I am puzzled at the following question: Let p be a polynomial of degree n and let $$E = \{x | e^x = p(x)\}.$$ Prove that the cardinality $|E|$ of E satisfies $|E| \le n+1$.

Known The hint in the question says to use the Generalized Mean Value Theorem, which states: Let $f$ and $g$ be continuous functions on $[a,b]$ and suppose that $f$ and $g$ are differentiable at least on $(a,b)$, with $g'(x) \not= 0$ on $(a,b)$. Then $\exists c \in (a,b)$ such that $$\frac{f(b)-f(a)}{g(b)-g(a)}=\frac{f'(c)}{g'(c)}.$$

Where I am confused I am confused as to were to apply the Mean Value Theorem (i.e. - which functions to pick for $f$ and $g$). I want to use a counting argument, but that is not working either. Any help, hints, guidance, etc. are well appreciated.

  • $\begingroup$ Hint: Use induction in $n$ - Show that $E$ is countable, and enumerate $E$ in increasing order as $\{a_i\}_{i \ge 1}$. Now use the MVT on $(a_i, a_{i+1})$. $\endgroup$ – stochasticboy321 Jan 29 '17 at 22:11
  • $\begingroup$ We don't need the notion of "countable" (note that a countable set often can't be put in increasing order anyway). $\endgroup$ – zhw. Jan 29 '17 at 22:30

$f(x)=e^x-p(x)$ is infinitely differentiable, and $f^{(n+1)}(x)= e^x\,$ since the $(n+1)^{th}$ derivative of an $n^{th}$ degree polynomial is identically zero $p^{(n+1)}(x) \equiv 0\,$.

If $\,|E| \ge n+2\,$, that would mean $f(x)$ had at least $\,n+2\,$ distinct zeros. Then, as a consequence of Rolle's theorem, $\,f^{(n+1)}(x)\,$ would need to have at least one real root (see here for example). But $\,f^{(n+1)}(x)= e^x \gt 0\,$, therefore $\,f(x)\,$ can have at most $\,n+1\,$ distinct roots $\iff |E| \le n+1\,$.


Hint: Don't use the GMVT; the MVT will do. Apply it to $e^x-p(x)$ and use induction as @stochasticboy321 suggests.


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