A simple function equation

Question

I come from a programming background and I can’t find a simple math function. The request might seem strange, but I needed it a graphical context to alter some points locations:

I need a function $f(x) = y$ defined for $x \ge 0$ such that:

• $f(x) \in [0, x)$
• $f(0) = 0$
• $f(x) \approx x$ as $x\to \infty$.
• It has to slowly grow at first — sort of like $x^2$ — and then get closer and closer to x.

The simplest equation form that satisfies this restrictions will do.

I tried to plot this so that I can make myself better understood:

http://www.wolframalpha.com/share/clip?f=d41d8cd98f00b204e9800998ecf8427eo8stqhpe1s

Actual values don’t matter, just the shape of the plot.

None of the basic functions (and combinations of them) that I tried were doing this (e.g. $x^2, \log x, \sqrt x, 1/x$).

Answer 1

This one fulfills your requirements: $$f(x)=\frac{x^2+x^3}{1+x+x^2}.$$ We have: $$\forall x>0,\ 0<f(x)<x$$ $$f(0)=0,$$ $$f(x)-x^2=-\frac{x^4}{1+x+x^2}$$ so that $f(x)$ and $x^2$ are very close for small values of $x$, and $$f(x)-x=-\frac{x}{1+x+x^2}$$ so that $f(x)$ and $x$ get closer and closer as $x\to+\infty$.

It's also cheap to compute with 2 additions, 2 multiplications and 1 division if you proceed thus:

• Compute $x^2$ (1 mult) and $x+x^2$ (1 addition); set $a=x+x^2$.
• Compute $x\times a$ (1 mult); set $b=x\times a$.
• Compute $1+a$ (1 addition); set $c=1+a$.
• Compute $b/c$ (1 division): that's $f(x)$.

Answer 2

Using your data, I found that $y=a x^b$ fits very well. My results are $a=0.008755$ and $b=3.134091$. The corresponding $R^2=0.999$.

This is a very flexible form.

Answer 3

You could use a common hyperbola $y = \sqrt{a^2 + x^2} - a$.

Example with $a = 10$:

Answer 4

Consider the derivative of your function. It has to be positive,increasing and $f'(x→inf)=1$ $f'(x)=tanh(x)$ fits.

$ \int \tanh(x) dx = \log (\cosh (x))+c$

Since we want $f(0)=0$

$\log (\cosh 0)+c=0\Leftrightarrow c=0$

Therefor $f(x)= \log (\cosh (x)) $ works.

Answer 5

You can use a weighting function $w(x)$ that allows you to create a mixture between the functions $x^2$ and $x$, as $\frac{x^2+w(x)x}{1+w(x)}$. Ensure $w(0)=0$ so that the initial behavior is $x^2$ and $w$ growing sufficiently fast that the term $x$ supersedes it. $$w(x)=x^3\to y=\frac{x^2+x^4}{1+x^3}.$$ $$w(x)=e^x-1\to y=\frac{x^2+x(e^x-1)}{e^x}.$$

Note that if you really want to reach $y=x$ (and not $y=x-c$), there must be an inflection point.

Answer 6

A simple function satisfying all above mentioned restrictions is:

$$f(x) = x \left( 1- \frac{1}{1+x} \right)$$

How to construct such a function?

1. $f(x)$ should behave like $x$ for $x \rightarrow \infty$ and for $x \gt 0$ it should map into the interval $[0,x)$ which mean $0 \leq f(x) \lt x$. So let's set up the function as $$f(x) = x \left( 1- g(x) \right)$$ To satisfy the above mentioned restriction we conclude: $$g(x) \rightarrow 0 \mbox{ for } x \rightarrow \infty \mbox{ and } g(0) = 1 \mbox{ and } 0 \lt g(x) \lt 1 \mbox{ for } x \gt 0$$
2. A simple function that satisfies all this is $$g(x) = \frac{1}{1+x}$$
3. Near $0$ the function $f(x)$ behaves like $x^2$ because for $|x| \lt 1$ we have $\frac{1}{1+x} = 1 - x + x^2 - x^3 +- \cdots$ (geometric series). So $$f(x) = x\left( 1 - ( 1 - x + x^2 - x^3 +- \cdots) \right) = x^2 - x^3 +- \cdots \approx x^2 \mbox{ for } x \mbox{ close to }0$$