how to use loop in this equation

sir how to use p loop in this equation 34

so for p llop we have to use eq 34

this is program

a(1)=1;

b(1)=1;

c(1)=1;

k2=1;

T=0.1;

r=2.8

k1=0.3;

for i=1:20

for p=1:i-1

a(i+1)=(1/(i+1))*(a(i)-b(i))*k2;

S=S+a(p)*c(i-p);

Z=Z+a(p)*b(i-p);

b(i+1)=(1/(i+1))*(r*a(i)-T*S-T*b(i));

c(i+1)=(1/(i+1))*(T*Z-k1*c(i));

end

end

Answer 1

[Torsten]

There are some mistakes in your article.

The code is based on

https://arxiv.org/pdf/2010.14993.pdf

  % Set parameters
  X0 = 0.0;
  Y0 = 1.0;
  Z0 = 0.0;

  N = 120;
  SIGMA = 10;
  R = 28;
  B = 8/3;
  
  % Initialize coefficient arrays
  a = zeros(N+1,1);
  b = zeros(N+1,1);
  c = zeros(N+1,1);
  a(1) = X0;
  b(1) = Y0;
  c(1) = Z0;
  
  % Compute Taylor coefficients
  for k = 1:N
    SB = 0.0;
    SC = 0.0;
    for i= 0:k-1
      SB = SB + a(i+1)*c(k-i);
      SC = SC + a(i+1)*b(k-i);
    end
    a(k+1) = (SIGMA*(b(k) - a(k)))/k; 
    b(k+1) = ((R*a(k) - b(k)) - SB)/k ;
    c(k+1) = (-B*c(k) + SC)/k ;
  end  

  % Evaluate Taylor series at T
  T = 0:0.01:0.2;
  
  X = zeros(size(T));
  Y = zeros(size(T));
  Z = zeros(size(T));
  for i = 1:numel(T)
    tau = T(i);
    x = a(N+1);
    y = b(N+1);
    z = c(N+1);
    for k = N:-1:1
      x = x*tau + a(k); 
      y = y*tau + b(k); 
      z = z*tau + c(k); 
    end
    X(i) = x;
    Y(i) = y;
    Z(i) = z;
  end
  figure(1)
  plot(T,X,'Color','red')
  hold on
  figure(2)
  plot(T,Y,'Color','red')
  hold on
  figure(3)
  plot(T,Z,'Color','red')
  hold on  
  
  % Compare with ODE solution
  fun = @(t,x)[SIGMA*(x(2)-x(1));x(1)*(R-x(3))-x(2);x(1)*x(2)-B*x(3)];
  x0 = [X0;Y0;Z0];
  tspan = T;
  options = odeset ("RelTol", 1e-8, "AbsTol", 1e-8);
  [TT,XX] = ode15s(fun,tspan,x0,options);
  figure(1)
  plot(TT,XX(:,1),'Color','blue')
  figure(2)
  plot(TT,XX(:,2),'Color','blue')  
  figure(3)
  plot(TT,XX(:,3),'Color','blue') 

[shiv gaur]

dear sir your approach is appreciable you are very close to my problem I have attach the paper of lorenz using power series using equation 32 to 34 very nearly to taylor so sir you are req to fig 5

[Torsten]

The code is the correct implementation of a power-series solution to the Lorenz equations.

So now it's up to you to do with the code what you like to do.

[shiv gaur]

but why 3d plot is not butterfly and t vs Y is not matching lorenz

[shiv gaur]

thanks for answer pl apply to this to my paper in later part

[Torsten]

Here is your butterfly:

% Set parameters
  Tstart = 0.0;
  Tend = 1000.0;
  Nt = 20000;
  dT = (Tend-Tstart)/Nt;
  X0 = 0.0;
  Y0 = 1.0;
  Z0 = 0.0;
  N = 20;
  SIGMA = 10;
  R = 28;
  B = 8/3;
%
% Initialize coefficient arrays
  T = zeros(Nt+1,1);
  X = zeros(Nt+1,1);
  Y = zeros(Nt+1,1);
  Z = zeros(Nt+1,1);
  a = zeros(N+1,1);
  b = zeros(N+1,1);
  c = zeros(N+1,1);
  
