1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Science Advisor
    Homework Helper

    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


    User Avatar
    Homework Helper

    [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


    User Avatar
    Science Advisor
    Homework Helper

    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


    User Avatar
    Science Advisor
    Homework Helper


    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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Taylor Series
  1. Taylor series (Replies: 3)

  2. Taylor series (Replies: 14)

  3. Taylor Series (Replies: 1)

  4. Taylor Series (Replies: 2)

  5. Taylor series (Replies: 15)