Results 1 to 12 of 12
Like Tree1Thanks
  • 1 Post By skeeter

Math Help - Power series

  1. #1
    Member
    Joined
    Aug 2011
    Posts
    117

    Power series

    I have a problem i cant seem to figure out. I am trying to find the power series expansion for the function

    f(x) = \frac {9}  {5 + x}
    c = 5

    Now, would i arrange the 5 + x to force a negative?

    f(x) = \frac {9}  {5 - (-x)}

    then, subtract C? I think this is where i get confused on how to handle C correctly!

    f(x) = \frac {9}  {-5 + 5 - (-x -5  ) }

    f(x) = \frac {9}  {-5 + 5 - (-x -5  ) }

    f(x) = \frac {9}  {0 - (-x -5 ) }
    But, somehow this isnt correct..
    Last edited by icelated; May 15th 2012 at 12:50 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,648
    Thanks
    1596
    Awards
    1

    Re: Power series

    Quote Originally Posted by icelated View Post
    I have a problem i cant seem to figure out.

    f(x) = \frac {9}  {5 + x}
    c = 5

    Now, would i arrange the 5 + x to force a negative?

    f(x) = \frac {9}  {5 - (-x)}

    then, subtract C? I think this is where i get confused on how to handle C correctly!

    f(x) = \frac {9}  {-5 + 5 - (-x -5  ) }

    f(x) = \frac {9}  {-5 + 5 - (-x -5  ) }

    f(x) = \frac {9}  {0 - (-x -5 ) }
    But, somehow this int correct..
    What is the point?
    What is the question?
    What are you trying do?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Aug 2011
    Posts
    117

    Re: Power series

    Sorry, i am trying to find the power series expansion for the function
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,690
    Thanks
    448

    Re: Power series

    f(x) = \frac{9}{10 - (-x+5)} = \frac{9}{10} \cdot \frac{1}{1 - \frac{5-x}{10}}

    \frac{9}{10} \left(1 - \frac{(x-5)}{10} + \frac{(x-5)^2}{10^2} - \frac{(x-5)^3}{10^3} + ... \right)
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Aug 2011
    Posts
    117

    Re: Power series

    Why 10 - (-x+5) can you show me how to properly add the C? i just dont see how you got 10
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,690
    Thanks
    448

    Re: Power series

    Quote Originally Posted by icelated View Post
    Why 10 - (-x+5) can you show me how to properly add the C? i just dont see how you got 10
    uhh ... 10 - 5 = ?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Aug 2011
    Posts
    117

    Re: Power series

    Sorry, still confused. I thought we are to follow this form (x-c) -c
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,690
    Thanks
    448

    Re: Power series

    Quote Originally Posted by icelated View Post
    Sorry, still confused. I thought we are to follow this form (x-c) -c
    no, you want the form k - (x-c)

    in this case, k - [-(x - c)] ... see why k = 10?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,693
    Thanks
    1466

    Re: Power series

    Another way to do that is to write
    \frac{9}{x+5}= \frac{9}{x-(-5)}
    and now divide both numerator and denominator by -5:
    \frac{-\frac{9}{5}}{1- \frac{x}{5}}

    If you let y= x/5 that becomes
    -\frac{9}{5}\frac{1}{1- y}
    which is, of course, the sum of the geometric series
    -\frac{9}{5}\sum y^n= -\frac{9}{5}\sum \frac{x^n}{5^n}
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,690
    Thanks
    448

    Re: Power series

    Quote Originally Posted by HallsofIvy View Post
    Another way to do that is to write
    \frac{9}{x+5}= \frac{9}{x-(-5)}
    and now divide both numerator and denominator by -5: <--- check this step ...
    \frac{-\frac{9}{5}}{1- \frac{x}{5}}
    ...
    Thanks from HallsofIvy
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Member
    Joined
    Aug 2011
    Posts
    117

    Re: Power series

    Quote Originally Posted by skeeter View Post
    no, you want the form k - (x-c)

    in this case, k - [-(x - c)] ... see why k = 10?
    I understand the form but we are to subtract c from x and then add c to k ? You seem to be forcing 10. I have not done this in the homework at all.
    So, would this be correct 9 / -5 + (x - 5) - 5 ??
    Last edited by icelated; May 15th 2012 at 07:33 PM.
    Follow Math Help Forum on Facebook and Google+

  12. #12
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,690
    Thanks
    448

    Re: Power series

    Quote Originally Posted by icelated View Post
    I understand the form but we are to subtract c from x and then add c to k ? You seem to be forcing 10. I have not done this in the homework at all.
    So, would this be correct 9 / -5 + (x - 5) - 5 ??
    exactly ... to get the denominator in the desired form to write a power series centered at 5


    no, 9/[-5 + (x-5) - 5] = 9/(x - 15)


    ... probably because you have not come across a problem (like this one) that does not follow a "cook-book" procedure. I recommend you speak w/ your instructor about it.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: September 29th 2010, 06:11 AM
  2. Replies: 0
    Last Post: January 26th 2010, 08:06 AM
  3. Formal power series & Taylor series
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: April 19th 2009, 09:01 AM
  4. Sequences and Series - Power Series Question
    Posted in the Calculus Forum
    Replies: 3
    Last Post: April 20th 2008, 07:32 PM
  5. Replies: 10
    Last Post: April 18th 2008, 10:35 PM

Search Tags


/mathhelpforum @mathhelpforum