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Homework Help: 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


    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
    \frac{1}{1-x} = \sum_{n=0}^{\infty} x^n (-3/5)(4x)^3n
  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:

    \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
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