Question

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

Answer 1

No, `g(q)` decays in proportion to the square of `q`.

Exponential decay is much faster (eventually).

Exponential functions are of the form `f(x) = k*a^x` for some `k in RR` and `a > 0, a != 1`.

Explanation:

Let `h(q) = 0.5^q`

Let's look at the first few values of `g(q)` and `h(q)`...

`((q, g(q), h(q)), (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))`

At first, `g(q)` seems to be decaying faster than `h(q)`, but eventually `h(q)` overtakes it.

In fact `h(q)` would eventually overtake any function of the form `g(x) = x^-n` with `n > 0` in its decay.