I agree with that remark.

On 12 March 2012 07:18, Aaron Meurer <asmeu...@gmail.com> wrote:
> 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?
I agree with that remark.

>
> Aaron Meurer
>
> On Thu, Feb 9, 2012 at 1:58 PM, krastanov.ste...@gmail.com
> <krastanov.ste...@gmail.com> wrote:
>> On 9 February 2012 21:39, Aaron Meurer <asmeu...@gmail.com> 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, krastanov.ste...@gmail.com
>>> <krastanov.ste...@gmail.com> 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, krastanov.ste...@gmail.com
>>>> <krastanov.ste...@gmail.com> 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, krastanov.ste...@gmail.com
>>>>> <krastanov.ste...@gmail.com> 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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