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Math Help - Taylor polynomial of a complex function

  1. #1
    MHF Contributor arbolis's Avatar
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    Taylor polynomial of a complex function

    Find Taylor's polynomial of the following function in z_0=-1 : f(z)=\frac{1}{z^2}.
    Determine the radius of convergence of the series.


    My attempt :
    f(-1)=\sum _{n=0}^{\infty} \frac{(-1)^n(n+1)}{z^{n+2}}(z+1)^n. Which answers the first question.

    Now trying to figure out the second question :
    So a_n=\frac{(-1)^n(n+1)}{z^{n+2}}.
    I apply the ratio test : \left| \frac{a_{n+1}}{a_n} \right| =\frac{(-1)(n+2)}{z} which tends to -\frac{1}{z} when n tends to \infty.
    Hence the radius of convergence is ...? It makes no sense at all.
    I'd like to know where is(are) my error(s). Thanks a lot!
    Last edited by arbolis; October 17th 2009 at 08:27 PM.
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  2. #2
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    Quote Originally Posted by arbolis View Post
    Find Taylor's polynomial of the following function in z_0=-1 : f(z)=\frac{1}{z^2}.
    Determine the radius of convergence of the series.


    My attempt :
    f(-1)=\sum _{n=0}^{\infty} \frac{(-1)^n(n+1)!}{z^{n+2}}(z+1)^n. Which answers the first question.

    [snip]
    No. This is definitely not a Taylor Series. Do you see why?

    Using the formula given in the theorem, the required Taylor Series is 1 + 2(z + 1) + 3(z + 1)^2 + 4(z + 1)^3 + .... = \sum_{n = 0}^{+ \infty}(n + 1) (z + 1)^n. Do you see why?
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  3. #3
    MHF Contributor arbolis's Avatar
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    Quote Originally Posted by mr fantastic View Post
    No. This is definitely not a Taylor Series. Do you see why?

    Using the formula given in the theorem, the required Taylor Series is 1 + 2(z + 1) + 3(z + 1)^2 + 4(z + 1)^3 + .... = \sum_{n = 0}^{+ \infty}(n + 1) (z + 1)^n. Do you see why?
    I made a mistake, I've edited my post. However I still don't reach the Taylor series you provided. I don't see why the series I found is not a Taylor series.

    I've used the fact that f(z)=\sum _{n=0}^{\infty} \frac{f^{(n)}(z-z_0)}{n!}.
    I've found out that f^{(n)}=\frac{(-1)^n (n+1)!}{z^{(n+2)}}.

    Now you make me doubt the formula holds for the function z \mapsto \frac{1}{z^2} due to the singularity at z=0.
    Edit : I understand the equality you gave, but I don't know how to reach it. I mean I don't know how to reach that the Taylor Series is 1 + 2(z + 1) + 3(z + 1)^2 + 4(z + 1)^3 + ...


    Edit 2 : Sorry, I just found my error. I forgot to evaluate f^{(n)} in z_0...
    I think I'll do it. Thanks a lot!
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  4. #4
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by arbolis View Post
    I made a mistake, I've edited my post. However I still don't reach the Taylor series you provided. I don't see why the series I found is not a Taylor series.

    I've used the fact that f(z)=\sum _{n=0}^{\infty} \frac{f^{(n)}(z-z_0)}{n!}.
    I've found out that f^{(n)}=\frac{(-1)^n (n+1)!}{z^{(n+2)}}.

    Now you make me doubt the formula holds for the function z \mapsto \frac{1}{z^2} due to the singularity at z=0.
    Edit : I understand the equality you gave, but I don't know how to reach it. I mean I don't know how to reach that the Taylor Series is 1 + 2(z + 1) + 3(z + 1)^2 + 4(z + 1)^3 + ...


    Edit 2 : Sorry, I just found my error. I forgot to evaluate f^{(n)} in z_0...
    I think I'll do it. Thanks a lot!
    Yes, \frac{(-1)^n}{z^{n+2}} evaluated at z=-1 always equals 1.
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  5. #5
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    The radius of convergence of a function will extend to the nearest point at which the function is not analytic. Here, the function is \frac{1}{z^2} which is not analytic at z= 0. Since the center of the region of convergence is at z_0= -1, the radius of convergence is |0-(-1)|= 1.

    This can be used even for real functions. For example, the function \frac{1}{1+ x^2} is analytic for all x except -i or -i. If it is expanded in a power series about x= 0, then its radius of convergence is |0- i|= 1.

    If it is expanded in a power series about x= 1, its radius of convergence is |1- i|= \sqrt{2}.
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  6. #6
    MHF Contributor arbolis's Avatar
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    Quote Originally Posted by HallsofIvy
    The radius of convergence of a function will extend to the nearest point at which the function is not analytic.
    So an entire function always has an infinite radius of convergence?
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