Finding the needed power without expanding

  • Thread starter Seydlitz
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  • #1
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Main Question or Discussion Point

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?
 

Answers and Replies

  • #2
pwsnafu
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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?
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}##
 
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  • #3
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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}##
Thanks pwsnafu! That is such a nice method, no more guesswork for me! :D
 

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