Symbolic solve giving wierd roots
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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.
采纳的回答
Matt Fig
2012-9-6
docsearch('rootof')
Look at the documentation of SOLVE, scroll down to rootof.
2 个评论
Matt Fig
2012-9-6
TUSHAR ANDRIYAS says,
"Hey Matt,
I tried the search and modified the function call to tp = solve(ff ,z,'MaxDegree',12)
since ff (ff=1/2 - (2*bn*x - z^4*(2*z^2 - 6*z + 5))^2/8;) is a 12th order polynomial. But matlab does not want to calculate the roots giving an error
_*Error using solve>processMaxDegree (line 324) Bad value for option 'MaxDegree'.
Error in solve>getEqns (line 259) [k, options] = processMaxDegree(k, options, v, nargin, varargin);
Error in solve (line 150) [eqns,vars,options] = getEqns(varargin{:});
Error in Potential_SOS (line 48) tp = solve(ff ,z,'MaxDegree',12)*_"
Matt Fig
2012-9-6
Tushar,
From the help:
"MaxDegree Do not use explicit formulas that involve radicals when solving polynomial equations of degree larger than the specified value. This value must be a positive integer *smaller than* 5. Default: 3"
Emphasis is mine. There is no explicit solution for a general polynomial of degree greater than 4. So it is not surprising that you get those results. What do you want to do with them, that is the question. Are you hoping to get a numerical value?
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