Square Root is of Exponential Order Epsilon

Theorem

The positive square root function:

$t \mapsto \sqrt t$

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

Proof

\(\ds \sqrt t\)

\(<\)

\(\ds K e^{a t}\)

an Ansatz

\(\ds \impliedby \ \ \)

\(\ds t\)

\(<\)

\(\ds \paren {K e^{a t} }^2\)

Square Root is Strictly Increasing

\(\ds \)

\(=\)

\(\ds K^2 e^{2 a t}\)

Exponential of Product

\(\ds \)

\(=\)

\(\ds K' e^{a' t}\)

$K^2 = K', 2 a = a'$

Recall from Identity is of Exponential Order Epsilon, $t < K'e^{a' t}$ for any $a' > 0$, arbitrarily small in magnitude.

Therefore the inequality $\sqrt t < K e^{a t}$ has solutions of the same nature.

$\blacksquare$