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 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. $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$ Any help is much appreciated, thanks in advance!
 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: $$\lim_{n \to \infty } \frac{c^n}{n!} = 0$$
 November 21st, 2012, 12:46 PM #3 Global Moderator   Joined: May 2007 Posts: 6,821 Thanks: 723 Re: Proof of Transcendence of constant e Your Latex is screwed up.
 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}
 November 22nd, 2012, 01:16 PM #5 Global Moderator   Joined: May 2007 Posts: 6,821 Thanks: 723 Re: Proof of Transcendence of constant e Have you tried reading it? I can't.

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