The below code takes much time to just compute till N=5. But I wanna know how to do it for N=100 or more a bit faster.
syms z;
N = 5; % Compute up to g_5′
alpha = -1; % α_n = -1 for all n
C = sym(‘C’, [1, N]); % Integration constants C0..CN
% Initialize
g_prime = cell(1, N+1);
g = cell(1, N+1);
g_prime{1} = z^2; % g_0′(z)
g{1} = z^3/3 % g_0(z)
% Iterate
for n = 1:N
% Step 1: Compute g_n'(z) = [ (g_{n-1} + C_{n-1})^2 / g_{n-1}’ ]^{α}
numerator = (g{n} + C(n))^2;
denominator = g_prime{n};
g_prime{n+1} = (numerator / denominator); % α = -1 ⇒ flip fraction
% Step 2: Simplify CRITICALLY here to prevent blowup
g_prime{n+1} = simplify(g_prime{n+1});
% Step 3: Compute g_{n}(z) for next step (if not last iteration)
if n < N
g{n+1} = int(g_prime{n+1}, z) + C(n+1);
g{n+1} = simplify(g{n+1}); % Simplify after integration
end
end
% Display fully simplified results
for k = 1:N
fprintf(‘g_%d”(z) = ‘, k);
disp(g_prime{k+1});
endsyms z;
N = 5; % Compute up to g_5′
alpha = -1; % α_n = -1 for all n
C = sym(‘C’, [1, N]); % Integration constants C0..CN
% Initialize
g_prime = cell(1, N+1);
g = cell(1, N+1);
g_prime{1} = z^2; % g_0′(z)
g{1} = z^3/3 % g_0(z)
% Iterate
for n = 1:N
% Step 1: Compute g_n'(z) = [ (g_{n-1} + C_{n-1})^2 / g_{n-1}’ ]^{α}
numerator = (g{n} + C(n))^2;
denominator = g_prime{n};
g_prime{n+1} = (numerator / denominator); % α = -1 ⇒ flip fraction
% Step 2: Simplify CRITICALLY here to prevent blowup
g_prime{n+1} = simplify(g_prime{n+1});
% Step 3: Compute g_{n}(z) for next step (if not last iteration)
if n < N
g{n+1} = int(g_prime{n+1}, z) + C(n+1);
g{n+1} = simplify(g{n+1}); % Simplify after integration
end
end
% Display fully simplified results
for k = 1:N
fprintf(‘g_%d”(z) = ‘, k);
disp(g_prime{k+1});
end syms z;
N = 5; % Compute up to g_5′
alpha = -1; % α_n = -1 for all n
C = sym(‘C’, [1, N]); % Integration constants C0..CN
% Initialize
g_prime = cell(1, N+1);
g = cell(1, N+1);
g_prime{1} = z^2; % g_0′(z)
g{1} = z^3/3 % g_0(z)
% Iterate
for n = 1:N
% Step 1: Compute g_n'(z) = [ (g_{n-1} + C_{n-1})^2 / g_{n-1}’ ]^{α}
numerator = (g{n} + C(n))^2;
denominator = g_prime{n};
g_prime{n+1} = (numerator / denominator); % α = -1 ⇒ flip fraction
% Step 2: Simplify CRITICALLY here to prevent blowup
g_prime{n+1} = simplify(g_prime{n+1});
% Step 3: Compute g_{n}(z) for next step (if not last iteration)
if n < N
g{n+1} = int(g_prime{n+1}, z) + C(n+1);
g{n+1} = simplify(g{n+1}); % Simplify after integration
end
end
% Display fully simplified results
for k = 1:N
fprintf(‘g_%d”(z) = ‘, k);
disp(g_prime{k+1});
end symbolic integration MATLAB Answers — New Questions
