Question

Analytic continuation for $\zeta(s)$ using finite sums?

score 431 · Answer 1 · 2016-03-12 22:51:38

431

Views

Analytic continuation for $\zeta(s)$ using finite sums?

Published on 12 Mar 2016 - 22:51

score 90 · Answer 2 · 2016-03-15 16:52:56

90

Views

Analytic continuation of $\sum_{n=0}^{\infty} \frac{z^{n+1}}{n+1}$

Published on 15 Mar 2016 - 16:52

score 1.9k · Answer 3 · 2016-04-13 20:00:39

1.9k

Views

How can I know the analytic continuation exists in certain cases?

Published on 13 Apr 2016 - 20:00

score 240 · Answer 4 · 2016-04-17 16:00:11

240

Views

Given an analytic continuation along $\gamma$ such that $R(t)\equiv \infty$ for some $t$, then $R(s)\equiv \infty$ for each $s\in [0,1]$

Published on 17 Apr 2016 - 16:00

score 2.8k · Answer 5 · 2016-04-17 17:15:54

2.8k

Views

Alternating series test for complex series

Published on 17 Apr 2016 - 17:15

score 174 · Answer 6 · 2016-04-18 17:27:29

174

Views

Let $U=B_1(0)$ and $ f(z)=\sum_{n=1}^{\infty}2^{-n^2}z^{2^n}$. Show that $f$ has not analytic extensions to any open set $G$ with $U\subsetneq G$.

Published on 18 Apr 2016 - 17:27

score 4k · Answer 7 · 2016-04-19 12:36:31

4k

Views

Show that $f(z)=\sum_{n=0}^{\infty}z^{2^n}$ can't be analytically continued past the unit disk.

Published on 19 Apr 2016 - 12:36

score 155 · Answer 8 · 2016-04-19 23:05:21

155

Views

Determine the largest open set to which $f(z)=\sum_{n=1}^{\infty}(-1)^n(2n+1)z^{n}$ can be analytically continued

Published on 19 Apr 2016 - 23:05

score 437 · Answer 9 · 2016-04-28 16:40:59

437

Views

Identity theorem for forms (on analytic manifolds)

Published on 28 Apr 2016 - 16:40

score 163 · Answer 10 · 2016-05-09 06:58:02

163

Views

Extending binomial identity $ \sum\limits_{k=0}^n\frac{(-1)^k}{k+x}\binom{n}{k}\binom{n+k}{k}=0$ to $0<x<1$

Published on 09 May 2016 - 6:58

score 192 · Answer 11 · 2016-05-12 17:51:15

192

Views

Showing that a function is the restriction of another holomorphic function

Published on 12 May 2016 - 17:51

score 73 · Answer 12 · 2016-05-13 17:35:41

73

Views

Analytic extension of $\sum_{k=1}^n\frac1k$ complex domain

Published on 13 May 2016 - 17:35

score 71 · Answer 13 · 2016-05-13 21:23:35

71

Views

The analytic extension of $\sum_{k=1}^n\frac1k$ and an induction

Published on 13 May 2016 - 21:23

score 150 · Answer 14 · 2016-05-21 17:44:18

150

Views

Analytic continuation of power series on the unit whose terms tends to 0

Published on 21 May 2016 - 17:44

score 56 · Answer 15 · 2016-06-03 03:14:59

56

Views

Analyitc Continuation of Partial $\zeta$ function

Published on 03 Jun 2016 - 3:14

List Question

Analytic continuation for $\zeta(s)$ using finite sums?

Analytic continuation of $\sum_{n=0}^{\infty} \frac{z^{n+1}}{n+1}$

How can I know the analytic continuation exists in certain cases?

Given an analytic continuation along $\gamma$ such that $R(t)\equiv \infty$ for some $t$, then $R(s)\equiv \infty$ for each $s\in [0,1]$

Alternating series test for complex series

Let $U=B_1(0)$ and $ f(z)=\sum_{n=1}^{\infty}2^{-n^2}z^{2^n}$. Show that $f$ has not analytic extensions to any open set $G$ with $U\subsetneq G$.

Show that $f(z)=\sum_{n=0}^{\infty}z^{2^n}$ can't be analytically continued past the unit disk.

Determine the largest open set to which $f(z)=\sum_{n=1}^{\infty}(-1)^n(2n+1)z^{n}$ can be analytically continued

Identity theorem for forms (on analytic manifolds)

Extending binomial identity $ \sum\limits_{k=0}^n\frac{(-1)^k}{k+x}\binom{n}{k}\binom{n+k}{k}=0$ to $0<x<1$

Showing that a function is the restriction of another holomorphic function

Analytic extension of $\sum_{k=1}^n\frac1k$ complex domain

The analytic extension of $\sum_{k=1}^n\frac1k$ and an induction

Analytic continuation of power series on the unit whose terms tends to 0

Analyitc Continuation of Partial $\zeta$ function

Trending Questions

Popular # Hahtags

Popular Questions