Get coeffients of symbolic expression
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I have a set of equations (derv_alpha) that I have arrived at by differentiating an expression and setting it to zero.
N=3; alphavec = sym('alpha',[1 N]); %N can take any value
derv_alpha = [ 0, 2*alphavec(3) - 2*alphavec(2) + 2, 2*alphavec(2) - 4*alphavec(3)];
Now I need to solve these simultaneous equations and come up with answers for alpha2, alpha3...alphaN. i.e. alphavec(2), alphavec(3)...alphavec(N).
I am trying to put this in matrix form A*x=b and solve for x. Is it possible? If so, HOW CAN I GET THE COEFFICIENTS OF EACH VARIABLE?
The solution will have to be irrespective of the position of the variable in the expression.
I did a quick search. collect() doesnt work. i dont think sym2poly can be used
OR, is there a better way to do it when working with symbols?
I am doing my first program using symbols this week and appreciate any help. Thanks again.
回答 (1 件)
Walter Roberson
2012 年 7 月 21 日
Generally,
solve(derv_alpha, alphavec)
However, derv_alpha(1) is the constant 0 and not an expression in the alpha variables, so the set of three equations in derv_alpha, in three variables, is really only two equations in three variables, which you aren't going to be able to solve for. This is going to be a problem even if you express as A*x=b
6 件のコメント
Mech Princess
2012 年 7 月 21 日
編集済み: Mech Princess
2012 年 7 月 21 日
Star Strider
2012 年 7 月 21 日
編集済み: Star Strider
2012 年 7 月 21 日
I am now completely lost as to what you are doing. What coefficients do you want to get? The ‘coeffs’ function will give you the coefficients, however it has to know the variables or expressions you want the coefficients of.
Meanwhile:
% alpha_sol=zeros(1:N);
% alpha_sol(2:N)=solve(derv_alpha, alphavec(2:N))
alpha_sol_12 = solve(derv_alpha(1) == 0, alphavec(2))
alpha_sol_13 = solve(derv_alpha(1) == 0, alphavec(3))
alpha_sol_22 = solve(derv_alpha(2) == 0, alphavec(2))
alpha_sol_23 = solve(derv_alpha(2) == 0, alphavec(3))
These produce:
alpha_sol_12 =
alpha3 + 1
alpha_sol_13 =
alpha2 - 1
alpha_sol_22 =
2*alpha3
alpha_sol_23 =
alpha2/2
Is this what you want to do?
I initially put that into a loop, but the Symbolic Math Toolbox objected and I don't have the energy tonight to argue with it.
Mech Princess
2012 年 7 月 21 日
編集済み: Mech Princess
2012 年 7 月 21 日
Walter Roberson
2012 年 7 月 21 日
With regard to four equations: if your code read
derv_alpha = [2*alphavec(3) -2*alphavec(2) + 2, 2*alphavec(2) -4*alphavec(3)];
with no space after the '-', then the '-' would be interpreted as unary minus and the space would be interpreted as the end of the term in the list. [A - B] is one expression but [A -B] is two, A and -B . You can ensure your desired groupings by adding ()
Mech Princess
2012 年 7 月 21 日
Mech Princess
2012 年 7 月 21 日
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