Oh, a solution is simple.
I got
(35) -> coefficients l0
(35) [0, 0, 0, 0]
So
(36) -> zero? removeZeroes l0
(36) true
Unfortunately, UnivariateTaylorSeries has no "removeZeroes". Maybe it
should be added?
(54) -> tx := x :: T
(54) x
Type: UnivariateTaylorSeries(Fraction(Integer),x,0)
(55) -> t := truncate(sin tx, 7)
1 3 1 5 1 7
(55) x - - x + --- x - ---- x
6 120 5040
Type: UnivariateTaylorSeries(Fraction(Integer),x,0)
(56) -> tt := t-t
(56) 0
Type: UnivariateTaylorSeries(Fraction(Integer),x,0)
(57) -> coefficients tt
(57) [0, 0, 0, 0, 0, 0, 0, 0]
Type: Stream(Fraction(Integer))
(58) -> zero?(tt)
(58) false
Type: Boolean
The following can, of course, be explained, but it looks strange to me.
(63) -> (tt =0)@Boolean
(63) true
And if I truncate at 20, then (tt =0)@Boolean also returns false.
Nevertheless, I don't think that it is a good idea that zero? and x=0
behave differently.
Ralf
On 10/21/20 9:51 PM, Ralf Hemmecke wrote:
> It is clear that in general there is no decidable zero test for power
> series, however, we have the function truncate: (%, INT) -> % that
> basically creates a finite series. I hoped that it were possible to use
> UnivariateLaurentSeries together with this truncate function to
> essentially work with Laurent polynomials. I ran into a problem that I
> did not expect.
>
> With the attached file I get.
>
> (9) -> zero? l0
>
> (9) false
>
> (13) -> zero? t0
>
> (13) false
>
> In fact, I consider this a bug. It is clearly possible to decide that l0
> and t0 are 0 since their corresponding stream should be finite and at
> most contain zeros.
>
> Waldek, I can look into it if you also want the above return true.
>
> Ralf
>
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