Results 1 to 3 of 3

Math Help - Power series representation????

  1. #1
    Newbie
    Joined
    Apr 2008
    Posts
    1

    Power series representation????

    Can anyone please help me???

    It asks to find the power series representation and radius of convergence for

    f(x) = ln(a - x) for a>0
    As an answer I got the series 1 to infinity of
    __x^n___
    a^n+1
    I don't know if it is right and the radius of convergence I got was a but I don't know if that is right

    f(x) = __x^3___
    (1-4x)^2
    I have no idea how to do this one...

    Please help me
    Last edited by Blackberry410; April 15th 2008 at 04:39 PM. Reason: Wrong formatting
    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 Blackberry410 View Post
    Can anyone please help me???

    It asks to find the power series representation and radius of convergence for

    f(x) = ln(a - x) for a>0
    As an answer I got the series 1 to infinity of
    __x^n___
    a^n+1
    I don't know if it is right and the radius of convergence I got was a but I don't know if that is right

    f(x) = __x^3___
    (1-4x)^2
    I have no idea how to do this one...

    Please help me
    Let

    f(x)=\ln(a-x) then
    n
    f'(x)=\frac{-1}{a-x}=-\frac{1}{a} \cdot \frac{1}{1-\frac{x}{a}}

    using the defintion of the geometric series we get

    f'(x)=-\frac{1}{a} \sum_{n=0}^{\infty} \left( \frac{x}{a}\right)^n= -\sum_{n=0}^{\infty} \frac{x^n}{a^{n+1}}

    integrating both sides we get

    f(x)= \ln(a-x)= \left( - \sum_{n=0}^{\infty} \frac{x^{n+1}}{a^{n+1}(n+1)} \right) +C

    To find the constant C we evaluate

     f(0)= \ln(a)= \left( - \sum_{n=0}^{\infty} \frac{0^{n+1}}{a^{n+1}(n+1)} \right) +C \iff \ln(a) = C

    f(x)= \ln(a-x)= \ln(a) - \sum_{n=0}^{\infty} \frac{x^{n+1}}{a^{n+1}(n+1)}

    Since this based on the geometric series it will converge when r < 1 so

    |\frac{x}{a}| < 1 \iff -1 < \frac{x}{a} < 1 \iff -a < x < a
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78

    Lightbulb hint for the 2nd part

    let g(x)=\frac{1}{4} \cdot \frac{1}{1-4x}=\sum_{n=0}^{\infty}4^{n-1} x^n

    note that:

     \frac{x^3}{(1-4x)^2} = x^3 \cdot g'(x)
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Power series representation
    Posted in the Calculus Forum
    Replies: 1
    Last Post: September 23rd 2010, 08:32 AM
  2. Power Series Representation
    Posted in the Calculus Forum
    Replies: 4
    Last Post: April 25th 2010, 02:19 PM
  3. power series representation!
    Posted in the Calculus Forum
    Replies: 5
    Last Post: April 19th 2009, 08:08 PM
  4. power series representation #2
    Posted in the Calculus Forum
    Replies: 3
    Last Post: April 12th 2009, 06:37 PM
  5. Power series representation
    Posted in the Calculus Forum
    Replies: 4
    Last Post: February 5th 2008, 01:17 PM

Search Tags


/mathhelpforum @mathhelpforum