MATLAB: Symbolic solve giving wierd roots

syms solve

Hi there,

I was trying to solve for a complicated but nevertheless a polynomial equation with the symbolic toolbox. I got a solution but can't make sense of it. I had input as

ff=1/2 - (2*bn*x - z^4*(2*z^2 - 6*z + 5))^2/8;
t = solve(ff,z);

with the output as

 RootOf(z1^6 - 3*z1^5 + (5*z1^4)/2 - bn*x + 2^(1/2)*Eo^(1/2), z1)
 RootOf(z1^6 - 3*z1^5 + (5*z1^4)/2 - 2^(1/2)*Eo^(1/2) - bn*x, z1)

I had read somewhere about this but can't find it now.

Best Answer

docsearch('rootof')

Look at the documentation of SOLVE, scroll down to rootof.

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MATLAB: Cannot find ALL solutions to simultaneous equation

Hi CZ

exp(x^2+y^2+z^2) = 9

then

x^2+y^2+z^2 = log(9)

which is a lot less challenging for symbolic evaluation.

clear all
syms x y z C
eq1 = x^2 + y^2 + z^2 == C;
eq2 = x + x*y - z == 1;
eq3 = x + y + z == 1;
[solx,soly,solz] = solve([eq1,eq2,eq3],[x y z])
solz = 
 root(z1^4 - 5*z1^3 - z1^2*(C - 7) + z1*((5*C)/2 - 1/2) - C/2 + C^2/4 + 1/4, z1, 1)
 root(z1^4 - 5*z1^3 - z1^2*(C - 7) + z1*((5*C)/2 - 1/2) - C/2 + C^2/4 + 1/4, z1, 2)
 root(z1^4 - 5*z1^3 - z1^2*(C - 7) + z1*((5*C)/2 - 1/2) - C/2 + C^2/4 + 1/4, z1, 3)
 root(z1^4 - 5*z1^3 - z1^2*(C - 7) + z1*((5*C)/2 - 1/2) - C/2 + C^2/4 + 1/4, z1, 4)

There are expressions that are twice as long for x and y. You can get numerical results by plugging in log(9) for C and using the roots function to get four solutions. I wish I knew how to get the coefficients in solx, soly and solz into the roots function in a convenient way, but I don't.

MATLAB: How to interprete output of solve function

syms x
double(solve ( x^2-(1+x)^2 + x^2*(1+x)^2 , 'MaxDegree', 4))

Best Answer

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MATLAB: Cannot find ALL solutions to simultaneous equation

MATLAB: How to interprete output of solve function

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