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Homework Help: Taylor Series

  1. Dec 20, 2005 #1
    [tex] \frac {1}{1-z} = \sum_{n=0}^{\infty} (\frac {(z-i)^n}{(1-i)^{n+1}}[/tex]
    we have to prove that, its our problem. So we start and we get,
    [tex] \frac {1}{1-z} = \frac {1}{(1-i)-(z-i)} = \frac {1}{1-i} (\frac {1}{1- \frac {z-i}{1-i}})[/tex]
    and then i think we're supposed to go to a power series or something, but i don't know, i'm not sure how to get to there, from here. help?
     
  2. jcsd
  3. Dec 20, 2005 #2

    Tide

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    You can use a Taylor series but it's also equivalent to a Binomial expansion which you should be able to handle.

    Here's a third way:

    [tex]\frac {1}{1-x} = \frac {1-x+x}{1-x} = 1 + \frac {x}{1-x}[/tex]

    and continue to replace the [itex]\frac {1}{1-x}[/itex] with [itex]1 + \frac {x}{1-x}[/itex] on the right side to obtain the desired expansion. Notice that -1 < x < 1 is required for convergence.
     
  4. Dec 20, 2005 #3
    does it matter then that z is a complex variable? it doesn't seem to make sense to say -1< z <1 if z is complex right? so then?

    and anyway, it says to use taylor series...
     
  5. Dec 20, 2005 #4

    benorin

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    [tex] \frac {1}{1-z} = \frac {1}{1-i} (\frac {1}{1- \frac {z-i}{1-i}})[/tex]

    Apply geometric series expansion to the right hand side. QED.
     
  6. Dec 21, 2005 #5

    shmoe

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    Then you'd talk about bounds on |z|

    have you tried directly finding the power series of 1/(1-z)? What point do you want to expand it about?
     
  7. Dec 21, 2005 #6
    Alright, so i guess, [tex]\frac {1}{1-i} (\frac {1}{1- \frac {z-i}{1-i}})= \frac {1}{1-i} (1 + \frac {z-i}{1-i} +({ \frac {z-i}{1-i})^2 + .... + (\frac{z-i}{1-i})^n ) = \frac {1}{1-i} \sum_{n=0}^{\infty}(\frac {1}{1- \frac {z-i}{1-i}})^n= \sum_{n=0}^{\infty} (\frac {(z-i)^n}{(1-i)^{n+1}}= \frac {1}{1-z} [/tex]

    look about right? i've never done series much before, so i was just a bit lost, but i think that seems right if i use the power series as mentioned.

    only question: are there restrictions on z? i'm not really sure how to find them if so.
     
    Last edited: Dec 21, 2005
  8. Dec 21, 2005 #7

    Tide

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    Gale,

    That is correct. It would simplify matters to replace [itex]\frac {z-i}{1-i}[/itex] with something like [itex]w[/itex] (call it whatever you want) so that you are expanding [itex]\frac {1}{1-w}[/itex] in your original equation. Then replace [itex]w[/itex] with [itex]\frac {z-i}{1-i}[/itex] when you're done.

    And, yes, there is a restriction on z which amounts to the absolute value of [itex]w[/itex] being less than 1.
     
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