- #1

- 263

- 4

The expansion for ##\sin x## is just ##(x-\frac{x^3}{3!}+...)##

For ##\ln |1+x|## it will be ##(x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+...)##

Then with substitution

$$(x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4})-\frac{1}{6}(x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4})^3$$

From the last term with the power of 3, I can immediately see that there's ##-\frac{x^3}{6}## and I take that into account to get.

$$x-\frac{x^2}{2}+\frac{x^3}{6}-\frac{x^4}{4}+...$$

But actually there's ##\frac{x^4}{4}## hidden in the last term group which can only be found after expansion.

My question is, is there anyway to know that there's a term with a specific power without expanding them so that I can get the needed Maclaurin series quickly?