On 9 February 2012 21:39, Aaron Meurer <[email protected]> wrote:
> I'm assuming you answered your own questions here with your symbolic
> nsolve.  Were there any issues that you didn't resolve?

Just one issue: the sympification of lambda x : x

Does it make sense for sympify(lambda x : x) to return an
implemented_function with an _imp = lambda x: x?

Otherwise the nsolve stuff is dealt with in the pull request for
symbolic nsolve.

>
> Aaron Meurer
>
> On Tue, Jan 10, 2012 at 9:34 AM, [email protected]
> <[email protected]> wrote:
>> The problem of "python functions evaluate immediately sympy functions don't"
>> is obvious when one tries to plot/lambdify something that contains both Expr
>> and some numerical routine. Try to imagine a way to plot
>> nsolve(..)*some_Expr for example.
>>
>> On 10 January 2012 17:30, [email protected]
>> <[email protected]> wrote:
>>>
>>> Or more likely it's better to use implemented_function that is imported
>>> from utilities.lambdify?
>>> It seems to me that implemented_function is quite important if one wants
>>> for example to have complicated numerical routines accessible as sympy
>>> expressions. In my opinion it's actually important enough to be mentioned in
>>> the tutorial/pitfalls. Because defining a function to be used in sympy is
>>> actually not as simple as defining a python function (python functions
>>> evaluate immediately, sympy functions do not).
>>>
>>> Another question that I need help with is what is Lambda used for. Is
>>> there something that Lambda does and implemented_function does not?
>>>
>>>
>>> On 10 January 2012 16:39, [email protected]
>>> <[email protected]> wrote:
>>>>
>>>> How can I represent an unevaluated call to nsolve as a sympy expression?
>>>> Is Lambda the best (and standard) solution?
>>>>
>>>> The expression I want to represent looks like nsolve(x-tanh(x-h),[x],0).
>>>> The free symbol is h. It's in the context of plotting phase transition
>>>> diagrams.
>>>
>>>
>>
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