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Bounds of summation

  1. Sep 24, 2010 #1
    can the lower bound of a summation(sigma) be any real number ?
    i.e ex: sigma(LB:sqrt(2) or (9/2) etc )
    Even a lower bound be a real number is possible or not can upper bound be any real number or is it a strict rule that '1' should be added to lower bound to get the consecutive number.?
    i.e. ex: LB + sqrt(2) or (9/2) etc.
     
  2. jcsd
  3. Sep 24, 2010 #2
    You mean something like

    [tex]\Sigma_{i=\pi}^{\pi^2}i[/tex]

    ?

    If so, no. In standard usage, the bounds are always integers.

    Of course, you could always choose a different definition. Many variations on Σ exist, like the Mobius function, which sums over the divisors of an integer.
     
  4. Sep 24, 2010 #3

    statdad

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    Homework Helper

    "Is it a strict rule that '1' should be added to lower bound to get the consecutive number.?"

    There are some exceptions: for instance, if we wanted to write an expression for the sum of the reciprocals of the prime integers (note the primes do not change by differences of 1) we write

    [tex]
    \sum_{j \text{ prime}} {\frac 1 j }
    [/tex]

    (or something similar). If you wanted the "jump" between successive terms in a sum to be [itex] \sqrt 2 [/itex], and start at [itex] \pi [/itex], you might do something like this:

    [tex]
    \sum_{j = 1}^n {(\pi + (j-1)\sqrt{2})}
    [/tex]
     
    Last edited: Sep 24, 2010
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