I'm looking for a way to create an inverted $S$ curve function. Below is an image of what I want. Left is a regular S curve, on the right is the kind of curve I want. Here are the allowed set of operations that I can use:
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How about $x^3$? – Rushabh Mehta Feb 23 '21 at 12:20
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With these vertical tangents and flat middle I would go for $x(a-\sqrt{1-x^2})$. With $a=1$ flat, with $a>1$ some slope. – zwim Feb 23 '21 at 12:29
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1Can you specify the domain and range? There are plenty of inverse sigmoid functions to pick from, but they all have vertical asymptotes, i.e. shoot up to infinifty. Please edit your question. – V.S.e.H. Feb 23 '21 at 12:31
3 Answers
Possible with Inverse function where function inversion is possible. The given graph indicates an odd function, straight forward function inversion solution. In polar coordinates $ r(\theta) = r(-\theta).$
In other cases absolute values to be taken and later on modified suitably.
$$ y= f(x)\to x= f(y) $$
$$ or$$
$$ y= f(x)\to y_{inv}= f^{-1}(x_{inv}) $$
In the above we have inverse tanh brown function solution graphed along with starting green tanh function. Similarly the sine function.
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Not exactly what I was looking for, I wanted a range for x of [0, 1] with the y value being nondecreasing from 0 to 1 in this range. I found a function which does this, I answered my own question. – porras Feb 24 '21 at 13:07
Using complex variables, we can transform the exact $S$-curve you started with very easily. Let us assume the inflection point passes through the origin and let the curve be expressed parametrically as $z(t)$. The first thing we want to do is to flip the curve about the $x$-axis, then rotate it $90^{\circ}$ counterclockwise, the the desired curve becomes
$$w(t)=z^*(t)e^{i\pi/2}=iz^*(t)$$
In Cartesian coordinates then
$$w(t)=i\big(x(t)-iy(t)\big)=y(t)+ix(t)$$
Which means that all you have to do is to switch $x$ and $y$. You can see as much by just flipping your original image about.
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