e is transcendental

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

μ

and

F

⁢

(

x

)

the sum of its

derivatives

,

F

⁢

(

x

)

:=

f

⁢

(

x

)

+

f

′

⁢

(

x

)

+

f

′′

⁢

(

x

)

+

…

+

f

(

μ

)

⁢

(

x

)

.

(1)

consider the

product

Φ

⁢

(

x

)

:=

e

-

x

⁢

F

⁢

(

x

)

.  The derivative of this is simply

Φ

′

⁢

(

x

)

≡

e

-

x

⁢

(

F

′

⁢

(

x

)

-

F

⁢

(

x

)

)

≡

-

e

-

x

⁢

f

⁢

(

x

)

.

Applying the mean value theorem (

http://planetmath.org/MeanValueTheorem

) to the

function

Φ

on the

interval

with

end points

0 and

x

gives

Φ

⁢

(

x

)

-

Φ

⁢

(

0

)

=

e

-

x

⁢

F

⁢

(

x

)

-

F

⁢

(

0

)

=

Φ

′

⁢

(

ξ

)

⁢

x

=

-

e

-

ξ

⁢

f

⁢

(

ξ

)

⁢

x

,

which implies that

F

⁢

(

0

)

=

e

-

x

⁢

F

⁢

(

x

)

+

e

-

ξ

⁢

f

⁢

(

ξ

)

⁢

x

.  Thus we obtain the

Lemma 1.

F

⁢

(

0

)

⁢

e

x

=

F

⁢

(

x

)

+

x

⁢

e

x

-

ξ

⁢

f

⁢

(

ξ

)

(

ξ

is between 0 and

x

)

When the polynomial

f

⁢

(

x

)

is expanded by the powers of

x

-

a

, one gets

f

⁢

(

x

)

≡

f

⁢

(

a

)

+

f

′

⁢

(

a

)

⁢

(

x

-

a

)

+

f

′′

⁢

(

a

)

⁢

(

x

-

a

)

2

2

!

+

…

+

f

(

μ

)

⁢

(

a

)

⁢

(

x

-

a

)

μ

μ

!

;

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

)

μ

are replaced respectively by the numbers 1!, 2!, …,

μ

!

.

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

+

…

+

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

≠

0

.

For any

positive integer

ν

, lemma 1 gives

F

(

0

)

e

ν

=

F

(

ν

)

+

ν

e

ν

-

ξ

ν

f

(

ξ

ν

)

(

0

<

ξ

ν

<

ν

)

.

(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

)

+

…

+

c

n

⁢

F

⁢

(

n

)

=

-

[

c

1

⁢

e

1

-

ξ

1

⁢

f

⁢

(

ξ

1

)

+

2

⁢

c

2

⁢

e

2

-

ξ

2

⁢

f

⁢

(

ξ

2

)

+

…

+

n

⁢

c

n

⁢

e

n

-

ξ

n

⁢

f

⁢

(

ξ

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

)

:=

x

p

-

1

(

p

-

1

)

!

⁢

[

(

x

-

1

)

⁢

(

x

-

2

)

⁢

⋯

⁢

(

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

)

=

1

(

p

-

1

)

!

⁢

[

(

-

1

)

n

⁢

p

⁢

n

!

p

⁢

x

p

-

1

+

A

1

⁢

x

p

+

A

2

⁢

x

p

+

1

+

…

]

where

A

1

,

A

2

,

…

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

)

=

1

(

p

-

1

)

!

⁢

[

(

-

1

)

n

⁢

p

⁢

n

!

p

⁢

(

p

-

1

)

!

+

A

1

⁢

p

!

+

A

2

⁢

(

p

+

1

)

!

+

…

]

=

(

-

1

)

n

⁢

p

⁢

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

)

n

⁢

p

⁢

n

!

p

.  On the other hand, the latter addend

p

⁢

K

0

is divisible by

p

.  Therefore:

(

α

)

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

-

ν

, putting

x

:=

ν

+

(

x

-

ν

)

.  Because

f

⁢

(

x

)

the factor

(

x

-

ν

)

p

, we obtain an of the form

f

⁢

(

x

)

=

1

(

p

-

1

)

!

⁢

[

B

p

⁢

(

x

-

ν

)

p

+

B

p

+

1

⁢

(

x

-

ν

)

p

+

1

+

…

]

,

where the

B

i

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

F

⁢

(

ν

)

=

1

(

p

-

1

)

!

⁢

[

p

!

⁢

B

p

+

(

p

+

1

)

!

⁢

B

p

+

1

+

…

]

=

p

⁢

K

ν

,

with

K

ν

a certain integer.  Thus:

(

β

)

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

)

+

p

⁢

K

with

K

an integer.  The factor

F

⁢

(

0

)

of the first addend is by (

α

) 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:

(

γ

) 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

ξ

1

, …,

ξ

n

all are positive (cf. (3)), so the

e

1

-

ξ

1

, …,

e

n

-

ξ

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

|

f

⁢

(

x

)

|

<

1

(

p

-

1

)

!

⁢

n

p

-

1

⁢

(

n

n

)

p

=

n

n

⋅

(

n

n

+

1

)

p

-

1

(

p

-

1

)

!

.

Because

ξ

1

, …,

ξ

n

all are between 0 and

n

(cf. (3)), we especially have

|

f

⁢

(

ξ

ν

)

|

<

n

n

⋅

(

n

n

+

1

)

p

-

1

(

p

-

1

)

!

∀

ν

=

1

,

2

,

…

,

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

+

…

+

n

)

⁢

c

⁢

e

n

⁢

n

n

⋅

(

n

n

+

1

)

p

-

1

(

p

-

1

)

!

=

n

⁢

(

n

+

1

)

2

⁢

c

⁢

(

e

⁢

n

)

n

⋅

(

n

n

+

1

)

p

-

1

(

p

-

1

)

!

.

But the limit of

(

n

n

+

1

)

p

-

1

(

p

-

1

)

!

is 0 as

p

→

∞

, 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

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