On 29 Nov., 23:42, smichr <[email protected]> wrote:
> In the calculation of roots_binomial a value of zeta is being
> calculated using Euler's formulation and expanded(complex=True). Is
> there a reason to do the expanding and is there a reason not to use
> the more compact "roots of -1" formulation?
>
> Also, after computing the root an expand(power_base=False) is done
>
> If the (-1) formulation and no expanding is done, a much more compact
> (and faster) result is attained. Is that expanding necessary?
>
> compare
>
>     h[2] >>> solve((x-1)**3 - 3)
>     [1 - (-3)**(1/3), 1 + I*(-3)**(1/3)*3**(1/2)/2 + (-3)**(1/3)/2,
>      1 - I*(-3)**(1/3)*3**(1/2)/2 + (-3)**(1/3)/2]
>
> to the unexpanded Euler representation:
>
>     [1 + 3**(1/3), 1 + 3**(1/3)*exp(2*pi*I/3), 1 + 3**(1/3)*exp(4*pi*I/
> 3)]
>
> or the "roots of -1" representation
>
>     [1 + 3**(1/3), 1 + (-1)**(2/3)*3**(1/3), 1 - (-3)**(1/3)]

I really like the "roots of -1" representation, because it is so
compact. Anyonre can expand it if he wants.

But, aren't there multiple possibilities for a complex root of -1?

Vinzent

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