error in multiplying symbolic variable in a matrix

i have an important problem in writing a program for calculating Lagrange. i want to assume z as parameter. then I face this error: Error in MuPAD command: DOUBLE cannot convert the input expression into a double array. If the input expression contains a symbolic variable, use the VPA function instead. this is my program:

 clear
clc
syms z
x=[0 300 1000 1500 1700 2000 3000 3300 4000 4500 5000]
y=[32.0 122.4 296.4 405.7 447.6 608.4 704.7 761.4 891.9 983 1072.6]
m=z-x
i=1
j=1
L=[1 1 1 1 1 1 1 1 1 1 1]
q=[1 1 1 1 1 1 1 1 1 1 1]
while j<12
    if j==1
        for i=2:11
             L(1)=L(1)-m(i)
            q(1)=q(1)*(x(j)-x(i))
        end
    else
              for i=1:(j-1)
              L(j)=L(j).*m(i)
              q(j)=q(j).*(x(j)-x(i))
              end
            for i=(j+1):11
            L(j)=L(j).*m(i)
            q(j)=q(j).*(x(j)-x(i))
            end
      end    
j=j+1
end
e=(y.*L)
o=e./q
k=1
f=0
while k<12
   f=f+o(k) 
   k=k+1
end
ezplot(f)

[Dyuman Joshi]

What is the final output you want?

Also, you can optimize your code -

1 - Club the 2 for loops under the else statement together -

for i = [1:(j-1) j+1:11]
    L(j) = L(j).*m(i);
    q(j) = q(j).*(x(j)-x(i));
end

After you combine them, you will see that you can vectorize them -

i = [1:(j-1) j+1:11];
L(j) = L(j)*prod(m(i));
q(j) = q(j)*prod(x(j)-x(i));

2 - replace the 2nd while loop by

f = sum(o(1:11));

You should also consider using semi-colons at the end of lines, to suppress the outcomes you don't want to get.

Answer 1

[VBBV]

 clear
clc
syms z L [1 11]
z , L

z = 

L = 

x=[0 300 1000 1500 1700 2000 3000 3300 4000 4500 5000];
y=[32.0 122.4 296.4 405.7 447.6 608.4 704.7 761.4 891.9 983 1072.6];
m=z-x

m = 

i=1;
j=1;
% L= [1 1 1 1 1 1 1 1 1 1 1];
q=[1 1 1 1 1 1 1 1 1 1 1];
while j<12
    if j==1
        for i=2:11
             L(i)=L(i)-m(i);
            q(i)=q(i)*(x(j)-x(i));
        end
    else
              for i=1:j-1
              L(j)=L(j).*m(i);
              q(j)=q(j).*(x(j)-x(i));
              end
            for i=j+1:11
            L(j)=L(j).*m(i);
            q(j)=q(j).*(x(j)-x(i));
            end
      end    
j=j+1;
end
e=(y.*L);
o=e./q;
k=1;
f=0;
while k<12
   f=f+o(k) ;
   k=k+1;
end
f

f =