what is the syntaxis for numerical solvers?
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Hi, for the last hour I am trying to start the numerical solver..
The copy-past from "help"
numeric::solve([sin(x) = y^2 - 1, cos(x) = y], [x, y]);
gives
Error: Unexpected MATLAB operator
for '::'
I can also use the other syntaxis option, which works fine in my script:
syms ha r2 positive; [h,rcone] = solve( hh(ha,r2) == 0, VV(ha,r2) == V0);
If use this, how do I define the range of ha and r2 in which I want the solutions?
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Accepted Answer
Petr
on 12 Dec 2012
3 Comments
Walter Roberson
on 13 Dec 2012
Note this from the numeric::solve documentation page:
Note: In contrast to the symbolic solver solve, the numerical solver does not react to properties of identifiers set via assume. To use these properties, call float ( hold( solve )(arguments)) instead.
More Answers (4)
Azzi Abdelmalek
on 12 Dec 2012
Edited: Azzi Abdelmalek
on 12 Dec 2012
sol=solve('sin(x) = y^2 - 1', 'cos(x) = y')
%or
sol=solve('sin(x) -y^2 + 1', 'cos(x) - y')
2 Comments
Walter Roberson
on 13 Dec 2018
Edited: Walter Roberson
on 13 Dec 2018
In older versions of MATLAB,
solve('x = exp(x+y)', 'x')
In more modern versions
syms x y
solve(x == exp(x+y), x)
However, MATLAB is not able to find a solution. A solution exists, and is
-lambertw(-exp(y))
but MATLAB is not strong on Lambert W processing.
Zuhaib
on 12 Dec 2012
it depend on your equation .. write function used in your equation for search in hepl of matlab
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Walter Roberson
on 12 Dec 2012
numeric::solve is a MuPAD call that cannot be directly used from MATLAB. Use
feval(symengine, 'numeric::solve', [sym('sin(x) = y^2 - 1'), sym('cos(x) = y')], [sym('x'), sym('y')])
If you have R2011b or later (I think it is), you can use
syms x y
feval(symengine, 'numeric::solve', [sin(x) = y^2 - 1, cos(x) = y], [x, y])
Chibuzo Chukwu
on 13 Dec 2018
How do I make x the subject of this equation.using matlab
x = exp(x+y)
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