substituting into a Sage/Python function

[john_perry_usm]

Hi!

When demonstrating stuff in class, I often want the variable to be an argument to a function:

def test_subs(f, a, x=x):
  print f(a)

This works if f is a function. If f is not a function, Sage issues a DeprecationWarning (rightly, IMHO). So we could try this:

def test_subs(f, a, x=x):
  print f(x=a)

...which sort of works, but not with a different variable:

sage: f = t^2 + 2
sage: test_subs(f, 1, t)
t^2 + 2

Another way to make it work with expressions is by dictionary substitution:

def test_subs(f, a, x=x):
  print f({x:a})

...and that now works with expressions, but not with functions:

sage: f(t) = t^2 + 2
sage: test_subs(f, 1, t)
...
TypeError: no canonical coercion from <type 'dict'> to Callable function ring with argument t

I can fix this with a try/except block, but when discussing it with a much wiser colleague he pointed out that it made the learning curve rather steep for kids who can barely do calculus.

Is there a smarter way to do substitution, so that a function doesn't have to worry about whether the input is an expression or an honest-to-goodness function?

[Ralf Stephan]

On Tuesday, June 14, 2016 at 10:07:30 PM UTC+2, john_perry_usm wrote:

sage: f(t) = t^2 + 2
sage: test_subs(f, 1, t)
...
TypeError: no canonical coercion from <type 'dict'> to Callable function ring with argument t

I am not sure why you didn't try:

sage: test_subs(f(t),1,t)
3

Of course it cannot work if you don't give a dependent var. 

[john_perry_usm]

I am not sure why you didn't try:

sage: test_subs(f(t),1,t)
3

I know that works! Part of the point is to demonstrate that we can't always count on a parameter having the type we assume. If the answer is simply, "Use functions. Period." then I would pass that advice on.

[john_perry_usm]

Actually, your answer stimulated a much better solution (IMHO).

def test_subs(f, a, x=x):
    f(x) = f
    print f(a)

Thanks for the stimulating thought.