transcendental – calculus + ε

We have already proved that $e$ is irrational (see the example in this post!)
The fact that $\pi$ is irrational, i.e. there are no integers $a,b$ such that $\displaystyle \pi=\frac{a}{b},$ was discovered a couple of hundreds of years ago and there are several proofs of this result; I like Ivan Niven’s proof better and that’s what I’m going to explain here.

Throughout this post, $f^{(k)}(a)$ is the $k$ -th derivative of a function $f(x)$ at $x=a.$

Problem 1. Let $P(x)$ be a polynomial of degree $2n.$ Show that

$\displaystyle \int_0^{\pi} P(x) \sin x \ dx = \sum_{k=0}^n (-1)^k(P^{(2k)}(0)+P^{(2k)}(\pi)).$

Solution. The proof is by induction over $n.$ If $n=0,$ then $P(x)=p$ is a constant and so $\displaystyle \int_0^{\pi} P(x) \sin x \ dx=p\int_0^{\pi} \sin x \ dx = 2p=P(0)+P(\pi).$
Now suppose the equality in the problem holds for polynomials of degree $2n$ and let $P(x)$ be a polynomial of degree $2n+2.$ Then using integration by parts with $P(x)=u$ and $\sin x \ dx = dv$ gives

$\displaystyle \int_0^{\pi} P(x) \sin x \ dx = P(0)+P(\pi)+\int_0^{\pi}P'(x)\cos x \ dx. \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$

Now, in $(1)$ , we use integration by parts again, this time with $P(x)=u$ and $\cos x \ dx = dv$ to get

$\displaystyle \int_0^{\pi} P(x) \sin x \ dx = P(0)+P(\pi)-(P'(0)+P'(\pi)) + \int_0^{\pi} P''(x) \sin x \ dx. \ \ \ \ \ \ \ \ \ \ \ \ (2)$

But since $P''(x)$ is a polynomial of degree $2n,$ we can use our induction hypothesis to write $\displaystyle \int_0^{\pi} P''(x) \sin x \ dx= \sum_{k=0}^n (-1)^k(P^{(2k+2)}(0)+P^{(2k+2)}(\pi))$ and so $(2)$ becomes

$\displaystyle \begin{aligned} \int_0^{\pi} P(x) \sin x \ dx = P(0)+P(\pi)-(P'(0)+P'(\pi)) + \sum_{k=0}^n (-1)^k(P^{(2k+2)}(0)+P^{(2k+2)}(\pi)) \\ = \sum_{k=0}^{n+1} (-1)^k(P^{(2k)}(0)+P^{(2k)}(\pi)). \ \Box \end{aligned}$

Problem 2. Let $n,a,b$ be integers with $n \ge 1$ and $b \ne 0.$ Let $\displaystyle c:=\frac{a}{b}$ and

$\displaystyle P(x):=\frac{x^n(a-bx)^n}{n!}.$

Show that $P^{(k)}(0)$ and $P^{(k)}(c)$ are integers for all $k \ge 0.$

Solution. Since $P(x)$ is a polynomial of degree $2n,$ we have $P^{(k)}(x)=0$ for all $x$ and $k > 2n$ and so there is nothing to prove for $k > 2n.$
Now, since $P(x)$ is a polynomial of degree $2n,$ we have

$\displaystyle P(x)=\sum_{k=0}^{2n} \frac{P^{(k)}(0)}{k!}x^k. \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$

On the other hand, using the binomial theorem, it is clear that

$\displaystyle P(x)=\frac{x^n(a-bx)^n}{n!}=\sum_{k=n}^{2n}\frac{p_k}{n!}x^k, \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)$

where $p_k$ are some integers.
So $(1)$ and $(2)$ together give $P^{(k)}(0)=0$ for $0 \le k \le n-1$ and $\displaystyle P^{(k)}(0)=\frac{k!}{n!}p_k$ for $n \le k \le 2n.$ Thus, since $\displaystyle \frac{k!}{n!}$ is an integer for $k \ge n,$ we have proved that $P^{(k)}(0)$ is an integer for all $k \ge 0.$
Finally, since $P(c-x)=P(x),$ we have $(-1)^kP^{(k)}(c-x)=P^{(k)}(x)$ for all $k \ge 0$ and thus $(-1)^kP^{(k)}(0)=P^{(k)}(c).$ So $P^{(k)}(c)$ is an integer because we have already proved that $P^{(k)}(0)$ is an integer. $\Box$

Problem 3. Show that $\pi$ is irrational.

Solution (Ivan Niven). Suppose, to the contrary, that $\pi$ is rational and put $\displaystyle \pi=\frac{a}{b},$ where $a,b > 0$ are integers. Let $n \ge 1$ be an integer and put

$\displaystyle P(x):=\frac{x^n(a-bx)^n}{n!}.$

Then by Problem 1 and 2, $\displaystyle \int_0^{\pi} P(x) \sin x \ dx$ must be an integer. But we are now going to prove that if $n$ is large enough, then $\displaystyle \int_0^{\pi} P(x) \sin x \ dx$ is not an integer and this contradiction proves that our assumption that $\pi$ is rational is false.
First note that, on the interval $[0,\pi],$ we have

$\displaystyle 0 \le P(x)\sin x \le P(x) =\frac{b^nx^n(\pi - x)^n}{n!} \le \frac{b^n \pi^{2n}}{n!}=\frac{(b \pi^2)^n}{n!}$

and thus

$\displaystyle 0 < \int_0^{\pi} P(x)\sin x \ dx \le \frac{(b \pi^2)^n}{n!} \pi. \ \ \ \ \ \ \ \ \ \ (*)$

(Note that $\displaystyle \int_0^{\pi} P(x)\sin x \ dx \ne 0$ because $P(x) \sin x$ is not identically zero on $[0, \pi]$ ).
But for every real number $r,$ we have $\displaystyle \lim_{n\to\infty} \frac{r^n}{n!}=0$ because the series $\displaystyle \sum_{n=0}^{\infty} \frac{r^n}{n!}$ is convergent (to $e^r$ ). So $\displaystyle \lim_{n\to\infty} \frac{(b \pi^2)^n}{n!} \pi=0.$ Hence, if $n$ is large enough, we will have $\displaystyle \frac{(b \pi^2)^n}{n!} \pi < 1$ and thus, by $(*),$

$\displaystyle 0 < \int_0^{\pi} P(x)\sin x \ dx < 1.$

Therefore $\displaystyle \int_0^{\pi} P(x)\sin x \ dx$ is not an integer if $n$ is large enough. $\Box$

Remark. Another way to say that a real number $t$ is irrational is to say that $t$ is not the root of a polynomial of degree one with integer coefficients. If $t$ is not a root of any polynomial with integer coefficients, then $t$ is called transcendental. Clearly every transcendental number is irrational. It is known that both $e$ and $\pi$ are in fact transcendental.

Exercise 1. Let $P(x)$ be a polynomial of degree $2n+1.$ Show that

$\displaystyle \int_0^{\pi} P(x) \sin x \ dx = \sum_{k=0}^n (-1)^k(P^{(2k)}(0)+P^{(2k)}(\pi)).$

Exercise 2. Show that the inequality $(*)$ in the solution of Problem 3 can be improved by proving that $\displaystyle \int_0^{\pi} P(x)\sin x \ dx \le \frac{(b \pi^2)^n}{4^n n!} \pi.$

Tag: transcendental

π is irrational