For 8 + 8/(1+p), apart() will give this form.
In general, it likely depends on the expression which function will do
this. You can use the count method to count how many times a variable
appears in an expression
In [65]: (8 + 8/(1+x)).count(x)
Out[65]: 1
In [66]: (8*(x + 3)/(x + 2)).count(x)
Out[66]: 2
You could use the measure option to simplify to tell it to minimize this:
In [67]: simplify(8 + 8/(1+x), measure=lambda i: i.count(x))
Out[67]:
8
8 + ─────
x + 1
You could also write your own function that tries various functions
and returns the result with the fewest occurrences of the variables.
Aaron Meurer
On Wed, May 18, 2016 at 5:49 PM, Timothy Wright <[email protected]> wrote:
> Hi
>
> I am using factor() to reduce the equation '8 + 8/(1+p)'
>
> I get 8*(p + 3)/(p + 2) because factor() must return a product. But I want
> a reduction that minimizes the number of each variable in the equation. In
> this case, my input referenced p once, running factor gives me a product
> that has two references to p.
>
> I need to generalize this to very long equations with 5 independent
> variables. I am replacing the use of mathematica with sympy, mathematica
> seems to be able to reduce the equation in this way
>
> I have looked at cancel(), simplify(), apart(), expand(), all don't do what
> I want.
>
> After reviewing the email history, I don't think this can be done out of the
> box. But there could be a way to write my own reducer to do this?
>
> Is this correct? I want to check if I am on the right path.
>
> Thanks
>
> Tim
>
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