Famous Theorems of Mathematics/e is transcendental

The mathematical constant $e=\sum _{n\,=\,0}^{\infty }{\frac {1}{n!}}=2.718281\ldots$ is a transcendental number.

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

Proof

Let us assume that $e$ is algebraic, so there exists a polynomial such that

$P(e)=a_{0}+a_{1}e+a_{2}e^{2}+\cdots +a_{n}e^{n}=0$

for $a_{0},\ldots ,a_{n}\in \mathbb {Z}$ and $a_{0}\neq 0$.

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 $G(x)=e^{-x}F(x)$. Taking its derivative yields:

$G'\!(x)=e^{-x}F'\!(x)-e^{-x}F(x)=e^{-x}{\bigl [}F'\!(x)-F(x){\bigr ]}=-e^{-x}f(x)$

Since $G(x)$ is differentiable, we shall apply the mean Value Theorem on the interval $[0,m]$ for $m\in \mathbb {N}$. So there exists a $x_{m}\in (0,m)$ such that

$G'\!(x_{m})={\frac {G(m)-G(0)}{m-0}}={\frac {e^{-m}F(m)-e^{-0}F(0)}{m}}=-e^{-x_{m}}f(x_{m})$

Now let:

${\begin{aligned}&F(m)-e^{m}F(0)=-m\,e^{m-x_{m}}f(x_{m})=A_{m}\\[6pt]&a_{m}F(m)-a_{m}e^{m}F(0)=a_{m}A_{m}\end{aligned}}$

Summing all the terms yields

${\begin{aligned}\sum _{m\,=\,1}^{n}a_{m}F(m)-\sum _{m\,=\,1}^{n}a_{m}e^{m}F(0)=\sum _{m\,=\,1}^{n}a_{m}A_{m}\\[6pt]\sum _{m\,=\,1}^{n}a_{m}F(m)-F(0)\sum _{m\,=\,1}^{n}a_{m}e^{m}=\sum _{m\,=\,1}^{n}a_{m}A_{m}\\[6pt]\sum _{m\,=\,1}^{n}a_{m}F(m)-F(0)(-a_{0})=\sum _{m\,=\,1}^{n}a_{m}A_{m}\\[6pt]\sum _{m\,=\,0}^{n}a_{m}F(m)=\sum _{m\,=\,1}^{n}a_{m}A_{m}\end{aligned}}$

Part 2

Let $f(x)$ be a polynomial with a root $x_{0}$ of multiplicity $d$. We will show that for all $0\leq k\leq d-1$ we get $f^{(k)}\!(x_{0})=0$.

Let us write $f(x)=(x-x_{0})^{d}Q_{0}(x)$, such that $Q_{0}(x)$ is a polynomial with $Q_{0}(x_{0})\neq 0$.

${\begin{aligned}f^{(1)}\!(x)&=d(x-x_{0})^{d-1}Q_{0}(x)+(x-x_{0})^{d}Q_{0}'(x)\\[5pt]&=(x-x_{0})^{d-1}{\Big [}d\,Q_{0}(x)+(x-x_{0})Q_{0}'(x){\Big ]}\\[5pt]&=(x-x_{0})^{d-1}Q_{1}(x)\\[5pt]Q_{1}(x_{0})&=d\,Q_{0}(x_{0})\neq 0\\\\[5pt]f^{(2)}\!(x)&=(d-1)(x-x_{0})^{d-2}Q_{1}(x)+(x-x_{0})^{d-1}Q_{1}'(x)\\[5pt]&=(x-x_{0})^{d-2}{\Big [}(d-1)Q_{1}(x)+(x-x_{0})Q_{1}'(x){\Big ]}\\[5pt]&=(x-x_{0})^{d-2}Q_{2}(x)\\[5pt]Q_{2}(x_{0})&=(d-1)Q_{1}(x_{0})\neq 0\\[5pt]&\,\,\,\vdots \\[5pt]f^{(d-1)}\!(x)&=2(x-x_{0})Q_{d-2}(x)+(x-x_{0})^{2}Q_{d-2}'(x)\\[5pt]&=(x-x_{0}){\Big [}2Q_{d-2}(x)+(x-x_{0})Q_{d-2}'(x){\Big ]}\\[5pt]&=(x-x_{0})Q_{d-1}(x)\\[5pt]Q_{d-1}(x_{0})&=2Q_{d-2}(x_{0})\neq 0\end{aligned}}$

with $Q_{0}'(x),Q_{1}'(x),\ldots ,Q_{d-1}'(x)$ all polynomials.

Part 3

Let us now define a polynomial

${\begin{aligned}f(x)&={\frac {1}{(p-1)!}}\,x^{p-1}{\bigl [}(1-x)(2-x)\cdots (n-x){\bigr ]}^{p}\\[5pt]&={\frac {(n!)^{p}}{(p-1)!}}x^{p-1}+\!\!\sum _{m\,=\,p}^{(n+1)p-1}\!\!\!{\frac {b_{m}}{(p-1)!}}x^{m}\quad :b_{m}\in \mathbb {Z} \end{aligned}}$

for prime $p$ such that $p>a_{0}$ and $p>n$. 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 $p$.

In other words, the sum $\sum _{m\,=\,0}^{n}a_{m}F(m)$ is an integer not divisible by $p$, and particularly non-zero.

Part 4

By part 1, for all $1\leq m\leq n$ we have $0<x_{m}<m\leq n$. Therefore,

${\begin{aligned}A_{m}&=-m\,e^{m-x_{m}}f(c_{m})=-{\frac {m\,e^{m-x_{m}}}{(p-1)!}}\,(x_{m})^{p-1}{\bigl [}(1-x_{m})(2-x_{m})\cdots (n-x_{m}){\bigr ]}^{p}\\[5pt]|A_{m}|&={\frac {m\,e^{m-x_{m}}}{(p-1)!}}\,(x_{m})^{p-1}{\Big |}(1-x_{m})(2-x_{m})\cdots (n-x_{m}){\Big |}^{p}\\[5pt]&\leq {\frac {ne^{n}}{(p-1)!}}\,n^{p-1}(1\cdot 2\cdots n)^{p}=e^{n}{\frac {(n\cdot n!)^{p}}{(p-1)!}}\end{aligned}}$

By the triangle Inequality we get:

$0<\left|\sum _{m\,=\,0}^{n}a_{m}F(m)\right|=\left|\sum _{m\,=\,1}^{n}a_{m}A_{m}\right|=\sum _{m\,=\,1}^{n}|a_{m}A_{m}|\leq \left(\sum _{m\,=\,1}^{n}|a_{m}|\right)e^{n}{\frac {(n\cdot n!)^{p}}{(p-1)!}}$

But $\lim _{p\to \infty }{\frac {(n\cdot n!)^{p}}{(p-1)!}}=0$, hence for sufficiently large $p$ we get $0<\left|\sum _{m\,=\,0}^{n}a_{m}F(m)\right|<1$. A contradiction.

$\blacksquare$