# Is g(q)=q^-2 an exponential function?

Jun 19, 2015

No, $g \left(q\right)$ decays in proportion to the square of $q$.

Exponential decay is much faster (eventually).

Exponential functions are of the form $f \left(x\right) = k \cdot {a}^{x}$ for some $k \in \mathbb{R}$ and $a > 0 , a \ne 1$.

#### Explanation:

Let $h \left(q\right) = {0.5}^{q}$

Let's look at the first few values of $g \left(q\right)$ and $h \left(q\right)$...

$\left(\begin{matrix}q & g \left(q\right) & h \left(q\right) \\ 1 & 1 & 0.5 \\ 2 & 0.25 & 0.25 \\ 3 & 0.111111 & 0.125 \\ 4 & 0.0625 & 0.0625 \\ 5 & 0.04 & 0.03125 \\ 6 & 0.027778 & 0.015625 \\ 7 & 0.020408 & 0.0078125\end{matrix}\right)$

At first, $g \left(q\right)$ seems to be decaying faster than $h \left(q\right)$, but eventually $h \left(q\right)$ overtakes it.

In fact $h \left(q\right)$ would eventually overtake any function of the form $g \left(x\right) = {x}^{-} n$ with $n > 0$ in its decay.