Prove that $e$ is transcendental.
I found a proof below, but I'm not sure if it's correct.
We prove by contradiction. Suppose $e$ is algebraic. Then $\exists c_0, c_1,…,c_n\in\mathbb{Z}$ such that $c_ne^n + c_{n-1}e^{n-1} + \dots + c_1e+c_0 = 0$. We may assume $c_n \neq 0$ and $c_0\neq 0$, since there must be at least one nonzero constant. We attempt to arrive at a contradiction by constructing an integer in $(0,1)$.
Pick a large prime $p > \max \{n, |c_0|\}$. Let $$f(x) = \dfrac{x^{p-1}}{(p-1)!}(1-x)^p (2-x)^p\dots(n-x)^p=\dfrac{n!}{(p-1)!}x^{p-1}+\dfrac{a_p}{(p-1)!}x^p+\dots+\dfrac{(-1)^n}{(p-1)!}x^{(n+1)p-1},$$ where $a_p$ is a constant.
Let $r:=\deg f(x) = (n+1)p-1$.
We claim that if $k \geq p \wedge j\in\mathbb{Z}$, $f^{(k)}(j) \equiv 0\pmod p$. $f^{(k)}(x) = \dfrac{a_p}{(p-1)!}p(p-1)\dots(p+1-k)x^{p-k}+\dots+\dfrac{a_l}{(p-1)!}l(l-1)(l-2)\dots(l+1-k)x^{l-k}\equiv 0\pmod p$ as the product of at least $p$ consecutive integers contains at least one multiple of $p$.
$f^{(k)}(j) = \displaystyle\sum_{l=p}^ra_l{l\choose {p-1}}(l+1-p)(l-p)\dots(l+1-k)j^{l-k}\equiv 0\pmod p$ as ${l\choose {p-1}}(l+1-p)$ is a product of $p$ consecutive integers and $(p-1)!$ does not contain any multiples of $p$.
We next claim that if $0\leq k < p,$ then $f^{(k)}(j) = \begin{cases}0&&\text{ , if $j\in \{1,2,…,n\}$ or $j=0$ and $k<p-1$}\\n!&&\text { , if $j=0$ and $k=p-1$}\end{cases}$.
For $i \leq j \leq n$, $f(x) = (x-j)^p (q_0(x)),$ where $q_0(x)$ is the required polynomial. $f'(x) = (x-j)^{p-1}q_1(x)$, $f''(x) = (x-j)^{p-2}q_2(x),$ and $f^{(p-1)}(x) = (x-j)q_{p-1}(x)$.
When $j=0$, $f(x) = \dfrac{x^{p-1}}{(p-1)!}n!+\dots$ and $f^{(p-1)}(x) = n!+\dots$.
Let $F(x) = \displaystyle\sum_{i=0}^r f^{(i)}(x)$, where $f^{(0)}(x):=f(x)$.
Let $G(x) = e^{-x}F(x)$. $G'(x) = e^{-x}(F'(x)-F(x)) = -e^{-x}f(x)$ as $f^{(r+1)}(x) =0 $.
For $k=1,2,…,n$, apply the MVT to $G(x)$ on $[0,k]$. $\dfrac{G(k)-G(0)}{k}=G'(x_k), x_k\in (0,k).$ Therefore, $e^{-k}F(k) – F(0) = -ke^{-x_k}f(x_k)\Rightarrow F(k)-e^{k}F(0) = -ke^{k-x_k}f(x_k).$
Hence $\displaystyle\sum_{k=j}^n c_k (F(k)-e^kF(0)) = \displaystyle\sum_{k=1}^n c_k F(k) – \displaystyle\sum_{k=1}^n (c_ke^{k})F(0) = \displaystyle\sum_{k=0}^nc_kF(k)\equiv c_0n!\pmod p$.
$F(k) = \displaystyle\sum_{j=0}^nf^{(j)}(k) \equiv 0\pmod p,1\leq k\leq n.$ $F(0) = \displaystyle\sum_{j=0}^{p-2}f^{(j)}(k) + f^{(p-1)}(k)+\displaystyle\sum_{j=p}^r f^{(j)}(k).$
Therefore, $\displaystyle\sum_{k=0}^n c_kF(k) = \displaystyle\sum_{k=1}^n -c_k(ke^{k-x_k}f(x_k))\leq
(\displaystyle\sum_{k=1}^n|c_k|ke^k)\dfrac{(n+1)^{(n+1)p}}{(p-1)!}$.
