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Power series

  1. Jan 30, 2012 #1
    [tex]z\in\mathbb{Z}[/tex]

    [tex]\frac{1}{1-(-z)}=\sum_{n=0}^{\infty}(-z)^n[/tex]

    [tex]\frac{1}{(z+2)^2}=\frac{d}{dz} \frac{-1}{1-(1-z)} = \frac{d}{dz} (1 + (1-z) + (1-z)^2+\cdots = 0 -1 -2(1-z)-3(1-z)^2 - \cdots = \sum_{n=0}^{\infty} ???[/tex]

    Not to sure about the second one.
     
    Last edited: Jan 30, 2012
  2. jcsd
  3. Jan 31, 2012 #2
    [tex]\frac{-1}{1-(1-z)}=\frac{-1}{z}[/tex]
    [tex]\frac{d}{dz}\frac{-1}{z}=\frac{1}{z^2}\neq\frac{1}{(z+2)^2}[/tex]
     
  4. Jan 31, 2012 #3
    There's a pretty well known power series for [itex]\frac{1}{1-x}[/itex]
    If you let x=-z, what do you get?
     
  5. Jan 31, 2012 #4
    What you can do is [tex] \frac{d}{dx} \frac{1}{(1+x)} = -\frac{1}{(1+x)^2} [/tex]
    If you just define y=x-1, you get
    [tex]\frac{1}{(2+y)^2}=-\frac{d}{dy}\sum_{n=0}^\infty (-(y+1)^n)[/tex]
    However, you still need to make sure that you are inside the radius of convergence, so |y+1|<1
     
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