Consider two functions of SymPy symbols e
and i
:
from sympy import Symbol, expand, Order
i = Symbol('i')
e = Symbol('e')
f = (i**3 + i**2 + i + 1)
g = (e**3 + e**2 + e + 1)
z = expand(f*g)
This will produce
z = e**3*i**3 + e**3*i**2 + e**3*i + e**3 + e**2*i**3 + e**2*i**2 + e**2*i + e**2 + e*i**3 + e*i**2 + e*i + e + i**3 + i**2 + i + 1
However, assume that e
and i
are both small and we can neglect both terms that are order three or higher. Using Sympy’s series tool or simply adding an O-notation Order
class can handle this:
In : z = expand(f*g + Order(i**3) + Order(e**3))
Out: 1 + i + i**2 + e + e*i + e*i**2 + e**2 + e**2*i + e**2*i**2 + O(i**3) + O(e**3)
Looks great. However, I am still left with mixed terms e**2 * i**2
. Individual variables in these terms are less than the desired cut-off so SymPy keeps them. However, mathematically small²·small² = small⁴. Likewise, e·i² = small·small² = small³.
At least for my purposes, I want these mixed terms dropped. Adding a mixed Order
does not produce the desired result (it seems to ignore the first two orders).
In : expand(f*g + Order(i**3) + Order(e**3) + Order((i**2)*(e**2)))
Out: 1 + i + i**2 + i**3 + e + e*i + e*i**2 + e*i**3 + e**2 + e**2*i + e**3 + e**3*i + O(e**2*i**2, e, i)
Question: Does SymPy have an easy system to quickly remove the n-th order terms, as well as terms that are (e^a)·(i^b) where a+b > n?
Messy Solution: I have found a way to solve this, but it is messy and potentially not general.
z = expand(f*g + Order((e**2)*i) + Order(e*(i**2)))
zz = expand(z.removeO() + Order(e**3) + Order(i**3))
produces
zz = 1 + i + i**2 + e + e*i + e**2 + O(i**3) + O(e**3)
which is exactly what I want. So to specify my question: Is there a way to do this in one step that can be generalized to any n? Also, my solution loses the big-O notation that indicates mixed-terms were lost. This is not needed but would be nice.