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Sigma Notation

  1. Nov 11, 2005 #1
    Show that:
    [tex]\sum_{k=0}^{N} a_k = \sum_{k=M}^{M+N} a_{k-M}[/tex]


    I did this, my answer is:
    [tex]\sum_{k=0}^{N} a_k = a_0 + a_1 + a_2 + ... + a_N[/tex]

    [tex]\sum_{k=M}^{M+N} a_{k-M} = a_{M-M} + a_{M+1-M} + a_{M+2-M} + ... + a_{M+N-M} = a_0 + a_1 + a_2 + ... + a_N[/tex]


    Now, I have to use this to prove that:
    [tex]\sum_{k=M}^{N} 2^{-k} = 2(2^{-M} - 2^{-N})[/tex]


    So, I tried expanding the sum:
    [tex]\sum_{k=M}^{N} 2^{-k} = 2^{-M} + 2^{-(M+1)} + 2^{-(M+2)} + ... + 2^{-N}[/tex]

    I factored out some 2's:
    [tex]\sum_{k=M}^{N} 2^{-k} = 2^{-M}2^{0} + 2^{-M}2^{-1} + 2^{-M}2^{-2} + ... + 2^{-M}2^{-N} = 2^{-M}(2^{0} + 2^{-1} + 2^{-2} + ... + 2^{-N})[/tex]

    Which I'm assuming is equal to some other sum:
    [tex]\sum_{k=M}^{N} 2^{-k} = 2^{-M}(\sum_{i=0}^{N} 2^{-i})[/tex]


    However, I don't know how to continue this. Any suggestions?
     
  2. jcsd
  3. Nov 11, 2005 #2

    benorin

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    Hint: [tex]\sum_{k=M}^{N} 2^{-k} = \sum_{k=0}^{N} 2^{-k}-\sum_{i=0}^{M-1} 2^{-i}[/tex]
     
  4. Nov 11, 2005 #3
    That doesn't seem to work.
    I get:
    [tex]\sum_{k=M}^{N} 2^{-k} = \sum_{k=0}^{N} 2^{-k}-\sum_{i=0}^{M-1} 2^{-i}[/tex]
    [tex]\sum_{k=M}^{N} 2^{-k} = (2^{0} + 2^{-1} + 2^{-2} + ... + 2^{-N}) - (2^{0} + 2^{-1} + 2^{-2} + ... + 2^{-M + 1}[/tex]
    [tex]= 2^{-N} - 2^{-M+1} = 2^{-N} + 2^{-M}2^{1}[/tex]

    Which will not equal to : [tex]2(2^{-M} - 2^{-N})[/tex]
     
    Last edited by a moderator: Nov 11, 2005
  5. Nov 11, 2005 #4

    benorin

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    [tex]a_{M}+a_{M+1}+\cdot\cdot\cdot+a_{N-1}+a_{N}=\left( a_{M}+a_{M+1}+\cdot\cdot\cdot+a_{N-1}+a_{N}\right) + \left[ \left( a_{0}+a_{1}+\cdot\cdot\cdot+a_{M-1}\right) - \left( a_{0}+a_{1}+\cdot\cdot\cdot+a_{M-1}\right) \right] [/tex]
    [tex] \left[ \left( a_{M}+a_{M+1}+\cdot\cdot\cdot+a_{N-1}+a_{N}\right) + \left( a_{0}+a_{1}+\cdot\cdot\cdot+a_{M-1}\right) \right] - \left( a_{0}+a_{1}+\cdot\cdot\cdot+a_{M-1}\right) = \left( a_{0}+a_{1}+\cdot\cdot\cdot+a_{N-1}+a_{N}\right) - \left( a_{0}+a_{1}+\cdot\cdot\cdot+a_{M-1}\right)[/tex]
     
    Last edited: Nov 11, 2005
  6. Nov 11, 2005 #5
    Wait, we posted at the same time.
    Read my previous post, I edited it.
     
  7. Nov 11, 2005 #6

    AKG

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    I think you mean:

    [tex]\sum _{k=M} ^N = 2(2^{-M}\mathbf{)} - 2^{-N}[/tex]

    Anyways, I don't know how you'd prove this using what you proved first, but you can prove this if you know that:

    2n - 1
    = 2n - 20
    = 2n + 0 + 0 + ... + 0 - 20
    = 2n + (-2n-1 + 2n-1) + (-2n-2 + 2n-2) + ... + (-21 + 21) - 20
    = (2n + 2n-1 + ... + 21) - (2n-1 + 2n-2 + ... + 20)
    = 2(2n-1 + 2n-2 + ... + 20) - (2n-1 + 2n-2 + ... + 20)
    = (2n-1 + 2n-2 + ... + 20)
     
