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I delete my file which I used to produce this graph. Does anybody have some idea how to produce it again?

enter image description here

Thanks for a while.

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This might be easier to answer if your picture were bigger. Do you have a bigger picture? –  MJD Jul 30 '12 at 20:47
    
Unfortunately I don't have. Maybe if you zoom it. I will try to find a bigger one. –  Sigur Jul 30 '12 at 20:51
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As it happens, $r^2\sin8\theta$ can be written as a rational function of $x$ and $y$, to wit: $8xy\big(x^6-7x^4y^2+7x^2y^4-y^6\big)\big/\big(x^2+y^2\big)^3$. –  Rahul Jul 30 '12 at 21:57
    
Have a look at the generalized monkey saddle‌​. –  J. M. Aug 1 '12 at 7:53
    
@J.M. Thanks for the msg, but I guess that my graph is not a saddle like this. –  Sigur Aug 5 '12 at 23:51

2 Answers 2

up vote 9 down vote accepted

For example : $$f(x,y)=\sqrt{x^2+y^2}\sin(8 \arctan(y/x))$$

enter image description here

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Isn't this roughly what Steven just posted? $\rho=\sqrt{x^2+y^2}$ and $\theta =\arctan \frac y x $. –  Pedro Tamaroff Jul 30 '12 at 21:03
    
@PeterTamaroff: don't know I was working! :-) Yes seems the same (coeff 1 at the end, his could be better!) –  Raymond Manzoni Jul 30 '12 at 21:04
    
OK, I guess that you got the answer. Thanks so much for all comments. For me, this is enough. Nice plot. Thanks. –  Sigur Jul 30 '12 at 21:11
    
@Sigur: Nice if it corresponds! I too thought that they were 7 oscillations at the start but it didn't work (it's better to count the white parts under the curves at the front). –  Raymond Manzoni Jul 30 '12 at 21:14
    
+1 for including the plot - this looks fantastic! –  Steven Stadnicki Jul 30 '12 at 21:18

A few key things that jump out looking at this graph:

  1. It's got a sort of 'radial symmetry' about it: if you look at a circular 'cross-section' centered around the origin then the shape looks roughly the same, just scaled. Similarly all of the 'radial' cross-sections along lines through the origin look roughly the same. This means that it's going to be best expressed as the plot of a function $z=f(r,\theta)$ where the 'base plane' is represented in polar coordinates. Moreover, the structure of it (a sequence of radial spires, of sorts) suggests that it's 'separable' in the sense that it can be written as $f(r,\theta) = g(r)\cdot h(\theta)$ for two individual functions $g$ and $h$.
  2. Going around the origin once seems (though it's unclear from the diagram) to encounter eight distinct peaks (possibly 7, but I think that's an artifact of the projection and there are actually 8), so the 'angular' part of the function, $h$, can be written as $h(\theta) = \sin(8\theta)$.
  3. Going out along any of the radial lines appears to be a straight line, so I'd say the radial portion of the function, $g$, can be written as $g(r) = c\cdot r$ for some small $c$, say $c=0.3$ or something on that order (but that's a function of the scale of the grap).

So, putting this together, the plot looks to be of an equation roughly like $z=.3r\sin(8\theta)$.

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Looks more like $r^2$ should go there. –  Pedro Tamaroff Jul 30 '12 at 20:57
    
@PeterTamaroff That's possible - I'm looking at the ridge lines along the peaks of the radial sine waves, and to me they look straight but it's very possible they're a quadratic with relatively small constant. –  Steven Stadnicki Jul 30 '12 at 21:00
    
I am trying to plot $z=r^2\sin(8\theta)$ with maple. But how to change to rectangular coordinates? I don't know how to use polar in maple. Thanks for the comments. –  Sigur Jul 30 '12 at 21:03
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@Sigur As you've no doubt gleaned from Raymond's plot, you can change to rectangular coordinates via the substitution $r=\sqrt{x^2+y^2}$ and $\theta = \mathrm{atan}(y/x)$. –  Steven Stadnicki Jul 30 '12 at 21:15
    
+1: you were first no doubt! –  Raymond Manzoni Jul 30 '12 at 21:15

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