Show that $e^{cx} \gt | P(x) |$ for some $c \gt 0$

Question

Let $P(x)$ be a polynomial, and let $c > 0$. Show that there exists a real number $N > 0$ such that $e^{cx} > |P(x)|$ for all $x > N$;

My work so far: $e^{cx}=\sum_{n=0}^{\infty} \frac{cx^n}{n!}$ and $|P(x)|=\sum_{n=0}^na_nx^n$

I would like to show that $e^{cx} - |P(x)| = \sum_{n=0}^{n} (\frac{c}{n!}-a_n) x^n > 0 $

I am not sure how to continue from here

Answer 1

By repeated application of L'Hopital's Rule you can show that $\frac {p(x)} {e^{cx}} \to 0$ as $x \to \infty$. Hence there exists $N$ such that $|\frac {p(x)} {e^{cx}}|<1$ for $x >N$.

Second proof: let $P(x)=\sum\limits_{k=0}^{N} a_ix^{i}$. From the series expansion of $e^{cx}$ we get $e^{cx} > \frac {c^{i+1}x^{i+1}} {(i+1)!}$. Hence $|c_ix^{i}| < \frac 1 {cx} {(i+1)!} e^{cx}$. From this it is clear that there exists a positive number $M_i$ such that $|c_ix^{i}| < \frac 1 {N+1} e^{cx}$ for $x >M_i$. [ In fact it is enough to take $M_i >\frac {(i+1)!(N+1)} c$]. Now let $M$ be the maximum of the numbers $M_0,M_1,...,M_N$. Then $x >M%$ implies that each term in the polynomial is less than $\frac 1 {N+1} e^{cx}$ in absolute value. There are $N+1$ terms in the polynomial. Hence $|P(x)| <e^{cx}$ if $x >M$.

Answer 2

(Ignoring some errors in the indexes)

You have for $x>0;$ $e^{cx} - |P(x)| = \sum_{n=0}^{\infty} (\frac{c}{n!}-a_n) x^n \ge \sum_{n=0}^{\mbox{deg}(P)+1} (\frac{c}{n!}-a_n) x^n$ where $a_k=0$ for all $k>\mbox{deg}(P)$.

Now this is comparing polynomials