It looks like srepr() doesn’t capture the real=True parameter to Symbol.
from sympy import *
s=Symbol('s')
f=Symbol('f',real=True)
SExp=Eq(s,2*pi*I*f)
srepr(SExp)
—> "Equality(Symbol('s'), Mul(Integer(2), I, pi, Symbol('f')))"
Is this another issue to be submitted?
In the mean time, I think it’s just going to be most straight forward if I
convert the iPython notebook example to .py, and attach it. Will that work for
these other issues?
Thanks—
Greg
> On Jul 16, 2015, at 13:32 , Aaron Meurer <[email protected]> wrote:
>
> Oh I didn't realize you had LaTeX strings in your Symbol names. It would be
> more useful to use srepr() then, so that it can just be copy-pasted.
>
> Aaron Meurer
>
> On Thu, Jul 16, 2015 at 12:33 PM, G.B. <[email protected]
> <mailto:[email protected]>> wrote:
> Will do. If I print the expression using str(), I get:
>
>
> 'I_{R_2}(s) == -C_{T}*s*(-R_{1}*V_{in}(s)/(R_{1} + 1/(1/(1/(C_{T}*s +
> 1/(L_{R}*s + R_{2} + 1/(C_{R}*s))) + 1/(C_{L}*s)) + 1/(L_{L}*s))) + V_{in}(s)
> - (-R_{1}*V_{in}(s)/(R_{1} + 1/(1/(1/(C_{T}*s + 1/(L_{R}*s + R_{2} +
> 1/(C_{R}*s))) + 1/(C_{L}*s)) + 1/(L_{L}*s))) +
> V_{in}(s))/(C_{L}*s*(1/(C_{T}*s + 1/(C_{R} + L_{R} + R_{2})) + 1/(C_{L}*s))))
> - (-R_{1}*V_{in}(s)/(R_{1} + 1/(1/(1/(C_{T}*s + 1/(L_{R}*s + R_{2} +
> 1/(C_{R}*s))) + 1/(C_{L}*s)) + 1/(L_{L}*s))) + V_{in}(s))/(1/(C_{T}*s +
> 1/(C_{R} + L_{R} + R_{2})) + 1/(C_{L}*s))’
>
> Is that usable or do I need to provide another means? Is that IPython doc
> preferable (or the same in straight Python form)?
>
> Thanks—
> Greg
>
>
>
>> On Jul 14, 2015, at 18:05 , Aaron Meurer <[email protected]
>> <mailto:[email protected]>> wrote:
>>
>> I think there is a bug. If you take the two expressions, subtract them, and
>> call simplify(), you get a result that isn't 0. equals() also says they are
>> different.
>>
>> simplify() returning False is also a bug.
>>
>> Can you open two issues for these things? Just put some minimal code to
>> create the expressions in the issue (i.e., just print the expressions using
>> str()).
>>
>> Aaron Meurer
>>
>> On Tue, Jul 14, 2015 at 12:51 PM, G.B. <[email protected]
>> <mailto:[email protected]>> wrote:
>> Hey Aaron—
>>
>> Maybe the best way to capture an example is an IPython notebook?
>>
>> Let me know if this doesn’t work properly…
>>
>> Cheers—
>>
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>>
>> > On Jul 13, 2015, at 10:25 , Aaron Meurer <[email protected]
>> > <mailto:[email protected]>> wrote:
>> >
>> > On Sun, Jul 12, 2015 at 7:46 PM, G B <[email protected]
>> > <mailto:[email protected]>> wrote:
>> >> What would case Eq.simplify() to return False? There isn't a lot of
>> >> computation time before the result.
>> >
>> > Can you show an example of what is doing this?
>> >
>> >>
>> >> Also why would using together() before substituting an expression lead to
>> >> a
>> >> numerically different result?
>> >
>> > How different? It's possible you are changing the numerical stability
>> > of the expression. Or it's possible there is a bug in together.
>> >
>> > Aaron Meurer
>> >
>> >>
>> >> I have a series of equations that I've been substituting in to each other
>> >> building up a final symbolic result. My first pass at this gave a result
>> >> that looks reasonable (though I haven't proven it correct) when I convert
>> >> it
>> >> to a numpy function after using
>> >> Eq.subs(dictionary of values for symbols).simplify().n()
>> >> and plot it.
>> >>
>> >> When I took the last symbolic equation to be substituted and use
>> >> Eq.together() on it before substituting to get a simpler result, the
>> >> Eq.subs().simplify() returns False. When I manipulate it differently to
>> >> get
>> >> it to successfully lambdify, the resulting plot is different (and appears
>> >> wrong).
>> >>
>> >> Thanks--
>> >>
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