Prove that $e$ is transcendental.

Solution 1:

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):

Theorem: If $h(x) = x^mg(x) / m! $ where $g(x) $ is a polynomial with integer coefficients and $m$ is a positive integer, then $h^{(j)} (0)$, the $j$-th derivative of $h(x) $ at $x=0$, is an integer for $j=0,1,2,\dots$. Moreover, with the possible exception of the case $j=m$, the integer $h^{(j)} (0)$ is divisible by $m+1$; no exception need be made in the case $j=m$ if $g(x) $ has the factor $x$: ie, if $g(0)=0$.

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.