Linear independence of the numbers $\{1,e,e^2,e^3\}$

Question

Does someone know a proof that $\{1,e,e^2,e^3\}$ is linearly independent over $\mathbb{Q}$?

The proof should not use that $e$ is transcendental.

$e:$ Euler's number.

$\{1,e,e^2\}$ is linearly independent over $\mathbb{Q}$

Any hints would be appreciated.

Answer 1

Since I've spent enough time thinking about this, yet not getting a proof, I might as well show what I've got. Others can comment on whether or not more can be done.

Your problem is solved if you can show that for any integers $a, b, c$, we have $$\sum^\infty_{n=0} \frac{1}{n!} (a + b 2^n + c 3^n)$$ irrational (using taylor series).

WLOG, assume $c>0$. Pick $N$ large so that $(a+b2^n+c3^n) > 0$ for all $N \geq 0$. Then our problem is equivalent to showing that the series with strictly positive terms $$\sum^\infty_{n=N} \frac{1}{n!} (a + b 2^n + c 3^n)$$ is irrational. Suppose it was not and equal to $p/q$. Now we try to mimic the proof of irrationality of $e$.

Define

$$x = q!\left(\frac{p}{q} - \sum_{n=N}^{q} \frac{1}{n!} (a + b 2^n + c 3^n) \right).$$ One easily sees by distributing that it is an integer, and because our original series contains only positive terms, $x>0$.

Note that we can also write $$x = \sum_{n=q+1}^\infty \frac{q!}{n!} (a + b 2^n + c 3^n).$$

Now if $b=c=0$, then using $q!/n! < 1/(q+1)^{n-q}$ gives a geometric series bound that gives $x < 1/q$. Then we can get $x<1$ which is a contradiction that $x$ is an integer.

The terms $2^n$ and $3^n$ grow too fast for this same trick to work. You'd get bounds of $2^q/q$ and $3^q/q$ respectively. Since $q!/n! < 1/(q+1)^(n-q)$ is not tight, it is still possible that we can get our sum under 1. Or maybe we can monkey with our original definition for $x$.

I think what really needs to be copied are proofs of the irrationality of $e^2$ and $e^3$, but I am not aware of such proofs. Googling, I found a very algebraic proof of the irrationality of $e^2$, but I didn't read it carefully. This suggests proofs of the irrationality of $e^2$ may not easily generalize, and hence you aren't really proving that $e$ is transcendental at the same time.

Answer 2

Below is my attempt which is too long for a comment and may be saveable, (doubt it).

Consider the differential equation $y^{(4)}-6y^{(3)}+11y''-6y'=\textbf 0$, where $\bf 0$ is the null function over some non-trivial interval $I$ containing $1$.

The theory of ODE tells us that a basis of solutions is $$\{\underbrace{x\mapsto 1}_{\large \varphi_0}, \underbrace{x\mapsto e^x}_{\large \varphi _1}, \underbrace{x\mapsto e^{2x}}_{\large \varphi _2}, \underbrace{x\mapsto e^{3x}}_{\large \varphi _3}\}$$

This implies that $$(\forall \lambda _0,\lambda _1, \lambda _2, \lambda _3\in \Bbb Q)\left[(\forall x\in I)\left(\sum \limits_{k=0}^3\lambda_k\varphi_k(x)=0\right)\implies \lambda _0=\lambda _1=\lambda _2=\lambda _3=0\right] \tag {*}$$

Now if we could somehow prove that $(*)$ would also hold for the intersection of all such intervals $I$ (containing $1$), what we want would follow. But I have no hope of this being doable.