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           (1)

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

displaystyle Phi

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

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

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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