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What power series represents 1/(1+x^2) on (-1,1)

  1. Mar 18, 2009 #1
    1. The problem statement, all variables and given/known data
    Word for word, the book says "Find a power series that represents 1/(1+x^2) on (-1, 1)


    2. Relevant equations
    It's in the chapter that talks about power series, so I think they want me to use the fact that 1/(1-x) is a power series with a=1 and r=x, but if I just substitute in -x^2 for x, that makes chain rule issues. The chapter also talks about term-by-term integration, which confuses me more than little bit.


    3. The attempt at a solution
    Uhm... I don't really have much yet. I think this is probably an extremely basic question and I'm just missing some underlying technique or trick to solve it.
    I know that the antiderivative of 1/(1+x^2) is arctan(x) of but we don't know the series for arctan(x), so that doesn't really help. The derivative of 1/(1+x^2) is -2x/(1+x^2)^2, which doesn't seem to help either.

    Thanks for any help :)
     
    Last edited: Mar 18, 2009
  2. jcsd
  3. Mar 18, 2009 #2

    HallsofIvy

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    What do you mean by "power series issues"? Of course, 1/(1-x), itself, is not a "power series", it is the sum of the geometric series [itex]\sum_{n= 0}^\infty x^n[/itex]. And, to make "[itex]1/(1- x)[/itex]" look like "[itex]1/(1+x^2)[/itex]" replace x by [itex]-x^2[/itex], not just [itex]x^2[/itex].

     
    Last edited: Mar 19, 2009
  4. Mar 18, 2009 #3
    Wow I'm sorry, for "power series issues" I actually meant to type "chain rule issues". Not sure what I was thinking there haha! Multiple bad typos in that, I'll go back and fix them in a second.

    I'm not quite understanding your answer - so I CAN just go back and sub in -x^2 (that's what I meant, I swear haha)? And it doesn't matter that x^2 is a different order from x?
     
  5. Mar 18, 2009 #4

    Dick

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    Yes. 1/(1+r)=sum (-r)^n for n=0 to infinity if |r|<1. Sub away.
     
  6. Mar 20, 2009 #5
    Thanks so much, Dick :)
    Unfortunately, the test still completely kicked my butt. Haha oh well.
     
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