Stuck on a step of a proof that $\pi$ is irrational.

Question

I'm trying to understand this proof that $\pi$ is irrational, yet I can't see why any derivative of the polynomial $$f(x)=\frac{x^n(a-bx)^n}{n!}$$ where $a,b\in\mathbb{Z}$ has integral values when $x=0$ or when $x=a/b$. I'd appreciate any help.

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Edit: to be somewhat clearer, I fail to see why

$$f^{(i)}(a/b)\in \mathbb{Z}$$

for any natural number $i$.

Answer 1

$f(x)=\frac{x^n(a-bx)^n}{n!}$. As Ivan Niven mentions in his paper

$$f(\frac{a}{b}-x)=\frac{\big(\tfrac{a}{b}-x\big)^n\big(a-b(\tfrac{a}{b}-x)\big)^n}{n!}=\frac{b^{-n}\big(a-bx\big)^n(bx)^n}{n!}=f(x)$$

Thus $f^{(j)}(x)=(-1)^jf^{(j)}(\tfrac{a}{b}-x)$ and so $f^{(j)}(0)=(-1)^nf^{(j)}(a/b)$.

Consequently, it suffices to check that each each $f^{(j)}(0)$ is an integer.

Using the binomial formula you obtain $$ f(x)=\sum^n_{k=0}\frac{\binom{n}{k}}{n!}a^{n-k} b^{k}x^{n+k}=\sum^{2n}_{k=n}\frac{\binom{n}{k-n}}{n!}a^{2n-k}(-b)^{k-n}x^k$$ Recall the formula for power series for analytic functions around $0$:

$$\phi(x)=\sum^\infty_{k=0}\frac{\phi^{(k)}(0)}{k!}x^k$$ As a polynomial in $x$, $f(x)$ coincides with its Taylor series around $0$. Hence

1. $f^{(k)}(0)=0$ for $1\leq k\leq n-1$ and $k>2n$
2. $f^{(k)}(0)=\frac{k!}{n!}\binom{n}{k-n}a^{2n-k}(-b)^{k-n}$ for $n\leq k\leq 2n$. Each of which is an integer.

Answer 2

Just take a couple of derivatives- $$f'(x) = \frac {nx^{n-1}(a-bx)^n - nbx^n(a-bx)^{n-1}}{n!}\\ f''(x) = \frac {n(n-1)x^{n-2}(a-bx)^n - 2nbx^{n-1}(a-bx)^{n-1} + n(n-1)b^2x^n(a-bx)^{n-2}}{n!}$$ Evaluate at $0, \frac {a}{b}$ to see that

$f^{(j)}(0) = 0, f^{(j)}(\frac {a}{b}) = 0$ if $j< n$ and $f^{(j)}(0), f^{(j)}(\frac {a}{b})$ will be an integer in all cases.

Does that help?

Answer 3

HINT.- An easy way to see clear your question is to consider your polynomial as a particular case of $f(x)=(P(x))^n$ (you have $P(x)=ax-bx^2$) Then you have $$f'(x)=nP^{n-1}(x)P'(x)=P^{n-1}(x)Q_0(x)\\f''(x)=n(n-1)P^{n-2}(x)(P'(x))^2+nP^{n-1}(x)P''(x)=P^{n-1}(x)Q_1(x)\\f^{(3)}(x)=(n-1)P^{n-2}(x)P'(x)Q_1(x)+P^{n-1}(x)Q_1'(x)=P^{n-1}(x)Q_2(x)$$ and so on where the $Q_k(x)$ are succesively appearing polynomials. You can verify that all the derivatives have $P(x)$ as a factor excepting the $(nm)^{th}$ derivative which is equal to the constant $\pm (2n)!b$ (according to the parity of $n$), in your case the $(2n)^{th}$ derivative because of the degree of $ax-bx^2$.

It follows that all your derivatives until the $(2n-1)^{th}$ when $x=0$ or $x=\dfrac ab$are equal to zero and the last one is equal to $\pm\dfrac{(2n)!b}{n!}$ which is in integer too.

Answer 4

The function $f(x)$ and all its derivatives are polynomials.

The value of a polynomial in $x$ at $x = 0$ is equal to the constant term.

We must have $n > 0,$ so the constant term of $f(x)$ is zero. The constant term of the $k$th derivative of $f(x)$ is $k! c_k,$ where $c_k$ is the coefficient of $x^k$ in $f(x).$ We know that $c_k = 0$ for $k < n$ (the term in $x^n$ is the non-zero term with the lowest power of $x$), so those derivatives are all zero, an integer. For $k \geq n,$ the coefficient $c_k$ in $f(x)$ is some integer divided by $n!.$ But $k!$ is a multiple of $n!,$ so $k!c_k$ is an integer.

That's the proof for $x = 0,$ which you apparently already understand.

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For $x = a/b,$ the observation is that $f\left(\frac ab - x\right) = f(x).$ Just plug $x \to \frac ab - x$ into the formula for $f(x)$ and watch as you get $f(x)$ back again.

Then $$ f'(x) = \frac{\mathrm d}{\mathrm dx} f(x) = \frac{\mathrm d}{\mathrm dx} f\left(\frac ab - x\right) = f'\left(\frac ab - x\right) \frac{\mathrm d}{\mathrm dx} \left(\frac ab - x\right) = -f'\left(\frac ab - x\right). $$ You can continue taking derivatives that way indefinitely, so in general for positive $k,$ $$ f^{(k)}(x) = \pm f^{(k)}\left(\frac ab - x\right). $$

In particular this is true when $x = 0$:

$$ f^{(k)}(0) = \pm f^{(k)}\left(\frac ab\right). $$

We already know all the values on the left are integers, so the values on the right are integers too.