I need to find the MacLaurin polynomial of the 8-th order for the function:
$$f(x)=\cos{(\log{(1+2x^2)})}$$
I tried some possible solutions but the math behind it was getting pretty lengthy and because of that I'm sure I'm not doing it right.
So my approach was to substitute the MacLaurin expansion of $\log{(1+2x^2)}$ into the one of $\cos{y}$. So the 8-th order of the $\cos{y}$ function would be:
$$\cos{y}=1-\frac{y^2}{2}+\frac{y^4}{4!}-\frac{y^6}{6!}+\frac{y^8}{8!}+o(y^8)$$
and
$$\log{(1+2x^2)}=2x^2-2x^4+\frac83x^6-4x^8+o(x^8)$$
Now, using the logarithm expansion above within the cosine one seems a little bit crazy since it would take too much time to actually find a solution, like how would I even evaluate in a 2 hour calculus 1 test where this is one of the 6 other problems:
$$\ldots+\frac{(2x^2-2x^4+\frac83x^6-4x^8+o(x^8))^8}{8} + o((2x^2-2x^4+\frac83x^6-4x^8+o(x^8))^8)$$
However I'm pretty sure that the key here is '8-th' order and that I should just ignore all Xs with a higher exponent of 8, thus, if I were able to anticipate which of these expressions would not return a lower exponent than 8 I could simplify the whole calculation. Before going any further so I'd like to understand if I got this part correct, and please correct me if I'm wrong:
Saying to find the MacLaurin expansion up to the n-th order of a composite function $f(g(x))$, using substitution means that I need to find the MacLaurin expansion of $g(x)$ up to the i-th order such that substituting it inside the MacLaurin expansion of $f(x)$ up to the j-th order returns a polynomial greater than $n$**
With that said, if that's correct, is there any trickery here I can use to simplify the steps to get to the solution?
The polynomial expansion solution should be:
$$1-2x^4+4x^6-\frac{20}{3}x^8+o(x^9)$$