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I need a function with a specific shape:

  • Quadratic/gaussian concave shape ($-x^2$ like)
  • Centered in $\frac{1}{2}$ where it reaches the max value 1
  • On 0 and 1 to become null

I first tried using a second-degree polynomial function and I failed.
By repeated tries on, after I deduced on paper that $c=0$, I got $-4x^2+4x$ to be what I need.

Then I found the function $e^{-x^2}$ which have a very nice shape.
But now I have to span it in $[-0.5, 0.5]$ and add an offset.
$e^{-(x-0.5)^2}$ is centered in $\frac{1}{2}$ where it reaches 1, but it does not descend down to zero in 0 and 1.

I'm interested in either the solution to this problem and in the process of reaching that solution. How is this usually done?

If Q above is unclear:

How do I model $e^{-x^2}$ to meet the requirements in bullets above?

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do you want it to drop sharply at $0$ and $1$ (like the quadratic does) or do you want it to go smoothly (i.e slope goes to zero at these points)? And are piecewise definitions okay? – Robert Mastragostino Apr 18 '13 at 16:46
I guess a 4th degree polynomial would do the trick with the right coefficients (and the right $y$ offset to make it have only 2 roots : $0$ and $1$) – Dolma Apr 18 '13 at 16:48
@RobertMastragostino: I need it to reach 0 (any slope). I'd prefer it in one piece. – Iulius Curt Apr 18 '13 at 16:49
@Dolma: I added a more concise and clear question – Iulius Curt Apr 18 '13 at 16:53
I'm not sure what you mean by modelling $e^{-x^2}$, but you might try a function like $$f(x) = \exp\left(4-\frac{1}{x(1-x)}\right).$$ Here's a graph:… – Abel Apr 18 '13 at 16:58
up vote 2 down vote accepted

Well you're almost there. You have your nice function centered on $x=\frac{1}{2}$:


Now you can compute $f(0)$ and $f(1)$ which will be equal because $f$ is symmetric around $x=\frac{1}{2}$:

$$\large f(0)=f(1)=e^{-\frac{1}{4}}$$

If you want $f(0)=f(1)=0$ you just have to write:


But now for $x=\frac{1}{2}$, your function is equal to $\large1-e^{-\frac{1}{4}}$

Just multiply it by $\large\frac{1}{1-e^{-\frac{1}{4}}}$ and here you go:

$$\large f(x)=\frac{e^{-(x-\frac{1}{2})^2}-e^{-\frac{1}{4}}}{1-e^{-\frac{1}{4}}}$$

Here's the plot of this function:

enter image description here

share|cite|improve this answer
Nice, thanks. It turned out to look an awful lot like the quadratic. :) – Iulius Curt Apr 18 '13 at 17:35

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