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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?

develarist
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2 Answers2

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

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

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. $

Adam Rubinson
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  • how do these compare to logarithmic decay? – develarist Jan 24 '21 at 13:12
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    @develarist since the logarithm grows way slower than any positive power of x, logarithmic decay would be slower – Kinheadpump Jan 24 '21 at 13:15
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    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 '21 at 13:17
  • See here for a proof: https://math.stackexchange.com/questions/55468/how-to-prove-that-exponential-grows-faster-than-polynomial – Adam Rubinson Jan 24 '21 at 13:24
  • @AdamRubinson, do these $h_p(x)$ functions have a commonly used name? I'm looking for something in the vein of "exponential decay", but clearly not exponential. – David Tarandek Sep 22 '22 at 13:13
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    @DavidTarandek I don't know, but I can't think of a name off the top of my head. But like I say, I don't know. – Adam Rubinson Sep 22 '22 at 17:38