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Double sum of same variable simplification help

  1. Mar 6, 2015 #1


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    Hello all,
    I have been meditating on this for a while, but can't seem to understand how this simplification came to be. Any help will be greatly appreciated.

    So, here is what we start with:
    ##\mathop{\sum_{k=0}^m\sum_{l=0}^n}_{m{\geq}n} x(k,l)##
    We also know that: l (lower case L) = n-m+k

    and here is what my book ends with:
    ##\sum\limits_{k=m-n}^m x(k,n-m+k)##

    Here is my attempt, using the fact that l = n-m+k and substituting, I get:
    ##\mathop{\sum_{k=0}^m\sum_{n-m+k=0}^n}_{m{\geq}n} x(k,n-m+k)##

    ## = \mathop{\sum_{k=0}^m\sum_{k=m-n}^n}_{m{\geq}n} x(k,n-m+k)##

    I then have no idea where to go from here : ( I tried substituting numbers for m and n and looked at what was happening, but that didn't help me much. Can someone please provide some insight please!!! I have been at this for a bit.

    Thank you for reading.
  2. jcsd
  3. Mar 6, 2015 #2

    Stephen Tashi

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    If that is true then for a given value of [itex]k [/itex] there is only a single value of [itex] l [/itex]. So for a given value of [itex]k[/itex], the expression [itex] \sum_{l=0}^{n} x(k,l) [/itex] doesn't make sense since it implies that for a given [itex] k [/itex] you can vary [itex] l [/itex] over several different values. If we adopt the convention that [itex] \sum _{l=0}^{n} x(k,l) [/itex] shall mean the single term [itex] x(k,n-m+k) [/itex] then you could get the desired result.

    When you substitute, you can't use [itex] k [/itex] as the name of the variable whose values are summed by the second summation [itex]
    \sum_{l=0}^n [/itex] because the name [itex] k [/itex] is already in use as the variable for the first summation [itex] \sum_{k=0}^m [/itex]. The first argument of [itex] x(k,l) [/itex] refers to the [itex] k [/itex] that is summed by [itex] \sum_{k=0}^m [/itex].
  4. Mar 6, 2015 #3


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    Hey, thank you for your reply. I still can't figure this out.

    I didn't think of this, but now that you mentioned it, wouldn't [itex] l [/itex] have infinite possible values? If you set [itex] k [/itex] to [itex] a [/itex] then you can have any value for [itex] n [/itex] and [itex] m [/itex] which results in [itex] l [/itex] having many values. What is wrong with this?

    I'm not sure what you mean here : ( I'm sorry, but I really am trying... I have to admit, the equation I supplied is modified from the original shown in the book. I will attach a copy of what I am struggling with maybe it will be better. Here it is (I understand everything until where it says "Thus for m ≥ n"):

    Hmmm, ok that kind of makes sense. So what then shall we do in such a situation?

    Thank you for your help : )
  5. Mar 7, 2015 #4

    Stephen Tashi

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    There's nothing to indicate that [itex] m [/itex] and [itex] n [/itex] are variables. They appear to denote constants.

    To make an analogy with calculus, if you are finding [itex] \int ( \int f(x,y) dy) dx [/itex] and you use integration by subsitution to find [itex] \int f(x,y) dy [/itex] then you can do a substitution like [itex] y = 3u^2 [/itex], but you can't do the substitution [itex] y = 3x^2 [/itex] because the variable [itex] x [/itex] is already in use to denote a quantity that is independent of [itex] y [/itex].

    In your example, the variable [itex] k [/itex] is already in use in the first summation. You can't introduce it as meaning a different variable in the second summation since the [itex] k [/itex] in [itex] x(k,l) [/itex] refers to the [itex]k [/itex] used in the first summation.

    The correct analogy to what is given in the text is the statement "[itex] x(k,l) = 0[/itex] except when [itex] l = n - m + k[/itex] " (which is different than asserting " [itex] l = n - m + k [/itex] ").

    Then [itex] \sum_{l=0}^n x(k,l) [/itex] has only one nonzero term, which is [itex] x(k,n-m+k) [/itex].

    Perhaps there is also information in the text that [itex] x(k,n-m+k) [/itex] is zero for [itex] k < n-m [/itex], but I'd have to think about it further to see why.
  6. Mar 7, 2015 #5


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    Thank you Mr.Tashi. I am still trying to figure out how they got those last two lines. I will report back if I find anything.

    PS: Your calculus analogy was helpful.
    Last edited: Mar 7, 2015
  7. Mar 7, 2015 #6


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    I think I may have understood how they achieved the second to last line. So we know ##l = n - m + k## (else the equation will be zero). And we also know from the second summation that ## 0 ≤ l ≤ n ##. So I did the following to kind of see what was going on:

    for ##l = 0, l = 0 = n-m+k => k = m-n## this will make the first summation to be ##\sum\limits_{k=m-n}^m x(k,n-m+k)##
    for ##l = n, l = n = n-m+k => k = m## this will make the first summation to be ##\sum\limits_{k=m}^m x(k,n-m+k)##
    for ##l = n-1, l = n-1 = n-m+k => k = m-1## this will make the first summation to be ##\sum\limits_{k=m-1}^m x(k,n-m+k)##

    So after viewing these summations, I see it is for ##l =0## in which the first summation takes into account all possible values (or has the greatest summation intervals).

    What do you think of this?

    EDIT1: I think the last line is achieved from the one before by using the fact that ##\sum\limits_{k=0}^m x(k) = \sum\limits_{k=0+n}^{m+n} x(k-n)##
    EDIT2: My first edit (EDIT1) may be wrong. I am still checking.
    Last edited: Mar 7, 2015
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