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Is there an inverse of Summation?

  1. Apr 20, 2014 #1
    Say for some general function f(x), and g(x) = ∑x=0 f(x) (assuming function is defined)
    Is there a way to find the zeroes of g(x)? Is there any relationship between the zeroes of f(x) and g(x)? Sorry if this question is poorly asked, i just began learning about summations and infinite series.
  2. jcsd
  3. Apr 20, 2014 #2
    This makes no sense.

    First a minor point, but your summation index is ##x##, which is an unusual choice. So I assume you sum over the naturals?

    Second, your summation index is ##x## so it shouldn't be used outside the summation. Setting ##g(x)## equal to this makes little sense to me.

    Indeed, by definition we can write

    [tex]\sum_{x=0}^{+\infty} f(x) = f(0) + f(1) + f(2) + f(3) + ...[/tex]

    So your equality

    [tex]g(x) = \sum_{x=0}^{+\infty} f(x)[/tex]


    [tex]g(x) = f(0) + f(1) + f(2) + f(3) + ...[/tex]

    which is probably not what you want.
  4. Apr 20, 2014 #3
    What if we assume N is being used in the function f(x)
    And we reset g(x) (I'm on mobile right now so I can't use symbols) to equal

    G(x) = Summation from N=0 to +Infinity of f(x)?
  5. Apr 20, 2014 #4
    Now ##f(x)## is independent from ##N##. So you're just adding a bunch of constants. Is this your intention?
  6. Apr 20, 2014 #5
    How about a function f(nx)?
    (Maybe I just need to go learn more about Infinite series and Functions)
  7. Apr 20, 2014 #6
    So you're considering

    [tex]g(x) = \sum_{n=0}^{+\infty} f(nx)[/tex]

  8. Apr 20, 2014 #7
    Say the function f(x) is (off the top of my head): x^3/(1-n)^x

    How would we go about finding the zeroes g(x) of the sum of From n=0 to +infinity? Or am I asking all the wrong questions?
  9. Apr 20, 2014 #8
    I don't really think there is one universal method. Some things work in one occasion but not in the other. I think the best you can do is to consider a specific function and try to work it out for that.
  10. Apr 20, 2014 #9
    OKay, Thanks!
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