You can do this by simply factoring the expression with keyword `deep`::
```
>>> var('a')
a
>>> factor(sqrt(((4*a**2 + 1)/(4*a**6 - 15*a**4 + 12*a**2 + 4))), deep=True)
1/Abs(a**2 - 2)
```
On Thursday, April 1, 2021 at 9:43:35 AM UTC-5 [email protected]
wrote:
> What is described above has worked well for me. But there is a further
> simplification step that I need help with.
>
> I have some long expressions containing terms contain terms which look
> like this example:
> sqrt(4*a**2 + 1)*sqrt(1/(4*a**6 - 15*a**4 + 12*a**2 + 4))
> How can I instruct sympy to combine such square roots and factor the
> arguments? In this example that would lead to:
>
> sqrt(factor((4*a**2 + 1)/(4*a**6 - 15*a**4 + 12*a**2 + 4)))
> =
> 1/Abs(a**2 - 2)
>
> On Wednesday, March 31, 2021 at 9:03:08 AM UTC+2 B A wrote:
>
>> Dear Chris,
>>
>> On 31.03.21 05:48, Chris Smith wrote:
>> > Oscar posted code at issue https://github.com/sympy/sympy/issues/19164
>> > for a interva-based Newton solver.
>>
>> Thank you, that's very useful. I didn't know about interval arithmetic.
>>
>> I just implemented the following, which works very well and helps to
>> increase my confidence that problems will be caught:
>>
>> d_abs={}
>> d_problems={}
>>
>> def MyAbs(x):
>> # check if this argument is already in dictionary
>> if x in d_abs:
>> return d_abs[x]
>> # see if there are any roots nearby
>> soln_list = nsolve_interval(x, 0.563, 0.583)
>> # nearby roots provide a warning message and get saved
>> if len(soln_list) > 0:
>> print('WARNING: ambiguous case found, argument of Abs() is', x)
>> d_problems[x]=soln_list
>> # Check sign, determine correct output
>> x1 = x.evalf(subs={a:0.573})
>> if x1 < 0.0:
>> out = -x
>> else:
>> out = x
>> # save into dictionary and return
>> d_abs[x] = out
>> return out
>>
>> Cheers,
>> Bruce
>>
>
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