Radius of convergence

[rinatoc]

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?

 

[quantumdude]

rinatoc, you seriously need to dust off the old calculus book and study it. You won't survive complex analysis if you don't know real calculus.

(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??

No, the radius of convergence is not (z-1)/2. The radius of convergence of a power series is a number.

(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?

Now if the ROC is infinite, how could the series possibly diverge anywhere? Think about what you are doing!

(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?

If the ROC is 0, then clearly the series converges only at the center.

 

[HallsofIvy]

Rinatoc, You seem to be confusing "radius of convergence" with the "ratio test".

 

[mathfied]

ok after a bit more studying, would this be correct:

(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|\]$

the k's go to infinity on the top and bottom so infinity/infinity = 1. Hence,

=$\[ \frac{{2^{ - 1} (z - 1)}}{{1}} = \frac{{(z - 1)}} {2} < 1 \]$

= $|z-1| < 2,$

So the rad. of convergence is 2?

(B)

Using Ratio Test:

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

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

Hence, is ROC is divergent everywhere except when z = e, so the radius of conv. is 0?

(C)
Using Ratio Test:
$\[ \begin{gathered} \mathop {\lim }\limits_{k \to \infty } \left| {\frac{{a_{k + 1} }} {{a_k }}} \right| = \mathop {\lim }\limits_{k \to \infty } \left| {\frac{{z^{k + 1} }} {{((k + 1)!)^2 }} \times \frac{{(k!)^2 }} {{z^k }}} \right| = \mathop {\lim }\limits_{k \to \infty } \left| {\frac{{z^{k + 1} }} {{z^k }} \times \frac{{(k!)^2 }} {{((k + 1)!)^2 }}} \right| = \mathop {\lim }\limits_{k \to \infty } \left| {z \times \frac{{(k!)^2 }} {{((k + 1)!)^2 }}} \right| = \mathop {\lim }\limits_{k \to \infty } \left| {z\frac{{(k!)^2 }} {{((k + 1)k!)^2 }}} \right| \hfill \\ = \mathop {\lim }\limits_{k \to \infty } \left| {\frac{z} {{(k + 1)^2 }}} \right| = 0 < 1 \hfill \\ \end{gathered} \]$

So is ROC infinity?

 

[HallsofIvy]

I'm sorry, what does "mathfied" have to do with "rinotoc"?

 

[mathfied]

oh hi, yes i had to open a new account "rinatoc" because my mathfied account for some reason was giving an error whenever i tried logging in. i thought maybe my account got disabled or something so i opened the other one.. turned out to be some ip problem..

but then my mathfied account is working fine now so i don't really need the rinatoc.. but i don't know how to delete the account.

im jus following up my queries with this original account now that everything is working fine.

many apologies for the double :)

i reattempted the questions there. are they ok?