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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?

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    $\begingroup$ 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". $\endgroup$ – The Chaz 2.0 Oct 18 '11 at 14:11
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    $\begingroup$ 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 $x^{2/3}$ $\endgroup$ – mathmath8128 Oct 18 '11 at 14:12
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    $\begingroup$ A related thread. $\endgroup$ – J. M. is a poor mathematician Oct 18 '11 at 14:16
  • $\begingroup$ Perhaps you found the image here: mathematische-basteleien.de/heart.htm (That's what came up when I used Google image search.) $\endgroup$ – PersonX Oct 18 '11 at 15:42
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    $\begingroup$ @Jeroen "[your text](the http address)" with the [ ] ( ) included $\endgroup$ – Dr. belisarius Oct 18 '11 at 21:20
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I did it in Maple...

heart

Vary b to change the picture.

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  • $\begingroup$ ok thanks it is perfect! and can i ask you also some little comment about the kind and analysis of this formula? $\endgroup$ – nkint Oct 31 '11 at 9:25
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$\sqrt{x^2}$ 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.

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