Re: [Reduce-algebra-developers] taylor and sqrt of full squares

[Alan B. <ala...@gm...>]

I note that using alternative power-series package tps, one gets:

1: ps(sqrt((x-y)^2), z,0);

|x - y|

which is an improvement, but of course this is only correct if x-y is 
real. However, after

on precise_complex;

both power-series packages produce (sqrt(x^2-2x*y+y^2) as the first term 
which is correct, but somewhat messy if x-y is real .

I think main problem lies with the simplication of sqrt((x-y)^2) not 
with the power series packages themselves, although at first sight the 
differnce in the behaviours of ps and taylor is somewhat puzzling.

Note by default

sqrt((x-y)^2);

produces |x - y| but with precise_complex ON one gets (sqrt(x^2-2x*y+y^2).

One might think that the declaration realvalued x,y; should cure the 
problem, but with precise_complex ON one still gets (sqrt(x^2-2x*y+y^2).

Regards Alan

On 17/03/2025 03:48, Andrey Ignatenko via Reduce-algebra-developers wrote:
> Hello,
>
> It appears that square root of full square of an expression
> under the taylor operator is replaced by the expression and
> not its absolute value as it should,
>
> 1: taylor(sqrt((x-y)^2), s, 0, 1);
>            2
> x - y + O(s )
>
> Is it a bug or a feature?
>
> In deciding which sign of the expression to choose taylor seems
> to rely on internal order of kernels
>
> 2: korder y, x;
> 3: taylor(sqrt((x-y)^2), s, 0, 1);
>               2
>  - x + y + O(s )
>
> But what should I do in the following case if I have (1-z)>0,
>
> 4: taylor(sqrt((1-z)^2), s, 0, 1);
>            2
> z - 1 + O(s )
>
> Is it possible to tell Reduce to put a kernel z after a
> literal constants 1? Is there a workaround?
>
> Thanks in advance.
>
> Best regards,
> Andrey
>
>
>
>
>
>
>
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