﻿ Napier's constant e is transcendental - Gonit Sora

## 14 May Napier's Constant "e" is Transcendental

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This theorem was first proved by Hermite in 1873. The below proof is near the one given by Hurwitz. We at first derive a couple of auxiliary results.

Let \$\$f(x)\$\$ be any polynomial of degree \$\$ mu\$\$ and \$\$ F(x)\$\$ the sum of its derivatives,

\$\$displaystyle F(x) := f(x)+f'(x)+f”(x)+ldots+f^{(mu)}(x).\$\$           (1)

consider the product \$\$ Phi(x) := e^{-x}F(x)\$\$. The derivative of this is simply

\$\$displaystyle Phi'(x) equiv e^{-x}(F'(x)-F(x)) equiv -e^{-x}f(x).\$\$

Applying the mean value theorem to the function \$\$Phi\$\$ on the interval with end points 0 and \$\$ x\$\$ gives

\$\$displaystyle Phi(x)-Phi(0) = e^{-x}F(x)-F(0) = Phi'(xi)x = -e^{-xi}f(xi)x,\$\$

which implies that \$\$ F(0) = e^{-x}F(x)+e^{-xi}f(xi)x\$\$. Thus we obtain the

Lemma 1. \$\$ F(0)e^x = F(x)+xe^{x-xi}f(xi)\$\$    (\$\$ xi\$\$ is between 0 and \$\$x\$\$)

When the polynomial \$\$ f(x)\$\$ is expanded by the powers of \$\$ x!-!a\$\$, one gets

\$\$displaystyle f(x) equiv f(a)+f'(a)(x!-!a)+f”(a)frac{(x!-!a)^2}{2!}+ldots+ f^{(mu)}(a)frac{(x!-!a)^{mu}}{mu!};\$\$

comparing this with (1) one gets the

Lemma 2. The value \$\$ F(a)\$\$ is obtained so that the polynomial \$\$ f(x)\$\$ is expanded by the powers of \$\$ x!-!a\$\$ and in this expansion the powers \$\$x!-!a\$\$, \$\$ (x!-!a)^2\$\$, …, \$\$ (x!-!a)^{mu}\$\$ are replaced respectively by the numbers 1!, 2!,…,\$\$mu !\$\$.

Now we begin the proof of the theorem. We have to show that there cannot be any equation

\$\$displaystyle c_0+c_1e+c_2e^2+ldots+c_ne^n = 0\$\$           (2)

with integer coefficients \$\$ c_i\$\$ and at least one of them distinct from zero. The proof is indirect. Let’s assume the contrary. We can presume that \$\$ c_0 neq 0\$\$.

For any positive integer \$\$nu\$\$, lemma 1 gives

\$\$displaystyle F(0)e^{nu} = F(nu)+nu e^{nu-xi_{nu}}f(xi_{nu}) quad(0 < xi_{nu} < nu)\$\$.         (3)

By virtue of this, one may write (2), multiplied by \$\$F(0)\$\$, as

\$\$displaystyle c_0F(0)!+!c_1F(1)!+!c_2F(2)!+!ldots!+!c_nF(n) = -[c_1e^{1-xi_1}f(xi_1)!+!2c_2e^{2-xi_2}f(xi_2)!+ldots+nc_ne^{n-xi_n}f(xi_n)].\$\$   (4)

We shall show that the polynomial \$\$ f(x)\$\$ can be chosen such that the left side of (4) is a non-zero integer whereas the right side has absolute value less than 1.

We choose

\$\$displaystyle f(x) := frac{x^{p-1}}{(p-1)!}[(x!-!1)(x!-!2)cdots(x!-!n)]^p,\$\$    (5)

where \$\$ p\$\$ is a positive prime number on which we later shall set certain conditions. We must determine the corresponding values \$\$ F(0)\$\$, \$\$F(1)\$\$, …, \$\$F(n)\$\$.

For determining \$\$ F(0)\$\$ we need, according to lemma 2, to expand \$\$ f(x)\$\$ by the powers of \$\$ x\$\$, getting

\$\$displaystyle f(x) = frac{1}{(p!-!1)!}[(-1)^{np}n!^px^{p-1}+A_1x^p+A_2x^p+1+ldots]\$\$

where \$\$ A_1,,A_2,,ldots\$\$ are integers, and to replace the powers \$\$ x^{p-1}\$\$, \$\$ x^p\$\$, \$\$ x^{p+1}\$\$, … with the numbers \$\$ (p!-!1)!\$\$, \$\$ p!\$\$, \$\$ (p!+!1)!\$\$, … We then get the expression

\$\$displaystyle F(0) = frac{1}{(p!-!1)!}[(-1)^{np}n!^p(p!-!1)!+A_1p! +A_2(p!+!1)!+ldots] = (-1)^{np}n!^p+pK_0,\$\$

in which \$\$ K_0\$\$ is an integer.

