fsolve doesn't give the expected answer

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Federico MegaMan on 12 Nov 2019
Commented: Federico MegaMan on 13 Nov 2019
Hello,
I done (with the help of a member in another post) a fsolve script. It works but it doesn't give the expected result, or else the output spikes a lot.
Before I create this post i tried to solve my problem with other discussions and I read that there are more solution for trascendental equation (like mine). But I don't uderstand how to solve my problem.
This is my code, i have 12 equation and 13 unknows. I have to impose one of the unknows (T21) and it varies. After i have to calcolate all other 12 unknows while T21 varies.
T21=zeros(1,36);
j=0;
for t=0:0.1:2.755
j=j+1;
T21(j)=0.333*cos(t)-0.245;
end
sol = zeros(12,numel(T21));
for k = 1:numel(T21)
% the results are saved in a row 1...12
% every value of T21 match a column
if T21<=0
sol(:,k) = fsolve(@(x)funzmia(x,T21(k)), 1:12);
else
sol(:,k) = fsolve(@(x)funzmia2(x,T21(k)), 1:12);
end
end
figure;
plot(1:1:28, sol);
title('sol in funzione di t');
xlabe1=('t');
ylabe1=('sol');
grid on;
function F=funzmia(x,T21)
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=0
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=funzmia2(x,T21)
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=0
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

Nadir Altinbas on 12 Nov 2019
%T21=0
this line should be
T21 =0;

Walter Roberson on 12 Nov 2019
No, the value is passed in as a parameter.
Federico MegaMan on 12 Nov 2019
Exactly, the value of T21 changes and it is described by the for.

Nadir Altinbas on 13 Nov 2019
error: 'funzmia2' undefined near line 7 column 29
error: __plt2vm__: matrix dimensions must match error: called from __plt__>__plt2vm__ at line 419 column 5 __plt__>__plt2__ at line 250 column 14 __plt__ at line 113 column 17 plot at line 223 column 10

1 Comment

Federico MegaMan on 13 Nov 2019
Mmh it works on my pc, the only error i found was the precalling of T21 (1,28 instead of 1,36).
Maybe it depend by the matlab version(i have the 2018)?
Try again now.
T21=zeros(1,28);
j=0;
for t=0:0.1:2.755
j=j+1;
T21(j)=0.333*cos(t)-0.245;
end
sol = zeros(12,numel(T21));
for k = 1:numel(T21)
%the results for x are saved in rows 1...12
% every value of T21 match a column
if T21<=0
sol(:,k) = fsolve(@(x)funzmia(x,T21(k)), 1:12);
else
sol(:,k) = fsolve(@(x)funzmia2(x,T21(k)), 1:12);
end
end
figure;
plot(1:1:28, sol);
title('sol in funzione di t');
xlabe1=('t');
ylabe1=('sol');
grid on;
function F=funzmia(x,T21)
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=0
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=funzmia2(x,T21)
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=0
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

Nadir Altinbas on 13 Nov 2019
if T21<=0
funzmia(T21)<=0;
???

Federico MegaMan on 13 Nov 2019
Generally no.
I put that condition because of the reference system, when T21 is less than 0 some equations vary (like T13+T22-pi/2 --> T13+T22-3*pi/2).
Nadir Altinbas on 13 Nov 2019
i calculated the data is as following;
funzmiaT21 = [0.088000, 0.086336, 0.081362, 0.073127, 0.061713, 0.047235, 0.029837, 0.0096924, -0.012997, -0.038004
syms var funzmia,funzmiab;
funzmia(T21) = zeros(1,28);
j=0;
for t=0:0.1:2.755
j=j+1;
funzmiaT21(j)=0.333*cos(t)-0.245;
end
sol = zeros(12,numel(T21));
for k = 1:numel(T21)
%the results for x are saved in rows 1...12
% every value of T21 match a column
funzmia(T21)<=0
sol(:,k) = fsolve(@(x)funzmia(x,T21(k)), 1:12);
sol(:,k) = fsolve(@(x)funzmiab(x,T21(k)), 1:12);
end
figure;
plot(1:1:28, sol);
title('sol in funzione di t');
xlabe1=('t');
ylabe1=('sol');
grid on;
function F=funzmia(x,T21)
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=0
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=funzmiab(x,T21)
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=0
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);
ans:
double('ans');
end
Federico MegaMan on 13 Nov 2019
It doesn't work if i don't modify something(precalling from funzmia(T21) to funzmiaT21, ans:, and delete funzmia(T21)<=0) and when i do it the solution it's the same. Am i wronging something?
Anyway i didn't understad why you didn't use "if" to say when T21>=0 use funzmia and when T21<0 use funzmia2.