e is transcendental

Theorem.

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 $\mu$ and $F(x)$ the sum of its derivatives,

$F(x):=f(x)+f^{\prime }(x)+f^{\prime \prime }(x)+\mathrm{\dots }+f^{(\mu )}(x).$(1)

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

$${\mathrm{\Phi }}^{\prime }(x)\equiv e^{-x}(F^{\prime }(x)-F(x))\equiv -e^{-x}f(x).$$

Applying the mean value theorem (http://planetmath.org/MeanValueTheorem) to the function $\mathrm{\Phi }$ on the interval with end points 0 and $x$ gives

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

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

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

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

$$f(x)\equiv f(a)+f^{\prime }(a)(x-a)+f^{\prime \prime }(a)\frac{{(x-a)}^2}{2!}+\mathrm{\dots }+f^{(\mu )}(a)\frac{{(x-a)}^{\mu }}{\mu !};$$

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 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

$c_0+c_{1}e+c_2{e}^2+\mathrm{\dots }+c_n{e}^n=\mathrm{\hspace{0.33em}0}$(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  $c_0\ne 0$.

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

(3)

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

$c_{0}F(0)+c_{1}F(1)+c_{2}F(2)+\mathrm{\dots }+c_{n}F(n)=-[c_1{e}^{1-{\xi }_1}f({\xi }_1)+2c_2{e}^{2-{\xi }_2}f({\xi }_2)+\mathrm{\dots }+n{c}_n{e}^{n-{\xi }_n}f({\xi }_n)].$(4)

We shall show that the polynomial $f(x)$ can be chosen such that the left has absolute value less than 1.

We choose

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

$$f(x)=\frac{1}{(p-1)!}[{(-1)}^{np}{n!}^p{x}^{p-1}+A_1{x}^p+A_2{x}^p+1+\mathrm{\dots }]$$

where $A_1,A_2,\mathrm{\dots }$ 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

$$F(0)=\frac{1}{(p-1)!}[{(-1)}^{np}{n!}^p(p-1)!+A_{1}p!+A_2(p+1)!+\mathrm{\dots }]={(-1)}^{np}{n!}^p+p{K}_0,$$

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 $p{K}_0$ 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 $f(x)$ the factor ${(x-\nu )}^p$, we obtain an of the form

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

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

$$F(\nu )=\frac{1}{(p-1)!}[p!B_p+(p+1)!B_{p+1}+\mathrm{\dots }]=p{K}_{\nu },$$

with $K_{\nu }$ a certain integer.  Thus:
($\beta$) $F(1)$, $F(2)$, …, $F(n)$ are integers all divisible by $p$.

So, the left hand of (4) is an integer having the form $c_{0}F(0)+pK$ 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  $p>|c_0|$,  then also the factor $c_0$ is indivisible by $p$, and thus likewise the whole addend $c_{0}F(0)$.  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 $|c_0|$, then the left of (4) is a nonzero integer.

We then examine the right hand of (4).  Because the numbers ${\xi }_1$, …, ${\xi }_n$ all are positive (cf. (3)), so the $e^{1-{\xi }_1}$, …, $e^{n-{\xi }_n}$ all are .  If  , then in the polynomial (5) the factors $x$,$x-1$, …, $x-n$ all have the absolute value less than $n$ and thus

Because ${\xi }_1$, …, ${\xi }_n$ all are between 0 and $n$ (cf. (3)), we especially have

If we denote by $c$ the greatest of the numbers $|c_0|$, $|c_1|$, …, $|c_n|$, then the right hand of (4) has the absolute value less than

$$(1+2+\mathrm{\dots }+n)c{e}^n{n}^n\cdot \frac{{(n^{n+1})}^{p-1}}{(p-1)!}=\frac{n(n+1)}{2}c{(en)}^n\cdot \frac{{(n^{n+1})}^{p-1}}{(p-1)!}.$$

But the limit of $\frac{{(n^{n+1})}^{p-1}}{(p-1)!}$ is 0 as  $p\to \mathrm{\infty }$, 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 $n$, $|c_0|$ and $p_0$ (which is possible since there are infinitely many prime numbers (http://planetmath.org/ProofThatThereAreInfinitelyManyPrimes)), then the having the absolute value .  The contradiction proves that the theorem is right.