# 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,

 $\displaystyle F(x)\;:=\;f(x)+f^{\prime}(x)+f^{\prime\prime}(x)+\ldots+f^{(\mu)% }(x).$ (1)

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

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

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

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

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

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

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

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

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 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

 $\displaystyle c_{0}+c_{1}e+c_{2}e^{2}+\ldots+c_{n}e^{n}\;=\;0$ (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

 $\displaystyle c_{0}F(0)\!+\!c_{1}F(1)\!+\!c_{2}F(2)\!+\!\ldots\!+\!c_{n}F(n)\;% =\;-[c_{1}e^{1-\xi_{1}}f(\xi_{1})\!+\!2c_{2}e^{2-\xi_{2}}f(\xi_{2})\!+\ldots+% nc_{n}e^{n-\xi_{n}}f(\xi_{n})].$ (4)

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

We choose

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

 $f(x)\;=\;\frac{1}{(p\!-\!1)!}[(-1)^{np}n!^{p}x^{p-1}+A_{1}x^{p}+A_{2}x^{p}+1+\ldots]$

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

 $F(0)\;=\;\frac{1}{(p\!-\!1)!}[(-1)^{np}n!^{p}(p\!-\!1)!+A_{1}p!+A_{2}(p\!+\!1)% !+\ldots]\;=\;(-1)^{np}n!^{p}+pK_{0},$

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

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

where the $B_{i}$’s are integers.  Using the lemma 2 then gives the result

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

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

So, the left hand 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  $0, then in the polynomial (5) the factors $x$, $x\!-\!1$, …, $x\!-\!n$ all have the absolute value less than $n$ and thus

 $|f(x)|\;<\;\frac{1}{(p\!-\!1)!}n^{p-1}(n^{n})^{p}=n^{n}\cdot\frac{(n^{n+1})^{p% -1}}{(p\!-\!1)!}.$

Because $\xi_{1}$, …, $\xi_{n}$ all are between 0 and $n$ (cf. (3)), we especially have

 $|f(\xi_{\nu})|\;<\;n^{n}\cdot\frac{(n^{n+1})^{p-1}}{(p\!-\!1)!}\quad\forall\,% \nu=1,\,2,\,\ldots,\,n.$

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

 $(1\!+\!2\!+\!\ldots\!+\!n)ce^{n}n^{n}\cdot\frac{(n^{n+1})^{p-1}}{(p\!-\!1)!}\;% =\;\frac{n(n\!+\!1)}{2}c(en)^{n}\cdot\frac{(n^{n+1})^{p-1}}{(p\!-\!1)!}.$

But the limit of $\frac{(n^{n+1})^{p-1}}{(p\!-\!1)!}$ is 0 as  $p\to\infty$, 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 $<1$.  The contradiction proves that the theorem is right.

## References

• 1 Ernst Lindelöf: Differentiali- ja integralilasku ja sen sovellutukset I.  WSOY, Helsinki (1950).
 Title e is transcendental Canonical name EIsTranscendental Date of creation 2013-03-22 15:10:32 Last modified on 2013-03-22 15:10:32 Owner pahio (2872) Last modified by pahio (2872) Numerical id 40 Author pahio (2872) Entry type Theorem Classification msc 26C05 Classification msc 11J82 Classification msc 11J81 Synonym $e$ is transcendental Synonym transcendence of e Related topic NaturalLogBase Related topic FundamentalTheoremOfTranscendence Related topic LindemannWeierstrassTheorem Related topic EIsIrrational Related topic ErIsIrrationalForRinmathbbQsetminus0 Related topic ExampleOfTaylorPolynomialsForTheExponentialFunction Related topic ProofThatEIsNotANaturalNumber