MATLAB: How to rename lumped constant terms within parenthesis

MATLABmatlab symbolic constant coefficients find replace

Hi Guys

I have some huge symbolic equations (multiple pages) in the variables (x,y,z), of the form:

Eq1(x,y,z) = x*y*(.....) + y*z*x*(....) ....(many pages)

where the brackets contain constant symbolic terms (e.g. they are just constant coefficients), e.g.:

(Q23*l5*l6*m4 + Q22*l5*m4*m6)

Is there anyway to automatically go through each equation and delete / abbreviate the bracketed coefficients to a single letter? Currently I am doing it manually but it takes so long..! Maybe using a find and replace function?

Thanks

Toby

Best Answer

The easiest way to do this is to go in to the symbolic engine level and use subsex

syms QQ
result = feval(symengine, 'subsex', Eq1(x,y,z), (Q23*l5*l6*m4 + Q22*l5*m4*m6), QQ);

Note that this will be an expression rather than a function. I am not sure if you would be able to execute this on the function returning a function, but you could try -- list Eq1 instead of E1(x,y,z) as the argument.

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MATLAB: Eig() gives wrong results for complex eigenvalues in comparison to eigs()

Hi Lennart,

It seems there are a few things that are not correct with the last part of the code:

1. Instead of li(qq), it should be li(qq,qq).

2. You should not use (.'), instead (') should be used. (.') computes the nonconjugate transpose of a matrix whereas (.) computes complex conjugate transpose. Since the eigenvalues and eigenvectors could be complex in nature even for a real matrix, the latter is the correct form to be used.

3. The code below gives you the desired results:

for qq = 1:length(L1)
  %





  right1(:, qq) = (A - L1(qq,qq) * B) * X1(:, qq);
  left1(qq, :) = Y1(:, qq)' * (A - L1(qq,qq) * B);
  %
  right2(:, qq) = (A - L21(qq,qq) * B) * X2(:, qq);
  left2(qq, :) = Y2(:, qq)' * (A - L22(qq,qq)' * B);
  %
  right3(:, qq) = (A - L3(qq,qq) * B) * X3(:, qq);
  left3(qq, :) = Y3(:, qq)' * (A - L3(qq,qq) * B);
  %
  right4(:, qq) = (A - L41(qq,qq) * B) * X4(:, qq);
  left4(qq, :) = Y4(:, qq)' * (A - L42(qq,qq)' * B);
  %
  right5(:, qq) = (A - L5(qq,qq) * B) * X5(:, qq);
  left5(qq, :) = Y5(:, qq)' * (A - L5(qq,qq) * B);
  %
  right6(:, qq) = (A - L61(qq,qq) * B) * X6(:, qq);
  left6(qq, :) = Y6(:, qq)' * (A - L62(qq,qq)' * B);
end
[left1 right1;
left2 right2;
left3 right3;
left4 right4;
left5 right5;
left6 right6]

Please note that I have changed Li2(qq,qq) to Li2(qq,qq)' for the computation of left2, left4 and left6 in the 2nd, 4th and 6th case. Here is why: As per the documentation of eig,

[Y2, L22] = eigs(A', B');

ensures that the solution satisfies the following:

A'*Y2 = B'*Y2*L22   % which gives  
Y2'*A = L22'*Y2'*B  % (taking transpose of both sides)
Y2'*A - L22'*Y2'*B = 0  
Y2'*(A-L22'*B) = 0

(since values of L22 are taken one by one in your calculations, so it can behave as a scalar and moving its position does not change the result). However, taking its (complex conjugate) transpose is crucial since it is a complex number. Instead of computing the residuals inside a loop, it can be written as one single statement (treating L22 as a matrix):

left2 = Y2'*A - L22'*Y2'*B;

The same reasoning applies for cases 4 and 6. Please note this reasoning does not apply to cases 1, 3 and 5 and the original form of equation stays valid.

I hope the above helps.

Nalini

MATLAB: Expression of the equations in the Symbolic Math Toolbox.

eq2 = cos(solve(eq1, Q))

Best Answer

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MATLAB: Eig() gives wrong results for complex eigenvalues in comparison to eigs()

MATLAB: Expression of the equations in the Symbolic Math Toolbox.

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