[GiNaC-list] series((x+x^2)^2,x,0) is broken

[Vitaly Magerya]

On 24/06/2023, Richard B. Kreckel <kreckel at in.terlu.de> wrote:
>> OK, sorry, that's not corect. Let me say it differently:
>>
>>      series(sin(x)-x, x, 2) == Order(x^2)
>>
>> This does not mean
>>
>>      1/series(sin(x)-x, x, 2) == Order(x^-2),
>>
>> because
>>
>>      series(1/(sin(x)-x), x, -2) == -6/x^3 + Order(x^-2)
>
> But sin(x)-x starts with (-1/6)*x^3 which is of Order(x^3).

Yes, but series(sin(x)-x,x,2) reports Order(x^2), it does not
try figure out the actual first non-zero term.

If we take Order(...) to be equivalent to the big-O notation, this
makes perfect sense, because f(x)=O(g(x)) only means that f(x)
is asymptotically bounded from above by g(x), not necceserily
from below, so x^3 is both O(x^3) and O(x^2).

In general it would not be practical to require Order(x^n) to
mean the big-Omega (i.e. bounded from above and from below),
because after the truncation of the series happened, it is no
longer possible to figure out if the next term is zero or not
without looking at the original expression, so for example while

    (x + bigO(x^2)) + x^2 = x + bigO(x^2)

the same can not be said about big-Omega,

    (x + bigOmega(x^2)) + x^2 =/= x + bigOmega(x^2)

because if the original series that generated "x + bigOmega(x^2)"
was actually "x - x^2 + x^7 + ...", then

    (x + bigOmega(x^2)) + x^2 = x + bigOmega(x^7),

but that's not possible to know just by looking at "x + bigOmega(x^2)".

So, Order(x^n) must be bigO(x^n), and it should not mean the next
non-zero term in the series is x^n, it can be x^(n+1), or higher.
This is why

    1/(Order(x^n))

is not

    Order(x^-n),

it can be

    Order(x^(-n-1)),

or lower.

On 24/06/2023, Richard B. Kreckel <kreckel at in.terlu.de> wrote:
> On 6/24/23 18:15, Vitaly Magerya wrote:
>>> If a series A starts with x^n, then the series 1/A *ends*, not starts,
>>> with 1/x^n.
>>
>> OK, sorry, that's not corect. Let me say it differently:
>>
>>      series(sin(x)-x, x, 2) == Order(x^2)
>>
>> This does not mean
>>
>>      1/series(sin(x)-x, x, 2) == Order(x^-2),
>>
>> because
>>
>>      series(1/(sin(x)-x), x, -2) == -6/x^3 + Order(x^-2)
>
> But sin(x)-x starts with (-1/6)*x^3 which is of Order(x^3).
>
>    -richy.
>
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