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I'm looking for a function that creates the blue curve shown in the picture. The point is that within a certain range (e.g. x=0 to x=0.8) the output of the function should be relatively close to 1.0. If possible it would be exactly 1 but "close" to 1 (within 10% or so) would be good enough. However, the first value produced by the function should be exactly 1. eventually, the output should drop sharply to y=0.

The available input value is x which linearly increases along the x-axis from 0 to 1. The function could rely on any other parameter necessary to achieve the illustrated curve.

Any ideas?

graph dropping from 1 to 0

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  • $\begingroup$ Try $y=(1-x^n)^{1/n}$ for, say, $n>8$. $\endgroup$ – g.kov Apr 27 at 10:03
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The first candidate that comes to mind is the function of the form

\begin{align} y(x)&=(1-x^n)^{1/n} . \end{align}

This is how the graph of this function looks for $n=5\dots12$:

enter image description here

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  • $\begingroup$ That's what I was looking for. The interesting property of that function is that it results in a straight line if n=1 and "bends" in the other direction if n<1. That might come in handy for my application. Thanks! $\endgroup$ – ackh Apr 27 at 11:22
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Another function you could try graphing is:

$$x^{2n} + y^{2n} = 1$$ As $n$ gets bigger and bigger, its shape resembles a square! For your purpose I would think $n=4$ and $n=6$ are good choices.

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