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[parent] e is transcendental (Theorem)

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 > \vert c_0\vert$, 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 $ \vert c_0\vert$, 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

$\displaystyle \vert f(x)\vert < \frac{1}{(p\!-\!1)!}n^{p-1}(n^n)^p = n^n\cdot\frac{(n^{n+1})^{p-1}}{(p\!-\!1)!}.$
Because $ \xi_1$, ..., $ \xi_n$ all are between 0 and $ n$ (cf. (3)), we especially have
$\displaystyle \vert f(\xi_{\nu})\vert < n^n\cdot\frac{(n^{n+1})^{p-1}}{(p\!-\!1)!} \quad\forall \,\nu = 1,\,2,\,\ldots,\,n.$
If we denote by $ c$ the greatest of the numbers $ \vert c_0\vert$, $ \vert c_1\vert$, ..., $ \vert c_n\vert$, then the right hand side of (4) has the absolute value less than
$\displaystyle (1\!+\!2\!+\!\ldots\!+\!n)ce^nn^n\cdot\frac{(n^{n+1})^{p-1}}{(p\!-\!1)!} = \frac{n(n\!+\!1)}{2}c(en)^n\cdot\frac{(n^{n+1})^{p-1}}{(p\!-\!1)!}.$
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$, $ \vert c_0\vert$ 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 $ \geqq 1$, whereas the right side having the absolute value $ < 1$. The contradiction proves that the theorem is right.

Bibliography

1
ERNST LINDELÖF: Differentiali- ja integralilasku ja sen sovellutukset I. WSOY, Helsinki (1950).



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See Also: natural log base, fundamental theorem of transcendence, Lindemann-Weierstrass theorem, e is not a quadratic irrational, $e^r$ is irrational for $r\in\mathbb{Q}\setminus\{0\}$, example of Taylor polynomials for the exponential function, proof that e is not a natural number

Other names:  $e$ is transcendental, transcendence of e

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Cross-references: contradiction, limit, right, factor, divisible, expression, expand, prime number, absolute value, positive, coefficients, integer, equation, numbers, expanded, implies, end points, interval, function, product, derivatives, sum, polynomial, near, transcendental, Napier's constant
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This is version 36 of e is transcendental, born on 2005-04-04, modified 2008-01-17.
Object id is 6929, canonical name is EIsTranscendental.
Accessed 14391 times total.

Classification:
AMS MSC: 11J81 (Number theory :: Diophantine approximation, transcendental number theory :: Transcendence )
 11J82 (Number theory :: Diophantine approximation, transcendental number theory :: Measures of irrationality and of transcendence)
 26C05 (Real functions :: Polynomials, rational functions :: Polynomials: analytic properties, etc.)
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