Looking for a function with $f(0) = 0$, $f'(0) = 1$ and $\lim\limits_{x\to\infty}f(x)=1$

Question

I'm looking for a monotonic, continuous function function with these properties:

$$f(0) = 0$$ $$f'(0) = 1$$ $$\lim\limits_{x\to\infty}f(x)=1$$

I also need a coefficient that allows me to configure the speed at which it reaches 1.

I've got the feeling that there is a well-known function that I am looking for but my brain just refuses to give it to me right now.

What is "sine-like" supposed to mean? What about $f(t)=\frac2\pi\arctan(\frac\pi2 t)$? Or in fact, what about $f(t)=\min{t,1}$? — Hagen von Eitzen, May 02 '13 at 19:56
@HagenvonEitzen I replaced sine-like with continuous. Your proposed function is fine. — Marcel Jackwerth, May 02 '13 at 20:04
@CameronBuie Assume that $x$ is a time-variable (should have named it $t$). So a coefficient to adjust for which x $f'(x) < 0.01$. — Marcel Jackwerth, May 02 '13 at 20:27

Ross B. · Answer 1 · 2013-05-02T20:20:31.940

11

Here are a variety of functions (taken from http://en.wikipedia.org/wiki/Sigmoid_function)!

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Many of these types of equations can be parameterized to control rate of convergence. A simple way would be to paramaterize a combination of two of these functions e.g. $f(x) = ag_1(x) + (1-a)g_2(x)$, where $g_1,g_2$ are relatively "fast" or "slow" converging functions.

edited May 02 '13 at 20:20

answered May 02 '13 at 20:09

Ross B.

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Robert Mastragostino · Accepted Answer · 2013-05-02T21:12:49.283

This might make things easier. Take any function such that $f(0)=0, f'(0)=1$. Then

$$\frac{f(x)}{1+f(x)}$$

automatically satisfies both of these and your third requirement if $\lim_{x\to\infty} f(x)=\infty$. It's also monotonically increasing when your original function is. If you have any single function $g$ such that $g(x)=0, g'(x)=1$, then $\frac 1n g(nx)$ will also have these properties, allowing you to generate a family of functions. For example, if we take $f(x)=\sinh(x)$, then our family of functions is $\frac 1n\sinh(nx)$ and our overall set of solutions is $$\frac{\sinh(nx)}{n+\sinh(nx)}$$

which you can verify have all of your needed properties and approach $1$ at varying rates. Plug in any increasing function with $f(x)=0$ and $f'(x)=1$ in place of $\sinh(x)$ and you'll have all the fine-tuning you need.

Ron Gordon · Answer 3 · 2013-05-02T21:04:53.297

4

$f(x) = 1-e^{-x}$ would seem to be a candidate.

Another one is

$$f(x) = 1-e^{-(e^x-1)}$$

In general:

$$f(x) = 1-e^{-g(x)}$$

where $g(0)=0$, $g'(0)=1$, and $\lim_{x \to \infty} g(x) = \infty$.

edited May 02 '13 at 21:04

answered May 02 '13 at 20:10

Ron Gordon

138,521

Clayton · Answer 4 · 2013-05-02T20:41:14.620

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I suggest $$f(x)=\frac{x\alpha}{x\alpha+1},\qquad\alpha>0.$$

edited May 02 '13 at 20:41

answered May 02 '13 at 20:03

Clayton

24,751

How about you add a parameter $\alpha$ with the $x$'s so that you can satisfy the requirement about "how fast it approaches 1"? – rschwieb May 02 '13 at 20:07
@rschwieb: Thanks for suggestion. I've edited it in. – Clayton May 02 '13 at 20:28
Err, heh, actually I was thinking $\alpha x$ would be a bit more predictable. – rschwieb May 02 '13 at 20:29
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I'm not sure your parametrization satisfies the derivative requirement. – Ross B. May 02 '13 at 20:33
$$f'(x) = \frac{\alpha x^{\alpha-1}}{(x^{\alpha}+1)^2}$$ – Ron Gordon May 02 '13 at 20:36

score 1 · Answer 5 · answered May 02 '13 at 20:30

1

$$f(x) = \left\{\begin{array}{l l} x & x\le 1\\ 1 & x > 1\end{array}\right.$$

answered May 02 '13 at 20:30

mez

10,497

Ok it doesn't have a parameter to adjust "speed" – mez May 02 '13 at 20:34

Looking for a function with $f(0) = 0$, $f'(0) = 1$ and $\lim\limits_{x\to\infty}f(x)=1$

5 Answers5

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