For variety, I switched from SageMath to Sympy since Sympy is a small more focused package. However, I’m struggling with it’s lack of seeming lack of equational reasoning. When you define an equation via Eq(lhs,rhs), it’s not a “first class” object – ie. you can’t divide it by 2 and expect both sides to be half as big, and you have to manually say “subs(titute) eq.lhs with eq.rhs” each time. Finally, I haven’t found good ways to say “substitute just this one occurrence of foo” – sympy wants to do them all. This is all totally anathema to how I think about maths – I’m used to pointing at some subexpression and saying “and use equation 2 to simplify that” etc.
To try out Sympy, I took up the challenge in Chapter 1 of Nahin’s Transients or Electrical Engineers book, in which he conjures up a second-order differential equation for a simple R/L/C ciruit and invites the reader to check he’s not lying. This was a decent workout, since 1) it involves first- and second-derivatives, 2) it involves plenty of equations and manipulating expressions. To be honest, I found this hard. I tried doing it on computer and got nowhere. I then did it by hand using paper and pen to figure out the steps. Then I went back to the computer and had to tackle several “how do I get the computer to do x” puzzles before I got anywhere.
In the end, I did indeed verify that Nahin is telling the truth (phew!) but it was hours and hour of headscratching work. It’s nice to have the computer verify I haven’t err’d in my derivation, but this was not “computer makes task easier” stuff – quite the opposite! I guess the nice bit comes later when I can get the computer to actually solve the differential equations for concrete cases. But this ‘abstract algebra manipulation’ stuff is painful.
In the end, the “proof” looks like..
from sympy import *
from sympy.abc import R,L,C,t
i = Function("i")(t)
i1 = Function("i1")(t)
i2 = Function("i2")(t)
v = Function("v")(t)
u = Function("u")(t)
# Chapter 1 of Transients For Electrical Engineers
# Equations which describe the given "R ser L par (R ser C)" circuit (ie. KVL/KCL)
eq1 = Eq(i, i1 + i2 )
eq2 = Eq(u, i*R + v)
eq3 = Eq(v, L * i1.diff())
eq4 = Eq(v, i2*R + (1/C) * Integral(i2,(t,0,t)))
# The book states a large differential equation with 'input' voltage u(t) + derivative on left,
# and current i(t) (+derivatives) on right
(u.diff(t,2) + (R/L) * u.diff() + 1/(L*C)*u)
# and challenges you to check the equality does actually follow from above 4 equations
# First step is easy; there's only one equation involving 'u' so let's apply that everywhere
_.subs( eq2.lhs, eq2.rhs).simplify()
# Now we have to be tactical; we have v and v' and v'' but we want to do different things to each
# I don't know how to tell sympy to "just substitutes the first instance, not all instances"
# So start by substituting for v'' (since that leaves v' and v alone for now) and use eq4 the capacitor eqn
_.subs(eq4.lhs.diff(t,2), eq4.rhs.diff(t,2)).simplify()
# And now we can tackle v' and then v - both use eq3 (the inductor one)
# This leaves us with just currents, albeit a mixture of i/i1/i2
_.subs(eq3.lhs.diff(t), eq3.rhs.diff(t)).simplify()
_.subs(eq3.lhs,eq3.rhs).expand().simplify()
# Now we want to group + collapse the i1's and i2's using eq1.
# But I don't know how to get sympy to "see" that we have lots of (i1+i2) patterns
# So as a hack, we can always subtract something then add something equal to it and preserve equality, right?
# So we use that to turn i1+i2 into i ...
(_ + diff(eq1.rewrite(Add),t)/C).expand()
# Getting there .. now we need to sweep up the twice-differentiated cases..
(_ + R*eq1.rewrite(Add).diff(t,2)).expand()
# w^5 - which was what we wanted
