ for all $n\ge2$, and hence that ${\theta}^{2n+1}I_n(\theta)=n!(P_n(\theta)sin(θ)+Q_n(\theta)cos(\theta))$ for some pair $P_n$ and $Q_n$ of polynomials of degree at most $2n$ with integer coefficients.

Deduce that $\pi$ is irrational.

I've done most of the question but I'm just struggling with the final deduction. I've set $\theta=\pi$ to obtain $\pi^{2n+1}I_n(\pi)+n!Q_n(\pi)=0$, this is clearly a polynomial in $\pi$ of degree $2n+1$ with integer coefficients. I was thinking it may help to show that $I_n(\pi)$ is irrational, but beyond this I'm not sure.

Thank you

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    $\begingroup$ Suppose $\pi/2 = a/b$ is rational, then set $\theta = \pi/2$. $\endgroup$ – Sameer Kailasa Aug 20 '16 at 21:16
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    $\begingroup$ for broadening your culture, see the (hopefully shortest) proof of irrationnality of $\pi$ in (projecteuclid.org/euclid.bams/1183510788) $\endgroup$ – Jean Marie Aug 20 '16 at 21:50
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This is Cartwright's proof of the irrationality of $\pi$. Put $\theta=\frac\pi2$ and obtain $$ \theta^{2n+1}I_n(\theta)=n!P_n(\theta).\tag1 $$ If $\theta=\frac ab$ is rational, then rearranging (1) gives $$ {a^{2n+1}\over n!}I_n(a/b) = b^{2n+1}P_n(a/b).\tag2 $$ Note that the RHS of (2) is an integer, since $P_n$ has degree at most $2n$. Now look at the LHS. Notice that $I_n(a/b)$ lies between $0$ and $2$ for any $n$ (since the integrand is positive and bounded by $1$), while the expression $a^{2n+1}/n!$ gets arbitrarily small as $n\to\infty$. Thus for large enough $n$ we have found an integer between $0$ and $1$, a contradiction.

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