# What is a slower decay rate than hyperbolic?

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?

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 $$\frac{1}{\ln(\ln(x))}$$

• are there names for those? log-hyperbolic? – develarist Jan 24 at 12:54
• Probably just logarithmic decay – Kinheadpump Jan 24 at 12:56

$$f(x) = \frac{1}{x}.$$

Reciprical square root,

$$g(x)=\frac{1}{\sqrt{x}}$$

or $$x^{-p}$$ for any $$0:

$$h_p(x)=\frac{1}{x^p}.$$

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.$$

• how do these compare to logarithmic decay? – develarist Jan 24 at 13:12
• @develarist since the logarithm grows way slower than any positive power of x, logarithmic decay would be slower – Kinheadpump Jan 24 at 13:15
• logarithmic decay is always a lot slower than $h_p(x).$ For example, $\frac{1}{\log_{10} (1,000,000)} = \frac16,\$ whereas $\frac{1}{1,000,000^{\frac13}} = \frac{1}{100}.$ – Adam Rubinson Jan 24 at 13:17
• See here for a proof: math.stackexchange.com/questions/55468/… – Adam Rubinson Jan 24 at 13:24