# 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^{},

$F(x):=f(x)+{f}^{\prime}(x)+{f}^{\prime \prime}(x)+\mathrm{\dots}+{f}^{(\mu )}(x).$ | (1) |

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

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

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

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

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

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

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

$$f(x)\equiv f(a)+{f}^{\prime}(a)(x-a)+{f}^{\prime \prime}(a)\frac{{(x-a)}^{2}}{2!}+\mathrm{\dots}+{f}^{(\mu )}(a)\frac{{(x-a)}^{\mu}}{\mu !};$$ |

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

${c}_{0}+{c}_{1}e+{c}_{2}{e}^{2}+\mathrm{\dots}+{c}_{n}{e}^{n}=\mathrm{\hspace{0.33em}0}$ | (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 ${c}_{0}\ne 0$.

For any positive integer $\nu $, lemma 1 gives

$$ | (3) |

By virtue of this, one may write (2), multiplied by $F(0)$, as

${c}_{0}F(0)+{c}_{1}F(1)+{c}_{2}F(2)+\mathrm{\dots}+{c}_{n}F(n)=-[{c}_{1}{e}^{1-{\xi}_{1}}f({\xi}_{1})+2{c}_{2}{e}^{2-{\xi}_{2}}f({\xi}_{2})+\mathrm{\dots}+n{c}_{n}{e}^{n-{\xi}_{n}}f({\xi}_{n})].$ | (4) |

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

We choose

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

$$f(x)=\frac{1}{(p-1)!}[{(-1)}^{np}{n!}^{p}{x}^{p-1}+{A}_{1}{x}^{p}+{A}_{2}{x}^{p}+1+\mathrm{\dots}]$$ |

where ${A}_{1},{A}_{2},\mathrm{\dots}$ 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

$$F(0)=\frac{1}{(p-1)!}[{(-1)}^{np}{n!}^{p}(p-1)!+{A}_{1}p!+{A}_{2}(p+1)!+\mathrm{\dots}]={(-1)}^{np}{n!}^{p}+p{K}_{0},$$ |

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 ${(-1)}^{np}{n!}^{p}$. On the other hand, the latter addend $p{K}_{0}$ 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

$$f(x)=\frac{1}{(p-1)!}[{B}_{p}{(x-\nu )}^{p}+{B}_{p+1}{(x-\nu )}^{p+1}+\mathrm{\dots}],$$ |

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

$$F(\nu )=\frac{1}{(p-1)!}[p!{B}_{p}+(p+1)!{B}_{p+1}+\mathrm{\dots}]=p{K}_{\nu},$$ |

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 $$. If $$, then in the polynomial (5) the factors $x$, $x-1$, …, $x-n$ all have the absolute value less than $n$ and thus

$$ |

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

$$ |

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

$$(1+2+\mathrm{\dots}+n)c{e}^{n}{n}^{n}\cdot \frac{{({n}^{n+1})}^{p-1}}{(p-1)!}=\frac{n(n+1)}{2}c{(en)}^{n}\cdot \frac{{({n}^{n+1})}^{p-1}}{(p-1)!}.$$ |

But the limit of $\frac{{({n}^{n+1})}^{p-1}}{(p-1)!}$ is 0 as $p\to \mathrm{\infty}$, 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 $$. 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 |

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Related topic | ExampleOfTaylorPolynomialsForTheExponentialFunction |

Related topic | ProofThatEIsNotANaturalNumber |