Finding the needed power without expanding

1. Oct 11, 2013

Seydlitz

Say I need to find Maclaurin Series of $\sin(\ln |1+x|)$ until I reach the forth power.

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?

2. Oct 11, 2013

pwsnafu

Yes, this is a partition question. The partitions of 4 are 4, 3+1, 2+1+1, 1+1+1+1. We want three terms so that means the expression we want is
$(-\frac{x^2}{2})(x)(x)$
times the multinomial coefficient (3 choose 2,1 which is $\frac{3!}{2!1!} = 3$) times the original $-\frac{1}{6}$ out the front, giving us $\frac{x^4}{4}$

Last edited: Oct 11, 2013
3. Oct 12, 2013

Seydlitz

Thanks pwsnafu! That is such a nice method, no more guesswork for me! :D

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook