# Thread: Proof of Transcendence of constant e

1. ## Proof of Transcendence of constant e

Hello all,

I'm a first year university student who is tasked with giving a presentation on the transcendence of e. I have a very basic grasp of the proof that we have been given to understand, however any explanations of ANY parts of the proof is very much appreciated as the math is way above my head (not quite sure if the proof itself is sound). Below is the pasted LATEX text of our proof as is right now. Please also list the topics that, in your opinion, must be covered in the 20-minute mathematical presentation of the topic.

To begin, we must assume that $e$ is algebraic. We suppose that $e$ is a root of the polynomial $g(x)=b_0+b_1x^1+...+b_rx^r$ where $b_i$ are in $\mathbb{Z}$ and $b_r$ doesn't not equal 0. $$I_p= \displaystyle\sum_{k=1}^{r} b_k \int^k_0 e^{k-x} f(x)dx$$
We will show that for $p$ large enough, $I_p$ is an integer, does not equal 0, but is arbitrarily close to 0.
Assume that $e$ is algebraic. Suppose that it is a root of the polynomial $g(x)=b_0+b_1x^1+...+b_rx^r$ where $b_i$ are in $\mathbb{Z}$ and $b_r$ doesn't not equal 0.
$$I_p= \displaystyle\sum_{k=1}^{r} b_k \int^k_0 e^{k-x} f(x)dx$$
We will show that for $p$ large enough, $I_p$ is an integer, does not equal 0, but is arbitrarily close to 0.
$$\int^t_0 e^{-x} f(x)dx = \int^t_0 (-e^{-x})' f(x)dx$$
Using integration by parts, we obtain the following expression.
$$-e^{-x} f(x) A - \int^t_0 (-e^{-x}) f'(x)dx$$
Plug in t,
$$-e^{-t}f(t) + f(0) + \int^t_0(-e^{-x}f'(x)dx$$
$$-e^{-t} \sum^n_{k=0} f^k (t) + \sum^n_k=0 f^k (0)$$
We will assume that $e$ is algebraic. The equation for the coefficients of the polynomial $e$ is $c_0 + c_1e+..+c_me^m=0$
$$f_p(x) = \frac{x^{p-1}(x-1)^p(x-2)^p...(x-k)^p}{(p-1)!}$$
The first p derivatives of $k$ $(x-1)^p g(x)$
$$f_p'(x) = p(x-1)^{p-1} g(x) + (x-1)^p g'(x)$$
Either will =0, or will cancel with (p-1)! and give integer.
$$I_p= \sum^m_k=0 c_k e^k \int^k_0 e^{-x} f_p(x) dx$$
$$= \sum^m_{k=0} c_k e^k \left( -e^{-k} \sum^n_{i=0} f^i (k) + \sum^n_{i=0} f^i (0)\right)$$
$$-\sum^m_{k=0} c_k (e^k)(e^{-k}) \sum^n_{i=0} f_p^i (k) + \left(\sum^m_{k=0} c_ke^k\right) \left(\sum^n_{i=0} f^i (0)\right)$$
$$= -\sum^m_{k=0} \sum^n_{i=0} c_k f_p^i (k)$$
Now, we will prove that this number does not equal 0 by showing that it is non-divisble by a prime number. Zero is of course divisible by any number, therefore a contradiction proves this point.
$$-\sum^m_{k=0} \sum^n_{i=p} c_k f_p^i (k) - \sum^n_{i=p-1} c_0 f^i (0)$$
Where $-\sum^m_{k=0} \sum^n_{i=p} c_k f_p^i (k)$ is divisble by p. If this part is differentiated $p$ times, everything can give you a nonzero that has a factor of p.
$$\sum^n_{i=p-1} c_0 ^{f(i)} (0) = c_0 f^{p-1} (0) + \sum^n_{i=p} c_0 f^{(i)} (0)$$
To get the nonzero, we will target $x^{p-1}$ times. Extra factor remains of p when you apply one derivative. $\sum^n_{i=p} c_0 f^i (0)$ is divisble by p.
So far, we have shown that $I_p \neq 0$, and that $I_p$ is and integer.
We will now justify that as p tends to infinity, $I_p$ tends to 0.
$$If\ you\ have\ a\ function \Rightarrow \left | g(x) \right |\leq M \ on \ \left [ a,b \right ],$$

