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Combining sums problem

  1. Oct 3, 2010 #1
    1. The problem statement, all variables and given/known data

    I'm having some confusion about combining sums. Our goal when combining these sums is to have the,

    [tex](x-c)^{\text{whatever}}[/tex]

    term to be the same in both sums.

    My confusion is better explained in an example. (see below)

    2. Relevant equations



    3. The attempt at a solution

    Let's say we have the following 2 sums and we want to simplify them into one sum,

    [tex]\sum_{n=0}^{\infty} (-1)^{n}2^{n}nx^{n+1} + \sum_{n=0}^{\infty} (-1)^{n}2^{n}nx^{n-1}[/tex]

    As you can see the,

    [tex](x-c)^{\text{whatever}}[/tex]

    terms are not identical, one is (n+1) and the other is (n-1).

    So if we wanted to make the two exponents identical for the first sum we would look as,

    [tex]n \rightarrow n-1[/tex],

    and plug in (n-1) where all the n's used to be in the first sum, and change the starting point of the sum to 1

    Now for the second sum, we would look as,

    [tex]n \rightarrow (n+1)[/tex],

    and plug in (n+1) where all the n's used to be in the second sum,

    ***Here's where I get confused***

    But my professor had mentioned to the class that this would not change the starting point of the sum to n= -1, it stays at n=0.

    Why is that? Can someone please clarify?

    Thanks again!
     
  2. jcsd
  3. Oct 3, 2010 #2
    Bump, still trying to figure this one out.
     
  4. Oct 3, 2010 #3
    [tex]
    \sum_{n=0}^{\infty} (-1)^{n}2^{n}nx^{n+1} + \sum_{n=0}^{\infty} (-1)^{n}2^{n}nx^{n-1} =
    [/tex]

    [tex]\sum_{n=0}^{\infty} (-1)^{n}2^{n}n\left(x^{n+1} + x^{n-1} \right) = \sum_{n=0}^{\infty} (-1)^{n}2^{n}nx^{n+1}\left(x^{-2}+1 \right) = \left(x^{-2}+1 \right) \sum_{n=0}^{\infty} (-1)^{n}2^{n}nx^{n+1} [/tex]

    Dunno what else to do..
     
  5. Oct 3, 2010 #4

    This doesn't really answer the question I had in the OP.
     
  6. Oct 3, 2010 #5
    Ok, then I don't know.

    Btw I don't even see any [tex] (x-c)^{\text{whatever}} [/tex] term in any of the sums unless you mean that c = 0. And I made it into one sum with one [tex] x^{\text{whatever}} [/tex] where whatever is a function of n.
     
  7. Oct 3, 2010 #6
    Ok. I think the reason it stays at zero is because when you replace n with n+1 and plugin n = -1 for the first term the term will be zero so it doesn't matter if you start your sum at 0 or -1, you will still get the same result.
     
  8. Oct 3, 2010 #7
    Can anyone else verify this? Is this actually why, or is it just a coincidence?
     
  9. Oct 3, 2010 #8
    Me again! ;)

    Well it works with all sums where the first term is zero which they are if you have a sum starting at n=0 of n*f(n,x) and f(n,x) is 'nice' in n=0.
     
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