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

1 Answer
Jun 19, 2015

Answer:

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.