Question

Let $f(z)$ be a function analytic in a domain containing the segment $[0,1]$ and satisfying $f(z+1)=azf(z)+p(z)$.

score 125 · Answer 1 · 2015-07-25 15:29:01

125

Views

Let $f(z)$ be a function analytic in a domain containing the segment $[0,1]$ and satisfying $f(z+1)=azf(z)+p(z)$.

Published on 25 Jul 2015 - 15:29

score 224 · Answer 2 · 2015-07-30 21:02:19

224

Views

Prove that the given condition implies analytic continuation

Published on 30 Jul 2015 - 21:02

score 1k · Answer 3 · 2015-07-31 00:58:25

1k

Views

Is there a holomorphic function $f$ on the unit disc such that $|f(z)|\rightarrow\infty$ as $|z|\rightarrow 1$?

Published on 31 Jul 2015 - 0:58

score 2.4k · Answer 4 · 2015-07-31 01:15:02

2.4k

Views

On every simply connected domain, there exists a holomorphic function with no analytic continuation.

Published on 31 Jul 2015 - 1:15

score 462 · Answer 5 · 2015-08-03 14:04:23

462

Views

If $|f| \le |g|$, does analytic continuation of $g$ imply analytic continuation of $f$?

Published on 03 Aug 2015 - 14:04

score 289 · Answer 6 · 2015-08-08 17:34:41

289

Views

Does there necessarily exist such a holomorphic function?

Published on 08 Aug 2015 - 17:34

score 204 · Answer 7 · 2015-08-16 19:56:08

204

Views

Evaluate the integral $\int_0^\infty x^{t-1}e^{-\beta x}dx$

Published on 16 Aug 2015 - 19:56

score 546 · Answer 8 · 2015-08-20 20:44:57

546

Views

Showing a function analytic on open unit disk can be analytically extended

Published on 20 Aug 2015 - 20:44

score 103 · Answer 9 · 2015-08-31 04:58:24

103

Views

There exists a $M$ such that $\mid f^k (0)\mid \leq k^4 M^k$. Show that $f$ can be extended analytic on $\Bbb C$.

Published on 31 Aug 2015 - 4:58

score 1.7k · Answer 10 · 2015-09-11 19:40:34

1.7k

Views

are there functions that have no derivative?

Published on 11 Sep 2015 - 19:40

score 111 · Answer 11 · 2015-09-23 11:07:43

111

Views

Analytic continuation vs. series convergence near convergence boundary

Published on 23 Sep 2015 - 11:07

score 892 · Answer 12 · 2015-09-28 21:06:00

892

Views

Why do $f(x)= \frac {{x²-1}}{{x-1} } $ and $g(x)=x+1$ not have the same domain of definition, but are the same?

Published on 28 Sep 2015 - 21:06

score 396 · Answer 13 · 2015-10-01 17:20:30

396

Views

analytic continuation for $(e^z)^{\frac{1}{3}}$

Published on 01 Oct 2015 - 17:20

score 166 · Answer 14 · 2015-10-10 01:12:49

166

Views

Summation continuation, over no common divisors

Published on 10 Oct 2015 - 1:12

score 103 · Answer 15 · 2015-10-13 07:09:45

103

Views

Analytic Continuation with Real Coefficients

Published on 13 Oct 2015 - 7:09

List Question

Let $f(z)$ be a function analytic in a domain containing the segment $[0,1]$ and satisfying $f(z+1)=azf(z)+p(z)$.

Prove that the given condition implies analytic continuation

Is there a holomorphic function $f$ on the unit disc such that $|f(z)|\rightarrow\infty$ as $|z|\rightarrow 1$?

On every simply connected domain, there exists a holomorphic function with no analytic continuation.

If $|f| \le |g|$, does analytic continuation of $g$ imply analytic continuation of $f$?

Does there necessarily exist such a holomorphic function?

Evaluate the integral $\int_0^\infty x^{t-1}e^{-\beta x}dx$

Showing a function analytic on open unit disk can be analytically extended

There exists a $M$ such that $\mid f^k (0)\mid \leq k^4 M^k$. Show that $f$ can be extended analytic on $\Bbb C$.

are there functions that have no derivative?

Analytic continuation vs. series convergence near convergence boundary

Why do $f(x)= \frac {{x²-1}}{{x-1} } $ and $g(x)=x+1$ not have the same domain of definition, but are the same?

analytic continuation for $(e^z)^{\frac{1}{3}}$

Summation continuation, over no common divisors

Analytic Continuation with Real Coefficients

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