Why does my Sympy code calculate the first order Taylor series approximation incorrectly?

Question

I have an expression like this

which is entered into Sympy like this (for the sake of a reproducible example in this question)

from sympy import *
expression = Add(Mul(Integer(-1), Float('0.9926375361451395', prec=2), Add(Mul(Float('0.33167082639756074', prec=2), Pow(Symbol('k1'), Float('-0.66666666666666674', prec=2)), Pow(Symbol('n1'), Float('0.66666666666666674', prec=2))), Mul(Float('0.97999999999999998', prec=2), exp(Mul(Integer(-1), Symbol('mu1'))))), Pow(Add(Mul(Float('0.97999999999999998', prec=2), Symbol('k1'), exp(Mul(Integer(-1), Symbol('mu1')))), Mul(Integer(-1), Symbol('k2')), Mul(Pow(Symbol('n1'), Float('0.66666666666666674', prec=2)), Pow(Mul(Symbol('k1'), exp(Mul(Integer(-1), Symbol('mu1')))), Float('0.33333333333333331', prec=2)))), Integer(-1))), Pow(Add(Mul(Float('0.97999999999999998', prec=2), Symbol('k0'), exp(Mul(Integer(-1), Symbol('mu0')))), Mul(Integer(-1), Symbol('k1')), Mul(Pow(Symbol('n0'), Float('0.66666666666666674', prec=2)), Pow(Mul(Symbol('k0'), exp(Mul(Integer(-1), Symbol('mu0')))), Float('0.33333333333333331', prec=2)))), Integer(-1)))

Eyeballing this expression, the first order Taylor approximation for any of the variables, eg k1 , around some nonzero value should be nonzero, but this code

x = symbol("x")
expression.series(k1, x0 = x, n = 1)

just returns 0 . This is a problem because I'm trying to (eventually) calculate a multivariate Taylor series approximation, in a similar vein to this answer , and if one of the series expansions mistakenly evaluates to zero, the whole thing breaks down.

Did I code something wrong, or is my basic calculus just THAT bad and this actually evaluates to zero? From the documentation on series , I'm fairly certain that I'm using it correctly.

Answer 1

I think this is a bug related to the way the addition operation handles Orders.

This bug applies only if you are calculating the zero order (n=1) of the Taylor series. To avoid it, you can do

next(expression.series(k1, x0=x, n=None))

which is equivalent to

expression.subs(k1, x0=x)

Here is a simple description of this bug:

from sympy import cos
from sympy.abc import x
cos(x).series(x, x0=1, n=2)

Results in

cos(1) - (x - 1)*sin(1) + O((x - 1)**2, (x, 1))

But

cos(x).series(x, x0=1, n=1)

Results in O(x - 1, (x, 1)) , instead of cos(1) + O(x - 1, (x, 1)) .

This bug results from a bug in Add

O(x).subs(x,x-1) + 1

Results in O(x - 1, (x, 1)) , instead of 1 + O(x - 1, (x, 1)) .