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Question about shifting the indexes [nth powers] and counting numbers of series?

  1. Nov 21, 2011 #1
    Question about shifting the indexes [nth powers] and counting numbers of series?
    When you multiply a series by x, therefore changing the nth power on x, is it mandatory that you change the number where you start counting? This is confusing me because some problems do both, where others explain it in two different separate steps.

    Basically,
    If I am multiplying a taylor series by x, then am I supposed to add one to the index AND count one lower? or is it only mandatory to change the power on x, not the counting number?

    If the index [nth power on x] is changed, is it mandatory to change where the series starts counting at?

    I keep seeing problems where they only change one, and not the others. I thought you're supposed to replace all n's by the new relation.

    I hope this makes sense, lol.
     
  2. jcsd
  3. Nov 21, 2011 #2

    HallsofIvy

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    If your series is [itex]\sum_{n=0}^\infty a_nx^n[/itex] and you multiply by x, you will have [itex]\sum_{n=0}^\infty a_nx^{n+1}[/itex]. You can, although you don't have to let j= n+1. Then n= j- 1 so [itex]a_n= a_{j-1}[/itex] and the lower limit n= j-1= 0 become j=1. That is the series changes to [itex]\sum_{j=1}^\infty a_{j-1}x^j[/itex].
     
  4. Nov 21, 2011 #3
    @HallsofIvy,

    Thanks for answering!

    I have a couple more questions, if you will...

    If you change the starting number of the series, then you don't have to change 'n' everywhere else? or not.

    If you change the nth power on 'x' w/o multiplying by another 'x', then you must / don't have to change the starting number of the series?


    I haven't had a problem with changing n everywhere before, but I'm working a problem where I'm not getting the right solution.

    It's a problem where I'm multiplying a power series of (x-1)^(n-1) by x, so I changed x to [1+ (x-1)], but it isn't working out right.
     
  5. Nov 21, 2011 #4

    SammyS

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    So that gives you x(x-1)(n-1) = [1+ (x-1)](x-1)(n-1) = (x-1)(n-1) + (x-1)(n).

    In this case it looks like shifting the index on either one of the series may simplify things.



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