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Power Series Question

  1. May 27, 2009 #1
    I need to find [tex] \frac{1}{i+z} [/tex] as a power series in z.

    I want to know if am doing this right.

    If i use the taylor series here by doing
    [tex]
    f(z) = z^i
    [/tex]

    [tex]
    f'(z) = i z^{-1} z^i
    [/tex]

    [tex]
    f''(z) = i (i-1) z^{-2} z^i
    [/tex]

    This taylor series is just for z= i +1, but i tried using it for my problem but i dont seem to get the right answer.
    this is the taylor series that i should be using but how do i find f(i) here?
    [tex]

    f(z) = f(i) + f'(i) (z-i) + f''(i) (z-i)^2 + \cdots

    [/tex]


    cheers
     
    Last edited: May 27, 2009
  2. jcsd
  3. May 27, 2009 #2

    Mark44

    Staff: Mentor

    If you needed to find the power series for 1/(1 - x), you could do it by getting the various derivatives, but an easier way would be just to do the long division. I would give that a shot first with the function you have. The thing about a power series is that it is unique, so however you get it, that's the one.
     
  4. May 27, 2009 #3
    oh yup thats cool.. i just had another question.. will i be finding the derivative of f(z) or f(i) for the taylor polynomial?
     
  5. May 27, 2009 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    I'm not sure you are getting it. f(i) is a constant. And you don't have to take any derivatives. As Mark44 said, you can expand 1/(i+z) in a power series without taking any derivatives. Write it as (1/i)*(1/(1-iz)). Now expand 1/(1-iz) as a geometric series. 1/(1-x)=1+x+x^2+x^3+... Does that ring a bell?
     
  6. May 27, 2009 #5
    You can do any of the short cuts given already (long division / geometric series), but if you want to grind the Taylor expansion machinery, then you are doing two things wrong.

    I read the quotation above to mean an expansion about z=0. What you are trying to do is an expansion about z=i. That is mistake number 1. (As an aside, I don't think you can do that because z=i is a singularity and I don't think your power series can be done around a singularity. However, I may not be 100% correct in this)

    Your second mistake is that the function f(z) = 1 / (i+z). Why did you use z^i? Take the derivatives of the full f(z) and plug in z=0 the get the series coefficients.

    If you use the function as given, and expand around z=0, then you should get the same thing as the short cut methods.
     
  7. May 28, 2009 #6

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    You could also use the fact that the sum of the geometric series
    [tex]\sum_{n= 0}^\infty r^n= \frac{1}{1- r}[/tex]
    That's similar to Mark44's suggestion.

    Oh, and
    [tex]\frac{1}{i+ x}= \frac{1}{i(1+ \frac{x}{i})}= -i\frac{1}{1- (-\frac{x}{i})}[/tex]
     
  8. May 28, 2009 #7
    sweet as.. got it.. cheers
     
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