Hi all,
I have this equation:
J = A1 + 2*A4 + 3*A2 - 4*A5 = 0
First I would like to arrange the symbols from A1 to A5 as follow:
J = A1 + 3*A2 + 2*A4 - 4*A5 = 0
Second step I would like to use their coefficients:
J = [1 3 2 -4] = 0
and in third step put zero instead of those variable which are not in the equation (here A3 is not in the equation ), for example :
J = [1 3 0 2 -4] = 0
and finally bring minus numbers to right hand side of the vector:
[1 3 0 2] = [-4]
Thank for your help
For the two first steps, try the code
syms A1 A2 A3 A4 A5
J = A1 + 2*A4 + 3*A2 - 4*A5
Jsorted = collect(J,[A1,A2,A3,A4,A5])
[c,t] = coeffs(J,[A1,A2,A3,A4,A5])
By the way I am not sure why you would need step one (it is not necessary for step two). For the other steps, there are several ways to do that. It depends in fact on what you really want to achieve.
Thanks Philippe,
You are right the step one is not necessary but in matlab 2011 which yesterday I tried didn't arrange the symbolic variables, in matlab 2014 works and with your code it works in matlab 2011 as well.
Actually I need to separate positive and negative coefficients to create two matrix (left hand side include zero and right hand side) exactly as I mentioned in step third and fourth.
Try the following code, it does the fours steps as described. Nevertheless, I still think there might be better ways to do that, depending on you overall problem.
syms A1 A2 A3 A4 A5
vars = [A1,A2,A3,A4,A5];
J = A1 + 2*A4 + 3*A2 - 4*A5;
Jsorted = collect(J,vars)
[c,t] = coeffs(J,vars)
index = 0;
newc = zeros(1,length(vars));
for k = 1:length(vars)
if any(t==vars(k))
index = index + 1;
newc(k) = c(index);
end
end
newc
newcplus = newc(find(double(newc)>=0))
newcminus = newc(find(double(newc)
Dear Philippe,
Sorry, I corrected the whole question. At the beginning I didn't want to make it complicated for the people. I hope I explain it clearly. I would be grateful if you please answer the question,
for example, I have this equation:
J_1 = A1 + 2*A4 + 3*A2 - 4*A5 +10*B1 - 2*A7
known_value = A5 A9 B1 B2
unknown_value = A1 A2 A3 A4 A6 A7 A8
First I would like to arrange the symbols from A1 to B1 as follow: (which you already did it)
J_1 = A1 + 3*A2 + 2*A4 - 4*A5 - 2*A7 + 10*B1
second step I would like to bring A5 and B1 (which are belong to known values) to the right hand side of the equation (which I already mistaked and mentioned minus values):
J_1 = A1 + 3*A2 + 2*A4 - 2*A7 = 4*A5 - 10*B1
third step I would like to use their coefficients:
J_1 = [1 3 2 -2] = [4 -10]
and in the fourth step put zero instead of those variable which are not in the left hand side of the equation (here A3 is not in the equation ) (known_value), for example :
J_1 = [1 3 0 2 0 -2 ] = [4 -10]
and finally, if I have 3 equations like this (J_1, J_2 and J_3), I need the matrix of left hand sides coefficient and right hand side as well:
M = [....J_1.... ; .....J_2.... ; .....J_3.....] ; % left side coefficients
N = [....J_1.... ; .....J_2.... ; .....J_3.....] ; % right side coefficients
Thanks for your help
Dear Ehsan,
As I said, there are many different ways to do what you want, it depends on your overall problem. Assuming your equations are linear , the following simple code does the trick. A question puzzles me, though: how come that in the first place you have symbolic equations and not arrays of coefficients (that is, how do you input the data into matlab?)?
syms A1 A2 A3 A4 A5 A6 A7 A8 A9 B1 B2
known = [A5 A9 B1 B2];
unknown = [A1 A2 A3 A4 A6 A7 A8];
J1 = A1 + 2*A4 + 3*A2 - 4*A5 +10*B1 - 2*A7
M = jacobian(J1,unknown)
N = -jacobian(J1,known)
NwithoutZero = N(find(not(double(N)==0)))
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