Shouldn't that be $F_n(x)$ ? \begin{eqnarray*} & F''(x) &=f^{2}(x)-f^{4}(x)+f^{6}(x) - ... \\ & F''(x) & + F(x) = f^{2}(x)-f^{4}(x)+f^{6}(x) - ... \\ &+& f(x)-f^{2}(x)+f^{4}(x)-f^{6}(x)+... \\ & = & f(x) \end{eqnarray*} No; $n$ is specified later in the proof to be a *sufficiently large* constant; the definition of $F(x)$ is simply in terms of this constant $n$. That is, assume $\pi$ is rational. Starting with $0<x<\pi$, we can raise to the $n$th power to get $0<x^n<\pi^n$. Starting again with $0<x<\pi$, multiply across by $-b$, add $a$, then raise to the $n$th power to get $a^n > (a-bx)^n>(a-b\pi)^n=0$. Multiplying both equations and dividing by $n!$ gets us $0 < \frac{x^n(a-bx)^n}{n!} < \frac{\pi^na^n}{n!}$. In particular, this will work for all $n$ large enough such that $\pi^na^n<n!$. (So $f$ and $F$ in fact depend on $a$, but there's no harm in that since $a$ has already been chosen to be arbitrary and fixed) The first rigorous proof that π is irrational is from Johann Heinrich Lambert in 1761. He proved that if $x \not= 0$ is rational, then $tan(x)$ must be irrational. Since $tan(π /4) = 1$ is rational, then π must be irrational. The simpler proof given here is due to Ivan Niven in 1947, at the time a young mathematician at the University of Oregon. It only assumes a knowledge of basic Calculus. nice clarification. We can see that $f(x)$ is positive over $(0, \pi)$, and since $\pi$ is the smallest positive zero of the sine function, $\sin x$ is positive there as well; thus their product also is. This is the first half of the inequality: $0 < f(x) \sin x$. For the other half, note that the zeroes of $f$ are at $0$ and $\pi$, and that the derivative of $f(x)$ is: $$f'(x) = \frac{n (a-2bx) \cdot x^{n-1} (a-bx)^{n-1}}{n!}$$ It has zeroes at $0$, $\pi/2$, and $\pi$. Since $f$ is positive over $(0, \pi)$, we can deduce that over the interval $[0, \pi]$, the graph of $f$ is a Bell curve: it goes up from $0$ at $x=0$, to $f(\pi/2)$ halfway, and back down from there to $0$ at $x=\pi$. Thus over the interval $(0, \pi)$, we conclude that $$f(x) < f(\pi/2) = \frac{\pi^n a^n}{4n!} < \frac{\pi^na^n}{n!},$$ completing the inequality. (Is there an easier way to show this half?) The idea now is to show that the integral of $f(x) sin(x)$ cannot always be an integer for all values of n, which will be in contradiction with $(1)$, thus proving that $\pi$ must be irrational. Consider a value of x that is between 0 and $\pi$. It's not difficult to see that $f(x)$ in this interval is always less than $\frac{\pi^{n}a^{n}}{n!}$. At the same time $sin(x)$ will always be less than 1 in the same interval and so $f(x)sin(x)<\frac{\pi^{n}a^{n}}{n!}$. As a consequence $\int_{0}^{\pi} f(x)sin(x)<\frac{\pi^{n+1}a^{n}}{n!}$. Now if we recall that $e^{\pi a}=\sum \frac{\pi^n a^n }{n!} $, and any term in a convergent series will have to tend to 0 as $n \rightarrow \infty $, we conclude that $\int_{0}^{\pi} f(x)sin(x)$ will have to tend to 0 for a sufficient large $n$, which contradicts the previous result stating that the integral must be an integer. By contradiction we can now conclude that $\pi$ needs to be irrational. To prove that $f(x)$ and its derivatives $f^{(j)}(x)$ have integral values for $x=0$ we should first note that $f(x)$ is a polynomial with terms in $x$ from $x^n...x^{2n}$. To do that we just need to expand the polynomial $(a-bx)^{n}$ using the binomial expansion and note that we end up with a polynomial with terms from $x^0...x^{n}$. Finally, when multiplied by $x^{n}$ we end up with terms ranging from $x^n...x^{2n}$. Now let's divide the derivatives in 3 groups: \begin{eqnarray*} & \text{1. } & j<n \\ & \text{2. } & n\leq j \leq 2n \\ & \text{3. } & j>2n \end{eqnarray*} For $j<n$ the terms in $f^{(j)}(x)$ will range from $x^{n-j}...x^{2n-j}$ and so $f^{(j)}(0) = 0 $. For $j>2n$ all the terms will vanish and again $f^{(j)}(0) = 0$. Finally for $n\leq j \leq 2n$, there will be at least a term that doesn't vanish for $x=0$ - the term in $x^{j}$ - and so $f^{(j)}(0) = C_{j}$, where $C_j$ is a binomial coefficient multiplied by $\frac{j!}{n!}$. The binomial coefficient is an integer as well as $\frac{j!}{n!}$, because $j>n$ and so their product ($C_j$) will also be an integer. We finally proved that $f(x)$ and its derivatives $f^{(j)}(x)$ have integral values for $x=0$, for all $j$. Now if we observe that: \begin{eqnarray*} f(a/b-x) & =& (a/b-x)^{n}(a-b(a/b-x))^{n}/n! \\ & = & (a/b-x)^{n}(bx)^{n}/n! \\ & = & (a-bx)^{n}x^{n}/n! \\ & = & f(x) \end{eqnarray*} We verify that the results derived for $x=0$ are also applicable for $x=a/b=\pi$.