Irrationality of $\pi$. Problem 7.32 of Apostol's analysis. The following problem in Apostol's Mathematical Analysis book gives an outline of Niven's proof that $\pi^2$ is irrational. It starts by letting

*

*$f_n(x)=\frac{1}{n!}x^n(1-x)^n$ over the interval $[0,1]$, which clearly satisfies $0\leq f_n(x)\leq \frac{1}{n!}$.

*Since $f_n$ is a polynomial of degree $2n$, using Taylor expansions we obtain that $f^{(k)}(0)=0\in\mathbb{Z}$ for all $0\leq k<n-1$ or $k>2n$, and $f^{(k)}=(-1)^k\binom{n}{k-n}\frac{k!}{n!}\in\mathbb{Z}$ for $n\leq k\leq 2n$. Since $f(x)=f(1-x)$, $f^{(k)}(1)=(-1)^kf^{(k)}(0)\in\mathbb{Z}$ for all $k\in\mathbb{Z}_+$.

The problem continues by  assuming that $\pi^2=a/b$, where $a,b\in\mathbb{N}$ and $(a,b)=1$, and then introducing
\begin{aligned}
F_n(x)&=b^n\sum^n_{k=0}(-1)^kf^{(2k)}_n(x)\pi^{2n-2k}\\
&= \sum^n_{k=0}(-1)^kf^{(2k)}_n(x)a^{n-k}b^k
\end{aligned}
3. From (2) it follows that $F_n(0)$ and $F_n(1)$ are integers.

Where I am stuck is in the subsequent parts 4 and 6 of the problem:


*Show that
$$ \pi^2a^n\sin \pi x = \frac{d}{dx}\big(F'_n(x)\sin \pi x - \pi F_n(x)\cos \pi x\big)$$

*Integration on the result in part (4) gives
$$F_n(0)+F_n(1)=\pi a^n\int^1_0 f(x)\sin\pi x\,dx$$

*Show $0<F_n(0)+F_n(1)<1$ for $n$ sufficiently large.

Part (6) would lead to a contradiction of (3).

Any help would be appreciated.
 A: For convenience, let's drop the index $n$.

Part (4) of your problem is straight forward:
$$
\begin{align}
(F'(x)\sin(\pi x) &- \pi F(x)\cos(\pi x)\big)'= F''(x)\sin(\pi x)+\pi F'(x)\cos(\pi x)\\
&\quad -\pi F'(x)\cos(\pi x) + \pi^2 F(x)\sin(\pi x)\\
&= \big(F''(x) +\pi^2 F(x)\big)\sin(\pi x)
\end{align}
$$
The term $F''+\pi^2 F$ is given by
$$
\begin{align}
b^n\Big(&\qquad \qquad\quad f^{(2)}(x)\tfrac{a^n}{b^n} - f^{(4)}(x)\frac{a^{n-1}}{b^{n-1}}+\ldots +(-1)^{n-1}f^{(2n)}(x)\frac{a}{b}+ (-1)^nf^{(2n+2)}(x)+\\
& f(x)\frac{a^{n+1}}{b^{n+1}}-f^{(2)}(x)\tfrac{a^n}{b^n} + f^{(4)}(x)\frac{a^{n-1}}{b^{n-1}}+\ldots -(-1)^{n-1}f^{(2n)}(x)\frac{a}{b}\Big)\\
&=a^nf(x)\frac{a}{b}
\end{align}
$$
Notice that the term with the derivative of order $2n+2$ vanishes since $f$ is a polynomial of degree $2n$.

As you pointed out in your description of the problem, integrating over $[0,1]$ leads to
$$
F(0)+F(1)=a^n\pi\int^1_0 f(x)\sin(\pi x)\,dx
$$
For the conclusion in part (6) use part (1), i.e. $0<f<\frac{1}{n!}$ for $0<x<1$ along with the fact that $\sin(\pi x)>0$ over $0<x <1$.

Notice that the identity in part (6) holds regardless of the index $n$. Choose $n$ large enough so that $\frac{a^n}{n!}<\frac12$.

Edit: According to this article in Wikipedia, the proof outlined above although closed to Niven's, it is actually  a problem in  Bourbaki's  Fonctions d'une variable réelle, chap. I–II–III, Actualités Scientifiques et Industrielles (in French), 1074, Hermann, pp. 137–138, year 1949.