$$then \left | \int_{b}^{a} g(x)dx \right |\leq M(b-a)$$
From here, we will estimate the integral of $\left | \int_{b}^{a} e^{-x}f_p(x)dx \right |$ where $0\leq a\leq m$
If $0\leq a\leq m$, then the difference $\left | x-j \right |$ is less than $m$.
$$\left | e^{-x} f_p(x) \right |\leq \left | f_p(x) \right | = \left | \frac{x^{p-1}(x-1)^p...(x-k)^p}{(p-1)!} \right |$$
$$\left | \frac{x^{p-1}(x-1)^p...(x-k)^p}{(p-1)!} \right |\leq \left | \frac{m^{p-1}m^p...m^p}{(p-1)!} \right |$$
$$=\frac{m^{(m+1)p+1}}{(p-1)!} < \frac{m^{(m+1)p}}{(p-1)!} = \frac{m^{(m+1)^p}}{(p-1)!}\rightarrow \bf0$$
Therefore, we can see that as $p$ tends to infinity, $I_p$ tends to 0. This limit tends to 0 because it is simply the limit $\lim_{n \to \infty } \frac{c^n}{n!} = 0$ in a more complicated form. This can be proved as follows:
Let $k$ be any natual number greater than $\ 2 \left | c \right |$. If $k>2 \left | c \right |$, then $\frac{ \left | c \right | } {k} < \frac{1} {2}$ and morever $\frac{\left | c \right |}{k}<\frac{1}{2}$ for $n\geq k$.
$$\frac{c^n}{n!} = \frac{c^k}{k!} \cdot \frac{c}{k} \cdot \frac{c}{k+1} \cdot ... \cdot \frac{c}{n} < \frac{c^k}{k!}\cdot \frac{1}{2} \cdot \frac{1}{2} \cdot...\cdot \frac{1}{2}$$
where $\frac{1}{2} \cdot \frac{1}{2} \cdot...\cdot \frac{1}{2} = \bf{n-k}$. Therefore the above equation equals:
$$\frac{c^k}{k!} \cdot \frac{1}{2^{n-k}} = 2^k\left ( \frac{c^k}{k!} \right )\frac{1}{2^{n}}$$
in which $\frac{1}{2^{n}}$ tends to 0. Therefore:
$$\lim_{n \to \infty } \frac{c^n}{n!} = 0$$

Any help is much appreciated, thanks in advance!

2. ## Re: Proof of Transcendence of constant e

Originally Posted by czar01
To begin, we must assume that $e$ is algebraic. We suppose that $e$ is a root of the polynomial $g(x)=b_0+b_1x^1+...+b_rx^r$ where $b_i$ are in $\mathbb{Z}$ and $b_r$ doesn't not equal 0. $$I_p= \displaystyle\sum_{k=1}^{r} b_k \int^k_0 e^{k-x} f(x)dx$$
We will show that for $p$ large enough, $I_p$ is an integer, does not equal 0, but is arbitrarily close to 0.
$$\frac{c^n}{n!} = \frac{c^k}{k!} \cdot \frac{c}{k} \cdot \frac{c}{k+1} \cdot ... \cdot \frac{c}{n} < \frac{c^k}{k!}\cdot \frac{1}{2} \cdot \frac{1}{2} \cdot...\cdot \frac{1}{2}$$
$$\frac{c^n}{n!} = \frac{c^k}{k!} \cdot \frac{c}{k} \cdot \frac{c}{k+1} \cdot ... \cdot \frac{c}{n} < \frac{c^k}{k!}\cdot \frac{1}{2} \cdot \frac{1}{2} \cdot...\cdot \frac{1}{2}$$ gives
$\frac{c^n}{n!} = \frac{c^k}{k!} \cdot \frac{c}{k} \cdot \frac{c}{k+1} \cdot ... \cdot \frac{c}{n} < \frac{c^k}{k!}\cdot \frac{1}{2} \cdot \frac{1}{2} \cdot...\cdot \frac{1}{2}$
Using the advanced menu toolbar the $\boxed{\Sigma}$ gives the  wrap. You can simply edit your post. Get rid of the \$ signs hightlight the code and click the $\boxed{\Sigma}$ .