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

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.

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