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Results tagged with sequences-and-series
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user 9025
for questions about sequences and series, e.g. convergence, closed form expressions, etc. Note that there is a different tag for spectral sequences, and also note that MathOverflow is not for homework. Please consider consulting the online encyclopedia for integer sequences, if you are trying to identify a given sequence that you have found in your research.
8
votes
Accepted
Is this series well known?
This is a slice of the taylor series for $\exp(t)$. The terms that dominate are those near $n=t^{1/2}$. Using Stirling's approximation, as $t\to\infty$ with $q$ more or less bounded, we have
$$\frac …
- 37.7k
4
votes
Asymptotics of a recurrence relation
I get a slightly different answer every time I look at it, so I hope the following is ok. I won't dot every last "i" in the error analysis.
Let $F(t)=\sum_{i=0}^\infty a_i(t)$, where $a_i(t)=t^{2^i}2 …
- 37.7k
4
votes
The coefficient of a specific monomial of the following polynomial
This answer is really just a convoluted edition of Fedor's answer. By Cauchy's theorem,
$$f_{abc} = \frac{1}{(2\pi)^3}\int_{-\pi}^\pi \int_{-\pi}^\pi \int_{-\pi}^\pi
\frac{(e^{i\theta_1}-e^{i\thet …
- 37.7k
7
votes
Lower/Upper bounds for $ \sum\limits_{i=0}^k \binom ni x^i $
Write $x=p/(1-p)$ and then
$$ \sum_{i=0}^k \binom ni x^i = (1-p)^{-n}\sum_{i=0}^k \binom ni p^k(1-p)^{n-k}.$$
The last sum is the cumulative binomial distribution, which has no exact formula (except a …
- 37.7k
3
votes
Defining $\{a_i\}$ as $(1+x+⋯+x^k)^n =\sum_{i=0}^{kn}a_ix^i$, then is the 'special' differen...
Yes, because convolutions of log-concave sequences are log-concave. Products of polynomials are convolutions of their coefficient sequence. Search on these keywords and you'll find tons of references …
- 37.7k
1
vote
what this type of series expansion is
It's a pretty ordinary asymptotic expansion for $$\frac{\log f(x)}{\log g(x)}.$$
- 37.7k
9
votes
Sum of square roots of binomial coefficients
For smallish $k$, we have
$$ \binom{n}{n/2+k} \approx \binom{n}{n/2} \exp(-2k^2/n). $$
So
$$\sum\sqrt{\binom{n}{n/2+k}}
\approx \sqrt{\binom{n}{n/2}} \int_{-\infty}^\infty e^{-k^2/n}\,dk
\approx …
- 37.7k
2
votes
Determining the asymptotic behavior of a series
Here is an elementary approach, which shows how to find the nature of $nf_n(t)$ as $n\to\infty$. But I'm not going to bound error terms or such so this remains an outline until those details are fill …
8
votes
Accepted
Inverting an asymptotic series
Since you say that you only want the first few terms, one way you can do this type of thing is by making a contraction mapping. As $x\to\infty$, inspection shows $y\sim x$, so rewrite
the equation as …
- 37.7k
9
votes
2
answers
353
views
Finding local patterns in a circular list
Consider a list $\boldsymbol{x}=x_0,x_1,\ldots,x_{n-1}$, which we consider to be circular by taking the subscripts modulo $n$. The entries in the list are distinct integers.
A local pattern is a Boo …
- 37.7k
2
votes
How to find the asymptotics of a linear two-dimensional recurrence relation
The book Analytic Combinatorics in Several Variables by Pemantle and Wilson covers problems like this extensively.
- 37.7k
6
votes
Convergency radius of the generating series for A93637
As noted in the OEIS entry, the sequence is bounded below by the number of unlabelled rooted trees, so the radius of convergence can be only equal or less. For unlabelled rooted trees, the radius of c …
- 37.7k
25
votes
Accepted
A conjectural infinite series for $\frac{\pi^2}{5\sqrt{5}}$
More generally, if $1\le k\le N-1$ is an integer, where $N$ is a positive interger,
$$S_{N,k} := \sum_{n=0}^\infty\biggl( \frac{1}{(N n+N-k)^2} + \frac{1}{(N n+k)^2}
\biggr) = \frac{\pi^2}{N^2\sin^ …
- 37.7k
2
votes
Is there a closed formula for the generating function of some trinomial coefficients?
For what it is worth (probably very little!),
$$\sum_{d\ge 0}\binom{3d}{d,d,d} x^d$$
is the coefficient of $y^0z^0$ in
$$\frac{1}{1-(yz+1/y+1/z)^3x},$$
and the original series is $\frac13$ of the deri …
- 37.7k
9
votes
Integrality of a sequence formed by sums
(More of a comment than an answer.) Maple's GuessRecurrence function finds that the following recurrence holds for at least 500 terms:
$$8(7n+8)(2n+5)(n+2)a_{n+2} - 6(252n^3+1233n^2+1930n+960)a_{n+1}
…
- 37.7k