Let $\dfrac{AB^p}{(p-1)!}$ represent the right hand side of this inequality.
Hence we have that $\exists x\in\mathbb{Z}$ such that $0 < x < \dfrac{AB^p}{(p-1)!}\to 0 $ as $p \to \infty$, which is a contradiction. Thus, $e$ is transcendental.
Edit: so it turns out this proof is similar to some popular proofs. So I have a few questions:
- I don't understand why we can't just say $f^{(k)}(x) = pa_p$ or $0$.
- Why does $f(x) = (x-j)^p(q_0(x))$?
- How is $\displaystyle\sum_{k=0}^n c_kF(k) = \displaystyle\sum_{k=1}^n -c_k(ke^{k-x_k}f(x_k))\leq
(\displaystyle\sum_{k=1}^n|c_k|ke^k)\dfrac{(n+1)^{(n+1)p}}{(p-1)!}$?
Best Answer
You had better take a look at Ivan Niven's book Irrational Numbers where this technique of Hermite is described in a simple manner and used to prove many results apart from transcendence of $e$.
He presents the following key result (Lemma 2.3 in his book on page 16):
Your $f(x) $ is $h(x) $ with $m=p-1$ with $p$ a prime and $$g(x) = (1-x)^p(2-x)^p\dots(m-x)^p$$ The above theorem then implies that all derivatives of $f$ at $0$ are integers and all of them are divisible by prime $p$ except possibly the $(p-1)$-th derivative. Further the value $p$ is chosen to ensure that $f^{(p-1)}(0)=(n!)^{p}$ is not divisible by $p$ (this is done by making $p>n$).
Further $$F(x) =\sum_{j=0}^{(n+1)p-1}f^{(j)}(x)$$ and $G(x) =e^{-x} F(x) $ so that $G'(x) =-e^{-x} f(x) $. Ivan Niven makes use of integrals from this point onwards but the same can be achieved by using mean value theorem (this is the approach in your question and by I N Herstein in his Topics in Algebra).
We have $$e^{-k} F(k) - F(0)=-ke^{-x_k}f(x_k)$$ or $$F(k) - e^kF(0)=-ke^{k-x_k}f(x_k)$$ Multiplying the above equation by $c_k$ and adding such equations for $k=1,2,\dots,n$ we get $$\sum_{k=1}^{n}c_kF(k)-F(0)\sum_{k=1}^{n}c_ke^k=-\sum_{k=1}^{n}kc_ke^{k-x_k}f(x_k)$$ By assumption we have $\sum_{k=1}^{n}c_ke^k=-c_0$ and therefore the last equation above can be written as $$\sum_{k=0}^{n}c_kF(k)=-\sum_{k=1}^{n}kc_ke^{k-x_k}f(x_k)\tag{1}$$ Now we do careful analysis of the expression on left side of the above equation and conclude that it is an integer not divisible by prime $p$ and is therefore non-zero (this also requires $p>|c_0|$).
The analysis of the term $c_0F(0)$ is a direct application of Niven's theorem and it is an integer not divisible by $p$ because $f^{(j)} (0)$ is divisible by $p$ for all $j\neq p-1$ and $f^{(p-1)}(0)$ is not divisible by $p$.
For analysis of $c_kF(k), k\neq 0$ note that if $p_k(x) =f(k-x) $ then $$p_k^{(j)} (0)=(-1)^{j}f^{(j)}(k)$$ and further $p_k(x) = f(k-x)$ can be written as $$\frac{x^{p-1}}{(p-1)!}xP_k(x)$$ for some polynomial $P_k(x) $ with integer coefficients. Therefore by Niven's theorem all the derivatives $p_k^{(j)} (0)$ and hence $f^{(j)} (k) $ are divisible by $p$. And therefore $c_kF(k) $ is an integer divisible by $p$.
Next step is the estimation of right side of $(1)$ and here one notices that $$|f(x_k)|\leq \frac{n^{(n+1)p-1}}{(p-1)!}$$ and therefore the right side of $(1)$ can be made arbitrarily small (in particular less than $1$) in absolute value by choosing large prime $p$ and we then get a contradiction.