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Homework Help: Is this a correct taylor series representation centered at 1

  1. May 27, 2010 #1
    f(x)=1/(1-x^2)^(1/2)
    1/x^(1/2)=1+ sum(( (-1)^n 1*3*5*7...(2n-1)(x-1)^n )/(2^n n! ) , n=1, infty )
    thus 1/(1-x^2)^(1/2) = 1+ sum(( 1*3*5*7...(2n-1)(x^2)^n )/(2^n n! ) , n=1, infty )
    is this a correct taylor series representation centered at 1
     
    Last edited: May 28, 2010
  2. jcsd
  3. May 27, 2010 #2

    cronxeh

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    Gold Member

    Re: series

    Are you sure its supposed to be centered at 1? And you are sure its 1/sqrt(1-x^2)? The answer you seek is in the imaginary domain.
     
    Last edited: May 27, 2010
  4. May 27, 2010 #3
    Re: series

    well ya for a=1 on D -1<x<1
     
  5. May 28, 2010 #4

    cronxeh

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    Gold Member

    Re: series

    So if f(x) = 1/sqrt(1-x^2)

    What is f(1) ?
     
  6. May 28, 2010 #5
    Re: series

    |x|<1
     
  7. May 28, 2010 #6

    Mark44

    Staff: Mentor

    Re: series

    So it can't be centered at 1, a number not in the domain.
     
  8. May 28, 2010 #7

    cronxeh

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    Gold Member

    Re: series

    Would make sense that it is centered at 0 then :biggrin:
     
  9. May 28, 2010 #8
    Re: series

    is the binomial series only way to go here
     
  10. May 28, 2010 #9

    Mark44

    Staff: Mentor

    Re: series

    No, but it is probably easier. Another approach is using the definition of the Maclaurin series (since you are expanding in powers of x): f(x) = f(0) + f'(0)x + f''(0)x^2/2! + ...
     
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