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Centre of power series

  1. Jul 20, 2014 #1
    Hello.
    I need someone to explain to me how to find the centre and radius of convergence of power series.
    I got the working and the answers but there are some things I don't understand.

    $$\sum_{n=0}^{\infty}\frac{(4i)^n(z-i)^n}{(n+1)(n+2)}$$

    Using the ratio test, we got
    $$\lim_{{n}\to{\infty}} \frac{4i(z-i)(n+1)}{n+3}$$=4i(z-i)

    Ok, in this part, why is the limit 4i(z-i)? Don't we have to divide all the terms by n?

    And the final answer is: $$R=1/4, z=i$$

    Why does the centre become i?
     
  2. jcsd
  3. Jul 20, 2014 #2
    A power series expansion is an infinite series
    \begin{equation*}
    \sum_{n=0}^{\infty} a_n (z-c)^n,
    \end{equation*}
    where ##a_n## are coefficients and ##c \in \mathbb{C}## is the center (the point we are expanding about).

    I think you can figure out why
    \begin{equation*}
    \lim_{n \rightarrow \infty} \frac{4i(z-i)(n+1)}{n+3} = 4i(z-i).
    \end{equation*}
    The easiest might be to substitute ##k = f(n)##, for some well-chosen function ##f##.
     
  4. Jul 20, 2014 #3

    HallsofIvy

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    This is incorrect. The "ratio test" only applies to series of non-negative real numbers. In order to apply the ratio test to more general series, you must take the absolute value
    [tex]\lim{m\to\infty} \frac{4|z- i|(n+1)}{n+ 3}= 4|z- i|\lim_{n\to\infty} \frac{n+1}{n+ 3}[/tex].

    Well, what do you get when you "divide all term" of [itex]\frac{n+1}{n+ 3}[/itex] by n?

    The "ratio" test says that series [tex]\sum a_n[/tex] converges if [tex]\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|< 1[/tex]. What is [tex]\lim_{\n to\infty}4|z- i|\frac{n+1}{n+ 3}[/tex]? For what values of z is that less than 1?
     
  5. Jul 20, 2014 #4
    The answer to your first question is 1 and for the second, z≤i. Right?
     
  6. Jul 21, 2014 #5

    HallsofIvy

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    "[itex]z\le i[/itex]" doesn't even make sense. The complex numbers are NOT an "ordered" field. You keep forgetting the absolute value!
     
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