Solving non-linear equation in vector form

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Tayyab Khalil
Tayyab Khalil 2021-4-28
评论： Tayyab Khalil 2021-4-28
Hi all, hope you are doing well.
Soi have a simple equation where the known value is a vector. So i need to get a vector as the solution.
The equation is very simple and can be easily caluclated by hand but i require it to be solved using Matlab.
Here is the code i have tried:
u2 = rand(1,1000);
syms t1
eq = t1.^2/64 == u2;
solve(eq, t1)
Any help would be appreciated, thanks.

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采纳的回答

Matt J
Matt J 2021-4-28
编辑：Matt J 2021-4-28
Ran in:
If you have the Optimization Toolbox,
u2=[1,4,9];
opts=optimoptions('fsolve','SpecifyObjectiveGradient',true,'OptimalityTolerance',1e-12);
t1=fsolve(@(t1)objfunc(t1,u2),u2,opts)
Equation solved, inaccuracy possible. The vector of function values is near zero, as measured by the value of the function tolerance. However, the last step was ineffective.
t1 = 1×3
8.0000 16.0000 24.0000
function [err, J]=objfunc(t1,u2)
err=t1.^2/64-u2;
J=speye(numel(u2))/32; %Jacobian
end
  1 个评论
Tayyab Khalil
Tayyab Khalil 2021-4-28
Yes that works fine, but is there not a easier way to solve this relatively simple problem?

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更多回答（2 个）

Matt J
Matt J 2021-4-28
编辑：Matt J 2021-4-28
Ran in:
syms u t
fun=matlabFunction( solve(t^2/64==u,t) );
u2=[1,4,9];
t1=fun(u2)
t1 = 2×3
-8 -16 -24 8 16 24
Matt J
Matt J 2021-4-28
u2 = rand(1,1000);
t1=8*sqrt(u2);
  3 个评论
显示 1更早的评论
Matt J
Matt J 2021-4-28
All the commands in my solution are Matlab commands...
Tayyab Khalil
Tayyab Khalil 2021-4-28
I meant that only putting the equation in its original form in Matlab and making matlab solve it for t1 rather than separting t1 ourself.

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