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

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

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

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

A coefficient is needed to configure the speed at which it reaches 1, or the actual shape of the curve

Could anyone give any advice? Thanks!

@AnuragA also, do you know anywhere can I plot these functions? I can manually plot them in python, but is there better approach? — ZK Zhao, Jul 12 '19 at 15:49
As soon as you have one such function $f(x)$, you can configure the speed by considering $f(ax)$ where $a > 0$. When $a > 1$, $f(ax)$ will reach $1$ "faster". — Ben Grossmann, Jul 12 '19 at 15:50
@Gae.S. in that question we have $f'(0) = 1$. In this question we require $f'(0) = 0$. — Ben Grossmann, Jul 12 '19 at 15:52
None of $\frac{x^2}{1+x^2}$, $\tanh(x)^2$, $\tanh(x^2)$ is monotonic as requested. But I think $\tanh(x^3)$ or similarly $\frac{2}{\pi} \tan^{-1}(x^3)$ should work. — Daniel Schepler, Jul 12 '19 at 15:58
Do you want it to be monotonic for all real $x$, or just $x > 0$? — Robert Israel, Jul 12 '19 at 15:58
@RobertIsrael Both is OK for my case. Since I can use f(|x|) . — ZK Zhao, Jul 12 '19 at 16:00
Not asked for, but a nice extension - $f(x) = \begin{cases} \operatorname{sgn}(x) e^{-1/x^2}, & x \ne 0 \ 0, & x = 0 \end{cases}$ is $C^\infty$, monotonic, has $f^{(n)}(0) = 0$ for all $n$, and $\lim_{x\to\infty} f(x) = 1$. — Daniel Schepler, Jul 12 '19 at 16:06
Differentiability is a local property, what does $f'(0) = 0$, $f(0) = 0$ have to do with $\lim_{x\to \infty} f(x) = 1$ ? — Jakobian, Jul 12 '19 at 16:31
But $f(|x|)$ will not be monotonic (unless it is constant). If it's increasing for $x > 0$, it's decreasing for $x < 0$. — Robert Israel, Jul 12 '19 at 18:45

score 3 · Accepted Answer · answered Jul 12 '19 at 16:17

I am just putting out my comment as the answer. Let $$ f(x)=\begin{cases} \frac{x^2}{1+x^2}, & \text{ for } x \geq 0\\ 0, & \text{ for } x <0. \end{cases} $$

Then $f(0)=0$ and $\lim_{x \to \infty}f(x)=1$. $$ f'(x)=\begin{cases} \frac{2x}{(1+x^2)^2}, & \text{ for } x \geq 0\\ 0, & \text{ for } x <0. \end{cases} $$ Thus $f'(0)=0$. Furthermore, $f$ is non-decreasing over $\Bbb{R}$.

Jürgen Sukumaran · Answer 2 · 2019-07-12T16:02:08.183

2

$$f(x) = \begin{cases}0 & x<1\\ x-1 & x\in[1,2]\\ 1 & x>2 \end{cases}$$

edit: here is a differentiable answer,

$$f(x) = \begin{cases}0 & x<1\\ \frac{2}{9}(x-1)^2 & x\in[1,\frac{3}{2}]\\ -\frac{2}{9}(x-2)^2 & x\in[\frac{3}{2},2]\\ 1 & x>2 \end{cases}$$

edited Jul 12 '19 at 16:02

answered Jul 12 '19 at 15:49

Jürgen Sukumaran

7,645

I think this is not continuous, right? – ZK Zhao Jul 12 '19 at 15:52
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@cqcn1991 it is continuous but not differentiable – Ben Grossmann Jul 12 '19 at 15:52
It certainly is continuous. It's not differentiable. – Jürgen Sukumaran Jul 12 '19 at 15:52
OK, I'll edit my question. Thanks! – ZK Zhao Jul 12 '19 at 15:53

score 0 · Answer 3 · answered Jul 12 '19 at 15:57

0

How about: $$ f(x)= \left( \frac{\arctan(x)}{\pi/2} \right)^2 $$

answered Jul 12 '19 at 15:57

Sambo

6,243

SlipEternal · Answer 4 · 2019-07-12T16:20:13.273

0

How about:

$$f(x) = \dfrac{x^3}{1+|x^3|}$$

$$f(0) = 0, f'(x) = \dfrac{3x^2}{\left(1+|x^3|\right)^2}, f'(0) = 0, \lim_{x \to \infty} f(x) = 1$$

This seems to satisfy all of your conditions. Its derivative is continuous. It is monotonically increasing.

edited Jul 12 '19 at 16:20

answered Jul 12 '19 at 16:04

SlipEternal

10,555
1
17
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score 0 · Answer 5 · answered Jul 13 '19 at 08:19

Although the above answers are complete, here’s another function that you may like.

Let $$ f(x)=\begin{cases} 1 - e^{-x^2}, & \text{ for } x \geq 0\\ e^{-x^2} - 1, & \text{ for } x <0. \end{cases} $$

Clearly, It is monotonous, $f(0) = 0$, $f’(0) = 0$, and, $lim_{x\to\infty} f(x) = 1$.

PS: This function is obtained by just an algebraic manipulation (really, adding constant 1) of a very famous curve.

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

5 Answers5