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Thread: radius of convergence

  1. #1
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    radius of convergence

    Hi please could you assist me: questions posted below:


    Assuming the function f is holomorphic in the disk $\displaystyle D(0,1) = \{ z \in \mathbb{C}:|z| < 1\}$, prove that $\displaystyle g(z) = \overline {f(\overline z )}$ is also holomorphic in D(0,1) and find its derivative?????



    Find the radii of convergence of the following series stating which result is being used.
    (a) $\displaystyle \sum\limits_{k = 0}^\infty {k^{113} 2^{ - k} (z - 1)^k } $

    (b)$\displaystyle \sum\limits_{n = 2}^\infty {n!(z - e)^{3n} } $

    (c)$\displaystyle \sum\limits_{k = 0}^\infty {\frac{{z^k }}{{(k!)^2 }}}$


    (A)
    Using Ratio Test:

    $\displaystyle \mathop {\lim }\limits_{k \to \infty } \left| {\frac{{a_{k + 1} }}
    {{a_k }}} \right| = \mathop {\lim }\limits_{k \to \infty } \left| {\frac{{(k + 1)^{113} 2^{ - (k + 1)} (z - 1)^{k + 1} }}{{k^{113} 2^{ - k} (z - 1)^k }}} \right| = $

    $\displaystyle
    \mathop {\lim }\limits_{k \to \infty } \left| {\frac{{(k + 1)^{113} 2^{ - k - 1} (z - 1)^{k + 1} }}{{k^{113} 2^{ - k} (z - 1)^k }}} \right| = \mathop {\lim }\limits_{k \to \infty } \left| {\frac{{(k + 1)^{113} 2^{ - 1} (z - 1)}}{{k^{113} }}} \right|$

    =$\displaystyle \frac{{(\infty + 1)^{113} 2^{ - 1} (z - 1)}}{{\infty ^{113} }} = \frac{{(z - 1)}}
    {2} < 1$

    So is rad. of convergence (z-1)/2 and converging since it is less than 1??



    (B)

    Using Ratio Test:

    $\displaystyle \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{a_{n + 1} }}
    {{a_n }}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{(n + 1)!(z - e)^{3(n + 1)} }}{{n!(z - e)^{3n} }}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {(n + 1)(z - e)^3 } \right|$

    $\displaystyle = (\infty + 1)(z - e)^3 = \infty$

    Hence, is ROC infinity and diverging since it is greater than 1?

    (C)
    Using Ratio Test:

    $\displaystyle \mathop {\lim }\limits_{k \to \infty } \left| {\frac{{a_{k + 1} }}
    {{a_k }}} \right| = \mathop {\lim }\limits_{k \to \infty } \left| {\frac{{z^{k + 1} }}{{((k + 1)!)^2 }} \times \frac{{(k!)^2 }}
    {{z^k }}} \right| = \mathop {\lim }\limits_{k \to \infty } \left| {\frac{{z^{k + 1} }}{{z^k }} \times \frac{{(k!)^2 }}
    {{((k + 1)!)^2 }}} \right|$

    $\displaystyle = \mathop {\lim }\limits_{k \to \infty } \left| {z \times \frac{{(k!)^2 }}{{((k + 1)!)^2 }}} \right| = \mathop {\lim }\limits_{k \to \infty } \left| {\sqrt z \frac{{k!}}{{k + 1!}}} \right|$

    $\displaystyle = \mathop {\lim }\limits_{k \to \infty } \left| {\frac{{\sqrt z }}
    {{k + 1}}} \right| = \frac{{\sqrt z }}{{\infty + 1}} = 0$

    So is ROC 0 and converging since it is less than 1?
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  2. #2
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    Quote Originally Posted by rinatoc View Post

    Assuming the function f is holomorphic in the disk $\displaystyle D(0,1) = \{ z \in \mathbb{C}:|z| < 1\}$, prove that $\displaystyle g(z) = \overline {f(\overline z )}$ is also holomorphic in D(0,1) and find its derivative?????

    using complex conjugate properties and this fact that $\displaystyle z \rightarrow a$ iff $\displaystyle \overline{z} \rightarrow \overline{a},$ we get:

    $\displaystyle \lim_{z\to a} \frac{g(z)-g(a)}{z-a}=\lim_{z\to a} \frac{\overline{f(\overline{z})} - \overline{f(\overline{a})}}{z-a}=\lim_{\overline{z}\to \overline{a}} \overline{\left(\frac{f(\overline{z})-f(\overline{a})}{\overline{z}-\overline{a}} \right)}$

    $\displaystyle =\overline{\left(\lim_{\overline{z}\to \overline{a}} \frac{f(\overline{z})-f(\overline{a})}{\overline{z}-\overline{a}} \right)} \ \ \ \ \ \text{since the map} \ \ z \rightarrow \overline{z} \ \ \text{is continuous}$

    $\displaystyle =\overline{f'(\overline{a})}.$ thus $\displaystyle g$ is holomorphic too and $\displaystyle g'(z)=\overline{f'(\overline{z})}. \ \ \ \square$


    Find the radii of convergence of the following series stating which result is being used.
    (a) $\displaystyle \sum\limits_{k = 0}^\infty {k^{113} 2^{ - k} (z - 1)^k } $

