Thanks, but this just shows how poorly I formulated my question. After we
solved some printing problems, I am very happy with the way it is working.
My problem is Function. With
q1F = Function('q1F')(t)
I get: q1F has value q1F(t) and type Q1F. With
q1F = Function('q1F')
I get nothing, even though this was preceded by
global q1F
I was previously using
q1F = Function('q1F')
and I think it was creating a global q1F with type Function, but I must
have broken something.
I am defining both q1S and q1F because I need a symbol for q1 for most
calculations, and I also need a function for q1 in order to take
derivatives.
if nDer == 0:
retVal = q1F#(t)
returns q1F with no derivative (nDer = number of derivatives), and
elif nDer == 1:
retVal = diff(qd(1, 0), t)
takes the first derivative of the above. (qd(qi=1, nDer) returns the
nDer-th derivative of qi=1, i.e. q1).
On Thursday, March 9, 2023 at 8:59:19 PM UTC+1 [email protected] wrote:
> If you want your code to match what is printed, you should write
>
> f = Function('f')
>
> instead of
>
> f = Function('f')(t)
>
> Alternatively you could use a special printer that prints things as
> you want. There is already a printer in the mechanics module that
> ignores the (t) part when printing a function (mprint()).
>
> Aaron Meurer
>
> On Thu, Mar 9, 2023 at 8:12 AM Thomas Ligon <[email protected]> wrote:
> >
> > 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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>
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