> I think that current state of 'rootSimp' is deliberate and
> you are opening a can of worms. Currently 'rootSimp' is
> purely algebraic operation, which does not touch transcendental
> kernels. 'exp(2)' or 'a^(2*b)' are transcendental so logically
> 'rootSimp does not change them.
The differen
oldk1331 wrote:
>
> >From docstring of rootSimp:
>
> rootSimp : F -> F
> ++ rootSimp(f) transforms every radical of the form
> ++ \spad{(a * b^(q*n+r))^(1/n)} appearing in f into
> ++ \spad{b^q * (a * b^r)^(1/n)}.
>
> Since rootSimp only deals with the rad
On Sat, May 20, 2017 at 9:09 PM, Bill Page wrote:
> On 20 May 2017 at 03:12, oldk1331 wrote:
>> From docstring of rootSimp:
>>
>> rootSimp : F -> F
>> ++ rootSimp(f) transforms every radical of the form
>> ++ \spad{(a * b^(q*n+r))^(1/n)} appearing in f into
>>
On 20 May 2017 at 03:12, oldk1331 wrote:
> From docstring of rootSimp:
>
> rootSimp : F -> F
> ++ rootSimp(f) transforms every radical of the form
> ++ \spad{(a * b^(q*n+r))^(1/n)} appearing in f into
> ++ \spad{b^q * (a * b^r)^(1/n)}.
>
> Since rootSimp onl
>From docstring of rootSimp:
rootSimp : F -> F
++ rootSimp(f) transforms every radical of the form
++ \spad{(a * b^(q*n+r))^(1/n)} appearing in f into
++ \spad{b^q * (a * b^r)^(1/n)}.
Since rootSimp only deals with the radicand which is a monomial,
I think
