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Taylor Series help?

  1. Jun 13, 2012 #1
    1. The problem statement, all variables and given/known data
    Have to find the Taylor series for (1-x)^(-0.5)
    Then use this to find the Taylor series for (1-x^2)^(-0.5)

    2. Relevant equations



    3. The attempt at a solution
    Was able to do the expansion for the first one quite easily, but not sure how to do the second one. My initial thought was to multiply the first expansion with the expansion for (1+x)^(-0.5)
    since (1-x^2) = (1-x)(1+x)
    But this is a one mark question, so this seems like too much working out to be the right way...
    Is there a trick to this question? :S
     
  2. jcsd
  3. Jun 13, 2012 #2

    SammyS

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    If [itex]\displaystyle f(x)=\frac{1}{\sqrt{1-x}}\,,[/itex] then what is [itex]f(x^2)\ ?[/itex]
     
  4. Jun 13, 2012 #3
    ah yes that's it, so I would just sub in x^2 to the expansion
    thanks!
    Would multiplying the two expansions have also worked?
     
  5. Jun 14, 2012 #4

    SammyS

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    Multiplying which two expansions ?
     
  6. Jun 14, 2012 #5

    Curious3141

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    Much, much more troublesome. You have to figure out which terms contribute to the coefficient of each term in the final expansion. For example, the [itex]x^3[/itex] term in the product is determined by the sum of the [itex]x^3[/itex] terms of both expressions, as well as the sum of the products of the [itex]x[/itex] term of one with the [itex]x^2[/itex] term of the other (and vice versa). The general term would involve lots of tedious algebra.

    It wouldn't be too bad if they just asked you to find the expression correct to the first few terms, though.
     
  7. Jun 14, 2012 #6

    HallsofIvy

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    No, squaring the original series would give [itex]\left(1/\sqrt{1- x}\right)^2= 1/\sqrt{1- x}[/itex], not [itex]1/(1- x^2)[/itex].
     
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