I am now using Symbol and Function where the name and the value match. That
was recommended for easier reading and then required for pickle. In order
to avoid confusing these, I have a different name for Symbol and for
Function, then I convert them just before printing. My problem is that I am
not sure how to use this when I take derivatives. When I look at q1S in the
debugger I see it has type Symbol, but q1F has value q1F(t) and type Q1F
(not Function).
The code looks like this:
def createSymbols():
global t, q1S, q2S, q1F, q2F, q1dotS, q2dotS, q1ddotS, q2ddotS
t = symbols('t')
q1S, q2S = symbols('q1S, q2S')
q1dotS, q2dotS = symbols('q1dotS, q2dotS')
q1ddotS, q2ddotS = symbols('q1ddotS, q2ddotS')
q1F = Function('q1F')(t)
q2F = Function('q2F')(t)
print('end of createSymbols HillCusp')
def symForLatex(expIn):
expOut = expIn
expOut = expOut.subs(q1S, symbols('q1'))
expOut = expOut.subs(q2S, symbols('q2'))
expOut = expOut.subs(q1F, symbols('q1'))
expOut = expOut.subs(q2F, symbols('q2'))
expOut = expOut.subs(q1dotS, symbols('{\dot{q}}_{1}'))
expOut = expOut.subs(q2dotS, symbols('{\dot{q}}_{2}'))
expOut = expOut.subs(q1ddotS, symbols('{\ddot{q}}_{1}'))
expOut = expOut.subs(q2ddotS, symbols('{\ddot{q}}_{2}'))
return expOut
Here is the code that calculates higher-order derivatives. The second-order
derivative is defined by Newton's equations (ODEs).
if qi == 1:
if nDer == 0:
retVal = q1F#(t)
elif nDer == 1:
retVal = diff(qd(1, 0), t)
elif nDer == 2:
retVal = 2*qd(2, 1) + 3*qd(1, 0) - qd(1, 0)*(qd(1, 0)**2 +
qd(2, 0)**2)**Rational(-3,2)
else:
retVal = diff(qd(1, nDer-1), t)
# replace q1dd & q2dd
#retVal = retVal.subs(Derivative(q1F(t), (t, 2)), qd(1, 2))
#retVal = retVal.subs(Derivative(q2F(t), (t, 2)), qd(2, 2))
retVal = retVal.subs(Derivative(q1F, (t, 2)), qd(1, 2))
retVal = retVal.subs(Derivative(q2F, (t, 2)), qd(2, 2))
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