My Math Forum  

Go Back   My Math Forum > College Math Forum > Applied Math

Applied Math Applied Math Forum


Reply
 
LinkBack Thread Tools Display Modes
November 21st, 2012, 07:22 AM   #1
Newbie
 
Joined: Nov 2012

Posts: 3
Thanks: 0

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.

Any help is much appreciated, thanks in advance!
czar01 is offline  
 
November 21st, 2012, 07:23 AM   #2
Newbie
 
Joined: Nov 2012

Posts: 3
Thanks: 0

Re: Proof of Transcendence of constant e

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,
$$-e^{-t}f(t) + f(0) + \int^t_0(-e^{-x}f'(x)dx$$
$$-e^{-t} \sum^n_{k=0} f^k (t) + \sum^n_k=0 f^k (0)$$
{\bf{Proof by contradiction}}
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:
$$\lim_{n \to \infty } \frac{c^n}{n!} = 0$$
czar01 is offline  
November 21st, 2012, 12:46 PM   #3
Global Moderator
 
Joined: May 2007

Posts: 6,607
Thanks: 616

Re: Proof of Transcendence of constant e

Your Latex is screwed up.
mathman is offline  
November 22nd, 2012, 06:01 AM   #4
Newbie
 
Joined: Nov 2012

Posts: 3
Thanks: 0

Re: Proof of Transcendence of constant e

\documentclass[11pt]{article}
\usepackage{amsmath}
\usepackage{amsfonts}


\setlength{\parindent}{0pt}
\setlength{\parskip}{12pt}




\begin{document}
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,

$$-e^{-t}f(t) + f(0) + \int^t_0(-e^{-x}f'(x)dx$$

$$-e^{-t} \sum^n_{k=0} f^k (t) + \sum^n_k=0 f^k (0)$$

{\bf{Proof by contradiction}}

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:

$$\lim_{n \to \infty } \frac{c^n}{n!} = 0$$



\end{document}
czar01 is offline  
November 22nd, 2012, 01:16 PM   #5
Global Moderator
 
Joined: May 2007

Posts: 6,607
Thanks: 616

Re: Proof of Transcendence of constant e

Have you tried reading it? I can't.
mathman is offline  
Reply

  My Math Forum > College Math Forum > Applied Math

Tags
constant, proof, transcendence



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Transcendence of Gelfond's-like Constants mathbalarka Number Theory 23 May 9th, 2013 05:28 AM
Conjecture regarding transcendence of a number mathbalarka Calculus 0 September 3rd, 2012 07:02 AM
Limit and Transcendence mathbalarka Applied Math 3 June 14th, 2012 12:43 PM
Is this a constant??? JamieStarr Algebra 3 November 24th, 2010 03:18 PM
Limit and Transcendence mathbalarka Number Theory 1 December 31st, 1969 04:00 PM





Copyright © 2018 My Math Forum. All rights reserved.