radius of convergence

**rinatoc** · July 8th 2008, 03:36 PM

Hi please could you assist me: questions posted below:

Assuming the function f is holomorphic in the disk $D(0,1) = \{ z \in \mathbb{C}:|z| < 1\}$ , prove that $g(z) = \overline {f(\overline z )}$ 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) $\sum\limits_{k = 0}^\infty {k^{113} 2^{ - k} (z - 1)^k }$

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

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

(A)
Using Ratio Test:

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

$ \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|$

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

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

(B)

Using Ratio Test:

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

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

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

(C)
Using Ratio Test:

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

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

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

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

**NonCommAlg** · July 8th 2008, 04:39 PM

Originally Posted by rinatoc

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

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

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

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

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

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

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

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

(A)
Using Ratio Test:

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

$ \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|$

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

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

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

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

(B)

Using Ratio Test:

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

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

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

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

(C)
Using Ratio Test:

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

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

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

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

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

$\lim_{k\to\infty} \left|\frac{a_{k+1}}{a_k} \right| = \lim_{k\to\infty} \frac{|z|}{(k+1)^2}=0 < 1, \ \ \forall z.$ so the radius of convergence is $\infty.$

**mathfied** · July 9th 2008, 10:49 AM

Originally Posted by NonCommAlg

so the series is divergent for all values of z except for $z=e.$ 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?

Originally Posted by NonCommAlg

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

$\lim_{k\to\infty} \left|\frac{a_{k+1}}{a_k} \right| = \lim_{k\to\infty} \frac{|z|}{(k+1)^2}=0 < 1, \ \ \forall z.$ so the radius of convergence is $\infty.$

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!!!

Thread: radius of convergence

LinkBack

Thread Tools

Display

radius of convergence

Similar Math Help Forum Discussions

investigate the convergence at the radius of convergence

Radius of convergence; Interval of convergence.

radius of convergence and interval of convergence of the series

Find the radius of convegence and the radius of convergence

series convergence and radius of convergence

Search Tags