  8. Nov 11, 2005 #7
    No, no... I wrote the sum correctly.
     
  9. Nov 11, 2005 #8

    benorin

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    AKG is right: [tex]\sum _{k=M} ^N 2^{-k}= 2(2^{-M}\mathbf{)} - 2^{-N}[/tex]
     
  10. Nov 11, 2005 #9

    AKG

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    Well it's false so good luck proving it:

    M = 3, N = 4

    [tex]\sum_{k=M}^{N} 2^{-k} = 2(2^{-M} - 2^{-N})[/tex]

    [tex]\sum_{k=3}^{4} 2^{-k} = 2(2^{-3} - 2^{-4})[/tex]

    [tex]2^{-3} + 2^{-4} = 2(2^{-3} - 2^{-4})[/tex]

    [tex]\frac{1}{8} + \frac{1}{16} = 2(\frac{1}{8} - \frac{1}{16})[/tex]

    [tex]\frac{3}{16} = 2(\frac{1}{16})[/tex]

    [tex]3 = 2[/tex]

    Good luck.
     
  11. Nov 11, 2005 #10

    benorin

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    Here's one way to do it:

    [tex]\sum_{k=M}^{N} 2^{-k} = \sum_{k=0}^{N} 2^{-k}-\sum_{i=0}^{M-1} 2^{-i}[/tex]

    the later sums are geometric... recall that the partial sum thereof is

    [tex]1+r+r^2+r^3+\cdot\cdot\cdot+r^n=\frac{1-r^{n+1}}{1-r}[/tex]
     
  12. Nov 11, 2005 #11
    This is going to be tricky... especially since the question I'm given has an error in it itself :S.
    I guess I will have to try your methods.

    Thank you AKG and benorin.

    I'll get back to you if I have any problems.
     
  13. Nov 11, 2005 #12
    Well, that is the partial sum for a positive n, but in my case-- there is a negative n.
    What is the partial sum for :
    [tex]\sum_{i=0}^{N} 2^{-i}[/tex]
     
  14. Nov 11, 2005 #13

    benorin

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    [tex]2^{-i}=\left( 2^{-1}\right)^{i}=\left( \frac{1}{2}\right)^{i}[/tex]

    hence [tex]\sum_{i=0}^{N} 2^{-i}= \sum_{i=0}^{N}\left( \frac{1}{2}\right)^{i} = \frac{1-\left( \frac{1}{2}\right)^{N+1}}{1-\frac{1}{2}}=2\left( 1-2^{-(N+1)}\right)[/tex]

    do likewise for [tex]\sum_{i=0}^{M-1} 2^{-i}[/tex] and subtract.
     
  15. Nov 11, 2005 #14
    Why am I doing likewise for [tex]\sum_{i=0}^{M-1} 2^{-i}[/tex] ?

    This was my last step: [tex]\sum_{k=M}^{N} 2^{-k} = 2^{-M}(\sum_{i=0}^{N} 2^{-i})[/tex]
     
  16. Nov 11, 2005 #15

    benorin

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    the last term of the sum on left is [tex]2^{-N}[/tex] on the right the last term is [tex]2^{-(N+M)}[/tex], so their not equal.
     
  17. Nov 11, 2005 #16

    benorin

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    [tex]2^{-i}=\left( 2^{-1}\right)^{i}=\left( \frac{1}{2}\right)^{i}[/tex]

    hence [tex]\sum_{i=0}^{N} 2^{-i}= \sum_{i=0}^{N}\left( \frac{1}{2}\right)^{i} = \frac{1-\left( \frac{1}{2}\right)^{N+1}}{1-\frac{1}{2}}=2\left( 1-2^{-(N+1)}\right)[/tex]

    do likewise for [tex]\sum_{i=0}^{M-1} 2^{-i}[/tex], that is

    [tex]\sum_{i=0}^{M-1} 2^{-i}= \sum_{i=0}^{M-1}\left( \frac{1}{2}\right)^{i} = \frac{1-\left( \frac{1}{2}\right)^{(M-1)+1}}{1-\frac{1}{2}}=2\left( 1-2^{-M}\right)[/tex]

    and subtract:

    [tex]\sum_{k=M}^{N} 2^{-k} = \sum_{k=0}^{N} 2^{-k}-\sum_{i=0}^{M-1} 2^{-i}= 2\left( 1-2^{-(N+1)}\right) - 2\left( 1-2^{-M}\right) = 2^{1-M}-2^{-N}[/tex]


    Sorry to be short, but gotta go...
     
  18. Nov 11, 2005 #17
    I don't quite understand.

    I realize that your method is correct, but I don't understand why.
     
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