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Math Help - Power Series (Complex Analysis)

  1. #1
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    Power Series (Complex Analysis)

    Find the radius of convergence of the given power series.

    (1) {\sum_{k=0}}^{\infty} \frac{(k!)^2}{(2k)!}(z-2)^k

    (2) {\sum_{j=0}}^{\infty} \frac{z^{3j}}{2^j}

    I'm not sure how to use the theorems given in my book to solve these.

    Theorem: Suppose that {\sum_0}^{\infty} a_n(z-z_0)^n is a power series with a positive or infinite radius of convergence R.

    (a) If lim_{n \to\infty} |a_{n+1}/a_n| exists, then \frac{1}{R} = lim_{n \to\infty}|\frac{a_{n+1}}{a_n}|

    (b) If lim_{n\to\infty}\sqrt[n]{|a_n|} exists, then \frac{1}{R} =lim_{n\to\infty}\sqrt[n]{|a_n|}

    Any help would be appreciated, thanks!
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  2. #2
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    try the ratio test
    so a(n+1)/a(n) = L
    and the series is only convergent when L < 1

    so if you can come up with an inequality and then solve for z
    Last edited by jbpellerin; September 21st 2008 at 08:14 PM. Reason: typo
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    (1) use ratio test:
    \lim_{n\to\infty}\frac{a_n}{a_{n+1}}=\lim\frac{(2n  +2)(2n+1)}{(n+1)^2}=4, so R=4 (I think you have the mistake in the theorem in part (a))

    (2) denote z^3=t. Then we have the series \sum_{j=1}^{\infty}t^j/2^j , so use (b):
    \lim a_n^{1/n}=\lim 1/2=1/2, so the series converge for any |t|<2. This means the original series converge for any |z|=|t|^{1/3}<2^{1/3}, i.e. R=2^{1/3}
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  4. #4
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    Quote Originally Posted by choovuck View Post
    (I think you have the mistake in the theorem in part (a))
    And why ?
    Isn't it equivalent to what you have done ? In any book talking about it, I've seen this rule stated this way... :/
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  5. #5
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    Oh, I think initially he wrote 1/R=\lim \frac{|a_n|}{|a_{n+1}|} rather than 1/R=\lim \frac{|a_{n+1}|}{|a_{n}|}. So either he edited his post after my reply, or I just got confused. in any case, the answer is right.
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    Quote Originally Posted by choovuck View Post
    (1) use ratio test:
    \lim_{n\to\infty}\frac{a_n}{a_{n+1}}=\lim\frac{(2n  +2)(2n+1)}{(n+1)^2}=4, so R=4 (I think you have the mistake in the theorem in part (a))

    (2) denote z^3=t. Then we have the series \sum_{j=1}^{\infty}t^j/2^j , so use (b):
    \lim a_n^{1/n}=\lim 1/2=1/2, so the series converge for any |t|<2. This means the original series converge for any |z|=|t|^{1/3}<2^{1/3}, i.e. R=2^{1/3}

    Why does R = 4 instead of \frac{1}{4} in problem (1)?
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  7. #7
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    I have a new problem I am stuck with too.

    \sum_{n = 0}^{\infty} 5^{(-1)n}z^n

    Seems simple enough, but I suck at power series...

    \frac{1}{R} = lim_{n \to \infty} \sqrt[n]{5^{(-1)}} = lim_{n \to \infty} (\sqrt[n]{5})^{(-1)}

    Right path so far? Way off? Help will be appreciated, thanks!
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  8. #8
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    Quote Originally Posted by shadow_2145 View Post
    Why does R = 4 instead of \frac{1}{4} in problem (1)?
    because 1/R is limit of a_{n+1}/a_n, so R is limit of a_n/a_{n+1}
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  9. #9
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    Quote Originally Posted by shadow_2145 View Post
    I have a new problem I am stuck with too.

    \sum_{n = 0}^{\infty} 5^{(-1)n}z^n
    I don't understand if you mean 5^{-n} or 5^{(-1)^n}
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  10. #10
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    Quote Originally Posted by choovuck View Post
    I don't understand if you mean 5^{-n} or 5^{(-1)^n}
    Oops, my bad. 5^{(-1)^n} is the correct one.
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  11. #11
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    Quote Originally Posted by choovuck View Post
    because 1/R is limit of a_{n+1}/a_n, so R is limit of a_n/a_{n+1}

    Oh, I see now. Hehe, duh!
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  12. #12
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    5^{(-1)^n} means that it's the sequence 1/5, 5, 1/5, 5, ..., and then \lim_{n\to\infty} \sqrt[n]{a_n}=1 (since both \lim_{n\to\infty} \sqrt[n]{5}=1 and \lim_{n\to\infty} \sqrt[n]{1/5}=1), so R=1
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