We now set for the prime \$\$ p\$\$ the condition \$\$ p > n\$\$. Then, \$\$n!\$\$ is not divisible by \$\$p\$\$, neither is the former addend \$\$ (-1)^{np}n!^p\$\$. On the other hand, the latter addend \$\$ pK_0\$\$ is divisible by \$\$p\$\$. Therefore:

(\$\$ alpha\$\$)    \$\$ F(0)\$\$ is a non-zero integer not divisible by \$\$ p\$\$.

For determining \$\$ F(1)\$\$, \$\$ F(2)\$\$, …, \$\$F(n)\$\$ we expand the polynomial \$\$ f(x)\$\$ by the powers of \$\$ x!-!nu\$\$, putting \$\$ x := nu!+!(x!-!nu)\$\$. Because \$\$f(x)\$\$ contains the factor \$\$ (x!-!nu)^p\$\$, we obtain an expansion of the form

\$\$displaystyle f(x) = frac{1}{(p!-!1)!}[B_p(x!-!nu)^p+B_{p+1}(x!-!nu)^{p+1}+ldots],\$\$

where the \$\$ B_i\$\$’s are integers. Using the lemma 2 then gives the result

\$\$displaystyle F(nu) = frac{1}{(p!-!1)!}[p!B_p+(p!+!1)!B_{p+1}+ldots] = pK_{nu},\$\$

with \$\$ K_{nu}\$\$ a certain integer. Thus:

(\$\$ beta\$\$)    \$\$F(1)\$\$, \$\$ F(2)\$\$, …, \$\$ F(n)\$\$ are integers all divisible by \$\$ p\$\$.

So, the left hand side of (4) is an integer having the form \$\$c_0F(0)+pK\$\$ with \$\$ K\$\$ an integer. The factor \$\$ F(0)\$\$ of the first addend is by (\$\$ alpha\$\$) indivisible by \$\$ p\$\$. If we set for the prime \$\$ p\$\$ a new requirement \$\$ p >mid c_0mid\$\$, then also the factor \$\$ c_0\$\$ is indivisible by \$\$ p\$\$, and thus likewise the whole addend \$\$ c_0F(0)\$\$. We conclude that the sum is not divisible by \$\$ p\$\$ and therefore:

(\$\$ gamma\$\$)    If \$\$ p\$\$ in (5) is a prime number greater than \$\$ n\$\$ and \$\$ mid c_0mid\$\$, then the left side of (4) is a non-zero integer.

We then examine the right hand side of (4). Because the numbers \$\$ xi_1\$\$, …, \$\$ xi_n\$\$ all are positive (cf. (3)), so the exponential factors \$\$e^{1-xi_1}\$\$, …, \$\$e^{n-xi_n}\$\$ all are \$\$ < e^n\$\$. If \$\$ 0 < x < n\$\$, then in the polynomial (5) the factors \$\$ x\$\$, \$\$ x!-!1\$\$, …, \$\$x!-!n\$\$ all have the absolute value less than \$\$ n\$\$ and thus

\$\$displaystyle mid f(x)mid < frac{1}{(p!-!1)!}n^{p-1}(n^n)^p = n^ncdotfrac{(n^{n+1})^{p-1}}{(p!-!1)!}.\$\$

Because \$\$ xi_1\$\$, …, \$\$xi_n\$\$ all are between 0 and \$\$ n\$\$ (cf. (3)), we especially have

\$\$displaystyle mid f(xi_{nu})mid < n^ncdotfrac{(n^{n+1})^{p-1}}{(p!-!1)!} quadforall ,nu = 1,,2,,ldots,,n.\$\$

If we denote by \$\$ c\$\$ the greatest of the numbers \$\$mid c_0mid\$\$, \$\$mid c_1mid\$\$, …, \$\$ mid c_nmid\$\$, then the right hand side of (4) has the absolute value less than

\$\$displaystyle (1!+!2!+!ldots!+!n)ce^nn^ncdotfrac{(n^{n+1})^{p-1}}{(p!-!1)!} = frac{n(n!+!1)}{2}c(en)^ncdotfrac{(n^{n+1})^{p-1}}{(p!-!1)!}.\$\$

But the limit of \$\$ frac{(n^{n+1})^{p-1}}{(p!-!1)!}\$\$ is 0 as \$\$ptoinfty\$\$, and therefore the above expression is less than 1 as soon as \$\$ p\$\$ exeeds some number \$\$ p_0\$\$.

If we determine the polynomial \$\$ f(x)\$\$ from the equation (5) such that the prime \$\$ p\$\$ is greater than the greatest of the numbers \$\$ n\$\$, \$\$mid c_0mid\$\$ and \$\$ p_0\$\$ (which is possible since there are infinitely many prime numbers), then the left side of (4) is a non-zero integer and thus \$\$geq  1\$\$, whereas the right side having the absolute value \$\$ < 1\$\$. The contradiction proves that the theorem is right.