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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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    @J.M. Thanks for the msg, but I guess that my graph is not a saddle like this.2012-08-05

2 Answers 2

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

enter image description here

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    :-) @Sigur: perhaps that the mathematica SE forum will help you more. For example this [thread)(http://mathematica.stackexcha$n$ge.com/questions/1542/exporting-graphics-to-pdf-huge-file). Cheers,2014-11-19
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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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    @J.M. Thanks for the link.2012-07-31