Polynomial is of Exponential Order Epsilon

Theorem
Let $P: \R \to \mathbb{F}$ be a polynomial, where $\mathbb{F} \in \left \{{\R,\C}\right\}$.

Then $P$ is of exponential order $\epsilon$ for any $\epsilon > 0$ arbitrarily small in magnitude.

Proof
If $P = 0$, the theorem holds trivially.

Let $P_n$ be a polynomial of degree $n$, where $n \ge 0$.

The proof proceeds by induction on $n$, where $n$ is the degree of the polynomial.

Basis for the Induction
Let $P_0$ be of degree zero.

Then $P_0$ is a constant polynomial.

By Constant Function is of Exponential Order Zero, $P_0 \in \mathcal{E}_0$.

and therefore is of exponential order $\epsilon$ as well.

This is the basis for the induction.

Induction Hypothesis
Fix $n \in \N$ with $n \ge 1$.

Assume:


 * $P_n \in \mathcal{E}_\epsilon$

That is,


 * $\left \vert {P_n\left({t}\right)}\right \vert < Ke^{\epsilon t}$

For some $K > 0$, for $\epsilon > 0$ arbitrarily small.

This is our induction hypothesis.

Induction Step
Let $P_{n+1}$ be of degree $n+1$.

By the Polynomial Factor Theorem,


 * $P_{n+1} = P_1 P_n$

for some polynomials of degree $1, n$.

But $P_1$ is of exponential order by the base case, and $P_n$ is of exponential order by the induction hypothesis.

Thus $P_1P_n$ is the product of functions of exponential order $\epsilon$, where $\epsilon > 0$ is arbitrarily small.

Thus $P_{n+1} = P_1 P_n$ is of degree $\epsilon$.

The result follows by the Principle of Mathematical Induction.