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I've been trying to come up with a one-variable function that roughly matches this graph:

enter image description here

The closest I've gotten is $ f(d) = \dfrac{\log(c \cdot (1 - \frac{d}{c}))}{\log c} $, but this approaches $y=0$ at $x=c-1$, and steadily declines from $x=0$, instead of sharply.

Are there any functions or tricks I should look into to develop this further?

Thank you.

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Have you tried piecewise-defined functions? This looks like you could approximate it well with a parabola near 0 (i.e., $1 - x^2$), a square root near $c$ (i.e., $\sqrt{c-x}$) and some thing like a third-degree polynomial or a rational function inbetween. – Johannes Kloos May 6 '12 at 17:11
Something like $\exp(-x^2)-1/(5-x) +1$ comes close. The exponential term gives you the "left part" of the graph (with the inflection point) and the $1/(5-x)$ term gives you the "right part". Now you can scale (replace $x$ with $ax$ to scale horizontally; multiply the whole thing by $a$ to scale vertically) this to obtain what you want. – David Mitra May 6 '12 at 17:15
That's beautiful, @DavidMitra! Thank you very much, and for the explanation. – theY4Kman May 6 '12 at 18:03
You could also try doing something in polar coordinates. – John Engbers Jun 7 '12 at 13:02

One thing might do is find a function $f$ whose graph is qualitatively the same as yours on the "left part", that is over $[0,C/2]$, say, and relatively constant on the right part, $[C/2, C]$.

Then find a function $g$ whose graph matches yours on the right part and is relatively constant on the left part.

Taking the sum $f+g$, you'll obtain a function whose graph is qualitatively the same as yours. The fact that $f$ and $g$ are nearly constant where they are is important, as this will insure that the sum $f+g$ will still have more or less the right shape.

However, the sum $f+g$ may not be 0 at $x=C$; you may need to vertically shift the function by adding a constant.

Here to get the bell shaped part on the left, you might use $f(x)=\exp(-x^2)$. For the shape of the right, the function $g(x)=.2|C-x|^{1/3}$ seems to work nicely.

So take $h(x)= \exp(-x^2) + .2|C-x|^{1/3}$. Even better would be $h(x)= \exp(-x^2) + .2|C-x|^{1/3} -\exp(-C^2)$, as this will insure that $h(C)=0$.

Well, that's almost right, except $h(0)$ is not $1$. But to ameliorate this, we can vertically scale, take $h(x)=a\bigl( \exp(-x^2) + .2|C-x|^{1/3} -\exp(-C^2)\bigr)$ where $a$ is chosen so that $h(0)=1$.

Actually, my choices of $f$ and $g$ are somewhat unsatisfactory (in particular, the function $g$ isn't nearly constant on the left part; perhaps the choice I made in my comment to your question ($g(x) = -{1\over C-x}$) would be a better choice). But I hope the method described is of help.

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Let's try it in polar coordinates (as suggested by John).

We will start with a four petals flower getting the polar expression $$\rho=\frac{m-1+\cos(4\theta)}m$$ $m$ is a parameter and $m\approx 5$ seems appropriate giving :


In your case the graph will be obtained by $$x=C\rho \cos(\theta),\ y=\rho \sin(\theta)$$ To rewrite it with just one parameter you may use $\ u:=\cos(\theta)$ getting : $$\rho=\frac{m+8(u^4-u^2)}m$$ $$x=C\;\rho\;u,\ y=\rho \sqrt{1-u^2}$$

But $y$ is not a simple function of $x$ and worse the bump at the left is a little too large and the middle part not smooth enough... (but it had to be tried!)

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