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Summation properties

  1. Oct 29, 2015 #1
    Hello, I can not find the way to solve the following equation:
    sum of k^2 f(k) from k=1 to n.

    In Particular, k^2 * (1/k)
     
  2. jcsd
  3. Oct 29, 2015 #2

    mfb

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    There is no general formula for arbitrary functions f.
    In your specific example, there is a very easy one that you'll find in every collection of sum formulas.
     
  4. Oct 29, 2015 #3
    I have found how to solve ∑{k=1,n} [k·f(k)], for example:

    ∑{k=1,n} [k·f(k)] = ∑{k=1,n} [ ∑{j=k,n} [f(j)] ]
    Dem:
    ∑{k=1,n} [k·f(k)] = 1·f(1) + 2·f(2) + 3·f(3) + ... + n·f(n) = f(1) + f(2) + f(2) +
    + f(3) + f(3) + f(3) + ... + f(n) + ... + f(n) = [f(1) + f(2) + f(3) + ... + f(n)] +
    + [f(2) + f(3) + ... + f(n)] + [f(3) + ... + f(n)] + ... + f(n) =
    = ∑{j=1,n} [f(j)] + ∑{j=2,n} [f(j)] + ∑{j=3,n} [f(j)] + ... + ∑{j=n,n} [f(j)] =
    = ∑{k=1,n} [ ∑{j=k,n} [f(j)] ]

    However, I need to solve ∑{k=1,n} [k^2·f(k)]
     
  5. Oct 29, 2015 #4

    mfb

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    That is not "solving", it is transforming one expression into another, here a sum into a double sum.

    You can do the same scheme here. Let g(k)=k*f(k) and apply the formula you found for k*g(k).
    Do the same thing again and you get a triple sum over f(m)
     
  6. Oct 29, 2015 #5
    I already tried that, but it didn't work.
    For example:

    ∑{k=1,n} [k·f(k)] = 1·f(1) + 2·f(2) + 3·f(3) + ... + n·f(n) = ∑{k=1,n} [ ∑{j=k,n} [f(j)] ]

    But,

    ∑{k=1,n} [k^2·f(k)] = 1·f(1) + 4·f(2) + 9·f(3) + ... + n^2·f(n) ≠ ∑{k=1,n} [ ∑{j=k,n} ∑{i=j,n}[[f(i)]] ]
     
  7. Oct 29, 2015 #6
    It seems to be that:
    Although: ∑{k=1,n} [k·f(k)] = ∑{k=1,n} [ ∑{j=k,n} [f(j)] ] holds,
    ∑{k=1,n} [ ∑{j=k,n} [j*f(j)] ] = ∑{k=1,n} [ ∑{j=k,n} ∑{i=j,n}[[f(i)]] ] does not hold.
     
  8. Oct 29, 2015 #7

    mfb

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    That's not what I suggested.

    Right, as you start the multiplications with j=k, not with 1. There is a way to avoid that.
     
  9. Oct 29, 2015 #8
    How can I solve: ∑{k=1,n} [k^2· 2^(-k)]
     
    Last edited: Oct 29, 2015
  10. Oct 29, 2015 #9

    Mark44

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    As you have already been told, you are not "solving" these summations. Instead, you are writing them in a different form.

    Please ask your question in such a way that makes it clear what you want to happen.
     
  11. Oct 29, 2015 #10
    I need to find an equivalent expression without the sigma notation of the expression: ∑{k=1,n} [k^2· 2^(-k)]
     
  12. Oct 30, 2015 #11

    mfb

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    It is possible with the ideas given in the thread.

    In case you are just interested in the result, WolframAlpha can calculate it - I copied your equation 1:1, didn't even need reformatting. I just changed the brackets so the forum displays the link correctly.
     
  13. Nov 7, 2015 #12

    mfb

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    Sigma (##\Sigma##) is the greek symbol used for sums. The question is about properties of sums.
     
  14. Nov 7, 2015 #13

    Mark44

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    I have changed the title of this thread to "Summation properties".
     
  15. Nov 7, 2015 #14

    fresh_42

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    What a pitty. ∑ properties would have been far more interesting.
     
  16. Nov 7, 2015 #15

    Mark44

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    Are you referring to sigma-algebras (##\sigma-\text{algebras}##) and the like? If so, lowercase ##\sigma## (sigma) is used.
    In any case, and interesting or not, the title now reflects what the OP asked.
     
  17. Nov 7, 2015 #16

    fresh_42

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    I meant hyperons and yes, it was meant to be both funny and a food for thought. But hey, I'm new and I'am trying to get used to the language here. Jokes don't seem to rank very high. Ok, lesson learned.
     
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