Limit at Infinity of Polynomial over Complex Exponential
Theorem
Let $n \in \N$.
Let $P_n \left({x}\right)$ be a real polynomial, of degree $n$.
Let $e^z$ be the complex exponential function, where $z = x + i y$.
Let $a \in \R_{>0}$.
Then:
- $\displaystyle \lim_{x \mathop \to +\infty} \frac {P_n \left({x}\right)}{e^{a z}} = 0$
Proof
Let $\epsilon > 0$.
By the definition of limits at infinity, we need to show that there is some $M \in \R$ such that:
- $\displaystyle x > M \implies \left \vert {\frac {P_n\left({x}\right)}{e^{a z}} - 0} \right\vert < \epsilon$
But:
| \(\displaystyle \left \vert {\frac {P_n\left({x}\right)}{e^{a z} } - 0} \right\vert\) | \(=\) | \(\displaystyle \frac {\left \vert {P_n\left({x}\right)} \right \vert}{\left \vert {e^{a z} } \right \vert}\) | Modulus of Product | ||||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \frac {\left \vert {P_n\left({x}\right)} \right \vert}{\left \vert {e^{a x \mathop + i a y} } \right \vert}\) | |||||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \frac {\left \vert {P_n\left({x}\right)} \right \vert}{\left \vert {e^{a x} } \right \vert}\) | Modulus of Exponential is Modulus of Real Part |
This means it is sufficient to find an $M \in \R$ such that:
| \(\displaystyle \frac {\left \vert {P_n\left({x}\right)} \right \vert}{\left \vert {e^{a x} } \right \vert}\) | \(<\) | \(\displaystyle \epsilon\) |
Exponential Dominates Polynomial doesn't say anything about the quotient. (Proof corrected Nov. 2016.)
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Let $\displaystyle P_n \left({x}\right) = \sum_{j \mathop = 0}^n \left({a_j x^j}\right)$ where $a_j \in \R$ for every $j$ in $\left\{{0, \ldots, n}\right\}$.
Let $\displaystyle M' = \frac 1 \epsilon \sum_{j \mathop = 0}^n \left\lvert{a_j}\right\rvert$.
We observe that $M' \ge 0$.
Also, for every $x > M'$:
- $x > 0$ as $M' \ge 0$
We have for every $x > M'$:
| \(\displaystyle M'\) | \(<\) | \(\displaystyle x\) | |||||||||||
| \(\displaystyle \iff \ \ \) | \(\displaystyle \frac 1 \epsilon \sum_{j \mathop = 0}^n \left\lvert{a_j}\right\rvert\) | \(<\) | \(\displaystyle x\) | ||||||||||
| \(\displaystyle \iff \ \ \) | \(\displaystyle \frac 1 x \sum_{j \mathop = 0}^n \left\lvert{a_j}\right\rvert\) | \(<\) | \(\displaystyle \epsilon\) | as $x \not = 0$ since $x > 0$ |
We have that for every $k \in \N$ there exists $N_k \in \N$ such that:
- $x^k < e^{a x}$ for all $x > N_k$ by Exponential Dominates Polynomial
Let $N = \max \left({N_1, \ldots, N_{n + 1}}\right)$.
We have for every $k \in \left\{{1, \ldots, n + 1}\right\}$:
| \(\displaystyle x^k\) | \(<\) | \(\displaystyle e^{a x}\) | for all $x > N$ | ||||||||||
| \(\displaystyle \iff \ \ \) | \(\displaystyle \frac {x^k} {e^{a x} }\) | \(<\) | \(\displaystyle 1\) | for all $x > N$ |
Let $M = \max \left({M', N}\right)$.
We get for every $x > M$:
| \(\displaystyle \frac {\left\lvert{P_n \left({x}\right)}\right\rvert} {\left\lvert{e^{a x} }\right\rvert}\) | \(=\) | \(\displaystyle \frac {\left\lvert{\sum_{j \mathop = 0}^n \left({a_j x^j}\right)}\right\rvert} {e^{a x} }\) | |||||||||||
| \(\displaystyle \) | \(\le\) | \(\displaystyle \frac {\sum_{j \mathop = 0}^n \left\lvert{a_j x^j}\right\rvert} {e^{a x} }\) | |||||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \frac {\sum_{j \mathop = 0}^n \left({\left\lvert{a_j}\right\rvert x^j}\right)} {e^{a x} }\) | as $x^j > 0$ since $x > 0$ | ||||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \frac 1 x \frac {\sum_{j \mathop = 0}^n \left({\left\lvert{a_j}\right\rvert x^\left({j + 1}\right)}\right)} {e^{a x} }\) | as $x \not = 0$ since $x > 0$ | ||||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \frac 1 x \sum_{j \mathop = 0}^n \left({\left\lvert{a_j}\right\rvert \frac {x^\left({j + 1}\right)} {e^{a x} } }\right)\) | |||||||||||
| \(\displaystyle \) | \(\le\) | \(\displaystyle \frac 1 x \sum_{j \mathop = 0}^n \left\lvert{a_j}\right\rvert 1\) | as $\dfrac {x^\left({j + 1}\right)} {e^{a x} } < 1$ for every $j$ in $\left\{ {0, \ldots, n}\right\}$ | ||||||||||
| \(\displaystyle \) | \(<\) | \(\displaystyle \epsilon\) |
$\blacksquare$
