Symbolic Math Toolbox™

subs - Symbolic substitution in symbolic expression or matrix

Syntax

R = subs(S) R = subs(S, new) R = subs(S,old,new)

Description

R = subs(S) replaces all occurrences of variables in the symbolic expression S with values obtained from the calling function, or the MATLAB^® workspace.

R = subs(S, new) replaces the default symbolic variable in S with new.

R = subs(S,old,new) replaces old with new in the symbolic expression S. old is a symbolic variable or a string representing a variable name. new is a symbolic or numeric variable or expression. That is, R = subs(S,old,new) evaluates S at old = new. The substitution is first attempted as a MATLAB expression resulting in the computation being done in double precision arithmetic if all the values in new are double precision. Convert the new values to sym to ensure symbolic or variable precision arithmetic.

If old and new are cell arrays of the same size, each element of old is replaced by the corresponding element of new. If S and old are scalars and new is an array or cell array, the scalars are expanded to produce an array result. If new is a cell array of numeric matrices, the substitutions are performed elementwise (i.e., subs(x*y,{x,y},{A,B}) returns A.*B when A and B are numeric).

If subs(s,old,new) does not change s, subs(s,new,old) is tried. This provides backwards compatibility with previous versions and eliminates the need to remember the order of the arguments. subs(s,old,new,0) does not switch the arguments if s does not change.

Note If A is a matrix, the command subs(S, x, A) replaces all occurrences of the variable x in the symbolic expression S with the matrix A, and replaces the constant term in S with the constant times a matrix of all ones. To evaluate S in the matrix sense, use the command polyvalm(sym2poly(S), A), which replaces the constant term with the constant times an identity matrix.

Examples

Single Input

Suppose a = 980 and C1 = 3 exist in the workspace.

The statement

y = dsolve('Dy = -a*y')

produces

y = C1*exp(-a*t)

Then the statement

subs(y)

produces

ans = 3*exp(-980*t)

Single Substitution

subs(a+b,a,4) returns 4+b.

Multiple Substitutions

subs(cos(a)+sin(b),{a,b},{sym('alpha'),2}) returns

cos(alpha)+sin(2)

Scalar Expansion Case

subs(exp(a*t),'a',-magic(2)) returns

[   exp(-t), exp(-3*t)]
[ exp(-4*t), exp(-2*t)]

Multiple Scalar Expansion

subs(x*y,{x,y},{[0 1;-1 0],[1 -1;-2 1]}) returns

     0    -1
     2     0