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Gaah! finding a power series out of a function

  1. Nov 13, 2007 #1
    1. Express the function as the sum of a power series by first using partial fractions. Find the interval of convergence.

    f(x)= (7x-1)/(3x^2 +2x-1)




    2. using the fact that 1/(1-x) =[tex]\sum[/tex] from [tex]\infty[/tex] to n=0 of x^n and the interval for convergence of that is (-1,1)



    3. I know that (7x-1)/(3x^2 +2x-1) can be factored to (7x-1)/(3x-1)*1/(1+x).


    So, using the idea of partial fractions and the method for changing functions into power series, 1/(1+x) would become 1/(1-(-x)).

    Let u=-x so that [tex]\sum[/tex] from infinity to n=0 is (-1)^n*x^n , interval of convergence is (-1,1).


    But I know don't know how to do (7x-1)/(3x-1). Is there some way to rewrite that? Like (1-7x)/(1-3x), but I still wouldn't know what to substitute...(-7/3)x???? At the end I would have the sum of two series, right?? Help...
     
  2. jcsd
  3. Nov 13, 2007 #2

    rock.freak667

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    Well you can use (7x-1)/(3x^2 +2x-1)

    to be equal to [tex]\frac{A}{3x-1} + \frac{B}{x+1}[/tex]
    [tex] \frac{A(x+1)+B(3x-1)}{(3x-1)(1+x)}=\frac{7x-1}{(3x-1)(1+x)}[/tex]

    then equate numerators and find the values for A and B
    then put the 2 partial fractions as power series and use the operations that you can do on a power series to put them into one complete series
     
  4. Nov 13, 2007 #3
    Okay, so I have 1/(3x-1) + 2/(x+1)

    when I went through and did all the work, I got this

    (1/2)[tex]\sum[/tex](-1)^n(x)^n - [tex]\sum[/tex](3)^n(x)^n

    Now when I put them together would I get ...

    (-1/2) [tex]\sum[/tex][(-1)^n -(3)^n]x^n ...?

    I'm not sure about that 1/2 thing... because I'm not sure if 2/(1+x) really = (1/2)[tex]\sum[/tex] (-x)^n. I want to know for sure because I have an exam in a couple days and my professor had to rush over the last few things we learned about series, which leaves me worried about making stupid small mistakes that could be easily avoided!
     
  5. Nov 13, 2007 #4

    rock.freak667

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    well if you can represent [tex]\frac{1}{1+x} = \sum_{n=0}^\infty (-1)^nx^n[/tex] then 2/(1+x) should be 2 by the power series
     
  6. Nov 13, 2007 #5
    oh duh! 2 not 1/2!! I don't know what I was thinking...
     
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