In MATLAB 2020a, how can I convert symfun to double?

8 vues (au cours des 30 derniers jours)
a a
a a le 6 Sep 2020
Commenté : Steven Lord le 6 Sep 2020
I have a symfun like this:
>> p
p(t) =
[ 0, -(800*pi^2*exp(10*log((489*pi)/50 + 5*sin((11*pi)/250)) - 10*log(10*pi))*(9*cos((11*pi)/250) + (489*pi*sin((11*pi)/250))/500 - 5*cos((11*pi)/250)^2 - 4))/((489*pi)/50 + 5*sin((11*pi)/250))^2, -(800*pi^2*exp(10*log((239*pi)/25 + 5*sin((11*pi)/125)) - 10*log(10*pi))*(9*cos((11*pi)/125) + (239*pi*sin((11*pi)/125))/250 - 5*cos((11*pi)/125)^2 - 4))/((239*pi)/25 + 5*sin((11*pi)/125))^2, -(800*pi^2*exp(10*log((467*pi)/50 + 5*sin((33*pi)/250)) - 10*log(10*pi))*(9*cos((33*pi)/250) + (467*pi*sin((33*pi)/250))/500 - 5*cos((33*pi)/250)^2 - 4))/((467*pi)/50 + 5*sin((33*pi)/250))^2, -(800*pi^2*exp(10*log((228*pi)/25 + 5*sin((22*pi)/125)) - 10*log(10*pi))*(9*cos((22*pi)/125) + (114*pi*sin((22*pi)/125))/125 - 5*cos((22*pi)/125)^2 - 4))/((228*pi)/25 + 5*sin((22*pi)/125))^2, -(800*pi^2*exp(10*log((89*pi)/10 + 5*sin((11*pi)/50)) - 10*log(10*pi))*(9*cos((11*pi)/50) + (89*pi*sin((11*pi)/50))/100 - 5*cos((11*pi)/50)^2 - 4))/((89*pi)/10 + 5*sin((11*pi)/50))^2, -(800*pi^2*exp(10*log((217*pi)/25 + 5*sin((33*pi)/125)) - 10*log(10*pi))*(9*cos((33*pi)/125) + (217*pi*sin((33*pi)/125))/250 - 5*cos((33*pi)/125)^2 - 4))/((217*pi)/25 + 5*sin((33*pi)/125))^2, ............
but when I try to convert it to doule I get(this used to work fine in MATLAB 2018b)
>> pp=double(p)
Error using symengine
Unable to convert expression into double array.
Error in sym/double (line 698)
Xstr = mupadmex('symobj::double', S.s, 0);

Réponse acceptée

Steven Lord
Steven Lord le 6 Sep 2020
You're not specifying a value for t. If any of the elements of the vector p returns when evaluated are functions of t, there's no way to convert that component to a double without specifying a value for t.
syms t p(t)
p(t) = sin(t);
double(p) % this will throw the error you received
p(pi/2) % this will return a sym that contains no symbolic variable
double(p(pi/2)) % this will return a double
  4 commentaires
a a
a a le 6 Sep 2020
Great answer! Thanks a lot!
Steven Lord
Steven Lord le 6 Sep 2020
One other thing I forgot to mention. If you're working symbolically because sin(pi) is not exactly 0 and you need to avoid the roundoff error:
>> y = sin(pi)
y =
1.22464679914735e-16
you may be able to avoid the need to work symbolically by using the sinpi function.
>> y = sinpi(1) % sin(1*pi)
y =
0

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