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Sigma notation problem

  1. Sep 9, 2006 #1
    Help!

    We need to solve the following problem:

    ∞ r-1
    find x if ∑ (3x/2) = 4
    r=1


    We know the answer is 1/2, but we have no idea how they got that answer. Can anyone help?

    Thanks,
    Sammie
     
  2. jcsd
  3. Sep 9, 2006 #2
    it looks like things shifted when we wrote our problem. It should be the summation of (3x/2) raised to the r-1 power = 4 where r is going from 1 to infinity.

    I hope this makes sense to someone!
     
  4. Sep 9, 2006 #3

    0rthodontist

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    Do you know the formula for an infinite geometric sum?
     
  5. Sep 9, 2006 #4
    Is it something like

    Sn = u1(r^n – 1)
    ---------- where Sn is really S sub n and u1 is really u sub 1?
    r-1
     
  6. Sep 9, 2006 #5

    0rthodontist

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    What does that formula mean? Identify u1, r, and especially n.
     
  7. Sep 9, 2006 #6
    o.k

    r is the constant common ratio and u1 is the first term of the series and I think n is a term in the series??
     
  8. Sep 9, 2006 #7

    0rthodontist

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    Well, first step is to find out what n is. You have to know what your formulas mean before you'll be able to use them.

    You might also have another formula that is more directly useful, but you should still understand that one.
     
    Last edited: Sep 9, 2006
  9. Sep 9, 2006 #8
    o.k. n is the number of terms we are adding up. In our case infinity???

    Right?
     
  10. Sep 9, 2006 #9

    0rthodontist

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    Right--though it's not "really" infinity, you take the limit as n goes to infinity. What you have is really the formula for a finite sum, which you could tell from the n. So--if you don't have it already, what do you think the formula for an infinite sum is? In an infinite sum, r will be less than 1, so what is the limit of r^n as n goes to infinity?

    Then it's just a matter of applying the formula to the series you gave and solving for x.
     
  11. Sep 9, 2006 #10
    i'm still confused. Is the formula for an infinite sum:

    u1/r-1 ?

    and is the limit 1 as r goes to infinity?
     
  12. Sep 9, 2006 #11

    0rthodontist

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    n goes to infinity, not r--typo on my part. You're close, but not quite. As n goes to infinity, what does r^n go to? Then what does r^n - 1 go to?
     
  13. Sep 9, 2006 #12
    wouldn't r^n go to 1, but never reach 1 since it is always <1?

    and if r^n goes to 1 then wouldn't r^n-1 go to 0?
     
  14. Sep 9, 2006 #13

    0rthodontist

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    Pick some r < 1 and find out what it does. What is r^10? What is r^20?
     
  15. Sep 9, 2006 #14
    o.k for example if we take r=.5 then as you increase the exponent r gets closer & closer to 1. right?
     
  16. Sep 9, 2006 #15

    0rthodontist

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    Well, does it?
     
  17. Sep 9, 2006 #16
    yes it does. as we increased the exponent the answer kept getting closer and closer to 1.
     
  18. Sep 9, 2006 #17

    0rthodontist

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    What's 0.5^2?
     
  19. Sep 9, 2006 #18
    .5 ^ 2 = .25
     
  20. Sep 9, 2006 #19

    0rthodontist

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    So can you guess what happens as you increase the exponent? Try a few more like .5^3 or .5^10.

    Now you may be able to find that limit.
     
  21. Sep 9, 2006 #20
    O.K. I think I finally got it.

    Sn = 4, u1=1 and r=3x/2, so

    4 = 1/(1-3x/2)

    solve for x and x = 1/2!

    FINALLY!
     
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