Create matrix from columns of variable (a)-for loop
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Hi I want to create a matrix TA from the columns EE generated in the last loop. I can display the wanted columns "disp (EE)' but when adding the matrix want to create from these columns, an error occurred, noted that the vectors/columns EE has a variable (a)
You may ignore the lines of the code before the line start with C = [I;A;B]; for getting the question
I=ones(20,2);
for i=0:20
theta=(pi/40)*i;
r=2.5;x1=r*cos(theta);
y1=r*sin(theta);s=i+1;
I(s,:)=[x1,y1];
end
A=ones(20,2);
for k=1:20
r=(k-1)/10;
A(k,:)=[1,r];
end
m=1:-(1/47):0;
B=ones(length(m),2);
B(:,1)=m';B(:,2)=2;
C = [I;A;B];
%%find the complex variable matrix, ZC connjgate Z
Z=ones(length(C),1);
for v=1:length(C)
Z(v)=C(v,1)+1i*C(v,2);end
ZC=conj(Z);
F=ones(89,2);syms Fb;
for ii=1:89
if ii<22
F(ii,1)=0;F(ii,2)=0;
elseif ii<42
F(ii,1)=1; F(ii,2)=0;
else
F(ii,1)=(89-ii)/(89-42);F(ii,2)=0;
end
end
FF=F.*Fb;
zzc=Z-ZC;
QQ=ones(length(Z),21);SS=ones(length(Z),21);ss=0;
T1=ones(length(Z),21);T2=ones(length(Z),21);syms a
for K=-6:14
ss=ss+1;
QQ(:,ss)=Z.^(2*K-1)+ZC.^(2*K-1);
SS(:,ss)=(2*ss-1)*(Z.^(2*ss-2));
T1(:,ss)=Z.^(2*K-2);T2(:,ss)=ZC.^(2*K-2);
end
%multiplying the T1,T2 BY THE FCTOR
Q=QQ.';S=SS.';
TA=ones(89,21);TB=ones(89,21);
Tz=sqrt(Z.^2-a^2); Tzc=sqrt(ZC.^2-a^2);
for j=1:21
CC=Tz.*T1(:,j);DD=Tzc.*T2(:,j);EE=CC+DD;
TA(:,j)=EE;
end
Accepted Answer
Walter Roberson
on 26 Sep 2018
syms a
so a is syms
Tz=sqrt(Z.^2-a^2); Tzc=sqrt(ZC.^2-a^2);
so Tz and Tzc are syms involving the unresolved variable a
CC=Tz.*T1(:,j);DD=Tzc.*T2(:,j);EE=CC+DD;
so CC and DD and EE all involve that unresolved variable a
TA=ones(89,21);
so TA can only store numeric values.
TA(:,j)=EE;
with TA only being able to store numeric values, the values in EE have to be converted from sym to numeric. But they cannot be converted, because they involve the unresolved symbolic variable a
Your EE entries are formulas, such as
(5209627982755595*(7990063086789097/36028797018963968 - a^2)^(1/2))/137438953472 + (5252908098854583*(6546788930781455/18014398509481984 - a^2)^(1/2))/4398046511104
You cannot store a formula in a numeric array.
You can use vpa() to convert the formulas from using rational numbers to using floating point numbers; for example vpa() of the above gives the result
1194.3730212020871022104984149337*(0.36341979041573407860354905096756 - 1.0*a^2)^(1/2) + 37905.032388193612860050052404404*(0.22176880017902014796682408359629 - 1.0*a^2)^(1/2)
but this is still a formula and would need a symbolic array to be stored into.
16 Comments
Stephen23
on 26 Sep 2018
Thanks, Walter for your answer, new things I have learned
I redefine all the matrices/arrays as symbolic. I did not use vap(). I am generating numerous linear equations to be solved at the end using least squares method.
Do you think I did not miss an important thing and if this is the right way for my purposes
Thanks
Walter Roberson
on 26 Sep 2018
Those equations are not linear equations: they have expressions of the form
sqrt(constant - x^2)
Wazy sky
on 26 Sep 2018
Walter Roberson
on 26 Sep 2018
Linear equations do not have squares of the variables. There are some quadratic solvers that can be good on equations containing the squares of variables (especially with no cross-terms of one variable multiplied by another). However, those solvers cannot deal with sqrt()
You will need to use nonlinear methods.
Wazy sky
on 27 Sep 2018
Walter Roberson
on 27 Sep 2018
They are non-linear equations with respect to the variables E, G, and a.
You cannot use a linear least squared approach for this -- not unless perhaps you are given one or more a values.
Walter Roberson
on 27 Sep 2018
Just putting on a constraint would not work. You would need to substitute a particular value of a to have a chance at the equations becoming linear.
Walter Roberson
on 10 Oct 2018
Sorry, I am not clear as to what your question is?
I am also unsure what your current code is.
Wazy sky
on 11 Oct 2018
Walter Roberson
on 11 Oct 2018
You have
for ij=1:29
Em1=0;Em2=0;
for jj=1:98
Em1=Em1+diff(Eq1(jj),E(ij,1));
Em2=Em2+diff(Eq1(jj),G(ij,1));nn=nn+1;
end
end
EmE(ij,1)=Em1;EmE(ij+29,1)=Em2;
end
The first "end" there finishes the "for jj". The second "end" there finishes the "for ij". The line after that uses the value of ij, which would at that point have the last value it was assigned in its for loop, which would be 29. The line of code looks more like it belongs inside the "for ij" loop than after it. The line after that, the "end" does not match anything and is a syntax error.
Walter Roberson
on 11 Oct 2018
The line
t_H=toc; tic;
attempts to do a "toc" before there has been any matching "tic". You never use the value of t_H so it is a mystery as to why that line is present.
Walter Roberson
on 11 Oct 2018
The line
Em2=Em2+diff(Eq1(jj),G(ij,1));nn=nn+1;
attempts to increment the undefined variable nn. you never use the value of nn so it is a mystery as to why that is present.
Stephen23
on 11 Oct 2018
"Your suggestions are so much appreacited"
Your code is very badly aligned. Badly aligned code is one way that beginners hide bugs in their code. In this case, your badly aligned code has hidden several bugs and made them harder to identify. Correctly aligned code makes identifying some kinds of bugs easier. You should align your code correctly.
I recommend using the MATLAB Editor's default alignment settings, which occur automatically as you type. You can also align existing code: select the code text, press ctrl+i.
Wazy sky
on 11 Oct 2018
More Answers (2)
Wazy sky
on 11 Oct 2018
Edited: Walter Roberson
on 11 Oct 2018
2 Comments
Walter Roberson
on 11 Oct 2018
If you dbstop at the vpasolve() and examine vpa(EmE) then you will see that it has a large number of real constants times real() or imag() of various variables, added together, with no obvious cross-products or exponentiation or exp() or the like. So each row is linear in real() and imag() of those variables. But each row also ends in a constant that has both a real and an imaginary coefficient. With all of those constants being real valued, and with real() and imag() also returning real-valued information, then the part involving the variables must be real-valued. That means there is no possibility of part of a term canceling out the imaginary part of the constant coefficient of the row. Therefore the rows cannot equal 0, because although you might balance the real portion, you are left with a non-zero imaginary portion.
Wazy sky
on 12 Oct 2018
Wazy sky
on 13 Oct 2018
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