All Questions
Tagged with divergent-series analytic-continuation
9 questions
21
votes
6
answers
1k
views
What is the relationship between $\sum_{n=0}^\infty f(n) x^n$ and $-\sum_{n=1}^\infty f(-n) x^{-n}$?
Background
Taking a relatively arbitrary combination of exponential and polynomial terms, for instance
$$\sum_{n=0}^\infty \left(n^{2}\sin\left(n\right)+n\cos\left(3n-2\right)\right)\cos\left(5n+1\...
- 1,730
12
votes
1
answer
1k
views
Divergent series summation beyond natural boundaries
I'm hoping to investigate the effects of divergent summation methods on series which cannot be analytically continued due to a dense set of singularities. At least a priori, it doesn't seem that a ...
- 1,730
9
votes
0
answers
313
views
Switching the order of a summation and replacing a series by its analytical continuation
Background
A useful trick when trying to analyze a series $\sum_{n=0}^\infty f(n)$ is to expand $f(n)$ as some kind of series, swap the order of summation, and then evaluate the inner infinite sum. ...
- 1,730
7
votes
0
answers
306
views
Gottfried Helms' tetra-eta series
Here Gottfried Helms introduces the following fascinating divergent series
$$ T_2(x)=- \sum_{n=1}^\infty (-1)^n n^{n^x}$$
The terms don't go to zero, so technically the series does not converge ...
- 1,730
6
votes
0
answers
2k
views
Do smooth cutoff functions analytically continue functions?
My goal is to prove (or disprove) that sufficiently smooth and quickly decaying cutoff functions being tacked on to a Taylor series correctly extend the radius of convergence to the analytic ...
- 1,730
4
votes
1
answer
249
views
Does the analytical continuation of $\sum f(n) x^n $ always have a branch cut if $f(z)$ has a pole?
I suspect the answer to the title question is 'no', but I'm hoping to find an explicit counterexample. Also, I am requiring that $\sum f(n) x^n $ has a finite radius of convergence, otherwise, the ...
- 1,730
4
votes
1
answer
401
views
How to correctly renormalize this function at the pole $x=1$? Evaluating: $\sum_{n=1}^{\infty} e^{e^n}$
So I was considering the divergent everywhere but 0 power series
$$ f(x) = \sum_{n=0}^{\infty} e^{e^n} x^n $$
Now one can do the following "questionable" manipulation
$$ f(x) = \sum_{n=0}^{\...
- 2,689
4
votes
1
answer
556
views
$\frac {f (0)}{2}+ \sum_{k=1}^{\infty}f (k)=\sum_{n=-\infty}^{\infty} \mathcal{L} \{ f \} (2 \pi i n)$
I obtained the very strange formula above and at begining I was just wanted know how to interpretate it. But now when I know what is this with help of @Carlo Beenakker, I am leaving it as a proof. BTW ...
- 161
3
votes
0
answers
358
views
New/useful method for summation of divergent series?
Questions
$$ S(n,x) = x+e^x + e^{e^x} + e^{e^{e^x}} + \dots \text{$n$ times}$$
Also obeys (see background for argument):
$$ \frac{1}{2 \pi i} \oint e^{S(k,x)} \frac{\partial \ln(\frac{\int_0^\...
- 237