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Power series expansion and largest disc of validity

  1. Oct 14, 2009 #1
    1. The problem statement, all variables and given/known data

    Find the power-series expansion about the given point for the function; find the largest disc in which the series is valid.

    f(z) = z^3 + 6z^2-4z-3 about z0=1.

    2. Relevant equations



    3. The attempt at a solution

    The series is fine. Since it's a polynomial, there are only three non-zero terms. I get:

    f(x) = 11(x-1) + 9(x-1)^2 + (x-1)^3.

    But I am very confused about the second part of the problem, the disc. I know that as a consequence of Cauchy's theorem, if f is analytic in a domain and there is a disk of some radius, say r, inside the domain, then f can be written as a (convergent) power series inside the disc. But I don't see how to find the disk! Unless it can be simply written as |x-x0| <= r. This kind of makes sense because you'd want the disc to be able to shrink about the point x0, where the expansion is going around.

    OR: Should be coefficients in the power series be something other than the normal I get from Taylor series? Is there some special technique that I'm not using?
     
  2. jcsd
  3. Oct 14, 2009 #2

    Dick

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    I think you are just confused because you are used to harder problems. Are there any values of x for which the 'series' doesn't converge?
     
  4. Oct 14, 2009 #3
    Hm, I suppose that if z=1, then each term would be zero. This would be bad, because there's another theorem that says if each term is zero then f(z) = 0 in the entire domain. Which can't be true because we know what f(z) is, and it's not 0! :)
     
  5. Oct 14, 2009 #4

    Dick

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    That's a mighty strange theorem. Can you quote it?
     
  6. Oct 14, 2009 #5
    Ok, I looked and I was mis-quoting something. The theorem says that if f is analytic in a domain D and further at some point zo in D we have f^(k)(z0), k=0,1,2,... Then f(z) = 0 for all z in D.

    Let's see - some if I understand, this says that if I pick a point in the domain D where all the derivatives of analytic f vanish, then f(z) = 0.
     
  7. Oct 14, 2009 #6

    Dick

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    That's fine, so it doesn't apply here, right?
     
  8. Oct 14, 2009 #7
    I suppose not *sigh*.

    What if z gets very large? (I meant to type z, not x, in my original post). I suppose if z were to get big, the whole thing would blow up and we'd lose convergence of the "series". Perhaps I am making it harder than it is :)
     
  9. Oct 14, 2009 #8

    Dick

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    You are right. You are making it harder than it is. The only 'point' where your series diverges is the 'point' at infinity. The way the question is phrased is also making it seem harder than it is.
     
  10. Oct 14, 2009 #9
    Ha! Well, it turns out that I am good at making things harder then they need to be. Thanks for the help! :-D
     
  11. Oct 15, 2009 #10

    HallsofIvy

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    To find the power series for [itex]f(z) = z^3 + 6z^2-4z-3[/itex] about [itex]z_0= 1[/itex]:

    1. Replace z in the polynomial by x+1.

    2. Replace x by z.

    The resulting polynomial is the power series you want. Any polynomial is analytic at any point. And any function, analytic everywhere (an "entire" function) has power series that converges for all x.
     
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