Sorry for bothering you with this. I have a serious issue and now im on clock to solve it, so here is my question.
I have an issue where I lambdify a quantity, but the result of the quantity differs from the ".subs" result, and sometimes it's way off, or it's a NaN, where in reality there is a real number (found by subs)
Here, I have a small MWE where you can see the issue! Thanks in advance for ur time
import sympy as sy
import numpy as np
##STACK
#some quantities needed before u see the problem
r = sy.Symbol('r', real=True)
th = sy.Symbol('th', real=True)
e_c = 1e51
lf0 = 100
A = 1.6726e-24
#here are some quantities I define to go the problem
lfac = lf0 2
rd = 4*3.14/4/sy.pi/A/lfac**2
xi = r/rd #rescaled r
#now to the problem:
#QUANTITY
lfxi = xi**(-3)*(lfac 1)/2*(sy.sqrt( 1 4*lfac/(lfac 1)*xi**(3) (2*xi**(3)/(lfac 1))**2) -1)
#RESULT WITH SUBS
print(lfxi.subs({th:1.00,r:1.00}).evalf())
#RESULT WITH LAMBDIFY
lfxi_l = sy.lambdify((r,th),lfxi)
lfxi_l(0.01,1.00)
##gives 0
CodePudding user response:
The issue is that your mpmath precision needs to be set higher!
By default mpmath uses prec=53
and dps=15
, but your expression requires a much higher resolution than this for it
# print(lfxi)
3.0256512324559e 62*(sqrt(1.09235114769539e-125*pi**6*r**6 6.74235013645028e-61*pi**3*r**3 1) - 1)/(pi**3*r**3)
...
from mpmath import mp
lfxi_l = sy.lambdify((r,th),lfxi, modules=["mpmath"])
mp.dps = 125
print(lfxi_l(1.00,1.00))
# 101.999... result
CodePudding user response:
Changing a couple of the constants to "modest" values:
In [89]: e_c=1; A=1
The different methods produce essentially the same thing:
In [91]: lfxi.subs({th:1.00,r:1.00}).evalf()
Out[91]: 1.00000000461176
In [92]: lfxi_l = sy.lambdify((r,th),lfxi)
In [93]: lfxi_l(1.0,1.00)
Out[93]: 1.000000004611762
In [94]: lfxi_m = sy.lambdify((r,th),lfxi, modules=["mpmath"])
In [95]: lfxi_m(1.0,1.00)
Out[95]: mpf('1.0000000046117619')