# Proving $\pi$ is irrational with recursive integrals

Given $$I_n = \int_{-1}^1 (1-x^2)^n\cos(\theta x)dx,$$ I have proven the relation $$\theta^2I_n = 2n(2n-1)I_{n-1} - 4n(n-1))I_{n-2}.$$ I am supposed to now notice that $$\theta^{2n+1}I_n = n!(P_n(\theta)\sin(\theta) + Q_n(\theta)\cos(\theta)),$$ where $$P_n$$ and $$Q_n$$ are polynomials of degree at most $$2n$$ with integer coefficients, then use this relation to prove that $$\pi$$ is irrational. I can sort of see the relation, but can't see a clear way to prove it besides ''the coefficients seem to decrease by one each time so the $$n!$$ makes sense, and $$I_1,I_0$$ have the desired $$\sin\theta$$ and $$\cos\theta$$, and the plus $$1$$ makes sense since we apply the relation either $$n$$ or $$n-1$$ times on the terms (giving us $$\theta^{2n}$$ or $$\theta^{2n-2}$$) but then $$I_1 = \frac{4}{\theta^3}(\sin\theta - \theta\cos\theta)$$ and $$I_0 = \frac{2}{\theta}\sin\theta$$''. So, I can vaguely see how all the parts could fit together but am not convinced with my argument and don't see an obvious way to prove it carefully by actually applying the relation by hand (I've tried expanding a few of the term by hand to see if a particularly nice pattern appears, but I don't see one). Can someone explain how to do this a bit more carefully? I see how to prove that $$\pi$$ is irrational given this relation (plug in $$\theta = \pi/2$$ and find an integer between $$0$$ and $$1$$ assuming $$\pi$$ is rational).

I got the problem from this practice sheet: [https://www.dpmms.cam.ac.uk/study/IA/AnalysisI/2020-2021/aI_4_21.pdf]

• Have you tried using strong induction on $n$ to prove it? Commented Aug 23, 2021 at 17:28
• I have now, it works very nicely, thanks! One thing I notice is that $P_{n+1}$ and $Q_{n+1}$ seem to have degree at most $2n$ and not $2(n+1)$, which seems a bit odd to me. Commented Aug 23, 2021 at 18:59
• It doesn't really matter since $\deg$ at most $2n$ implies $\deg$ at most $2(n+1)$. Indeed, I think the result still holds if you say $P_n,Q_n$ are polynomials with $\deg$ at most $2(n-1)$ for $n\gt 1$, are constant polynomials for $n=0$ and for $n=1$, $P_1$ is a constant polynomial while $Q_1$ is a degree $1$ polynomial, but it is easier to simply write at most $2n$ for all $n\ge 0$, even though it introduces a bit of redundancy. Commented Aug 23, 2021 at 19:28
• Checking on W|A does seem to confirm this, eg., n=2, n=3. Commented Aug 23, 2021 at 19:34

$$\theta^2 I_n = 2n(2n-1)I_{n-1} - 2n(2n-2)I_{n-2}\tag 1$$

You can do it directly by iteratively multiplying both sides of $$(1)$$ by $$\theta^2$$ and reducing it, but that seems tedious.

Looks like it will be easier to prove it using strong induction on $$n$$.

Show the base case holds. Assume the induction hypothesis for $$k\le n$$, ie,

$$\forall k\le n\colon\quad\theta^{2k+1}I_k = k!(P_k(\theta)\sin(\theta) + Q_k(\theta)\cos(\theta))\tag{\dagger}$$

Now, show the inductive step for $$n+1$$, ie,

\begin{align}\theta^{2(n+1)+1}I_{n+1}&=\theta^{2n+1}\theta^2 I_{n+1}\\&\overset{(1)}{=}\theta^{2n+1}\bigl((2n+2)(2n+1)I_n-(2n+2)(2n)I_{n-1}\bigr)\\&=(2n+2)(2n+1)[\theta^{2n+1}I_n]-(2n+2)(2n)\theta^2[\theta^{2(n-1)+1}I_{n-1}]\\&\overset{(\dagger)}{=}(2n+2)(2n+1)n!\bigl(P_n(\theta)\sin\theta+Q_n(\theta)\cos\theta\bigr)-\theta^2 (2n+2)(2n)\bigl((n-1)!P_{n-1}(\theta)\sin\theta+Q_{n-1}\cos\theta\bigr)\\&=2(2n+1)(n+1)!\bigl(P_n(\theta)\sin\theta+Q_n(\theta)\cos\theta\bigr)-4(n+1)!\bigl(\theta^2 P_{n-1}(\theta)\sin\theta+\theta^2 Q_{n-1}\cos\theta\bigr)\\&=(n+1)!(P_{n+1}(\theta)\sin\theta+Q_{k+1}(\theta)\cos\theta)\end{align}

where $$P_{n+1}(\theta):=2(2n+1)P_n(\theta)-4\theta^2 P_{n-1}(\theta)$$ and $$Q_{n+1}(\theta):=2(2n+1)Q_n(\theta)-4\theta^2 Q_{n-1}(\theta)$$

and we can finish noting that $$P_{n+1}$$ and $$Q_{n+1}$$ are both polynomials with integer coefficients and degree at most $$2n+2=2(n+1)$$