  T(1) = 0.0;
  X(1) = X0;
  Y(1) = Y0;
  Z(1) = Z0;
  

  for j = 2:Nt+1
    % Compute Taylor coefficients      
    a(1) = X(j-1);
    b(1) = Y(j-1);
    c(1) = Z(j-1);
    for k = 1:N
      SB = 0.0;
      SC = 0.0;
      for i= 0:k-1
        SB = SB + a(i+1)*c(k-i);
        SC = SC + a(i+1)*b(k-i);
      end
      a(k+1) = (SIGMA*(b(k) - a(k)))/k; 
      b(k+1) = ((R*a(k) - b(k)) - SB)/k ;
      c(k+1) = (-B*c(k) + SC)/k ;
    end  
    
    % Evaluate Taylor series at actual T
    x = a(N+1);
    y = b(N+1);
    z = c(N+1);
    for k = N:-1:1
      x = x*dT + a(k); 
      y = y*dT + b(k); 
      z = z*dT + c(k); 
    end
    
    %x = a(1);
    %y = b(1);
    %z = c(1);
    %for k = 2:N+1
    %  x = x + a(k)*dT^(k-1);
    %  y = y + b(k)*dT^(k-1);
    %  z = z + c(k)*dT^(k-1);
    %end
    
    % Prepare for T = T + dT
    T(j) = T(j-1) + dT;
    X(j) = x; 
    Y(j) = y; 
    Z(j) = z; 
  end
  figure(1)
  plot(T,X,'Color','red')
  hold on
  figure(2)
  plot(T,Y,'Color','red')
  hold on
  figure(3)
  plot(T,Z,'Color','red')
  hold on  
  figure(4)
  plot3(X,Y,Z,'Color','red')
  hold on
 
  % Compare with ODE solution 
  fun = @(t,x)[SIGMA*(x(2)-x(1));x(1)*(R-x(3))-x(2);x(1)*x(2)-B*x(3)];
  x0 = [X0;Y0;Z0];
  tspan = [Tstart,Tend];
  options = odeset ("RelTol", 1e-7, "AbsTol", 1e-7, "InitialStep", 1e-9);
  [TT,XX] = ode45(fun,tspan,x0,options);
  figure(1)
  plot(TT,XX(:,1),'Color','blue')
  figure(2)
  plot(TT,XX(:,2),'Color','blue')  
  figure(3)
  plot(TT,XX(:,3),'Color','blue') 
  figure(4)
  plot3(XX(:,1),XX(:,2),XX(:,3),'Color','blue') 

[Torsten]

thanks for answer pl apply to this to my paper in later part

As I mentionned, there are errors in your paper.

You'd better use the one I gave you the link for.

[shiv gaur]

sir how I use u=summation (a(i)*tau^i) so from calculating the taylor coeff is same as cauchy product

which you have calculate and finally my equation is

u=u+ a(i)*tau^i;

v=v+b(i)*tau^i;

w=w+c(i)*tau^i;

so plor3d(u,v,w) looks like butterfly as in fig5

[shiv gaur]

this is last help

[Torsten]

sir how I use u=summation (a(i)*tau^i) so from calculating the taylor coeff is same as cauchy product

which you have calculate and finally my equation is

u=u+ a(i)*tau^i;

v=v+b(i)*tau^i;

w=w+c(i)*tau^i;

so plor3d(u,v,w) looks like butterfly as in fig5

% Evaluate Taylor series at actual T
x = a(N+1);
y = b(N+1);
z = c(N+1);
for k = N:-1:1
    x = x*dT + a(k); 
    y = y*dT + b(k); 
    z = z*dT + c(k); 
end

is the part of the code where you do

x = x + a(i)*tau^i

y = y + b(i)*tau^i

z = z + b(i)*tau^i

It's done with Horner's scheme for polynomials because it uses less multiplications.

You might want to compare with

x = a(1);
y = b(1);
z = c(1);
for k = 2:N+1
  x = x + a(k)*dT^(k-1);
  y = y + b(k)*dT^(k-1);
  z = z + c(k)*dT^(k-1);
end

Both methods should give more or less the same result, but yours is less efficient.

[shiv gaur]

in this result is not mat

[Torsten]

Pardon ?

[shiv gaur]

in this result is not matching so pl need ful

[shiv gaur]

[shiv gaur]

thanks very much