Thank you very much for your reply, i learned a lot
My initial intention when asking the prior question was to define a
function where some evaluations would return itself, while other
evaluations would not. Such as
```
def f(x):
if x%2==0:
return hold(f(x))
else:
return x^2+1
```
In fact, I am trying to define the total differential in the following
way, this is my final intention.
```
def Dt(expression1):
if len(expression1.args()) == 1 \
and expression1.args()[0]==expression1:
# if the expression has only one variable
# and that variable is the entire expression
print("a")
with hold:
held = Dt(SR(expression1))
return held
else:
return sum(map(lambda
arg:diff(expression1,arg)*Dt(arg),expression1.args()))
```
How can I define this as a symbolic function? Perhaps multiple dispatch?
On 25/08/2024 23:33, Nils Bruin wrote:
Dear erentar,
There are two types of objects involved here: "symbolic functions"
that live as objects in SR and "python functions" which are part of
python itself. "hold" is only a directive that applies to symbolic
expressions, whereas "def" defines a python function. Hence, the
"hold" directive never applies.
Looking into what you are trying to compute, it would seem you want to
replace g(x) by g(SR(x)). Replacing that another time would get you
g(SR(SR(x))). However, no matter what x is, if SR(x) succeeds
properly, then y=SR(x) will be an element of the symbolic ring and
applying SR to an element of the symbolic ring should return exactly
the same thing, so SR(y=SR(SR(x))=SR(x).
But then g(SR(SR(SR(x))))=g(SR(x)) and g(SR(SR(SR(SR(x)))))=g(SR(x)) .
So by the properties of the system, applying SR repeatedly is not
going to make a difference.
Is it that you want to define a symbolic function `g` *and* a
convenience python function that forces the argument into SR? I don't
think you need that, since symbolic functions will try to put their
argument into SR when called (because SR arguments are the only thing
they can work on. Hence, if I'm right in guessing what you want, the
following would already do the job:
sage: g = function('g') ## on top-level you could also just use
`function('g')` for convenience
sage: a=10
sage: parent(a) #a is not an object in SR
Integer Ring
sage: E = g(a); E
g(10)
sage: E.operands()[0]
10
sage: parent(E.operands()[0]) #but the argument in the expression E is
Symbolic Ring
Note that g is not a python function:
sage: type(g)
<class
'sage.symbolic.function_factory.function_factory.<locals>.NewSymbolicFunction'>
(quite a mouth-full, but notably different from the type of an object
that results from a "def" statement)
On Sunday 25 August 2024 at 13:17:16 UTC-7 erentar wrote:
Greetings,
I am trying to return a function call from the function itself
with the
context `hold`
The following example does not behave as i'd want:
```
def g(x):
with hold:
held = sqrt(x)
return held
g(4)
> 2
```
To get around this, i make the change SR(x):
```
def g(x):
with hold:
held = sqrt(SR(x))
return held
g(4)
> sqrt(4)
```
This is what i want.
Now, i want to return the function call `g(4)` rather than `sqrt(4)`.
This is what i've tried:
```
def g(x):
with hold:
held = g(SR(x))
return held
```
Sadly it reaches recursion limit and exits, which means it does
not hold
the function call.
How can i return the function call from the function itself?
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