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I'm sorry for the ambiguity here but I've recently discovered a function which plots, what seems to be either a fractal or simply noise in a selected area. Can anyone explain this function:

$\sqrt{x^2+y^2} = \frac{1}{(\cos(\tan^{-1}(x/y)+\tan^{-1}(y/x)))}$

Graph it and see what you make of it.

I was trying to find the locus of a square, but instead found the equation of parallel lines through $abs(x) = a$

$\sqrt{x^2+y^2} = \frac{a}{\cos(\tan^{-1}(x/y))}$

and then added in an extra $\tan^{-1}(y/x)$, the reciprocal of $\tan^{-1}(x/y)$ and thats how I discovered this strange graph.

I'm in only in high school, so I'm sorry if my question is a a bit simple.

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How did you plot it? Via MATLAB, etc? – Gautam Shenoy Nov 30 '12 at 8:48
I must confess, there is a certain similarity on different scales - though I guess this is due to the number of roots of the trigonometric functions. – S.D. Nov 30 '12 at 8:59
I used grapher on mac – Jordan Brown Nov 30 '12 at 9:00
a screenshot would have been welcome, to help interpret it. – comprehensible Mar 19 '14 at 12:19
up vote 3 down vote accepted

Observe the following:

$$\tan^{-1}(x) + \tan^{-1}(y) = \tan^{-1}\left(\frac{x+y}{1-xy}\right)$$

Now squaring both sides of your equation yields

$$ x^2 +y^2 = \sec^2(\tan^{-1}(x/y)+\tan^{-1}(y/x))$$ $$ = 1 + \tan^2\left(\tan^{-1}\left(\frac{x/y + y/x}{1-1}\right)\right) = \infty$$ In the last step, I have implicitly taken a left hand limit to arrive at the result. Now This represents(as a locus) a circle with infinite radius which some would say technically is a line. Just like a circle with zero radius(radius tends to 0) is a point. So the software you are using could be giving weird results due to this anomaly. Unfortunately I do not know how the software plots these functions...

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