Sum of Functions of Exponential Order
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Theorem
Let $f, g: \R \to \F$ be functions, where $\F \in \set {\R, \C}$.
Suppose $f$ is of exponential order $a$ and $g$ is of exponential order $b$.
Then $f + g: t \mapsto \map f t + \map g t$ is of exponential order $\max \set {a, b}$.
Proof
Let $t$ be sufficiently large so that both $f$ and $g$ are of exponential order on some shared unbounded closed interval.
By the definition of exponential order:
\(\ds \size {\map f t}\) | \(<\) | \(\ds K_1 e^{a t}\) | ||||||||||||
\(\ds \size {\map g t}\) | \(<\) | \(\ds K_2 e^{b t}\) | ||||||||||||
\(\ds \size {\map f t} + \size {\map g t}\) | \(<\) | \(\ds K_1 e^{a t} + K_2 e^{b t}\) | Real Number Inequalities can be Added | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \size {\map f t + \map g t}\) | \(<\) | \(\ds K_1 e^{\max \size {a, b} t} + K_2 e^{\max \size {a, b} t}\) | Triangle Inequality for Real Numbers, Exponential is Strictly Increasing | ||||||||||
\(\ds \) | \(=\) | \(\ds K' e^{\max \size {a, b} t}\) | $K' = K_1 + K_2$ |
$\blacksquare$