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Summation Question

  1. Feb 25, 2014 #1
    1. The problem statement, all variables and given/known data

    I need a summation where the answer is 1 2 2 2 2 2 2 2

    2. Relevant equations

    a(0) + sum(2*a(1) + 2*a(2) +2*a(3))

    3. The attempt at a solution

    I unfortunately have no idea where to start, basically it is taking a symmetrical function from 0 to N-1. where the function is symmetric from the center point.

    I have an answer for it which is

    a(N-1/2) + sum_n=1^(N-1)/2 (2*a[((N-1)/2) - n])

    However I wanted to put it all inside the summation
     
  2. jcsd
  3. Feb 25, 2014 #2

    haruspex

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    I've no idea what you mean.
    In what sense does a summation have an answer like "1 2 2 2 2 2 2 2" rather than a single number?
    What does a(0) + sum(2*a(1) + 2*a(2) +2*a(3)) mean? Do you mean a(0) + sum(2*a(1) + 2*a(2) +2*a(3) + ....)? If so, how is that different from 2(∑an) - a0?
    From the mention of symmetry, I'm guessing this comes from ##\Sigma_{n=-N}^Na_n##, where a-n = an, but I still don't understand what you wish to achieve.
    What quantity does your final equation represent?
     
  4. Feb 25, 2014 #3
    Sorry, the "1 2 2 2 2 ..." bit was basically was how I wanted each part of the summation to equate to, so essentially it would be "1 + 2 + 2 + 2 ..."

    "2(∑an) - a0" is the exact same as "a0 + 2(∑an)" with the exception that n goes from 0->N for the first one and 1->N for the second one.

    Basically I want to achieve this, "2(∑an) - a0", without having to subtract the a0 at the end.
     
  5. Feb 25, 2014 #4

    haruspex

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    Still unclear.. are you looking for an algebraic function f(n) that takes the values f(0) = 1, f(n) = 2 for n > 1? So that you can write ∑n>=0anf(n)?
     
  6. Feb 26, 2014 #5
    yes that's exactly what i'm looking to do
     
  7. Feb 26, 2014 #6

    LCKurtz

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    You want ##f(n) = 2-\textrm{sinc}(n)##, where sinc is the normalized sinc function:$$
    \textrm{sinc} (x) = \begin{cases}
    \frac {\sin(\pi x)}{\pi x}&x \ne 0\\
    1 & x = 0
    \end{cases}$$
     
  8. Feb 26, 2014 #7

    haruspex

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    Cute, but I don't see the value in this. As you note, that function has to have a separate definition at x=0. So why not use Kronecker delta? I see nothing in the thread to suggest a continuous function is required.
    JeeebeZ, why do you want such a function? What ultimately are you trying to achieve?
     
  9. Feb 26, 2014 #8

    LCKurtz

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    Agreed. But then, I don't see any point to the original question either.
     
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