A computer algebra system written in pure Python http://sympy.org/ . To get started to with contributing https://github.com/sympy/sympy/wiki/Introduction-to-contributing
smichr on master
point2D now will subclass Evalf… fixed failing tests test evalf for entities Also, … and 1 more (compare)
solve([x - 5], x)
simply return [5]
?
Hi everyone, can someone give me some recommendation to the next problem, I have a large expression lambdified, the problem is that has Nan results when I evaluate. I have found one term which causes this behavior and given by:
from sympy import init_printing,symbols,log,sqrt
init_printing()
m = symbols('m')
f = 0.5*log(1 - 3.13400648e+22/(1.56700324e+22*sqrt(1 - 1.5976993129893e-11*m)*sqrt(1.5976993129893e-11*m + 1) + 1.56700324e+22))
Then, when I substitute in $m=1e-3$
f.subs(m,1e-3)
Which give $\tilde{\infty}$. However, if I use evalf command I obtain
f.evalf(subs={m:1e-3})
-32.4607738174251 + 1.5707963267949*I
Then, I don't know what result is correct. Is this a problem of the precision of subs command?.
lambdify
, you are going to run into this same problem, because you are evaluating it naively. You will need to rewrite the expression somehow to avoid the massive cancellation.
log1p
.
@asmeurer Thanks for the suggestions and the explanation. My expression is too large because I obtained after some SymPy calculations and I do not how to rewrite the log function following your suggestions of multiply for a big number inside of log. I have tried to create a new log function like log_rewrite(x) = log(x*y) - log(y) with y a big number and after using lambdify command to the f function
from numpy import log as nplog
def log_rewritted(x):
y = 1e30
return nplog(x*y)-nplog(y)
fnp = lambdify([m],f,modules=[{'log':log_rewritted}])
However, I obtained the same result when I evaluate f,
fnp(1e-3)
-inf.
I found some unusual behaviour in randprime and I try to fix it.
from sympy import randprime
print(randprime(-2,0))
ValueError: no preceding primes
Can someone please review this PR : #20949