I now have a test showing that I can write this either way, and also take
the derivative either way. But, it is still irritating not to see q2F in
the list of global variables in the debugger. Maybe there is a bug in the
debugger. After all I can assign q2F(t) to the variable C, which wouldn't
work if it weren't defined. Here is the code:
global t, q1F, q2F
t = symbols('t')
q1F = Function('q1F')(t)
q2F = Function('q2F')
#A = q1F(t) # 'q1F' object is not callable
A = q1F
print(A) # q1F(t)
B = diff(A, t)
print(B) # Derivative(q1F(t), t)
C = q2F(t)
print(C) # q2F(t)
D = diff(C, t)
print(D) # Derivative(q2F(t), t)
I am using Python inside of Visual Studio and looking for the global
variables in the Locals window (with {}Globals expanded). Could someone try
this in a different IDE?
On Friday, March 10, 2023 at 5:31:57 PM UTC+1 Thomas Ligon wrote:
> 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 asme...@gmail.com 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 <thomas...@gmail.com> 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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