I'm not sure about that.  Doesn't that assume that a lambda expression
is going to return a floating point (or at least numeric) value?

Aaron Meurer

On Thu, Feb 9, 2012 at 1:58 PM, [email protected]
<[email protected]> wrote:
> 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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