Use vpasolve to solve iterations of the same function when the number of iterations is N
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Hi,first time asking a question here, I did search through but couldn't find an answer to this posted already.
I'm trying to cut down this code so instead of running it for each iteration of F(z), i can input the amount of iterations I want and it will compute just that.
You can see in the code that sols3 or sols6 are calculating F^N(z) - z for N = 3 and 6 respectively. I want to just use N to input this and not write code for sols3/4/5/6 etc.
The problem is that when N = 1 I want to vpasolve(F(z) - z, z)
but for N = 2 I want vpasolve(F(F(z)) - z, z)
N = 3, vpasolve(F(F(F(z))) - z, z)
if that makes sense.
I tried to make g = F^N(z) and vpasolve(g(z) - z, z)
but obviously thats wrong.
I need to do the same for the derivatives in the later part of the code too but I think if it's explained for the first part I should be able to do that part trivially.
Thank you very much :)
clear variables
% Define symbolic variables r and phi.
% Also need to define a numerical value for complex constant c...
syms z; r = 0.9; Omega = (sqrt(5)+1)/2; phi = 2*pi*Omega;
F = @(z) z^2 + r*exp(1i*phi)*z;
dFdz = @(z) 2*z + r*exp(1i*phi);
% To find N-periodic points
N = 7;
%g = F^N(z);
% Obtain numerical approximations to the roots using symbolic algebra...
% PERIOD-N POINTS (i.e., the fixed points of the map F^(N))
% solsN = vpasolve (g(z)- z,z);
% disp(['PERIOD-N POINTS:']); pN_points = vpa(solsN)
% size(pN_points,1)
% PERIOD-3 POINTS [i.e., the fixed points of the map F^(3)]
sols3 = vpasolve(F(F(F(z))) - z,z);
disp(['PERIOD-3 POINTS:']); p3_points = vpa(sols3)
size(p3_points,1)
% PERIOD-6 POINTS [i.e., the fixed points of the map F^(6)]
sols6 = vpasolve(F(F(F(F(F(F(z)))))) - z,z);
disp(['PERIOD-6 POINTS:']); p6_points = vpa(sols6)
size(p6_points,1)
% Check for instability in a period-3 orbit...
disp([' ']); disp([' CHECKING PERIOD-3 ORBIT STABILITY.'])
a3 = p3_points(3); F3_prime = dFdz(F(F(a3)))*dFdz(F(a3))*dFdz(a3);
fprintf(" For selected point: z = %s \n",string(a3));
fprintf(" |dF^(3)/dz| = %s \n",string(abs(F3_prime)));
if abs(F3_prime) > 1
disp([' ==> selected period-3 point is UNSTABLE!'])
elseif abs(F3_prime) < 1
disp([' ==> period-3 point is STABLE!'])
else
disp([' ==> period-3 point is MARGINALLY STABLE!'])
end
% Check for instability in a period-6 orbit...
disp([' ']); disp([' CHECKING PERIOD-6 ORBIT STABILITY.'])
a6 = p6_points(6); F6_prime = dFdz(F(F(F(F(F(a6))))))*dFdz(F(F(F(F(a6)))))*dFdz(F(F(F(a6))))*dFdz(F(F(a6)))*dFdz(F(a6))*dFdz(a6);
fprintf(" For selected point: z = %s \n",string(a6));
fprintf(" |dF^(6)/dz| = %s \n",string(abs(F6_prime)));
if abs(F6_prime) > 1
disp([' ==> selected period-6 point is UNSTABLE!'])
elseif abs(F6_prime) < 1
disp([' ==> period-6 point is STABLE!'])
else
disp([' ==> period-6 point is MARGINALLY STABLE!'])
end
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回答(1 个)
Voss
2024-10-31,15:25
One way is to define a function that takes F, N, and z as arguments and iterates N times to construct F(F(...F(z)...))-z. Then call that function with the appropriate value of N.
% function definition
function out = FN(F,N,z)
in = z;
for ii = 1:N
in = F(in);
end
out = in - z;
end
% testing set-up
syms z;
r = 0.9;
Omega = (sqrt(5)+1)/2;
phi = 2*pi*Omega;
F = @(z) z^2 + r*exp(1i*phi)*z;
% compare original approach vs new approach with N = 3
sols3_original = vpasolve(F(F(F(z))) - z,z);
sols3_new = vpasolve(FN(F,3,z),z);
isequal(sols3_original,sols3_new)
% compare original approach vs new approach with N = 6
sols6_original = vpasolve(F(F(F(F(F(F(z)))))) - z,z);
sols6_new = vpasolve(FN(F,6,z),z);
isequal(sols6_original,sols6_new)
4 个评论
Walter Roberson
2024-10-31,17:49
Note that you are passing in a function handle to fold not a symbolic function. But still fold() gets the job done when used as you indicate.
Voss
2024-10-31,17:58
You're correct. I was thinking you meant a function handle that accepts symbolic variables, rather than a symbolic function. My mistake. I've edited the terminology I used in my comment.
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