Results 1 to 4 of 4

Math Help - Exploiting geometric series with power series to find Taylors series

  1. #1
    Member
    Joined
    Jan 2010
    Posts
    144

    Question Exploiting geometric series with power series to find Taylors series

    I'm confused between some formulae so I'm going to give some examples and you can let me know if what I'm writing is correct.

    Find the Taylor series for....

    EXAMPLE 1:
    f(x) = \frac{1}{1- (x)} around x = 2

    Then,

     \frac{1}{1-(x)} = \frac{1}{3-(x+2)} = \frac{1}{3} \left( \frac{1}{1 - \frac{(x+2)}{3}} \right) = \frac{1}{3} \sum_{n=0}^{\infty} (\frac{x+2}{3})^{n} =  \sum_{n=0}^{\infty} \frac{(x+2)^{n}}{3^{n+1}} provided that | \frac{x+2}{3} | < 1 or,  -5 < x < 1


    Does this look correct?

    EDIT: Now if I do the same question in another fashion...

    \frac{1}{1-x} = \frac{1}{1 + (-x)} = \frac{1}{3 + (-x -2)} = \frac{1}{3} \left( \frac{1}{1 + (\frac{-x-2}{3})} \right) = \frac{1}{3} \sum_{n=0}^{\infty}  (\frac{-x-2}{3})^{n} = \frac{1}{3} \sum_{n=0}^{\infty} (-1)^{n} \frac{(x+2)^{n}}{3^{n}} = \sum_{n=0}^{\infty} (-1)^{n} \frac{(x+2)^{n}}{3^{n+1}} provided that | \frac{(-x - 2)}{3} | < 1 or,  -5 < x < 1

    They aren't equal, in this one I have a (-1)^{n} kicking around in my sum.

    What am I doing wrong here?
    Last edited by jegues; September 28th 2010 at 05:50 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Quote Originally Posted by jegues View Post
    I'm confused between some formulae so I'm going to give some examples and you can let me know if what I'm writing is correct.

    Find the Taylor series for....

    EXAMPLE 1:
    f(x) = \frac{1}{1- (x)} around x = 2

    Then,

     \frac{1}{1-(x)} = \frac{1}{3-(x+2)} = \frac{1}{3} \left( \frac{1}{1 - \frac{(x+2)}{3}} \right) = \frac{1}{3} \sum_{n=0}^{\infty} (\frac{x+2}{3})^{n} =  \sum_{n=0}^{\infty} \frac{(x+2)^{n}}{3^{n+1}} provided that | \frac{x+2}{3} | < 1 or,  -5 < x < 1


    Does this look correct?

    EDIT: Now if I do the same question in another fashion...

    \frac{1}{1-x} = \frac{1}{1 + (-x)} = \frac{1}{3 + (-x -2)} = \frac{1}{3} \left( \frac{1}{1 + (\frac{-x-2}{3})} \right) = \frac{1}{3} \sum_{n=0}^{\infty}  (\frac{-x-2}{3})^{n} = \frac{1}{3} \sum_{n=0}^{\infty} (-1)^{n} \frac{(x+2)^{n}}{3^{n}} = \sum_{n=0}^{\infty} (-1)^{n} \frac{(x+2)^{n}}{3^{n+1}} provided that | \frac{(-x - 2)}{3} | < 1 or,  -5 < x < 1

    They aren't equal, in this one I have a (-1)^{n} kicking around in my sum.

    What am I doing wrong here?
    The 2nd series is not centered at x=2 it is centered at x=-2

    Notice that

    \displaystyle \frac{1}{3} \left( \frac{1}{1 + (\frac{-x-2}{3})} \right)=\frac{1}{3} \left( \frac{1}{1 - (\frac{x+2}{3})} \right)
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Jan 2010
    Posts
    144
    Quote Originally Posted by TheEmptySet View Post
    The 2nd series is not centered at x=2 it is centered at x=-2

    Notice that

    \displaystyle \frac{1}{3} \left( \frac{1}{1 + (\frac{-x-2}{3})} \right)=\frac{1}{3} \left( \frac{1}{1 - (\frac{x+2}{3})} \right)
    But that exact term is also in my 1st derivation of the question.

    Are they both centered at x = -2?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,412
    Thanks
    1328
    They are both centered at x= -2 but there is a missing sign in your second calculation.

    For a geometric series, \displaytype\sum_{n=0}^r ar^n= \frac{a}{1- r}.

    In your first calculation, you have \frac{\frac{1}{3}}{1- \frac{x+2}{3}} so a= \frac{1}{3} and r= \frac{x+2}{3}. Your calculations are correct.

    In your second calculation, you have \frac{\frac{1}{3}}{1+ \frac{-x-2}{3}} and that is NOT of the correct form for the geometric series. You have \frac{a}{1+ r}, not \frac{a}{1- r}

    You could then say \frac{a}{1+r}= \frac{a}{1-(-r)}= \sum_{n=0}^\infty a(-r)^n and replacing "r" with "-r" cancels that incorrect (-1)^n
    Last edited by HallsofIvy; September 29th 2010 at 07:21 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 0
    Last Post: January 26th 2010, 08:06 AM
  2. geometric power series
    Posted in the Calculus Forum
    Replies: 4
    Last Post: December 8th 2009, 09:30 AM
  3. Replies: 11
    Last Post: April 1st 2008, 12:06 PM
  4. taylors series
    Posted in the Calculus Forum
    Replies: 4
    Last Post: May 23rd 2007, 01:05 PM
  5. find the sum of the series (taylors fomula)
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 8th 2007, 09:48 AM

Search Tags


/mathhelpforum @mathhelpforum