Why would using together() change the solution and why would simplify() return False?

[G B]

What would case Eq.simplify() to return False?  There isn't a lot of computation time before the result.

Also why would using together() before substituting an expression lead to a numerically different result?

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--

[Aaron Meurer]

On Sun, Jul 12, 2015 at 7:46 PM, G B <g.c.b....@gmail.com> 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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[G.B.]

Hey Aaron—

Maybe the best way to capture an example is an IPython notebook?

Let me know if this doesn’t work properly…

Cheers—

[Aaron Meurer]

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

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[G.B.]

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

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[Aaron Meurer]

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

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[G.B.]

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

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[Aaron Meurer]

This is already fixed in master:

In [1]: srepr(Symbol('x', real=True))

Out[1]: "Symbol('x', real=True)"

Aaron Meurer

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