Famous Theorems of Mathematics/e is transcendental

The mathematical constant $\text{e}=\sum _{n=0}^{\mathrm{\infty }}\frac{1}{n!}=2.718281\dots$ is a transcendental number (or inalgebraic).

In other words, it is not a root of any polynomial with integer coefficients.

Proof

Let us assume that $\text{e}$ is algebraic, so there exists a polynomial

$P(x)=a_0+a_{1}x+a_2{x}^2+\cdots +a_n{x}^n\in \mathbb{Z}[x]$

such that $P(\text{e})=0$.

Part 1

Let $f(x)$ be a polynomial of degree $d$. Let us define $F(x)=\sum _{k=0}^d{f}^{(k)}(x)$. Taking its derivative yields:

$F^{\prime }(x)=\sum _{k=0}^d{f}^{(k+1)}(x)=\sum _{k=1}^d{f}^{(k)}(x)=F(x)-f(x)$

Let us define $G(x)={\text{e}}^{-x}F(x)$. Taking its derivative yields:

$G^{\prime }(x)={\text{e}}^{-x}{F}^{\prime }(x)-{\text{e}}^{-x}F(x)={\text{e}}^{-x}[F^{\prime }(x)-F(x)]=-e^{-x}f(x)$

By the fundamental theorem of calculus, we get:

$\begin{array}{cc}G(x)-G(0)={\text{e}}^{-x}F(x)-{\text{e}}^{-0}F(0)& =-\underset{0}{\overset{x}{\int }}{\text{e}}^{-t}f(t)dt\\ F(x)-{\text{e}}^{x}F(0)& =-\underset{0}{\overset{x}{\int }}{\text{e}}^{x-t}f(t)dt\end{array}$

Now let:

$\begin{array}{c}F(m)-{\text{e}}^{m}F(0)=-\underset{0}{\overset{m}{\int }}{\text{e}}^{m-t}f(t)dt=A_m\\ a_{m}F(m)-a_m{\text{e}}^{m}F(0)=a_m{A}_m\end{array}$

Summing all the terms yields

$\begin{array}{c}\sum _{m=1}^n{a}_{m}F(m)-\sum _{m=1}^n{a}_m{\text{e}}^{m}F(0)=\sum _{m=1}^n{a}_m{A}_m\\ \sum _{m=1}^n{a}_{m}F(m)-F(0)\sum _{m=1}^n{a}_m{\text{e}}^m=\sum _{m=1}^n{a}_m{A}_m\\ \sum _{m=1}^n{a}_{m}F(m)-F(0)(-a_0)=\sum _{m=1}^n{a}_m{A}_m\\ \sum _{m=0}^n{a}_{m}F(m)=\sum _{m=1}^n{a}_m{A}_m\end{array}$

Part 2

Lemma: Let $f(x)$ be a polynomial with a root $x_0$ of multiplicity $d\ge 1$. Then $f^{(j)}(x_0)=0$ for all $0\le j\le d-1$.

Proof: By strong induction.

Let us write $f(x)=(x-x_0{)}^{d}Q(x)$, with $Q(x)$ a polynomial such that $Q(x_0)\ne 0$.

For $d=1$ we get:

$f^{(0)}(x_0)=f(x_0)=0$

Assume that for all $1\le d\le k$ the claim holds for all $0\le j\le d-1$.
We shall prove that for $d=k+1$ the claim holds for all $0\le j\le k$:

$\begin{array}{cc}f(x)& =(x-x_0{)}^{k+1}Q(x)\\ f^{(1)}(x)& =(k+1)(x-x_0{)}^{k}Q(x)+(x-x_0{)}^{k+1}{Q}^{(1)}(x)\\ & =(x-x_0{)}^k[(k+1)Q(x)+(x-x_0)Q^{(1)}(x)]\\ & =(x-x_0{)}^{k}R(x)\end{array}$

The Template:Font color is of multiplicity $k\ge 1$, with $R(z)$ a polynomial such that $R(x_0)\ne 0$.
Hence their product satisfies the induction hypothesis.

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Part 3

Let us now define a polynomial

$\begin{array}{cc}f(x)& =\frac{1}{(p-1)!}{x}^{p-1}[(1-x)(2-x)\cdots (n-x){]}^p\\ & =\frac{(n!{)}^p}{(p-1)!}{x}^{p-1}+\sum _{m=p}^{(n+1)p-1}\frac{b_m}{(p-1)!}{x}^m:b_m\in \mathbb{Z}\end{array}$

and $p$ is a prime number such that $p>\max \{a_0,n\}$. We get:

$f^{(k)}(x)=\sum _{m=k}^{(n+1)p-1}\frac{k!}{(p-1)!}(\genfrac{}{}{0ex}{}{m}{k})b_m{x}^{m-k}:p\le k\le (n+1)p-1$

hence for all $k\ge p$, the function $f^{(k)}(x)$ is a polynomial with integer coefficients all divisible by $p$.

By part 2, for all $1\le m\le n$ we get:

$F(m)=\sum _{k=0}^{(n+1)p-1}{f}^{(k)}(m)=\sum _{k=p}^{(n+1)p-1}{f}^{(k)}(m)$

Therefore $F(m)$ is also an integer divisible by $p$.

On the other hand, for $m=0$ we get:

$F(0)=\sum _{k=0}^{(n+1)p-1}{f}^{(k)}(0)=\sum _{k=p-1}^{(n+1)p-1}{f}^{(k)}(0)$

but $f^{(p-1)}(0)=(n!{)}^p$, and $n,a_0$ are not divisible by $p$. Therefore $a_{0}F(0)$ is not divisible by $p$.

Conclusion: $N=\sum _{m=0}^n{a}_{m}F(m)$ is an integer not divisible by $p$, and particularly $N\ne 0$.

Part 4

By part 2, by the triangle inequality for integrals we get:

$\begin{array}{cc}|A_m|& =|\underset{0}{\overset{m}{\int }}{\text{e}}^{m-t}f(t)dt|\le \underset{0}{\overset{m}{\int }}|{\text{e}}^{m-t}||f(t)|dt\le {\text{e}}^m\underset{0}{\overset{m}{\int }}\frac{n^{p-1}}{(p-1)!}(n!{)}^{p}dt\le {\text{e}}^m\underset{0}{\overset{n}{\int }}\frac{n^{p-1}}{(p-1)!}(n!{)}^{p}dt={\text{e}}^m\frac{(n\cdot n!{)}^p}{(p-1)!}\end{array}$

By the triangle Inequality we get:

$0<|N|=|\sum _{m=0}^n{a}_{m}F(m)|=\left|\sum _{m=1}^n{a}_m{A}_m\right|\le \sum _{m=1}^n|a_m{A}_m|\le (\sum _{m=1}^n|a_m|{\text{e}}^m)\frac{(n\cdot n!{)}^p}{(p-1)!}$

But $\underset{p\to \mathrm{\infty }}{\lim }\frac{(n\cdot n!{)}^p}{(p-1)!}=0$, hence for sufficiently large $p$ we get $0<|N|<1$. A contradiction.

$◻$

Conclusion: $\text{e}$ is transcendental.

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