>
> ... This model has the advantage that (sqrt(1+t)^2 -1)/t == 1 returns
> true, as one would expect mathematically. (insisting equality up to the
> default precision would have to lead to false, because the arithmetic on
> the LHS leads to precision loss)
>
Your example is convincing and I'
On Wednesday, January 15, 2020 at 12:45:10 PM UTC-8, Sebastian Oehms wrote:
>
> My expectation is that two power series are considered to be equal if
> there polynomials agree, but also their individual precisions as long as
> one of them is below the default precision of the ring. Accordingly
>
> For example, suppose you want to do some long computation and determine
> whether you get 1 at the end. At the end of your computation, you find the
> answer 1 + O(t^33). I think it's more useful to say that this is equal to
> 1 than that it's not.
>
Sorry, that I was not clear enought a
On Wednesday, January 15, 2020 at 7:09:29 AM UTC-8, David Roe wrote:
>
>
> I'm not sure what kind of equality you would imagine, but if we defined
> equality by something like "equal absolute precision and equal value modulo
> that absolute precision" then you no longer have additive and
> multi
We only store power series and p-adics to finite precision, so it's
impossible to tell whether the mathematical objects underlying two elements
in Sage are actually equal. In practice, the goal of a p-adic or power
series computation is to determine some result approximately, since that's
all that
Dear all,
obviously Sage does not distinguish between precisions of power series:
sage: R. = PowerSeriesRing(ZZ)
sage: O(t).is_zero()
True
sage: O(t) == O(t**2)
True
Similarly for p-adics:
sage: O(3).is_zero()
True
sage: O(3) == O(3^2)
True
sage:
This seems to be explicitly intended:
def
