Hi Aaron !
I was thinking in context with the following problem,where I was looking
into the pattern matching. I found this :
>>> pattern = (a+b*sin(c+d*x))**n
>>> term = (1 + 3 * sin(5 + 4 * y))** 10
>>> pattern.matches(term)
{a_: 1, n_: 10, d_: 4, c_: 5, b_: 3, x_: y}
>>> pattern = (a*sin(b+c*x))**n
>>> term = (3*sin(5+4*y))**3
>>> pattern.matches(term)
>>>
Same goes, with :
>>> pattern2 = (a+ b*log(x))**n
>>> term2 = (3 + 2*log(x))**2
>>> pattern2.matches(term2)
{a_: 3, n_: 2, b_: 2}
But,
>>> pattern3 = (b*log(x))**n
>>> term3 = (3*log(x))**2
>>>
This is due to the fact,that the term, (3*log(x))**2 is now 9*log(x_)**2.
(Is it a bug? or an improper implementation of pattern matching? If, yes,
how can this be tackled?)
Due to this, I think special cases have to be considered(while implementing
integration rules) where such case is occurring, during pattern matching.
Please let me know your views on it.
Thanks,
Ankit
On Wednesday, March 8, 2017 at 11:24:21 PM UTC-5, Aaron Meurer wrote:
>
> You can use Pow(10*x, 2, evaluate=False).
>
> Aaron Meurer
>
> On Wed, Mar 8, 2017 at 11:06 PM, Ankit <[email protected]
> <javascript:>> wrote:
> > Hello ! I have this question !
> >
> > Python automatically evaluates 10**2 to 100 or say (10*10) to 100 .
> >
> > Same is the case where our expression is , for example: (2*log(x))**2,
> it
> > will become 4*log(x_)**2
> >
> > or
> >
> > (10*x)**2 becomes 100*x_**2 (Also, how is x different
> from
> > x_ ?)
> >
> > Is there anyway,we can stop this evaluation and write these expressions
> in
> > non-evaluated form.
> >
> > What I thought of was to take them as Strings, and use Evaluate= False,
> with
> > sympify . But, if this expression is written in Sympy, like expr=
> (10*x)**2,
> > it will automatically get evaluated and expr = 100*x_**2.
> >
> > Please let me know the possible solution !
> >
> > Thanks,
> > Ankit
> >
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