How to use options in fsolve

Question

Federico MegaMan 2019-11-14

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/491109-how-to-use-options-in-fsolve

评论： Matt J 2019-11-14

采纳的回答： Matt J

在 MATLAB Online 中打开

Hello,

I'm trying to use the options in fsolve but i didn't understand how i can do...

I know that i have to do fsolve(@(x) myfunz,x0,options) but I don't know what i have to write, option=..?

This is my code:

j=0;
xsol=zeros(13,28);
T21=zeros(1,28);
for t=0:0.1:2.7
    j=j+1;
    T21(j)=0.333*cos(t)-0.245;
    jT21=T21(j);
%jT21=0;
jT22=pi/2;%87°
jr23=3.1623;
jT23=4.391248;%248.5°
jT31=0;%0.226;%13°
jr32=3.1623;
jT32=1.8919369;%106°
jT33=1.5*pi;
jT11=1.106538;%62
jT12=0;
jT13=0;%01.5*pi+1.517
jT14=0;%0.226;
jT15=5.176646;%316°
for k=1:numel(T21)
    
x0=[jT11,jT12,jT13,jT14,jT15,jT22,jT23,jr23,jT31,jT32,jT33,jr32,jT21];
xsol=fsolve(@(x) funzz(x), [kT11,kT12,kT13,kT14,kT15,kT22,kT23,kr23,kT31,kT32,kT33,kr32,jT21]);
kT11=jxsol(1);
kT12=jxsol(2);
kT13=jxsol(3);
kT14=jxsol(4);
kT15=jxsol(5);
kT22=jxsol(6);
kT23=jxsol(7);
kr23=jxsol(8);
kT31=jxsol(9);
kT32=jxsol(10);
kT33=jxsol(11);
kr32=jxsol(12);
end
end
function F=funzz(x)
T11=x(1);
T12=x(2);
T13=x(3);
T14=x(4);
T15=x(5);
T22=x(6);
T23=x(7);
r23=x(8);
T31=x(9);
T32=x(10);
T33=x(11);
r32=x(12);
T21=x(13);
F(1)=(5^(0.5))*cos(T11)+cos(T12)+4*cos(T13)+cos(T14)+(5^(0.5))*cos(T15)-8;
F(2)=(5^(0.5))*sin(T11)+sin(T12)+4*sin(T13)+sin(T14)+(5^(0.5))*sin(T15);
F(3)=cos(T21)+3*cos(T22)+r23*cos(T23);
F(4)=sin(T21)+3*sin(T22)+r23*sin(T23);
F(5)=cos(T31)+r32*cos(T32)+3*cos(T33);
F(6)=sin(T31)+r32*sin(T32)+3*sin(T33);
F(7)=T12-T21;
F(8)=T31-T14;
F(9)=T13+T22-pi/2;
F(10)=T33-T22-pi;
F(11)=r23-(1+9+6*sin(T12-T13))^(0.5);
F(12)=r32-(1+9+6*cos(T14-T33))^(0.5);
end

But when I run it, it says:

Equation solved, fsolve stalled.

fsolve stopped because the relative size of the current step is less than the

default value of the step size tolerance and the vector of function values

is near zero as measured by the default value of the function tolerance.

Warning: Trust-region-dogleg algorithm of FSOLVE cannot handle non-square systems; using Levenberg-Marquardt

algorithm instead.

> In fsolve (line 310)

In wsadfg (line 30)

I saw that Levenberg-Marquardt is linked to options(is it right?), but i didn't understand how i can do.

Another thing, how can i make xsol a xsol(k)? Because matlab give me error if i put it.

Thank you in advance.

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Answer 1

Matt J 2019-11-14

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链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/491109-how-to-use-options-in-fsolve#answer_401527

编辑：Matt J 2019-11-14

在 MATLAB Online 中打开

I know that i have to do fsolve(@(x) myfunz,x0,options)

I don't think so. The exit message says "Equation solved".

Another thing, how can i make xsol a xsol(k)?

for k=1:numel(T21)
    
xsol(:,k)=fsolve(@(x) funzz(x), [kT11,kT12,kT13,kT14,kT15,kT22,kT23,kr23,kT31,kT32,kT33,kr32,jT21]);
end

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Federico MegaMan 2019-11-14

编辑：Federico MegaMan 2019-11-14

在 MATLAB Online 中打开

Thank you for the answer.

