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Suppose that for some two variable rational polynomial that evaluation of the second variable at $e^x$ gives 0, then the evaluation at any $y$ also gives 0.

I've seen a proof that uses a calculus-based limiting argument, but I was wondering if there was some sort of algebraic proof, maybe using the transcendence of $e$ over $\mathbb{Q}$?

Any ideas? Thanks!

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Suppose that $p$ is nonzero. Then, there exists $n\in\mathbb{N}$ such that $p(n,y)$ is a nonzero element of $\mathbb{Q}[y]$. Now, the equation $p\left(n,\text{e}^n\right)=0$ implies that $\text{e}^n$ is algebraic over $\mathbb{Q}$, which is a contradiction.

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  • $\begingroup$ Well, that was deceptively simple. The argument I saw was about factoring out powers of $y$ and using the domination of $e^x$ over a polynomial, etc. Not very enlightening for an algebra problem. But this is a very nice proof, bravo. $\endgroup$ – walkar Aug 13 '15 at 18:43
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Allow me to provide the argument I've referenced a few times. This is taken from a page of qualifying exam solutions at the University of Missouri.

Let $n$ be the largest exponent of $y$ in $p(x,y)$. Then we can write $0 = p(x,e^x) = e^{nx}p_n(x) + \cdots + e^xp_1(x) + p_0(x)$, where each $p_i \in \mathbb{Q}[x]$. If $n=0$, then there is nothing to prove, so for $n \ge 1$, we have $p_n(x) = -(e^{-x}p_{n-1}(x)+ \cdots + e^{-nx}p_0(x))$. Since $e^x$ always dominates polynomials in $x$, we have if we take the limit as $x \rightarrow \infty$, we get 0. So $p_n(x) = 0$. But this says $y^n$ does not appear in $p(x,y)$, a contradiction.

I like @Batominovski's answer much better.

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