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Change a variable to transform a series into a power series

  1. Sep 5, 2016 #1
    1. The problem statement, all variables and given/known data
    The following series are not power series, but you can transform each one into a power
    series by a change of variable and so find out where it converges.

    0 ((3n(n+1)) / (x+1)n

    2. Relevant equations

    a power series is a series of the form:

    a0 + a1x + a2x^2 .... + ...

    3. The attempt at a solution

    What exactly does it mean by transforming a power series by changing a variable? the only thing I could look up and find was converting to another coordinate plane in calculus 3 which would be further along than this problem focuses on... but i am still confused as i can't see how a series can be in a separate plane like that
     
  2. jcsd
  3. Sep 5, 2016 #2

    Krylov

    User Avatar
    Science Advisor
    Education Advisor

    Introduce a new variable ##y## as a function of ##x##, such that if you write your series in terms of ##y## it takes the form of a power series. Then determine for which values of ##y## that power series converges. Finally, translate these values back to values of ##x##.
     
  4. Sep 5, 2016 #3
    Let me see if i understand this then...

    Let Y = Y(x),
    Where Y(x) = 1 / (x - 1)

    Then we would have:

    3n(n+1) / ( 1/ (x -1 +1)n)
    = 3n(n+1) (x)n

    this would in turn be a power series, yes?
     
  5. Sep 5, 2016 #4

    Mark44

    Staff: Mentor

    No, this isn't the substitution. There is another that should be much more obvious.
    You have a series whose general term is ##3^n(n + 1) \cdot \frac 1 {(x + 1)^n}##. You'd like to end up with ##3^n(n + 1) \cdot y^n##. What's the most obvious substitution to make?
     
  6. Sep 5, 2016 #5
    how is mine not obvious enough? It removes the added 1 and puts y back in the numerator.
     
  7. Sep 5, 2016 #6
    Sorry, I meant for my original definition of y to be:
    x = 1/y -1
    then you would have the relationship I intended to have:

     
  8. Sep 5, 2016 #7

    Mark44

    Staff: Mentor

    That works, but why not write it as ##y = \frac 1 {x + 1}##?
    Then ##y^n = \frac 1 {(x + 1)^n}##
     
  9. Sep 5, 2016 #8
    So, essentially, my answer could work but you could also simply say "Let y = all of the business that makes this difficult"?
     
  10. Sep 5, 2016 #9

    Mark44

    Staff: Mentor

    No, not really.

    When you define something (in this case, y), you generally start it with "Let y = <whatever> ". What you wrote is the inverse of the function that defines y.
     
  11. Sep 5, 2016 #10
    Yeah, I noticed that. I guess it makes sense. Thank you!
     
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