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Write a closed form expression for the approximation y(nC)

  1. Apr 16, 2012 #1
    [tex]y(4C) ≈ 7.3 + C + \frac{C}{3^{10C}} + \frac{C}{3^{20C}} + \frac{C}{3^{30C}}[/tex]

    Would:

    [tex]y(nC) ≈ 7.3 + C\sum_{n = 0}^{\infty}{\frac{1}{3^{10Cn}}}[/tex]

    Be an acceptable answer? If not, what am I doing wrong here?
     
  2. jcsd
  3. Apr 16, 2012 #2
    Any help?
     
  4. Apr 16, 2012 #3

    Dick

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    The sum part is a geometric series r^n where r=1/(3^(10C).
     
  5. Apr 16, 2012 #4
    Oh, ok. So should I be relating this to [tex]a\frac{1-r^n}{1-r}[/tex]
    In particular, will my solution look something like:

    [tex]y(nC) ≈ 7.3 + C\frac{1-{\frac{1}{3^{10C}}}^n}{1-\frac{1}{3^{10C}}}[/tex]
    ?
     
  6. Apr 16, 2012 #5

    Dick

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    Depends. What are you really trying to do here? Why are you writing approximately equal to? (1+r+r^2+...+r^n) is actually (1-r^(n+1))/(1-r).
     
  7. Apr 16, 2012 #6
    Well, because y(nC) is an approximation of some function, I assume. I just realized that there should have been a "..." in the initial y(4C).
     
  8. Apr 16, 2012 #7

    Dick

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    Now why would you do that? Then the first sum is the same as second sum. You can sum a geometric series like that exactly. As you've basically already said. If n is large and r=1/(3^(10C) is small you can approximate it by the infinite sum. I don't see any other role for 'approximately equal' here.
     
  9. Apr 17, 2012 #8
    I just assumed that I couldn't use the sigma notation since they specified a "closed form expression." But, then again, I'm not entirely what that means.
     
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