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Cesaro Sum. Understanding the sequence.

  1. Nov 27, 2009 #1
    1. The problem statement, all variables and given/known data
    I have a finite sequence Z=(z1,...,zn). The Cesaro sum of Z is [tex]\frac{(B1+B2+...+Bn)}{n}[/tex]

    BC=z1+z2+...zC (1[tex]\leq[/tex]C[tex]\leq[/tex]n)

    Lets say the problem asks "The Cesaro sum of the 99th term sequence of (z1,...,z99) is 2000, what is the Cesaro sum of the 100 term sequence (1, z1,...,z99)?

    2. Relevant equations



    3. The attempt at a solution
    I read about Cesaro sum on wikipedia but it didn't elaborate much. Here is where I'm at:

    2000=[tex]\frac{(B1+B2+...+B99)}{99}[/tex]

    But, honestly, I have no idea how to solve this because I can't find any info on it.
     
  2. jcsd
  3. Nov 27, 2009 #2

    LCKurtz

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    Call z1 + z2 +...+z99 = S

    You are given that S/99 = 2000.

    Now you are asked to calculate (1 + S)/100. It isn't that tough...
     
  4. Nov 27, 2009 #3
    So, putting 1 in front of the sequence allows z99 to become the 100th term? But it isn't the 100th nth term right? With what you said it would just be 1980.01 as the answer? I am trying to fundamentally understand this. I understand it's just an average but when they start saying the value of C could be lesser or equal to n and all of that I lose the concept.
     
  5. Nov 27, 2009 #4

    LCKurtz

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    Yes, that is the correct answer. Add a 1 to the sequence means there are now 101 terms to average.

    The Cesaro sum of a sequence gives you a new sequence which gives cumulative average of the given sequence. One place they are used is in the study of divergent sequences. For example, the sequence 1, -1, 1, -1, 1,... diverges. But, informally, you might say its "average value" is 0. And that is exactly what the Cesaro sum sequence converges to.
     
  6. Nov 27, 2009 #5
    Big "oh..." moment. I think I misread the problem. It asks "what is the Cesaro sum of the 100th term sequence (1, z1, ..., z99)?" I realize now that it is the whole 100 term sequence and I don't actually have to go up to z100 right? Because the 1 in front makes it 100 terms.

    I understand this. It's just the technical definition that got me. Thanks!
     
  7. Nov 27, 2009 #6

    LCKurtz

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    Yes. It's 100 terms including the 1. I mistyped 101 earlier I see. I think you've got it now.
     
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