# Seeking a function based on its level set

I'm trying to create a function for a research project, but I fear my math knowledge is insufficient to derive it from the attached diagram I've created showing its desired behavior. I'm hoping someone will recognize the behavior and provide some leads on similar functions to lead me in the right direction.

$\hskip 1.1in$

For the axes we have $X \in [0,1]$ and $Y \in (0, \infty)$. The dotted lines are isometric lines of the Z axis. The Z values and exact shape of the isolines are not as important as the monotonicity of the graph w.r.t. both X and Y. I'm not against having a piecewise function for $Y>1$, $Y=1$, and $Y<1$ if necessary but a single function would be a bonus.

-Jeff

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Sorry, I should have been more specific. It's a 3 dimensional function with the Z axis being perpendicular to the screen. The lines are the "points" at which the function crosses those values. For example, at the line marked 1, $Z<1$ when $Y>1$ and $Z>1$ when $Y<1$. It's also referred to as a contour line, where Z equals that value at all points along the line. – Jeff Apr 13 '12 at 23:27
Oh, contour lines, of course. – anon Apr 13 '12 at 23:29

Try something like $(x,y) \mapsto 1+e^{-(x^2+y^2)}-e^{-(x^2+(y-2)^2)}$? May need a little scaling...

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That is fantastic. I apologize for my ignorance, but what parameters would I change to shift the contour lines at y=2 and y=0 to the left a little bit? – Jeff Apr 14 '12 at 6:41
Replace $x$ by something like $1.1 x$, ie, $(x,y) \mapsto 1+e^{-((1.1 x)^2+y^2)}-e^{-((1.1 x)^2+(y-2)^2)}$. Some experimentation will be needed! – copper.hat Apr 14 '12 at 6:50
Thank you so much. This captures exactly what I was looking for! – Jeff Apr 14 '12 at 16:52

It looks like it ought to be something like $$z = y^{-g(x)}$$ where $g$ is some strictly decreasing function $[0,1] \to (0,\infty)$.

The rough sketch does not give much clue to the exact choice of $g$; you could try something like $g(x) = 10 - 9x$ as a starting point.

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This is definitely the right direction, but I'm not sure about the behavior as Y approaches $0$. I was hoping for a derivative that decreases monotonically when $Y<0$. – Jeff Apr 14 '12 at 0:25
@Jeff, you specified $Y\in(0,\infty)$, so what's that about behavior when $Y<0$? – Henning Makholm Apr 14 '12 at 0:27
When $y<1$ rather. – Jeff Apr 14 '12 at 0:30

After some thought, to me this looks like it could be an implicit relationship between $x$, $y$, and $z$:

$$y = 1+(1-z)\textrm{e}^{(1-z)^2x}$$

Solving this equation for $z$ explicitly in terms of $x$ and $y$ is not possible with elementary functions though.

In case it helps, I found this by subtracting $1$ from your contour values, so that they are balanced around $0$ at the horizontal contour line. Then I negated them, so that positives would naturally correspond to contours above the horizontal line. The actual shape of your contour lines reminds me of $$y=1+k\mathrm{e}^{k^2x}$$ where the $k$ in the exponent is really only squared to make it positive. $k$ ranges from $-1$ to $\infty$. $k$ plays the role of the transformed contour values, and then we have my guess at the relationship.

(And maybe replacing $\mathrm{e}$ by some other base would make contour lines more precisely in line with yours.)

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Does making it a piecewise function and solving for only $y<1$ or $y>1$ simplify the relationship enough? It's starting to look like that's the best (if not the only) option at this point. – Jeff Apr 14 '12 at 0:38
No, that wouldn't simplify this particular stab at your problem. With $z$ both "down below" and up in the exponent, it will not be possible to solve for $z$ explicitly using usual functions. However, if you are willing to work with the product log function, then breaking up piecewise does help and you have the two functions defined here. I'd be curious if this works - let me know. – alex.jordan Apr 14 '12 at 21:49

The simplest thing I can think of is $$z=\frac{1−y}{1+4x^2}+1.$$ Its contour lines are parabolae.

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This is excellent, and I like the simplicity of it. I accepted copper.hat's example because it exactly matched the shape I was looking for but I may use this instead. – Jeff Apr 14 '12 at 17:02
No fair, you said "the exact shape of the isolines are not as important"! :) I'm glad you found what you were looking for, in any case. – Rahul Apr 15 '12 at 19:45