So do you think the code is right? Because the output are all costant...

I did some changes for T21 <0 and T21>0 and add the plot, but the output are all costant...

%jT21=0;
kT22=pi/2;%87°
kr23=3.1623;
kT23=4.391248;%248.5°
kT31=0;
kr32=3.1623;
kT32=1.8919369;%106°
kT33=1.5*pi;
kT11=1.106538;%62
kT12=0;
kT13=0;
kT14=0;
kT15=5.176646;%316°
    
j=0;
xsol=zeros(13,28);
T21=zeros(1,28);
for t=0:0.1:2.7
    j=j+1;
    T21(j)=0.333*cos(t)-0.245;
    jT21=T21(j);
end
x0=[kT11,kT12,kT13,kT14,kT15,kT22,kT23,kr23,kT31,kT32,kT33,kr32,jT21];
for k=1:numel(T21)
    
    if T21<=0
xsol(:,k)=fsolve(@(x) funz1(x), [kT11,kT12,kT13,kT14,kT15,kT22,kT23,kr23,kT31,kT32,kT33,kr32,T21(k)]);
    else
        xsol(:,k)=fsolve(@(x) funz2(x), [kT11,kT12,kT13,kT14,kT15,kT22,kT23,kr23,kT31,kT32,kT33,kr32,T21(k)]);
    end
        jxsol=xsol(:,k);
kT11=jxsol(1);
kT12=jxsol(2);
kT13=jxsol(3);
kT14=jxsol(4);
kT15=jxsol(5);
kT22=jxsol(6);
kT23=jxsol(7);
kr23=jxsol(8);
kT31=jxsol(9);
kT32=jxsol(10);
kT33=jxsol(11);
kr32=jxsol(12);
end
figure;
plot(1:1:28, xsol);
title('xsol in funzione di t');
xlabe1=('t');
ylabe1=('xsol');
grid on;
function F=funz1(x)
T11=x(1);
T12=x(2);
T13=x(3);
T14=x(4);
T15=x(5);
T22=x(6);
T23=x(7);
r23=x(8);
T31=x(9);
T32=x(10);
T33=x(11);
r32=x(12);
T21=x(13);
F(1)=(5^(0.5))*cos(T11)+cos(T12)+4*cos(T13)+cos(T14)+(5^(0.5))*cos(T15)-8;
F(2)=(5^(0.5))*sin(T11)+sin(T12)+4*sin(T13)+sin(T14)+(5^(0.5))*sin(T15);
F(3)=cos(T21)+3*cos(T22)+r23*cos(T23);
F(4)=sin(T21)+3*sin(T22)+r23*sin(T23);
F(5)=cos(T31)+r32*cos(T32)+3*cos(T33);
F(6)=sin(T31)+r32*sin(T32)+3*sin(T33);
F(7)=T12-T21;
F(8)=T31-T14;
F(9)=T13+T22-pi/2;
F(10)=T33-T22-pi;
F(11)=r23-(1+9+6*sin(T12-T13))^(0.5);
F(12)=r32-(1+9+6*cos(T14-T33))^(0.5);
end
function F=funz2(x)
T11=x(1);
T12=x(2);
T13=x(3);
T14=x(4);
T15=x(5);
T22=x(6);
T23=x(7);
r23=x(8);
T31=x(9);
T32=x(10);
T33=x(11);
r32=x(12);
T21=x(13);
F(1)=(5^(0.5))*cos(T11)+cos(T12)+4*cos(T13)+cos(T14)+(5^(0.5))*cos(T15)-8;
F(2)=(5^(0.5))*sin(T11)+sin(T12)+4*sin(T13)+sin(T14)+(5^(0.5))*sin(T15);
F(3)=cos(T21)+3*cos(T22)+r23*cos(T23);
F(4)=sin(T21)+3*sin(T22)+r23*sin(T23);
F(5)=cos(T31)+r32*cos(T32)+3*cos(T33);
F(6)=sin(T31)+r32*sin(T32)+3*sin(T33);
F(7)=T12-T21;
F(8)=T31-T14;
F(9)=T13+T22-3*pi/2;
F(10)=T33-T22-pi;
F(11)=r23-(1+9+6*sin(T12-T13))^(0.5);
F(12)=r32-(1+9+6*cos(T14-T33))^(0.5);
end