Quantcast
Channel: blog化学
Viewing all articles
Browse latest Browse all 2255

Newton/Julia Fractals (1677)~(1689)

$
0
0
Newton/Julia Fractal (1677)

f(z) = z - (1+z^2+z^4+z^6/20)/(2.475/37*z^5+7*z^2+1)+(-.26942824+6.30800331E-3)
 
イメージ 1

  
Sonate K3023
イメージ 2



Newton/Julia Fractal (1678)

f(z) = z - (1+z^2+z^4+z^6/20)/(2.475/37*z^5+7*z^2+1)+(-.27942824+6.30800331E-3)
イメージ 3

    
Newton/Julia Fractal (1679)

f(z) = z^2+z*SQR(z)+(-0.157571966+2.7918228E-2*i)  
イメージ 4

  
Sonate K3026  
イメージ 5

 
Sonate K3027
イメージ 6



Newton/Julia Fractal (1680)

f(z) = z^2+z*SQR(z)+ (-.159274923+4.56837331E-2*i)
 
イメージ 7

  
Sonate K3029
イメージ 8


Sonate K3030
イメージ 9



Newton/Julia Fractal (1681)

f(z) = z^2+z*SQR(z)+ (-.158574923+4.56837331E-2*i)
イメージ 10


Sonate K3032
イメージ 11



Newton/Julia Fractal (1682)

f(z) = z^2+z*SQR(z)+ (-.162274923+4.56837331E-2*i)
イメージ 12


Sonate K3034
イメージ 13



Newton/Julia Fractal (1683)

f(z) = z^2+(0.0001+0.1*i)/z
イメージ 14



Newton/Julia Fractal (1684)

f(z) = z^2+(0.0001+0.4*i)/z
イメージ 15



Newton/Julia Fractal (1685)

f(z) = z^2+(0.007-0.23*i)/z
イメージ 16



Newton/Julia Fractal (1686)

f(z) = z^2+(0.0001+0.05*i)/z+(0.0001+0.05*i)
イメージ 17



Sonate K3039
イメージ 18



Newton/Julia Fractal (1687)

f(z) = z^2+(0.0001+0.05*i)/z+(0.0001-0.05*i)
イメージ 19



Newton/Julia Fractal (1688)

f(z) = z^2+(0.007-0.137*i)/z
イメージ 20



Sonate K3042
イメージ 21



Newton/Julia Fractal (1689)

f(z) = (z^2+0.001/z)/(0.95-0.31225*i)
イメージ 22

By Dr. M.Becker
イメージ 23
f(z)=(z3+c)/(dz) with c=0.001 and d=0.95-0.31225i, shown on [-1.5;1.5]×[-1.5;1.5].
d is a complex number with absolute value approx. 1. It makes the triangle knots on the last picture turn a bit. So their three branches cannot meet each other straightly, but run into spirals.



Viewing all articles
Browse latest Browse all 2255

Trending Articles