Finding a variable value using no fzero etc.
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Hi I have another problem that I cannot using fzero or solve or any method that uses initial answers because It can produce the exact answer
Y(t,j)=[0,0,1,0;1,0,0,0;0,0,0,1;0,1,0,0];
A(i,j)=[1,4,6,2;4,5,2,3;5,3,2,4];
now I have K(i,j) which needs to be calculated as follow
sigma(Y(1,j)*(K(1,j)-A(1,j)))=0
this part is simple I could easily calculate K(1,j) after having K(1,j) I need to use it in the following equation to calculate K(t,j)
sigma( sigma(Y(t,j)*(K(1,j)-(sigma(Y(t-1,f)*K(1,f)+A(1,j)))))=0
I calculated the answer K(1,3) = 6 from equation 1
from equation 2 we have K(1,1)=7, K(1,4)=9, K(1,2)=13
it can be easily calculated because it is gradually like this Y(K-A)=0 and we have y that is one and A
The problem is : how can I code the second equation that I would be able to calculate K(t,j) and use it in another equation? no fzero, solve etc. can anyone help?
1 个评论
Matt J
2014-8-12
because It can produce the exact answer
Do you mean "can", or did you really mean "can't"? No solver will produce an exact answer. You always have inexact calculations in finite precision computers.
回答（1 个）
Matt J
2014-8-12
Your equations are linear, so you just have to build an appropriate matrix equation
P*X=Q
and solve as X=P\Q
5 个评论
alexaa1989
2014-8-13
You have equations of the form
sigma_j Pij *Kj = Qi
This is equivalent to
P*K=Q
where Pij are the entries of the matrix P and Qi are the entries of the vector Q.
alexaa1989
2014-8-13
Matt J
2014-8-13
There shouldn't be a fundamental difference. Both summations are equivalent to weighted sums of the K(i) with different sets of coefficients. Just put the coefficients into the rows of a matrix, one row per equation.
alexaa1989
2014-8-14
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