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