1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Finding the needed power without expanding

  1. Oct 11, 2013 #1
    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. jcsd
  3. Oct 11, 2013 #2

    pwsnafu

    User Avatar
    Science Advisor

    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
  4. Oct 12, 2013 #3
    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