Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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 at 12:19

1 Answer 1

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...

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.