# Famous Theorems of Mathematics/e is transcendental

The mathematical constant ${\text{e}}=\sum _{n\,=\,0}^{\infty }{\frac {1}{n!}}=2.718281\ldots$ 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:
    $F'\!(x)=\sum _{k\,=\,0}^{d}f^{(k+1)}\!(x)=\sum _{k\,=\,1}^{d}f^{(k)}\!(x)=F(x)-f(x)$
Let us define $G(x)={\text{e}}^{-x}F(x)$. Taking its derivative yields:
    $G'\!(x)={\text{e}}^{-x}F'\!(x)-{\text{e}}^{-x}F(x)={\text{e}}^{-x}{\biglF'\!(x)-F(x){\bigr ]}=-e^{-x}f(x)$
By the fundamental theorem of calculus, we get:
    ${\begin{aligned}G(x)-G(0)={\text{e}}^{-x}F(x)-{\text{e}}^{-0}F(0)&=-\!\int \limits _{0}^{x}{\text{e}}^{-t}f(t)dt\\F(x)-{\text{e}}^{x}F(0)&=-\!\int \limits _{0}^{x}{\text{e}}^{x-t}f(t)dt\end{aligned}}$
Now let:
    ${\begin{aligned}F(m)-{\text{e}}^{m}F(0)=-\!\int \limits _{0}^{m}\!{\text{e}}^{m-t}f(t)dt=A_{m}\\[5pt]a_{m}F(m)-a_{m}{\text{e}}^{m}F(0)=a_{m}A_{m}\end{aligned}}$
Summing all the terms yields
    ${\begin{aligned}\sum _{m\,=\,1}^{n}a_{m}F(m)-\sum _{m\,=\,1}^{n}a_{m}{\text{e}}^{m}F(0)=\sum _{m\,=\,1}^{n}a_{m}A_{m}\\[5pt]\sum _{m\,=\,1}^{n}a_{m}F(m)-F(0)\sum _{m\,=\,1}^{n}a_{m}{\text{e}}^{m}=\sum _{m\,=\,1}^{n}a_{m}A_{m}\\[5pt]\sum _{m\,=\,1}^{n}a_{m}F(m)-F(0)(-a_{0})=\sum _{m\,=\,1}^{n}a_{m}A_{m}\\[5pt]\sum _{m\,=\,0}^{n}a_{m}F(m)=\sum _{m\,=\,1}^{n}a_{m}A_{m}\end{aligned}}$

### Part 2

Lemma: Let $f(x)$ be a polynomial with a root $x_{0}$ of multiplicity $d\geq 1$. Then $f^{(j)}\!(x_{0})=0$ for all $0\leq j\leq d-1$.
Proof: By strong induction.
Let us write $f(x)=(x-x_{0})^{d}Q(x)$, with $Q(x)$ a polynomial such that $Q(x_{0})\neq 0$.
For $d=1$ we get:
    $f^{(0)}\!(x_{0})=f(x_{0})=0$
Assume that for all $1\leq d\leq k$ the claim holds for all $0\leq j\leq d-1$. We shall prove that for $d=k+1$ the claim holds for all $0\leq j\leq k$:
    ${\begin{aligned}f(x)&=(x-x_{0})^{k+1}Q(x)\\[5pt]f^{(1)}\!(x)&=(k+1)(x-x_{0})^{k}Q(x)+(x-x_{0})^{k+1}Q^{(1)}\!(x)\\[5pt]&={\color {blue}(x-x_{0})^{k}}{\color {red}{\bigl(k+1)Q(x)+(x-x_{0})Q^{(1)}\!(x){\bigr ]}}\\[5pt]&={\color {blue}(x-x_{0})^{k}}{\color {red}R(x)}\end{aligned}}$
The blue part is of multiplicity $k\geq 1$, with $R(z)$ a polynomial such that $R(x_{0})\neq 0$. Hence their product satisfies the induction hypothesis.
$\square$

### Part 3

Let us now define a polynomial
    ${\begin{aligned}f(x)&={\frac {1}{(p-1)!}}\,x^{p-1}{\bigl(1-x)(2-x)\cdots (n-x){\bigr ]}^{p}\\[5pt]&={\frac {(n!)^{p}}{(p-1)!}}x^{p-1}+\!\!\sum _{m\,=\,p}^{(n+1)p-1}\!\!\!{\frac {b_{m}}{(p-1)!}}x^{m}\quad :b_{m}\in \mathbb {Z} \end{aligned}}$
and $p$ is a prime number such that $p>\max\{a_{0},n\}$. We get:
    $f^{(k)}\!(x)=\!\!\sum _{m\,=\,k}^{(n+1)p-1}\!\!\!{\frac {k!}{(p-1)!}}{\binom {m}{k}}b_{m}x^{m-k}\quad :p\leq k\leq (n+1)p-1$
hence for all $k\geq p$, the function $f^{(k)}\!(x)$ is a polynomial with integer coefficients all divisible by $p$.
By part 2, for all $1\leq m\leq n$ 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 $N\neq 0$.

### Part 4

By part 2, by the triangle inequality for integrals we get:
    ${\begin{aligned}|A_{m}|&={\Bigg |}\int \limits _{0}^{m}\!{\text{e}}^{m-t}f(t)dt\,{\Bigg |}\leq \int \limits _{0}^{m}\!|{\text{e}}^{m-t}|{\bigr |}f(t){\bigl |}dt\leq {\text{e}}^{m}\!\int \limits _{0}^{m}\!{\frac {n^{p-1}}{(p-1)!}}\,(n!)^{p}dt\leq {\text{e}}^{m}\!\int \limits _{0}^{n}\!{\frac {n^{p-1}}{(p-1)!}}\,(n!)^{p}dt={\text{e}}^{m}{\frac {(n\!\cdot \!n!)^{p}}{(p-1)!}}\end{aligned}}$
By the triangle Inequality we get:
    $0<|N|=\left|\sum _{m\,=\,0}^{n}\!a_{m}F(m)\right|=\left|\sum _{m\,=\,1}^{n}\!a_{m}A_{m}\right|\leq \sum _{m\,=\,1}^{n}\!|a_{m}A_{m}|\leq \left(\sum _{m\,=\,1}^{n}\!|a_{m}|{\text{e}}^{m}\right)\!{\frac {(n\!\cdot \!n!)^{p}}{(p-1)!}}$
But $\lim _{p\to \infty }{\frac {(n\!\cdot \!n!)^{p}}{(p-1)!}}=0$, hence for sufficiently large $p$ we get $0<|N|<1$. A contradiction.
$\square$
Conclusion: ${\text{e}}$ is transcendental.
$\blacksquare$