1. Not finding help here? Sign up for a free 30min 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!

Power series

  1. Nov 14, 2005 #1
    Hi
    I"m having truoble with fnding the power series of the following::frown:

    1+(x^3)/3+(x^6)/18+(x^9)/162+...

    Can anyone give me a hand?

    Also, any hints in finding a power series?

    Thanks !
     
  2. jcsd
  3. Nov 14, 2005 #2

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    That IS a power series...
     
  4. Nov 14, 2005 #3

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Hahaha... or did you mean, you have trouble putting it into compact [itex]\sum[/itex] notation?
     
  5. Nov 14, 2005 #4

    Tide

    User Avatar
    Science Advisor
    Homework Helper

    I am guessing that overseastar is looking for a closed form of the sum.
     
  6. Nov 15, 2005 #5
    Looks like:

    [tex]\sum_{n=0}^{a}\frac{x^{3n}}{3^n3!}[/tex]
     
  7. Nov 15, 2005 #6

    Why do you use the factorial when 3 ! = 6 which is constant ?
     
  8. Nov 15, 2005 #7
    That fails to work when the denominator is 3.
     
  9. Nov 15, 2005 #8
    I think he means [tex]\sum_{n=0}^{\infty}\frac{x^{3n}}{3^nn!}[/tex]
     
  10. Nov 15, 2005 #9
    Thankyou Moo Of Doom, that is what I meant to write.
     
  11. Nov 16, 2005 #10

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    And [tex]\sum_{n=0}^{\infty}\frac{x^{3n}}{3^nn!}[/tex]
    is equal to
    [tex]\sum_{n=0}^{\infty}\frac{1}{n!}\left(\frac{x^3}{3}\right)^n[/tex]
    Anyone recognize THAT??
     
  12. Nov 16, 2005 #11
    No I don't, please enlighten me.
     
  13. Nov 16, 2005 #12
    [tex]\sum_{n=0}^{\infty}\frac{1}{n!}\left(\frac{x^3}{3}\right)^n=\exp{\left(\frac{x^3}{3}\right)}[/tex]
     
  14. Nov 19, 2005 #13
    oh wow, yes, that's what i meant, sorry
     
  15. Nov 19, 2005 #14
    And thank you for all your help !
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Power series
  1. Power Series Problem (Replies: 8)

  2. Complex power series (Replies: 2)

  3. Power Series? (Replies: 16)

  4. The Power Series (Replies: 1)

  5. Power series (Replies: 1)

Loading...