problem by using symengine

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

[Walter Roberson]

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]

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