Taylor expansion with $o(x^{2n})$ or $o(x^{2n+1})$

Question

I notice some books worte taylor expansion for Trigonometric and Hyperbolic functions with $o(x^{2n})$ or $o(x^{2n+1})$

for example

$$\cos \,x=\sum _{k=0}^{n }{(-1)^k\frac {x^{2k}}{(2\,k)!}}+o(x^{2n})$$

and others

$$\cos \,x=\sum _{k=0}^{n }{(-1)^k\frac {x^{2k}}{(2\,k)!}}+o(x^{2n+1})$$

Could someone explain to me why

Answer 1

Both are true, though the second formula says a little more. The next term in the Taylor expansion is $(-1)^{n+1}x^{2n+2}/(2n+2)!$, and so we see that the Taylor remainder term is $O(x^{2n+2})$ as $x\to 0$. But if the remainder term is $O(x^{2n+2})$ as $x\to 0$, then it is also $o(x^{2n+1})$ and $o(x^{2n})$.

Similar reasoning holds for the other asymptotic expressions: they are consequences of the big-O asymptotic expression for each Taylor approximation.