# Matrix inversion and summation

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BeeTiaw el 29 de Jun. de 2018
Comentada: BeeTiaw el 30 de Jun. de 2018
Hi all, can anyone help me to explain the mistake in my code below? The problem is as follow:
• The following parameter is assumed,
lma = 3;
ma = zeros(1,lma);
ma(2) = -0.2;
ma(lma) = -0.01;
• The parameter is then used to calculate the coordinate z using the following equation,
where z = x + i*y (complex number)
dumZ = 0;
sumdumZ = 0;
theta = linspace(0,2*pi,37);
xi = exp(1i.*theta);
for ii=1:length(theta)
for jj=1:lma
dumZ = ma(jj).*xi(ii).^jj;
sumdumZ = sumdumZ + dumZ;
end
zcon(ii) = 1./xi(ii)+sumdumZ;
dumZ = 0;
sumdumZ = 0;
end
zcon=zcon';
The following profile is obtained by plotting zcon,
• Now, using the hypothetical data above, i.e. the zcon, I want to back-calculate the variable "ma" or "alpha" as shown in the equation above. To do this, I first created a matrix A, which essentially a matrix containing all the constants. The following code is used to generate matrix A with 370 terms,
lx = length(zcon);
N = 10*37;
A = zeros(lx,N);
for ii=1:lx
for jj=1:N
A(ii,jj) = xi(ii).^jj;
end
end
A=[1./xi' A]; % insert a new colum to the left containing the 1/xi(ii) into the existing matrix A;
• Then the variable "ma" or "alpha" can be easily obtained using the following operation in Matlab,
maiter = A\zcon;
which give me a [370 x 1] matrix. To validate this, I again calculate the "predicted" z using the following code:
zdirect = A*maiter;
Which results an exactly the same curve (as expected) as shown below,
Now, coming to my question. Instead of using a direct calculation as shown above, i.e. the code "zdirect = A*maiter, instead I would like to manually use summation to calculate the "z" for each "theta" using the code below. Somehow, I could not get it right and cannot find the mistake. Below is the code that I used to calculate "z"
dumZ = 0;
sumdumZ = 0;
pow = [-1 1:1:N];
for ii=1:length(theta)
for jj=1:length(maiter)
dumZ = maiter(jj).*xi(ii).^pow(jj);
sumdumZ = sumdumZ + dumZ;
end
ziter(ii) = sumdumZ;
dumZ = 0;
sumdumZ = 0;
end
Instead, this is what I got,
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David Goodmanson el 29 de Jun. de 2018
Hi Bee,
Are you plotting the right variable? I ran your code, and zcon, zdirect and ziter all reproduce the same curve. Only the simple exponential variable xi gives the red circle in the second plot.
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David Goodmanson el 30 de Jun. de 2018
Hi Bee,
Below are the first few rows of [zcon zdirect ziter]. The only change I made to your code was
zcon = zcon.' instead of zcon = zcon'
so that the complex conjugate of zcon is not taken. That means that the table I get is the complex conjugate of yours. I also added
ziter = ziter.'
so that ziter is a column vector. Of course neither of these changes/additions affects the shape of the plot, 'triangle' vs. circle.
0.7900 + 0.0000i 0.7900 + 0.0000i 0.7900 + 0.0000i
0.7882 - 0.2471i 0.7882 - 0.2471i 0.7882 - 0.2471i
0.7815 - 0.4792i 0.7815 - 0.4792i 0.7815 - 0.4792i
0.7660 - 0.6832i 0.7660 - 0.6832i 0.7660 - 0.6832i
0.7363 - 0.8484i 0.7363 - 0.8484i 0.7363 - 0.8484i
0.6862 - 0.9680i 0.6862 - 0.9680i 0.6862 - 0.9680i
0.6100 - 1.0392i 0.6100 - 1.0392i 0.6100 - 1.0392i
0.5039 - 1.0633i 0.5039 - 1.0633i 0.5039 - 1.0633i
0.3666 - 1.0446i 0.3666 - 1.0446i 0.3666 - 1.0446i
0.2000 - 0.9900i 0.2000 - 0.9900i 0.2000 - 0.9900i
0.0093 - 0.9077i 0.0093 - 0.9077i 0.0093 - 0.9077i
-0.1975 - 0.8061i -0.1975 - 0.8061i -0.1975 - 0.8061i
-0.4100 - 0.6928i -0.4100 - 0.6928i -0.4100 - 0.6928i
-0.6167 - 0.5741i -0.6167 - 0.5741i -0.6167 - 0.5741i
-0.8058 - 0.4545i -0.8058 - 0.4545i -0.8058 - 0.4545i
BeeTiaw el 30 de Jun. de 2018
Hi David Goodmanson,
I got it correct now after change it as per your suggestion. Thanks a lot!
It seems that it has something to do with consistency of the conjugate/non-conjugate of the complex number. Although I still do not clearly understand it.
But I would consider the problem is solved now.

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