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Let $P$ be a polynomial function , then we need to show that the equation :

$P(x)=e^x$ cannot have infinitely many solutions .

I thought about the nth-derivative of $g(x)=P(x)-e^x$ , if there are infinitely many solutions then $g$ takes the value $0$ more than $n=deg(P)+1$ times . Thus the n-th derivative of $g$ vanishes at some point , which is absurd . Because the n-th derivative of $g$ is $e^x$.

(My intuition tells me that there can be a proof using the variation of $P$ and $exp$) is there any idea to solve it besides the idea of the n-th derivative ?

Oussama Akar
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  • If $P(x)=1$, identically, then $e^x=1$ has infinitely many solutions. Or did you mean to specify real solutions? – lulu Jan 07 '22 at 12:23
  • Why do you think it has a finite number of solutions, when $\sum_{0}^\infty e^x=\frac{1}{n!}x^n$? – Superunknown Jan 07 '22 at 12:25
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    If you meant real, then note that between each two roots of $P(x)-e^x$ there must be a root of $P'(x)-e^x$, so you can work inductively. – lulu Jan 07 '22 at 12:25
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    The function $g(x)=P(x)-e^x$ is analytic and non-constant, so its roots are isolated. To conclude, you have to show that for $|x|$ large enough, $g$ has no root. For this, use that for such $x$ you have $|P(x)|\sim C|x|^{\mathrm{deg}(P)}$. – user47274 Jan 07 '22 at 12:26

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An alternative solution would be to first note that, the set of solutions must be bounded.

This is because for some $M_1$, you have $e^x>P(x)$ for all $x>0$, and some $M_2$ such that $P(x)>1$ for $x<M_2$ (because the limit of $P$, as $x\to-\infty$, is $\pm \infty$.

Because the set of solutions is bounded, it must have a convergence point. This is not possible because the function $g(x)=P(x)-e^x$ is analytic and non-constant, and the roots of such function are always isolated.

5xum
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