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$\begingroup$

i don't remember where exactly, i found in internet this image: beautiful curve

i tried to replicate the formula with python and i tried this:

b = 0.9 y = 2*b + sqrt(x*x) + sqrt((x+b)*(3*b-x)) y1 = 2*b + sqrt(x*x) - sqrt((x+b)*(3*b-x)) plot(x,y, x, y1) 

where sqrt is the square root!

but my curve is not very similar to the picture.. mayebe i'm not able to read it because of it is handwritten. some help?

  • 3
    I'm fairly certain that it's not $\sqrt{x^2}$... the root and the exponent are probably different so as to give the desired "cusp".2011-10-18
  • 3
    The first radical appears to not be a square root, but an $n$th root, which is pretty hard to make out. It'd be silly to use $\sqrt{x^2}$, which is $|x|$, since there's probably a nice function for that in Python. Try experimenting with some other roots, maybe $x^{2/3}$2011-10-18
  • 1
    [A related thread.](http://math.stackexchange.com/questions/12098)2011-10-18
  • 0
    Perhaps you found the image here: http://www.mathematische-basteleien.de/heart.htm (That's what came up when I used Google image search.)2011-10-18
  • 0
    This reminds me of the famous batman equation (http://math.stackexchange.com/questions/54506/is-this-batman-equation-for-real) @ J.M., how can you link your text in comment mode?2011-10-18
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    [Parametric Heart](http://www.wolframalpha.com/input/?i=parametric%20plot%20%28%20abs%28tan%28t%29%29%5E%281/%28abs%28tan%28t%29%29%29%29%20cos%28t%29,%20abs%28tan%28t%29%29%5E%281/%28abs%28tan%28t%29%29%29%29%20sin%28t%29%20%29,%20%20t%20=%200%20to%20pi) thanks to Dwight Boddorf (MAA's American Mathematical Monthly Volume 115 Number 2 (February 2008) page 113)2011-10-18
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    @Jeroen "[your text](the http address)" with the [ ] ( ) included2011-10-18
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    What python library are you using?2012-09-11
  • 0
    @asmeuere: numpy2012-09-12

2 Answers 2

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I did it in Maple...

heart

Vary b to change the picture.

  • 0
    ok thanks it is perfect! and can i ask you also some little comment about the kind and analysis of this formula?2011-10-31
1

$\sqrt{x^2}$ is the same thing as $|x|$, the absolute value of $x$, whose graph has a sharp corner. When I plot exactly the first equation you wrote above, what I get is quite similar to the part of the graph above the two left and right vertical tangents. But it doesn't have a vertical tangent at the cusp in the middle, although it does have a sharp corner there.