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 $0$.

Plot of hyperbolic and vs. exponential decay

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

  • $\begingroup$ are there names for those? log-hyperbolic? $\endgroup$ – develarist Jan 24 at 12:54
  • $\begingroup$ Probably just logarithmic decay $\endgroup$ – 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 $0<p<1$:

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

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. $

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

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