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problem by using symengine

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Basem Nouh
Basem Nouh 2020-9-14
编辑： Walter Roberson 2020-9-14
Hallo,
i am trying to program this code but when i try to use jacobian by A matrix i became wrone msg that error by using symengine
but it works with B matrix !!
the msg also says "unable to convert experssion into double array"
%aufgabestellung3
clear all;clc;close all;
L=[0 1 4 9 0 1 2 3]';[n,~]=size(L);
r=4; n=8; u=2;
Lsym=sym('Lsym',[n,1]);
v0=zeros(n,1); vsym=sym('vsym',[n,1]);
X0=[0 0]';
[u,~]=size(X0);
Xsym=sym('Xsym',[u,1]);
r=4;
%sto Modell
Qll=eye(n);
P=inv(Qll);
%funk Modell % Ansatz: L=[x;y]; 0=ax+b-y; X=[a b];
% A=[x ones(r,1)]; B=[X0(1,1)*eye(r) -1*eye(r)];
for i =1:r
f(i,1)=Xsym(2,1)+Xsym(1,1)*(Lsym(r+i,1)+vsym(r+i,1))-(Lsym(i,1)+vsym(i,1));
end
it=0;abr=1;
while it<10 & abr>10^-20
B=double(subs(jacobian(f,vsym),[vsym;Xsym],[v0;X0]));
% A=[x+v0(1:r,1) ones(r,1)]
A=double(subs(jacobian(f,Xsym),[vsym;Xsym],[v0;X0])); % here i become the wrong !
w0=double(subs(f,[Lsym;Xsym],[L;X0]));
w=-B*v0+w0
% Loesungsalgorithmus (Niemeier S. 178 Gl. 5.2.30)
iBQB=inv(B*inv(P)*B'); % Substitution
Q22=-inv(A'*iBQB*A);
Q12=-iBQB*A*Q22;
Q21=Q12';
Q11=iBQB*(eye(r)-A*Q21);
% Gleichung 5.2.31
k=-Q11*w;
xd=-Q21*w;
Xd=X0+xd
Qll=inv(P);
v=Qll*B'*k;
Ld=L+v;
wd=zeros(r,1)
wd=double(subs(f,[vsym;Xsym],[v;Xd]));
abr=max(abs(wd))
it=it+1
X0=Xd
end
  2 个评论
Walter Roberson
Walter Roberson 2020-9-14
Your code is assuming that the jacobian of f with respect to Xsym gets rid of all Lsym variables, but it does not. You have sub-expressions that are Lsym times something involving Xsym variables, so the Lsym remain in the jacobian.
Basem Nouh
Basem Nouh 2020-9-14
编辑：Walter Roberson 2020-9-14
i edited it i have to insert L as a vector not as sympolic variable , i have no idea why
but the Prof at the uni didnt use jacobian for the Partial differentiation he inserted the both matrix A,B direct
A=[x+v0(1:r,1) ones(r,1)]
B=[X0(1,1)*eye(r) -1*eye(r)]
but i thought by jacobian or using diff its faster and safer
but there is a problem by Partial differentiation
by me B at the final is
B =
0 0 0 0 -1 0 0 0
0 0 0 0 0 -1 0 0
0 0 0 0 0 0 -1 0
0 0 0 0 0 0 0 -1
by prof B =
3.2418 0 0 0 -1.0000 0 0 0
0 3.2418 0 0 0 -1.0000 0 0
0 0 3.2418 0 0 0 -1.0000 0
0 0 0 3.2418 0 0 0 -1.0000
A =
0 1
1 1
2 1
3 1
A =
0.3838 1.0000
0.7524 1.0000
1.6843 1.0000
3.1795 1.0000
it means there is no correction happened

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