Hyperbolic decay
$$f_\alpha(x)=\frac{1}{\alpha x + 1} $$
is slower than exponential decay
$$f(x) = e^{-x}$$
where $\alpha > 0$ is a scaling factor. The larger that $\alpha$ is, the steeper the descent towards $0$.
What would be a slower decay function than hyperbolic?
$\endgroup$2 Answers
$\begingroup$Any reciprocal of a function that grows slower than linear. E. g. a much slower decay rate than hyperbolic would be $$\frac{1}{\ln(x)}$$ or even $$\frac{1}{\ln(\ln(x))}$$
$\endgroup$ 2 $\begingroup$$$f(x) = \frac{1}{x}.$$
Reciprical square root,
$$g(x)=\frac{1}{\sqrt{x}} $$
or $x^{-p}$ for any $0<p<1$:
$$h_p(x)=\frac{1}{x^p}. $$
These are much slower than $\frac{1}{x}$ as $x \to \infty$, since
$\large{\frac{f(x)}{h_p(x)} = \frac{\frac{1}{x}}{\frac{1}{x^p}} = \frac{x^p}{x} =}$ $x^{p-1}\to 0 $ as $ x \to \infty\ $ because $p-1<0. $
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