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 representation

  1. Dec 15, 2009 #1
    1. Find a power series for F(x)= 3/4x^3-5, where c=1
    2. power series = 1/a-r



    3. What I did was take a derivative to get a similar function that was easier to solve. I used 1/x^3-1. Then I found a series for that function. Which I got [tex]\sum[/tex](x^3)^n. Then I added back from my orginal problem and got [tex]\sum[/tex](4/3x)^3n does this look at all right?
     
  2. jcsd
  3. Dec 15, 2009 #2

    Mark44

    Staff: Mentor

    Without parentheses it's difficult to tell what your function F is. It should be interpreted as (3/4)*x^3 - 5, but I don't think that's what you had in mind.
     
  4. Dec 15, 2009 #3
    Doesn't seem right to me. How does the 5 affect your result for instance?

    From the geometrical series you know that

    [tex]\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n[/tex]

    Your function

    [tex]\frac{3}{4x^3 - 5}[/tex]

    is already almost in the same form, all you have to do is manipulate it a little bit.
     
  5. Dec 15, 2009 #4
    Ok if I rewrite I need to make the first term of the denominator 1. So I divide by -5. So then does the sum from 0 to infinity of (-3/5)(x)^3n look better?
     
  6. Dec 15, 2009 #5
    Where did the 4 go?
     
  7. Dec 15, 2009 #6
    oops. Im meant to write [infinity]\sum[/n=0] (-3/5)(4x)^3n.
     
  8. Dec 15, 2009 #7
    [tex]
    \frac{1}{1-x} = \sum_{n=0}^{\infty} x^n (-3/5)(4x)^3n
    [/tex]
     
  9. Dec 15, 2009 #8
    does that look ok?
     
  10. Dec 15, 2009 #9
    When you divide by five you should do that to the whole denominator. And you have too much x's in there though I suspect that's a typo.
     
  11. Dec 15, 2009 #10
    Yes that was a typo. So.....[tex]\sum[/tex] (from 0-infinty) (-3/5)(4/5X)^3n would be ok? Sorry this is a new topic for me. Thanks you so much for helping!!!
     
  12. Dec 15, 2009 #11
    Almost, except that 4/5 is not cubed so it's power is n:

    [tex]
    \frac{3}{4x^3 - 5} = -\frac{3}{5} \frac{1}{1-\frac{4x^3}{5}} = -\frac{3}{5}\sum_{n=0}^{\infty} (\frac{4x^3}{5})^n
    [/tex]
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Power series representation
Loading...