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

1 Answer
George C.
Jun 19, 2015

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.

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