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# Unable to perform assignment because the left and right sides have a different number of elements.

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alok ranjan on 14 May 2021
Commented: Image Analyst on 15 May 2021
clear
syms X1 X2 X3 X4
xx=[X1,X2,X3,X4]
X1 = xx(1);
d = length(xx);
term1 = (X1-1)^2;
sum = 0;
for ii = 2:d
xi = xx(ii);
xold = xx(ii-1);
new = ii * (2*xi^2 - xold)^2;
sum = sum + new;
end
xc1(1)=1;
xc2(1)=1;
xc3(1)=1;
xc4(1)=1;
err=10^(-2)
%yb4 = @(X1,X2,X3,X4) (term1 + sum)
yb4=(term1+sum)
dyb4_x1=diff(yb4,X1)
dyb4_x2=diff(yb4,X2)
dyb4_x3=diff(yb4,X3)
dyb4_x4=diff(yb4,X4)
J4=[subs(dyb4_x1,[X1,X2,X3,X4],[xc1(1),xc2(1),xc3(1),xc4(1)]) subs(dyb4_x2,[X1,X2,X3,X4],[xc1(1),xc2(1),xc3(1),xc4(1)]) subs(dyb4_x3,[X1,X2,X3,X4],[xc1(1),xc2(1),xc3(1),xc4(1)]) subs(dyb4_x4,[X1,X2,X3,X4],[xc1(1),xc2(1),xc3(1),xc4(1)])]
S4=-(J4)
i=1;
while norm(J4)>err
I4=[xc1(i),xc2(i),xc3(i),xc4(i)]';
syms h;
gr=subs(yb4,[X1,X2,X3,X4],[xc1(i)+h*S4(1),xc2(i)+h*S4(2),xc3(i)+h*S4(3),xc4(i)+h*S4(4)])
dgr_h=diff(gr,h)
h=vpasolve(dgr_h==0,h,[-Inf,Inf])
i=i+1
xc1(i+1)=I4(1)+h*S4(1)
xc2(i+1)=I4(2)+h*S4(2)
xc3(i+1)=I4(3)+h*S4(3)
xc4(i+1)=I4(3)+h*S4(4)
end
Error in dixon (line 38) xc1(i+1)=I4(1)+h*S4(1)
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### Answers (2)

DGM on 14 May 2021
The result of the vpasolve() call is a vector of multiple solutions (3). The result of operations on h with scalars is still a 3-element vector. The LHS of the assignment is a scalar. You're trying to put a vector into a scalar.
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DGM on 15 May 2021
The equation dgr_h==0 has multiple solutions. If you expect only one, you're going to have to decide which one is meaningful to you. I can't tell you that if you don't know it yourself. I don't know anything about your task at a conceptual level.
If you want to work with all solutions, you're going have to write your loop such that it can handle vectors of multiple solutions -- possibly vectors of unequal length if the number of solutions is not always 3. Since you're using xc1, etc as inputs to the solver, such an approach could get complicated if not impractical. Maybe someone else can offer a better recommendation.

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Image Analyst on 14 May 2021
Try this (before the error line) and tell us what it says.
whos h
whos S4
whos I4
Chances are h is not a single scalar number, like it needs to be if you're going to stick it in the (i+1) element of xc1.
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Image Analyst on 15 May 2021
@alok ranjan, perhaps you overlooked the part where I said "tell us what it says". I'm sure MATLAB did not report "sir, still not working" into the command window. Read my answer again.

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