Sum of Functions of Exponential Order

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

\(\ds \)

\(=\)

\(\ds K' e^{\max \size {a, b} t}\)

$K' = K_1 + K_2$

$\blacksquare$