symbolic variable and assignment operator

2 Ansichten (letzte 30 Tage)
yogeshwari patel
yogeshwari patel am 21 Aug. 2024
Kommentiert: Steven Lord am 21 Aug. 2024
syms x
syms t
b0=0.05
a3=0.1
b3=0.3;
A=(2*b3-1)/(2*a3-1)
B=(1/2)*b0*((4*a3*b3-1)/2*a3-1);
U=zeros(1,2,'sym');
V=zeros(1,2,'sym');
A=zeros(1,2,'sym');
B=zeros(1,2,'sym');
C=zeros(1,2,'sym');
D=zeros(1,2,'sym');
series1(x,t)=sym(zeros(1,1));
series2(x,t)=sym(zeros(1,1));
U(1)=0.05*(1-tanh(B*(20*(x-0.5))));
V(1)=b0*(A-tanh(B*(20*(x-0.5))))
Unable to perform assignment because the left and right sides have a different number of elements.
Error in sym/privsubsasgn (line 1168)
L_tilde2 = builtin('subsasgn',L_tilde,struct('type','()','subs',{varargin}),R_tilde);
Error in indexing (line 999)
C = privsubsasgn(L,R,inds{:});
  1 Kommentar
Shantanu Dixit
Shantanu Dixit am 21 Aug. 2024
Hi Yogeshwari, the error occurs because you're trying to assign a vector (1x2) to a single element of the symbolic array U and V.
U(1) = 0.05 * (1 - tanh(B * (20 * (x - 0.5))));
V(1)=b0*(A-tanh(B*(20*(x-0.5))));
%% results in a 1x2 vector on the right side, which cannot be assigned to a single element on the left
Also double-check your implementation as 'B' is initially defined as a symbolic variable and is later changed to a symbolic vector making the initial declaration redundant.
B=(1/2)*b0*((4*a3*b3-1)/2*a3-1);
B=zeros(1,2,'sym');

Melden Sie sich an, um zu kommentieren.

Akzeptierte Antwort

Walter Roberson
Walter Roberson am 21 Aug. 2024
B=zeros(1,2,'sym');
B will be a symbolic vector of size 1 x 2
U(1)=0.05*(1-tanh(B*(20*(x-0.5))));
On the right hand side, all of the symbolic vector B is used, so the right hand side will be a 1 x 2 result. But you are attempting to assign that 1 x 2 result into a single location, U(1)
  1 Kommentar
Steven Lord
Steven Lord am 21 Aug. 2024
If you'd eliminated the line of code that overwrites the scalar you'd assigned to B first with the symbolic vector, it would give you an answer. I'll leave it to you to determine if that's the answer you expected.
syms x
syms t
b0=0.05
b0 = 0.0500
a3=0.1
a3 = 0.1000
b3=0.3;
% These assign scalars to A and B
A=(2*b3-1)/(2*a3-1)
A = 0.5000
B=(1/2)*b0*((4*a3*b3-1)/2*a3-1)
B = -0.0261
U=zeros(1,2,'sym');
V=zeros(1,2,'sym');
% Commenting these two lines out (so A and B aren't overwritten) ...
% A=zeros(1,2,'sym');
% B=zeros(1,2,'sym');
C=zeros(1,2,'sym');
D=zeros(1,2,'sym');
series1(x,t)=sym(zeros(1,1));
series2(x,t)=sym(zeros(1,1));
% ... allows these two lines to work
U(1)=0.05*(1-tanh(B*(20*(x-0.5))))
U = 
V(1)=b0*(A-tanh(B*(20*(x-0.5))))
V = 

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (0)

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by