On 24/06/2023, Richard B. Kreckel <kreckel@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@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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