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Problem Solving Symbolic Inequalities

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Ben
Ben 2012-3-14
评论： filston Rukerandanga 2020-7-14
I'm trying to use Matlab to solve inequalities like the example below, but only have partial sucess, with other times getting the result shown below.
EDU>> solution=solve('((k1^2 + 1080.0*k1 - 2948400.0)/(k1 - 4660.0))>0')
solution = matrix([[solve([0.0 < (k1^2 + 1080.0*k1 - 2948400.0)/(k1 - 4660.0)], [k1])]])
I know that the solutions for this example are -2340<k1<1260 & k1>4660, is there something that I can do differently to make this work in Matlab? Thanks.

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采纳的回答

Stefan Wehmeier
Stefan Wehmeier 2012-3-19
Note that by default solve is in complex mode, i.e., you are looking for all solutions within the complex numbers. Try
solution=feval(symengine, 'solve', '((k1^2 + 1080.0*k1 - 2948400.0)/(k1 - 4660.0))>0', 'k1', 'Real')
  2 个评论
Alexander
Alexander 2012-3-19
|solve| also supports the option |real|, so you don't need |feval|:
solution = solve('((k1^2 + 1080.0*k1 - 2948400.0)/(k1 - 4660.0))>0', 'Real', true)
filston Rukerandanga
filston Rukerandanga 2020-7-14
Confirmed, the option 'real', solved my problem. Before it was giving me a warning like :
Warning: Unable to find explicit solution. For options, see help.
> In solve (line 317)
% So here is my working code
syms n
eq1 = -10*log10(abs(1/(1 + (.25)^(2*n))))<=0.05;
eq2 = -10*log10(abs(1/(1 + (2)^(2*n))))>10;
eq1 = rewrite( -10*log10(abs(1/(1 + (.25)^(2*n))))<=0.05,'log');
eq2 = rewrite(-10*log10(abs(1/(1 + (2)^(2*n))))>10, 'log');
soln = solve(eq1,eq2, n, 'IgnoreAnalyticConstraints',1,'real',1);
n = vpa(soln)

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更多回答（1 个）

Walter Roberson
Walter Roberson 2012-3-14
Symbolic solvers are notoriously poor at inequalities. All except the long-gone Axiom: it was supposedly good.
In the particular case above, Maple 15 gives the solution as
RealRange(Open(-2340), Open(1260))
RealRange(Open(4660), infinity)
In general, though, what I usually end up doing is transforming the inequality in to an equality by introducing a variable that I add constraints on to:
syms k1
syms c positive
solve( ((k1^2 + 1080.0*k1 - 2948400.0)/(k1 - 4660.0)) - c, k1)
Since the assumed-positive value c needs to be subtracted for the expression to equal 0, then that is equivalent to saying that the result of the expression (without the "- c") must be positive.
There have been a fair number of expressions in Maple that I could not get anywhere on until I substituted a particular number (symbolic) as the difference and made the expressions in to equalities.

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