# The transcendence of e (part 2)

This is the second part in a 3-part blog post in which we prove that ${e}$ is transcendental.

Three-step proof that $e$ is transcendental
Step 1
Step 2
Step 3

Recall that in step 1 we proved the following lemma.

Lemma 1. Suppose ${e}$ is a root of the polynomial ${\varphi(x)=c_{0}+c_{1}x+c_{2}x^{2}+\cdots+c_{n}x^{n}}$. Let ${f}$ be a polynomial and ${F(x)=\sum_{i=0}^{\infty}f^{(i)}(x)}$. Then there exist ${\alpha_{1},\ldots,\alpha_{n}\in(0,1)}$ such that ${c_{0}F(0)+\cdots+c_{n}F(n)=c_{1}\beta_{1}+\cdots+c_{n}\beta_{n}}$, where ${\beta_{k}=-ke^{k(1-\alpha_{k})}f(k\alpha_{k})}$.

Step 2.

Now we are ready pick a specific polynomial ${f}$ for Lemma 1. Since there are infinitely many primes, we can find a prime number ${p}$ greater than both ${n}$ and ${c_{0}}$. Let $\displaystyle f(x)=\frac{x^{p-1}(1-x)^{p}(2-x)^{p}\cdots(n-x)^{p}}{(p-1)!}.$

Our goal for this step is to show that ${c_{0}F(0)+\cdots+c_{n}F(n)}$ is a nonzero integer. In fact we will show is that it is an integer not divisible by ${p}$, which implies that it is nonzero.

We will accomplish this by proving that for ${m=1,\ldots n}$, ${F(m)}$ (and hence ${c_{m}F(m)}$) is an integer divisible by ${p}$, but that ${c_{0}F(0)}$ is an integer that is not divisible by ${p}$.

First we state but do not prove the following lemma. (This is a homework problem in Herstein.)

Lemma 2. If ${g}$ is a polynomial with integer coefficients and ${h(x)=g(x)/(p-1)!}$, then for ${i\ge p}$, ${h^{(i)}}$ is a polynomial with integer coefficients, each of which is divisible by ${p}$.

Applying Lemma 2 to our function ${f}$ we see that when ${i\ge p}$, ${f^{(i)}(x)}$ is a polynomial with integer coefficients, each of which is divisible by ${p}$. In particular, when ${i\ge p}$ and ${m}$ is any integer, ${f^{(i)}(m)}$ is an integer that is divisible by ${p}$.

Now let us restrict our attention to ${m\in\{1,\ldots,n\}}$. From the definition of ${f}$ we see that ${m}$ is a root of ${f}$ with multiplicity ${p}$. The following lemma tells us that ${f(m)=f^{(1)}(m)=\cdots=f^{(p-1)}(m)=0}$. (We leave the proof of this lemma as an exercise for the reader.)

Lemma 3. Suppose ${x_{0}}$ is a root of a polynomial ${g}$ with multiplicity ${m}$. Then ${g^{(k)}(x_{0})=0}$ for ${k=0,\ldots,m-1}$.

So, $F(m)=f(m)+f^{(1)}(m)+\cdots+f^{(r)}(m)$ $=f^{(p)}(m)+\cdots+f^{(r)}(m).$

We know from above that each of the terms ${f^{(p)}(m),\cdots,f^{(r)}(m)}$ is an integer divisible by ${p}$. Thus ${F(m)}$ is an integer divisible by ${p}$.

Now let us consider ${m=0}$. By the definition of ${f}$, ${0}$ is a root with multiplicity ${p-1}$. By Lemma 3, ${f(0)=f^{(1)}(0)=\cdots=f^{(p-2)}(0)=0}$.

Thus, $F(0)=f(0)+f^{(1)}(0)+\cdots+f^{(r)}(0)$ $=f^{(p-1)}(0)+f^{(p)}(0)+\cdots+f^{(r)}(0).$

From the result above, we know that ${f^{(p)}(0),\cdots,f^{(r)}(0)}$ are all divisible by ${p}$. What about ${f^{(p-1)}(0)}$?

If we multiply out the terms in ${f}$, we obtain $\displaystyle f(x)=\frac{(n!)^{p}}{(p-1)!}x^{p-1}+\{\text{higher order terms in }x\}$

So, $\displaystyle f^{(p-1)}(x)=(n!)^{p}+\{\text{higher order terms in }x\},$

and hence ${f^{(p-1)}(0)=(n!)^{p}}$. Since ${p>n}$, ${p}$ does not divide ${f^{(p-1)}(0)}$. Therefore ${p}$ does not divide ${F(0)}$. Moreover, ${p>c_{0}}$, so ${p}$ does not divide ${c_{0}}$. We conclude that ${p}$ does not divide ${c_{0}F(0)}$.

Finally this implies that ${c_{0}F(0)+\cdots+c_{n}F(n)}$ is an integer not divisible by ${p}$. In particular, it is a nonzero integer.

We are almost finished with the proof. In the next step we will show that ${|c_{1}\beta_{1}+\cdots+c_{n}\beta_{n}|<1}$, thus it cannot be a nonzero integer and we will obtain our sought-after contradiction.

1. David Smith says:

Thanks for posting this, it’s very helpful. I have one question though, I know I’m being stupid but I can’t seem to prove lemma 2. I’d really appreciate any help on this. Thank you.

1. For simplicity, suppose $g(x)=x^n$. Then $\displaystyle h^{(i)}(x)=\frac{n(n-1)\cdots(n-i+1)}{(p-1)!}x^{n-i}$. Rewritten, the coefficient is $\displaystyle \frac{n!}{(p-1)!(n-i)!}=(i(i-1)\cdots p)\frac{n!}{i!(n-i)}=(i(i-1)\cdots p){{n}\choose{i}}$. Seen in this way, it is clearly an integer divisible by $p$.

2. David Smith says:

Thanks a lot for your help, I really should have been able to work that out myself!

3. David Smith says:

Sorry to bother you again, but I’ve been trying to prove that $e^{m/n}$ is transcendental, where $m>0$ and $n$ are integers (it’s an exercise in Herstein on p.178), but I’ve had no luck. Any help would be much appreciated. Thanks.