Friendship of $e$ and $\pi$

A number which doesn't satisfy any algebraic equation of undivided (solid) rational coefficients is called transcendental.

I will here prove that $e$ is transcendental:

I'll start from the formula achieved from integration by parts:

$\large\int^{r}_{0}e^{-x}f(x)dx = - [e^{-x}f(x)]^r_0 + \int^{r}_{0}e^{-x}f'(x) dx$ .

If $f(x)$ is a polynomial of $x N$ th degree, we get:

$F(x) = f(x)+f'(x)+f''(x)+...+f^{N}(x)$

$\int^{r}_{0}e^{-x}f(x) dx = F(0) - F(r) e^{-r}$

From which:

$F(r) = F(0) e^r - e^r \int^{r}_{0} e^{-x} f(x)$

Let's now suppose that $e$ satisfies an algebraic equation: $a_0 e^n + a_1e^{n-1} +...+ a_{n-1}e + a_n = 0$ ,

and let's define for $f(x)$ : $f(x) = \frac{1}{(p-1)!} x^{p-1} (1-x)^p (2-x)^p...(n-x)^p$ .

So the equation: $F(r) = F(0) e^r - e^r \int^{r}_{0} e^{-x} f(x)$ implies, that if $r = n$ , so $r = n-1, r = n-2 ,... r = 0$ there are totally $n+1$ equations, which we multiply by $a_0, a_1, a_2, ..., a_n$ and sum together.

If we then substitute the $F(0)$ on the right hand, we get:

$a_0 F(n) + a_1 F(n-1) +...+a_{n-1} F(1) + a_n F(0) = - \sum^{n}_{k=0} a_k e^{n-k} \int^{n-k}_{0} e^{-x} f(x) dx$ .

Which is impossible for $e$ , as:

If we plug in $f(x)$ any of the numbers $1, 2, ...,n$ we get either $0$ or an integer divisible by $p$ . So for $x = 0$ , we get $f(0) = f'(0) = f''(0) = ...f^{(p-2)}(0) = 0$ ;

Further $f^{(p-1)}(0) = (1 \times 2 \times 3 ...\times n)^p$ ; $f^{(p)}(0); f^{(p+1)}(0)$ are integers, which again are divisible by $p$ . So if $p$ is a prime number $p>n$ ,

$F(0) = n \in \mathbb Z$ not divisible by $p$ .

Let also $p> \vert a_n \vert$ , then the polynomial:

$a_0 F(n) + a_1 F(n-1) +...+ a_{n-1} F(1) + a_n F(0)$

is a whole number not divisible by $p$ .

From $- \sum^{n}_{k=0} a_k e^{n-k} \int^{n-k}_{0} e^{-x} f(x) dx$ we assume from the Intermediate value theorem, that:

$\vert \int^{n-k}_{0} f(x) e^{-x} dx \vert = \vert f(\xi) \vert (1 - e^{-n+k}) < \vert f(\xi) \vert < \frac{n^{(n+1)p-1}}{(p-1)!}$ .

So let

$M = \vert a_0 \vert e^n + \vert a_1 \vert e^{n-1} + \vert a_2 \vert e^{n-2} +...+ \vert a_n \vert$ , then

$- \sum^{n}_{k=0} a_k e^{n-k} \int^{n-k}_{0} e^{-x} f(x) dx < M \frac{n^{(n+1)p-1}}{(p-1)!}$ in absolute value.

However the prime number $p$ is defined only by conditions $p > \vert a_n \vert, p > n$ , so it can be of any size. So if they are large enough, the expression: $M \frac{n^{(n+1)p-1}}{(p-1)!}$ becomes arbitrarily small, for example we could choose a $p$ large enough, so that the expression is smaller than $1$ , then the equation

$\vert a_0 F(n) + a_1 F(n-1) +...+a_{n-1} F(1) + a_n F(0) = - \sum^{n}_{k=0} a_k e^{n-k} \int^{n-k}_{0} e^{-x} f(x) dx\ \vert$ must equal at least $1$ in it's absolute value equal to a number $<1$ , which is impossible since an absolute value can't be negative. That means that $e$ doesn't satisfy the equation $a_0 e^n + a_1e^{n-1} +...+ a_{n-1}e + a_n = 0$ , which means, that $e$ $\boxed{\text{is transcendental}}$ .

Therefore the answer to the claiming is $\boxed{\text{false} }$