Find the polynomial of minimal degree such that
Using the Taylor expansion of we know as
we have
(This is where we see how nice -notation can be. All of the terms in the polynomials larger than
will get absorbed into the
. This simplifies computations tremendously when we don’t care about the higher order terms.) Therefore, we have
Why does the expression inside the little o not change when you replace x in the sin expansion? Is it because the new sin expression approaches 0 as x approaches 0 both before and after the substituion?
The expression inside little o can be arbitrary and in this case the problem statement implied that we want to have a solution specifically with o(x). It could also be o(x-x^2) instead [and then we could stop the calculations at the first line as it’s clearly would be o((x-x^2)^6)].
This is also my problem with this approach. I think this topic is poorly explained in Apostol’s book and little o can mean a lot of things at the same time. Rant aside, I got the same answer by differentiating sin(x-x^2) six times (using maxima) and evaluating the Taylor polynomial of degree 6 which gives us o(x^6) without problems.