How to put value in symbolic variable in differential equation (Newton Raphson Method)?

hi

i need help in the following code. i require "df" value in numeric form but here i get the complete expressions due to syms alpha1 alpha2 alpha3. Kindly help me and guide me how i get results of "df" and "inv(df)" in numeric format. kindly help me please :)

with best regards Mudasir

alpha1=0.2;            
alpha2=0.4;
alpha3=0.6;

v1desired=10;

T=[v1desired 0 0]';

f1=[ cos(alpha1)+cos(alpha2)+cos(alpha3)         
     cos(5*alpha1)+cos(5*alpha2)+cos(5*alpha3)   
     cos(7*alpha1)+cos(7*alpha2)+cos(7*alpha3)];

   syms alpha1 alpha2 alpha3

            f=[ cos(alpha1)+cos(alpha2)+cos(alpha3)         
              cos(5*alpha1)+cos(5*alpha2)+cos(5*alpha3)   
              cos(7*alpha1)+cos(7*alpha2)+cos(7*alpha3)];

df=jacobian(f,[alpha1;alpha2;alpha3])  

dalpha=(inv(df))*(T-f1);

[Walter Roberson]

Mudasir Ahmed commented

Dear sir, i want to delete this question. because i post here my actual code of research. so i will be highly thankful to you if you delete this question please :) best regards mudasir

[Walter Roberson]

Musadir, your code contains algorithm that are in common use, nothing new or confidential. We will leave the question so that other people using the same kind of algorithm can learn from it.

Answer 1

[Walter Roberson]

v_alpha1=0.2;            
v_alpha2=0.4;
v_alpha3=0.6;

v1desired=10;

T=[v1desired 0 0]';

syms alpha1 alpha2 alpha3

f1 = [ cos(alpha1)+cos(alpha2)+cos(alpha3)         
     cos(5*alpha1)+cos(5*alpha2)+cos(5*alpha3)   
     cos(7*alpha1)+cos(7*alpha2)+cos(7*alpha3)];

f = [ cos(alpha1)+cos(alpha2)+cos(alpha3)         
     cos(5*alpha1)+cos(5*alpha2)+cos(5*alpha3)   
     cos(7*alpha1)+cos(7*alpha2)+cos(7*alpha3)];

df = jacobian(f,[alpha1;alpha2;alpha3])  

dalpha_sym = (inv(df))*(T-f1);

dalpha = double( subs(dalpha_sym, [alpha1, alpha2, alpha3], [v_alpha1, v_alpha2, v_alpha3]));

[Walter Roberson]

The inv() will fail when det(df)) is 0. You can solve(det(df),alpha1) and you will be given back a bunch of conditions. For example, the inv() will fail if alpha1 = alpha2. There is nothing you can do about that as long as you continue to use inv(). This is why inv() is not used in real problems.

You should consider substituting in numeric values first and then using pinv() instead of inv()