Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Series expression

  1. Apr 25, 2012 #1
    Hey, I'm going over series expansions and was wondering if someone could check my work and tell me if my work is correct. If not, could you explain it to me? I couldn't find any example like this problem in my book so I'm posting it online. Here it is,

    The closed form series expansion for cos(x) is Ʃ[(-1)^n(x)^2n]/(2n)!. Use this series to find a series expression for [cos(x)-1]/x^2.

    Okay here's what I did:

    [cos(x)-cos(pi)]/(x)^(2)

    (x)^(-2)*Ʃ[(-1)^(n)(x)^(2n)-(pi)^(2n)]/(2n)!

    Ʃ[(-1)^(n)(x)^(2n-2)-(pi)^(2n)

    Is that the way I'm supposed to do it? Thanks!
     
  2. jcsd
  3. Apr 25, 2012 #2
    This step is wrong, it should be ##x^{-2} \cdot \displaystyle \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}+(-1)^n \pi^{2n}}{(2n)!}##

    Remember, ##\cos{\pi} = -1##, and you also forgot that extra factor of the ##(-1)^n##. Also, you didn't multiply your ##x^{-2}## term in the sum properly in your attempt, make sure you do so this time!


    Alternatively, you could break up your expression to find the series expansion of ##\displaystyle \frac{\cos{x}}{x^2} - \frac{1}{x^2}##.
     
  4. Apr 26, 2012 #3
    Okay thanks!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook