Forward and backward ( inverse) iteration of complex quadratic polynomial
Function :
 f_{c}(z) = z*z +c; Gives one value ( image)
 f_{c}^{1}(z) =sqrt( z c); Gives 2 values (preimages). It is multivalued function.
Iteration :
 forward : z_{n+1} = f_{c}(z_{n}); is simple
 backward : z_{n1} = f_{c}^{1}(z_{n}); is not simple. It is impossible to choose good preimage without extre informations !
Orbit :
 forward : list of points {z_{0}, z_{1}, z_{2}, z_{3}... , z_{n}}
 backward : binary tree of preimages of z_{n}; One can't choose good path in such tree without extra informations.
Where orbit goes ?
 forward : to attractor ( for example infinity is always superattracting fixed point for polynomials)
 backward : to repelling point ( which are in Julia set = IIM/J) or preperiodic points
Example :
external angle : t = 1/3 in turns
radius : r = abs(z) = 2
c = 0
z0 = r*exp(i*t) = 1.732050807568877*%i1.0
Forward iteration of z0 :
z0 = 1.732050807568877*%i1.0 = 2*exp(i*1/3)
z1 = 3.464101615137754*%i2.0 = 4*exp(i*2/3)
z2 = 13.85640646055102*%i8.0 = 16*exp(i*1/3)
so forward orbit = {1/3, 2/3, 1/3 }
Backward iteration of z2 :
 z_{2} = 13.85640646055102*%i8.0 = 16*exp(i*1/3)
 2 preimages :

z2m1pv = z_{21} = 3.464101615137754*%i+2.0 = 4*exp(i*1/6) ( principal value of arg )
 z2m2ppv = z_{22} = %i+1.732050807568877 = 2*exp(i*1/12)
 z2m2psv = z_{22} = %i1.732050807568877 = 2*exp(i*7/12)

z2m1sv = z_{21} =3.464101615137754*%i2.0 = 4*exp(i*2/3)
 z2m2spv = z_{22} = 1.01.732050807568877*%i = 2*exp(i*10/12)
 z2m2ssv = z_{22} = 1.732050807568877*%i1.0 = 2*exp(i*4/12)
So here is 4 preimages of z2 with backward orbits :
 {1/3, 1/6, 1/12} = {4/12, 2/12, 1/12} = {z2, z2m1pv,z2m2ppv}
 {1/3, 1/6, 7/12} = {4/12, 2/12, 7/12} = {z2, z2m1pv, z2m2psv}
 {1/3, 2/3, 5/6} = {4/12, 8/12, 10/12} = {z2, z2m1sv, z2m2spv}
 {1/3, 2/3, 1/3} = {4/12, 8/12, 4/12} = {z2, z2m1sv, z2m2ssv} = {z2,z1,z0}
so :
z2m1sv is our z1 with angle = 8/12 = 2/3
z2m2ssv is our z0 with angle = 4/12 = 1/3
and here are 4 new points which are preperiodic points falling into period 2 cycle
z2m1pv with angle = 1/6 with forward orbit ={1/6, (2/6, 4/6)}= {1/6, (1/3, 2/3)}
z2m2ppv with angle = 1/12 with forward orbit ={ 1/12 , 2/12 ,( 1/3 , 2/3 )}
z2m2psv with angle = 7/12 with forward orbit ={ 7/12 , 2/12 ,( 1/3 , 2/3 )}
z2m2spv with angle = 5/6 with forward orbit = { 5/6 , ( 2/3 , 1/3 )}
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Autor: Adam Majewski adammaj1ato2dotpl
Feel free to email me. (:))
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