All Questions
48
questions
0
votes
0
answers
33
views
Maximizing the radius of convergence around a point for an analytic function
Let $f:\Omega\longrightarrow \mathbb{C}$ be analytic, and let $z_0\in\Omega$ s.t.
$$f (z)=\sum_{n\geq 0} a_n(z-z_0)^n,\quad\forall \left|z-z_0\right|<r,$$
for some $r>0$, and some complex-valued ...
- 67
1
vote
1
answer
94
views
Meaning of "$f$ has a power series expansion around $p$"
In Complex Analysis by Donald Marshall (page 29), there is an exercise problem that starts with "Suppose $f$ has a power series expansion at $0$ which converges in all of $\mathbb{C}$. " ...
- 1,268
0
votes
0
answers
25
views
If $f$ is analytic on $B(z_0, R)$, does that imply $f$ has a power series expansion centered at $z_0$ with radius of convergence $R$? [duplicate]
I know that if a power series $f:=\sum_{n=0}^{\infty}a_n(z-z_0)^n$ has radius of convergence $R$, then $f$ is analytic on $B(z_0,R)$. I wonder if the converse of this statement is true. That is, ...
- 1,268
0
votes
0
answers
77
views
Series of Analytic Functions is Analytic
Let $0 \in \mathbf{N}$. Let $P_m(x): [0,1] \to \mathbf{C}$ be bounded analytic functions for every $m\in \mathbf{N}$. Formally, define
$$
f(x) = \sum_{m\in \mathbf{N}}c_m P_m(x)\overline{P_m}(x),
$$
...
- 147
2
votes
0
answers
61
views
Complex Analysis Qualifying Exam Problem Regarding Taylor Series and Normal Convergence
I am studying for a qualifying exam on Gamelin's Complex Analysis Chapters 1-11 and am stuck on the following past exam question:
Let $\phi(n): \mathbb{N} \to \mathbb{R}$ such that $\lim_{n\to \infty} ...
- 149
0
votes
1
answer
105
views
Is univalent polynomials dense in $\mathbb S$?
Let $\mathbb S$ be the collection of all univalent and analytic functions defined on unit disc $\mathbb D$ such that for each $f \in \mathbb S,$ we have $f(0)=0, f'(0)=1.$
Let $P$ be the subsets of $\...
- 751
1
vote
0
answers
27
views
Prove that $f$ is a polynomial function of $z$ [duplicate]
Given that $f$ is a holomorphic function on a domain $D \subset \mathbb{C}$, and $ \forall a \in D$ we have a $n_a \in \mathbb{N}$ such that $f^{n_a}(a) = 0$. We have to use this to show that $f$ is a ...
0
votes
0
answers
31
views
Sum of an infinite series [duplicate]
I want to calculate
$$ \sum_{k=1}^{\infty} \frac{k^2}{2^k} $$
Usually (assuming it was $k$ instead of $k^2$) I would consider this as a special case of the series $ f(x) =\sum_{k=1}^{\infty} k \cdot z^...
- 473
0
votes
1
answer
106
views
How can I show that an analytic function can be bounded from below as follows?
I have the following problem:
Let $f$ be a nonzero analytic function on $B:=\{|z|<1\}$. Show that there exists $c>0$, $r>0$ and $k=0,1,2,...$ such that $$|f(z)|\geq c|z|^k$$when $|z|<r$.
...
- 2,935
1
vote
1
answer
681
views
Prove that the radius of convergence of $\log(1+z)$ is precisely $1$
On Ahlfors' Complex Analysis, he states that the logarithmic series $$\log(1+z)=z-\frac{z^2}{2}+\frac{z^3}{3}-\frac{z^4}{4}+\frac{z^5}{5}-...$$ centered at the origin must have a radius of $1$ because ...
- 3,800
1
vote
3
answers
58
views
Struggling to compute a power series for a complex value function
I am struggling to compute the power series expansion of $$f(z) = \frac{1}{2z+5}$$ about $z=0$, where $f$ is a complex function. I tried comparing it to the geometric series as follows,$$ f(z) = \frac{...
- 145
0
votes
0
answers
39
views
if $f$ is holomorphic in $\Bbb D$ then it is equal to its power series in all $\Bbb D$?
I saw a solution to some question somewhere and this was implicitly used.
I know holomorphic functions are given LOCALLY by their taylor series, but from the fact that $f\in Hol(\Bbb D)$ can we ...
- 746
1
vote
1
answer
169
views
Prove $\sum_{n=1}^\infty \frac{nz^n}{1-z^n}=\sum_{n=1}^\infty \frac{z^n}{(1-z^n)^2}$ [closed]
Prove that
$$\sum_{n=1}^\infty \frac{nz^n}{1-z^n}=\sum_{n=1}^\infty \frac{z^n}{(1-z^n)^2}$$
and give the region where is an holomorphic function.
I already know that are holomorphic functions in the ...
- 589
2
votes
1
answer
75
views
Convergence of power series at the boundary
Consider the following complex power series
$$
\sum_{n\geq 1}{\frac{ni^n}{2^n}{z^{n-1}}}
$$
By the root test, I have concluded that the disc of convergence is $D:=D(0,2)$. Then, I would like to study ...
- 195
1
vote
0
answers
66
views
Analyticity at infinity: Laurent series
Suppose that the series $$A_0 + A_1(z-z_0)^{-1} + \cdots + A_n (z-z_0)^{-n}+ \cdots$$ converges to $f(z)$ for all $z$ such that $r<|z-z_0|<\infty$ and let $f^*(\xi)=f(1/\xi), f^*(0)=A_0$. Show ...
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