Prove: $e^x$ is transcendental over the polynomials with coefficients in $\mathbb{R}$

Question

I have to prove the following for my math study:

Prove: $e^x$ is transcendental over the polynomials with coefficients in $\mathbb{R}$.

So far, I've done this:

It's enough to prove that if $a_n(x),\dots,a_1(x), a_0(x)$ are polynomials with coefficients in $\mathbb{R}$ such that $a_n(x)e^{nx} + a_{n-1}(x)e^{(n - 1)x} + \dots + a_1(x)e^x + a_0(x) = 0$, then $a_i(x) = 0$ $\forall i$
To prove this I assumed that $\exists i : a_i(x) \neq 0$. Then take the smallest $n$ such that $a_n(x)e^{nx} + a_{n-1}(x)e^{(n - 1)x} + ... + a_1(x)e^x + a_0(x) = 0$.

I don't know how I have to complete the proof from here. Could you explain it to me?

Answer 1

An approach not using limits. Let $f(x) = \sum_{k=0}^n a_k(x)e^{kx}$ and call $a_0$ the lowest coefficient. Then $f'(x)$ has a similar form and only the the degree of the lowest coefficient decreases, other coefficients remain of the same degree. Take the derivative repeatedly until the lowest coefficient vanishes.

The same approach also works for the difference operator $(\Delta f)(x)=f(x+1)-f(x)$.

Answer 2

Clearly, as $n>0$, $$ \lim_{x\to-\infty}-a_0(x)= \lim_{x\to-\infty} \bigl(a_n(x)e^{nx} + a_{n-1}(x)e^{(n - 1)x} + ... + a_1(x)e^x\bigr)= 0 $$ So…

(Note: the only needed property is $\lim_{x\to-\infty}x^he^{kx}=0$ for $h\ge0$, $k>0$.)

Answer 3

You can do it, well, by using calculus.

Suppose $$\sum_{i=0}^{n}a_i(x)e^{ix}=0$$

with $n$ being smallest possible, i.e. $a_n(x)\neq 0$.

In particular, $$\lim_{x\rightarrow \infty}a_n(x) \neq 0.$$

Then it follows that for any $\alpha \in \mathbb{R},$

$$\lim_{x \rightarrow \infty} \frac{\sum_{i=0}^{n}a_i(x)e^{ix}}{e^{\alpha x}}=0.$$

Also, for positive $\alpha$ and any polynomial $a(x)$, we have $$\lim_{x\rightarrow \infty}\frac{a(x)}{e^{\alpha x}}=0.$$

Putting $\alpha=n$, one has that $$0 \neq \lim_{x \rightarrow \infty}a_n(x)=\lim_{x \rightarrow \infty} \frac{\sum_{i=0}^{n}a_i(x)e^{ix}}{e^{\alpha x}}=0,$$

a contradiction.