How to Solve for x for 20 functions

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Meteb Mejbel
Meteb Mejbel on 26 Aug 2021
Commented: Meteb Mejbel on 26 Aug 2021
Hi,
I have 4 column vectors (C,V,B and N), each has 20 intial values and they represent 20 spatial points ([C(i), V(i), B(i), N(i)], [C(i+1),V(i+1), B(i+1), N(i+1)]) and so on. I want to solve for x for each point using this function -> K=[(C(i)+x(i))*(V(i)+x(i))]/[(B(i)-x(i))*(N(i)-x(i))] (K is constant) and then get new values for each row in each vector ( C(i_new)=C(i)+x(i), V(i_new)=V(i)+x(i).....etc) and sort them again in new vectors (C_new, V_new, B_new and N_new)
1- So how can I construct this model? I always get confused when I want to do for loops by numel.
2- I was thinking to use vpasolve to get all x. However, I get 2 answers of x since the functons is second order in x. How can I choose first answer of x and use it later to find my next solutions?
Thnaks

Accepted Answer

Wan Ji
Wan Ji on 26 Aug 2021
You can get symbolic solution by
syms K C V B N X
eq = K*(B-X)*(N-X)-(C+X)*(V+X);
x = solve(eq, X)
and the final process is
K = 1;
% There are two solutions, x1 is the first solution
x1 = @(C,V,B,N) (C + V - (B.^2.*K.^2 + 2*B.*C.*K - 2*B.*K.^2.*N + ...
4*B.*K.*N + 2*B.*K.*V + C.^2 + 2*C.*K.*N + 4*C.*K.*V - 2*C.*V + ...
K.^2.*N.^2 + 2*K.*N.*V + V.^2).^(1/2) + B.*K + K.*N)./(2*(K - 1));
% x1 is the second solution
x2 = @(C,V,B,K) (C + V + (B.^2.*K.^2 + 2*B.*C.*K - 2*B.*K.^2.*N + ...
4*B.*K.*N + 2*B.*K.*V + C.^2 + 2*C.*K.*N + 4*C.*K.*V - 2*C.*V + ...
K.^2.*N.^2 + 2*K.*N.*V + V.^2).^(1/2) + B.*K + K.*N)./(2*(K - 1));
C = rand(20,1); % randomly initialized data
V = rand(20,1);
B = rand(20,1);
N = rand(20,1);
X = x1(C,V,B,N); % choose the first answer of x, you can also choose the second
C_new = sort(C + X);
V_new = sort(V + X);
B_new = sort(B + X);
N_new = sort(N + X);
  7 Comments
Meteb Mejbel
Meteb Mejbel on 26 Aug 2021
Great, I just tried it. It did work and yes without symbolic process. Thank you so much

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