Power series

  • Thread starter fauboca
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  • #1
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[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.
 
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Answers and Replies

  • #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]
 
  • #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?
 
  • #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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