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Function represented as a power series

  1. Aug 3, 2010 #1
    1. The problem statement, all variables and given/known data

    [url]http://img.photobucket.com/albums/v257/gamer567/powerseries.png[/url]

    2. Relevant equations



    3. The attempt at a solution

    [url]http://img.photobucket.com/albums/v257/gamer567/cramster.png[/url]


    I am getting lost from the transition in the 1/x-2 to 1/(x-2)^2. Could someone explain to me what happened? I think I could finish this problem and several others if I could understand this. Also is there a better approach? In other problems I took the derivative and put into a power series and then took the integral. This doesn't seem to be working on this one(?)
     
  2. jcsd
  3. Aug 3, 2010 #2
    Apologies I am unable to edit, the equation in the picture needs to be represented as a power series if I didn't make it clear.
     
  4. Aug 3, 2010 #3
    Before differentiating, the right-hand side can be rewritten as

    [tex] \frac{-1}{2} \sum \frac{x^n}{2^n} = - \sum \frac{x^n}{2^{n+1}} [/tex]

    Thus, we differentiate to get:

    [tex] \frac{d}{dx}[\frac{1}{x-2}] = \frac{d}{dx} [ - \sum \frac{x^n}{2^{n+1}}] [/tex]

    [tex] \frac{-1}{(x-2)^2} = - \sum \frac{n}{2^{n+1}}x^{n-1} [/tex]

    The reason we do this is because now we can multiply both sides by -x to give us:

    [tex] \frac{x}{(x-2)^2} = \sum \frac{n}{2^{n+1}}x^{n}. [/tex]

    Do you see how this is helpful in trying to get to the power series of f(x)?
     
  5. Aug 3, 2010 #4
    As far as I know, this is the standard and easiest approach. You should learn and understand this method because I wouldn't be surprised if it pops up in some other problems.
     
  6. Aug 3, 2010 #5

    Awesome that helps :) Got confused when they rearranged stuff while saying they are taking derivative.

    No doubt this would be a nice mean test question :( Shall be redoing this one a couple times.
     
  7. Aug 3, 2010 #6
    cramster2.png

    ^^ New problem

    If you dont mind me asking one more question, I am stuck as to how they got the last line of this solution to another problem.

    From what I can tell, they plug in n=0 and got 1/(x^2) and then changed the power series to n=1

    1/(x^2)-1/(x^2) cancels out and disappears. The thing I don't understand is how (-1)^n becomes (-1)^(n+1)

    What am I missing here? The answer in the back of the book has it as well.
     
  8. Aug 3, 2010 #7
    There's a negative sign in front of [tex] tan^{-1}x. [/tex] They bring it into its power series formula, thus raising the exponent on -1 to n + 1.
     
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