# How to make something that will cap on 20?

A user on the chat asked how could he make something that would cap when it gets a specific value like 20. Then the behavior would be as follows:

$f(...)=...$

$f(18)=18$

$f(19)=19$

$f(20)=20$

$f(21)=20$

$f(22)=20$

$f(...)=20$

He said he would like to perform it with a regular calculator. Is it possible to do this?

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Could you better explain your problem? – Sigur Aug 20 '12 at 12:47
Your use of "=" has so many people rolling in their graves now I think you have triggered the mathematician zombie apocalypse. Do you mean something like capping a sequence $f(1),f(2),f(3)\dots$? – rschwieb Aug 20 '12 at 12:49
From here: $$\frac{x+20-|x-20|}{2}$$ – J. M. Aug 20 '12 at 12:49
@GustavoBandeira You shouldn't put "=" between quantities that aren't equal. If you do, you have a false statement. Writing "22=20" is as absurd as writing "0=1". – rschwieb Aug 20 '12 at 13:07
@GustavoBandeira Yeah, that's a much better version :) People will understand you more quickly. – rschwieb Aug 20 '12 at 17:17

$x \mapsto \min ( x , 20 )$
We can also get a bit (unnecessarily) fancier: $$f(x) = x + (20 - x) \int\limits_{-\infty}^{x-20} \delta(t)\ dt$$ where $$\int\limits_{-\infty}^{x-20} \delta(t)\ dt = \begin{cases} 0 & x < 20 \\ 1 & x \ge 20 \end{cases}$$ (See Heaviside step function.)
While we are at fancy expressions, what about $$20-\lim_{n\to\infty}\frac1n\ln\left(1+\mathrm e^{n(20-x)}\right)$$