    (b)$\displaystyle \sum\limits_{n = 2}^\infty {n!(z - e)^{3n} } $

    (c)$\displaystyle \sum\limits_{k = 0}^\infty {\frac{{z^k }}{{(k!)^2 }}}$


    (A)
    Using Ratio Test:

    $\displaystyle \mathop {\lim }\limits_{k \to \infty } \left| {\frac{{a_{k + 1} }}
    {{a_k }}} \right| = \mathop {\lim }\limits_{k \to \infty } \left| {\frac{{(k + 1)^{113} 2^{ - (k + 1)} (z - 1)^{k + 1} }}{{k^{113} 2^{ - k} (z - 1)^k }}} \right| = $

    $\displaystyle
    \mathop {\lim }\limits_{k \to \infty } \left| {\frac{{(k + 1)^{113} 2^{ - k - 1} (z - 1)^{k + 1} }}{{k^{113} 2^{ - k} (z - 1)^k }}} \right| = \mathop {\lim }\limits_{k \to \infty } \left| {\frac{{(k + 1)^{113} 2^{ - 1} (z - 1)}}{{k^{113} }}} \right|$

    =$\displaystyle \frac{{(\infty + 1)^{113} 2^{ - 1} (z - 1)}}{{\infty ^{113} }} = \frac{{(z - 1)}}
    {2} < 1$

    So is rad. of convergence (z-1)/2 and converging since it is less than 1??[/tex]

    you should say that $\displaystyle \lim_{k\to\infty} \frac{(k+1)^{113}}{k^{113}} = 1,$ rather than putting $\displaystyle k=\infty$!!!

    so you got $\displaystyle |z-1| < 2,$ which means that the radius of convergence is 2.


    (B)

    Using Ratio Test:

    $\displaystyle \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{a_{n + 1} }}
    {{a_n }}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{(n + 1)!(z - e)^{3(n + 1)} }}{{n!(z - e)^{3n} }}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {(n + 1)(z - e)^3 } \right|$

    $\displaystyle = (\infty + 1)(z - e)^3 = \infty$

    Hence, is ROC infinity and diverging since it is greater than 1?

    so the series is divergent for all values of z except for $\displaystyle z=e.$ thus the radius of convergence is 0.


    (C)
    Using Ratio Test:

    $\displaystyle \mathop {\lim }\limits_{k \to \infty } \left| {\frac{{a_{k + 1} }}
    {{a_k }}} \right| = \mathop {\lim }\limits_{k \to \infty } \left| {\frac{{z^{k + 1} }}{{((k + 1)!)^2 }} \times \frac{{(k!)^2 }}
    {{z^k }}} \right| = \mathop {\lim }\limits_{k \to \infty } \left| {\frac{{z^{k + 1} }}{{z^k }} \times \frac{{(k!)^2 }}
    {{((k + 1)!)^2 }}} \right|$

    $\displaystyle = \mathop {\lim }\limits_{k \to \infty } \left| {z \times \frac{{(k!)^2 }}{{((k + 1)!)^2 }}} \right| = \mathop {\lim }\limits_{k \to \infty } \left| {\sqrt z \frac{{k!}}{{k + 1!}}} \right|$

    $\displaystyle = \mathop {\lim }\limits_{k \to \infty } \left| {\frac{{\sqrt z }}
    {{k + 1}}} \right| = \frac{{\sqrt z }}{{\infty + 1}} = 0$

    So is ROC 0 and converging since it is less than 1?

    where did $\displaystyle \sqrt{z}$ come from??! since $\displaystyle (k+1)!=(k+1)k!,$ you'll get:

    $\displaystyle \lim_{k\to\infty} \left|\frac{a_{k+1}}{a_k} \right| = \lim_{k\to\infty} \frac{|z|}{(k+1)^2}=0 < 1, \ \ \forall z.$ so the radius of convergence is $\displaystyle \infty.$
    Last edited by NonCommAlg; Jul 8th 2008 at 04:57 PM.
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  3. #3
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    Quote Originally Posted by NonCommAlg View Post
    so the series is divergent for all values of z except for $\displaystyle z=e.$ thus the radius of convergence is 0.
    i understand your method here, thanks for that. but why is it 0.
    is there a general rule saying that the ROC is 0 if the function is divergent everywhere except for one point?



    Quote Originally Posted by NonCommAlg View Post
    where did $\displaystyle \sqrt{z}$ come from??! since $\displaystyle (k+1)!=(k+1)k!,$ you'll get:

    $\displaystyle \lim_{k\to\infty} \left|\frac{a_{k+1}}{a_k} \right| = \lim_{k\to\infty} \frac{|z|}{(k+1)^2}=0 < 1, \ \ \forall z.$ so the radius of convergence is $\displaystyle \infty.$
    lol i was trying to square root the z so that it would be easier to work with the k's but your method is more appropriate.
    as before though, why is the ROC infinity. is there a general rule saying that if you get the value as 0 then the ROC is infinity?
    how did you conclude these answers?

    thanx very much for the help btw, very much appreciated!!!
    Last edited by mathfied; Jul 9th 2008 at 11:02 AM.
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