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Questions tagged [analytic-continuation]

For questions related to analytic continuation

Analytic continuation is a technique used in complex analysis to expand the domain of an analytic function.

If $f$ is analytic on some open $U\subseteq \Bbb{C}$, then it can usually be extended to an analytic function $F$ on a larger connected domain. That extension, when it exists, is unique.

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Easy explanation of analytic continuation

Today, as I was flipping through my copy of Higher Algebra by Barnard and Child, I came across a theorem which said, The series $$ 1+\frac{1}{2^p} +\frac{1}{3^p}+...$$ diverges for $p\leq 1$ and converges for $p>1$. But later I found out that the…
kodyv
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Show that $f(z)=\sum_{n=0}^{\infty}z^{2^n}$ can't be analytically continued past the unit disk.

I'm reading the problems of Stein and Shakarchi's Complex Analysis, Chapter 2 Problem 1 asks to show that $$f(z)=\sum_{n=0}^{\infty}z^{2^n}$$ cannot be analytically continued past the unit disk. (Hint: Suppose $\theta =\frac{2\pi p}{2^k}$ for…
Mike
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Analytic continuation of $\Phi(s)=\sum_{n \ge 1} e^{-n^s}$

(After 3 bounties I've also posted on mathoverflow). While discussing theta functions, I thought: $\zeta(s)=\sum n^{-s}=1+2^{-s}+3^{-s}+ \cdot\cdot\cdot$ and $\Phi(s)=\sum e^{-n^s}=e^{-1}+e^{-2^s}+e^{-3^s}+\cdot\cdot\cdot $ What is the analytic…
J. Zimmerman
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On every simply connected domain, there exists a holomorphic function with no analytic continuation.

I am working on a question that requires me to prove that on every simply connected open subset of $\mathbb{C}$, there exists a holomorphic function that cannot be extended to a holomorphic function on a larger connected open set. I know an example…
Chuwei Zhang
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Analytic continuation of double integral $\int_0^\infty \int_0^\infty f(x,y)^{-s} dx dy$

Consider the following integral $$F(s) = \int_0^\infty \int_0^\infty [xy(x+1) (y+1) (y+2) (x+y+1) (2 x+y+2 )(x+y+2))]^{-s} dxdy$$ It converges and thus defines an analytic function when $1/4<\Re(s)<1$, it can be meromorphically continued to…
pisco
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Analytic continuation of harmonic series

Is there an accepted analytic continuation of $\sum_{n=1}^m \frac{1}{n}$? Even a continuation to positive reals would be of interested, though negative and complex arguments would also be interesting. I don't have a specific application in mind, but…
Richard Burke
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Can we use analytic continuation to obtain $\sum_{n=1}^\infty n = b, b\neq -\frac{1}{12}$

Intuitive question It is a popular math fact that the sum definition of the Riemann zeta function: $$\zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s} $$ can be extended to the whole complex plane (except one) to obtain $\zeta(-1)=-\frac{1}{12}$. The right…
Jens Renders
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Fourier transform of meromorphic function

Suppose that I have a function $f(z)$ which is meromorphic on the entire complex plane, meaning holomorphic everywhere except for a discrete set of poles. I then take a vertical slice of this function, which we can denote $f(s+it)$ (for constant…
Mike Battaglia
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Expressing Zeta function using Gamma series

Motivated by Gautschi double inequality, $$ \frac{n^{s}}{n^{\small1}}\ge\frac{\Gamma(n+s)}{\Gamma(n+1)}\ge\frac{(n+1)^{s}}{(n+1)^{\small1}}\ge\frac{\Gamma(n+1+s)}{\Gamma(n+1+1)}\ge\,\cdots \quad\colon\,0\lt{s}\lt1\tag{1} $$ From the main…
Hazem Orabi
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Intuition behind $\zeta(-1)$ = $\frac{-1}{12}$

When I first watched numberphile's 1+2+3+... = $\frac{-1}{12}$ I thought the sum actually equalled $\frac{-1}{12}$ without really understanding it. Recently I read some wolframalpha pages and watched some videos and now I understand (I think), that…
Brian010515
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Analytic extension of $\text{Li}_0^{(1,0)}(z):=-\displaystyle\sum_{n=1}^{\infty}\ln(n)z^{n}$ for $|z|>1$

I would like to extend the domain of the following function: $$\text{Li}_0^{(1,0)}(z):=-\sum_{n=1}^{\infty}\ln(n)z^{n}\qquad\text{where }|z|<1$$ The part in red is the series, the part in green is a hypothetical extension I'll start by saying that…
Math Attack
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A Family of Limits Leading to an Interesting Function

A while back I got very interested in limits of the form $$ \lim_{n\to\infty} (2A)^n \left (A-\underbrace{\sqrt{a+\sqrt{a+\ldots\sqrt{a+z}}}}_{n\textrm{ radicals}} \right )=f_a^{-1}(z) $$ Where $A$ is the positive solution of $A^2=a+A$. As notated,…
NMK
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Reference request for a book that covers analytic continuation in great detail starting from basics

I have earlier self studied Tom M. Apostol's Introduction to Analytic Number Theory after doing a course in complex analysis, but my instructor at university didn't even mention analytic continuation. Although I self studied from Complex Variables…
user775699
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Is the “sum of all natural numbers” unique?

A while ago, there was a great hype about the “identity” $$\sum_{n=1}^{\infty} n = -\frac{1}{12}.$$ Apart from some series manipulations where the validity seems to be at least questionable, the derivation of this always goes through the zeta…
celtschk
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Exercise 2 from Terry Tao's blog on Euler-Maclaurin, Bernouilli numbers, and the zeta function

In the blog post The Euler-Maclaurin formula, Bernoulli numbers, the zeta function, and real-variable analytic continuation, Terry Tao looks at the commonly-cranked 'absurd' formulae $$\begin{align} \sum_{n \geq 1} 1 &= -1/2 \tag{1} \\ \sum_{n \geq…
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