## 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 be any polynomial of degree and the sum of its derivatives,

(1)

consider the product . The derivative of this is simply

Applying the mean value theorem to the function on the interval with end points 0 and gives

which implies that . Thus we obtain the

Lemma 1. ( is between 0 and )

When the polynomial is expanded by the powers of , one gets

comparing this with (1) one gets the

Lemma 2. The value is obtained so that the polynomial is expanded by the powers of and in this expansion the powers , , ..., 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

(2)

with integer coefficients and at least one of them distinct from zero. The proof is indirect. Let's assume the contrary. We can presume that .

For any positive integer , lemma 1 gives

. (3)

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

(4)

We shall show that the polynomial 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

(5)

where is a positive prime number on which we later shall set certain conditions. We must determine the corresponding values , , ..., .

For determining we need, according to lemma 2, to expand by the powers of , getting

where are integers, and to replace the powers , , , ... with the numbers `\( (p!-!1)\)`

, `\( p\)`

, `\( (p!+!1)\)`

, ... We then get the expression

in which is an integer.

We now set for the prime the condition n" />. Then, `\(n\)`

is not divisible by , neither is the former addend . On the other hand, the latter addend is divisible by . Therefore:

() is a non-zero integer not divisible by .

For determining , , ..., we expand the polynomial by the powers of , putting . Because contains the factor , we obtain an expansion of the form

where the 's are integers. Using the lemma 2 then gives the result

with a certain integer. Thus:

() , , ..., are integers all divisible by .

So, the left hand side of (4) is an integer having the form with an integer. The factor of the first addend is by () indivisible by . If we set for the prime a new requirement mid c_0mid" />, then also the factor is indivisible by , and thus likewise the whole addend . We conclude that the sum is not divisible by and therefore:

() If in (5) is a prime number greater than and , then the left side of (4) is a non-zero integer.

We then examine the right hand side of (4). Because the numbers , ..., all are positive (cf. (3)), so the exponential factors , ..., all are . If , then in the polynomial (5) the factors , , ..., all have the absolute value less than and thus

Because , ..., all are between 0 and (cf. (3)), we especially have

If we denote by the greatest of the numbers , , ..., , then the right hand side of (4) has the absolute value less than

But the limit of is 0 as , and therefore the above expression is less than 1 as soon as exeeds some number .

If we determine the polynomial from the equation (5) such that the prime is greater than the greatest of the numbers , and (which is possible since there are infinitely many prime numbers), then the left side of (4) is a non-zero integer and thus , whereas the right side having the absolute value . The contradiction proves that the theorem is right.

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