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Taylor series of z^i at z=i

  1. Apr 26, 2005 #1
    I need to find the first three terms of this series.

    Am I correct in saying z^i = exp(i*Log(z)), then using the taylor series for e^z, giving me:

    (i*Log(z)) - 1/2*(Log(z))^2 - i/6*(Log(z))^3 + 1/24*(Log(z))^4 + ....

    I haven't worked it out, but this seems to mean that the coefficients for every term in the Taylor series for z^i is actually an infinite series, found by combining terms in the Log(z) Taylor series.

    This seems like I could be way off though, so I don't want to do the calculations if I don't have to.

    Thanks in advance.
     
  2. jcsd
  3. Apr 26, 2005 #2

    HallsofIvy

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    I do not understand this at all! A "Taylor's series" of a function f(z) is, by definition,
    a power series in z. You said in your title that this was to be "at z= i" so your series should have (z- i), (z- i)2, etc. The coefficients would be, of course,
    f(i), f '(i)/2, etc. which are f(i)= ii, f '(i)= i(ii-1)= ii,
    f"(i)= i(i-1)ii-2= ii-2ii-1 etc so that the Taylor's series is ii+ ii(z-i)+ ((ii- 2ii-1)/2)(z-i)2+ . . .
     
  4. Apr 26, 2005 #3

    mathman

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    zi=(i+w)i, where w=z-i
    Use the binomial expansion to get
    ii+i*i(i-1)(z-i)+0.5*i(i-1)i(i-2)(z-i)2
     
  5. Apr 26, 2005 #4

    Hurkyl

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    Why not just apply the definition of Taylor series?

    [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]
    ...

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

    ...
     
  6. Apr 26, 2005 #5
    Firstly, I made a mistake in my OP. It should be at [tex]z = 1 + i[/tex], but that shouldn't make a huge difference. The problem is that terms such as [tex](z)^i[/tex] aren't allowed. We need to restate them as [tex]e^i Log(z)[/tex]. So in your last line, it would have to be:

    [tex]f(z) = e^{i Log(i))} + i e^{i Log(i)} (z- i- 1) +\cdots[/tex]

    Does that sound right? When my professor asks for the "first three terms" is it just the first three of these?
     
  7. Apr 26, 2005 #6

    Hurkyl

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    [itex]i^i[/itex] isn't a term of the form [itex]z^i[/itex]. :biggrin: And, as you point out, one usually defines [itex]i^i = \exp(i \mathop{Log} i)[/itex], so one doesn't really need to convert. By the way, you can simplify that expression quite a bit!

    Anyways, you can't replace (z-i) with (z-i-1) in this Taylor series any more than you can replace z with (z-1) to get the (wrong) Taylor series: [itex]e^z = 1 + (z-1) + (z-1)^2/2! + (z-1)^3/3! + \cdots[/itex].
     
  8. Apr 26, 2005 #7
    Ok, basically I'm clearly being an idiot today. I've just not been thinking this through. The first three terms of the correct Tayler series expansion for [tex]z^i[/tex] about [tex]z = 1+i[/tex] should be (I hope) something along the lines of:

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

    That look ok? I realize it can be simplified, I can do that on my own (hopefully).
     
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