0
votes
0answers
9 views

Asymptotic approximate

How would you approximate the following sum in terms of n: $$\sum_{k=1}^{n} \binom{n}{k} (k-1)(a - n + k)^{-a + n - k - 1/2}(-1 - a - k + 2)^{1 + a + k - 5/2} e^{-n + 2-\frac{(n - k)}{12*a*(a - n + ...
1
vote
0answers
18 views

Approximating the max value of a function containing the half-sum of binomial series

I recently met a problem and I have been finding related materials about it. However it is still hard to handle. Precisely, I want to maximize a function related to $$S_n = \sum_{i=0}^{\lfloor ...
0
votes
1answer
62 views

Is this possible or hopeless to try to prove?

If I have $x_1, ..., x_k=o(n)$ and $j=O(1)$. Is it possible to prove something like: $$\sum_{i=1}^k {n \choose j} \left(\frac{x_i}{n}\right)^j \left(1-\frac{x_i}{n}\right)^{k-j} \sim {n \choose j} ...
1
vote
0answers
23 views

How to get a nice approximation of $f(N,s)=\sum_{k=0}^{N}{N \choose k}{k \choose s-k+N}$ when $N>>1$ and $|s|<<N$?

I need to approximate the above sum in order to calculate $\mathbb{E}(s^2)$, which is the expectation value determined by the probability density function $f$ and the position $s$. Any idea?
0
votes
0answers
10 views

Confidence intervals for bernoulli trials over a cyclic time series

I have a time series with observations of 0 or 1 observed yearly for approx. 20 years. The time series is cyclic and I want to find a CI for the probability p over the cycle (mean). Unfortunately I ...
1
vote
1answer
118 views

Asymptotics of sum of Binomial Coefficients (Binomial distribution) - Poisson approximation?

Let $$f(n):=\sum_{i=k}^n {n \choose i } p^i (1-p)^{n-i}$$ where $k\geq 2$ is a fixed Parameter and $p=p(n) \in (0,1]$ depends on $n$ where $np\leq 1$. We consider $n \rightarrow \infty$. I've found ...
1
vote
2answers
50 views

Is $n\binom{\epsilon n}{t}>t\binom{n}{t}$ for large $n$ and fixed $\epsilon$ and $t$

Let $\epsilon$ and $t$ be fixed numbers with $t$ and integer. I came across the following inequality in a counting problem. $$n\binom{\epsilon n}{t}>t\binom{n}{t}.$$ I want to show that for $n$ ...
0
votes
0answers
28 views

Approximation of sum with binomial summands

I am new here, so hopefully my question will be understood correctly. I have a function (originating from expected untility theory in economics) that looks the following: ...
2
votes
2answers
174 views

Sums of central binomial coefficients

Are there closed forms for $$\sum^n_{i=0} \binom{2i}{i}$$ and $$\sum^n_{i=0} \binom{2i}{i}^2$$? Also, how can these sums be approximated?
2
votes
0answers
99 views

Good upper bound on a binomial sum

What is a good upper bound on the following binomial sum: $$\sum_{i,j< \frac{m}{n}} {m \choose i}{m-i \choose j} z^{i+j}$$ where $z = \frac{1}{n-2}$?
2
votes
0answers
93 views

Weighted sum of ratio of partial sum of binomial coefficients

I would like to approximate the following sum when $n \rightarrow \infty$ and $n \gg k$, $$\sum_{x = k}^n \sum_{y > x}^n \frac{\sum_{m = 0}^{k - 1} {y - 2 \choose m}}{\sum_{m = 0}^k {y - 1 \choose ...
4
votes
1answer
89 views

How to calculate (or approximate) “trimmed” (a+b)^n?

$a^n + C_n^{1}a^{n-1}b + ... C_n^{n-1}a^{1}b^{n-1}+b^n = (a+b)^n$ But how to calculate (maybe approximately) $a^n + C_n^{1}a^{n-1}b + ... C_n^{i}a^{n-i}b^{i} = ?$ For info, the underlying problem ...
1
vote
0answers
51 views

Approximation of factorial - Stirling formula [duplicate]

Possible Duplicate: Elementary central binomial coefficient estimates How can I prove that $$ \binom{n}{n/2} = \Theta\left(\frac{2^n}{\sqrt n}\right) $$ I tried with Stirlings ...
1
vote
1answer
386 views

Using binomial theorem find general formula for the coefficients

Using binomial thaorem (http://en.wikipedia.org/wiki/Binomial_theorem) find the general formula for the coefficients of the expantion: $$ ...
3
votes
1answer
89 views

is the approximation of the sum true?

Someone commented under my question Calculation of the moments using Hypergeometric distribution that $$ \sum_{k=0}^l\frac{{l \choose k}{2n-l \choose n-k}(2k-l)^q}{{2n\choose n}}\sim \sum_{k=0}^l ...
3
votes
1answer
286 views

Surprising approximation of weighted sum of binomial coefficients

The following sum appeared in connection to the problem addition of angular momentum in physics: $$ \frac{1}{2^{n+3}}\sum_{k=0}^n \left(\frac{n-2k-1}{\sqrt{k+1}}+\frac{n-2k+1}{\sqrt{n-k+1}}\right)^2 ...
0
votes
0answers
111 views

Calculation of sum

I am wondering if it is possible to calculate or approximate the following sum $$ \sum_{k=0}^l\frac{(l-2k)^p(2l+k(k-1))l^{k-1}}{(k+3)(k+2)} $$here $p \geq 2$. Thank you.
6
votes
2answers
415 views

Elementary bound of binomial coefficient

I'm working my way through an Erdős paper from the sixties and some of the elementary bounds he claims seem to be just beyond my reach. The expression looks horrendous but maybe there is a clever ...
1
vote
0answers
286 views

Calculation of a 'double' sum

Let $n \in N$ and $q\geq 2$. I am trying to calculate the following sum: $$ \sum_{i=0}^{\sqrt n/2}\sum_{j= i \sqrt n }^{(i+1)\sqrt n}\frac{(-1)^q2^q(\frac{n}{2}-j)^q}{(n-j)!j!} $$ Any help will be ...
4
votes
1answer
203 views

Evaluating a limit involving binomial coefficients.

If $N_c=\lfloor \frac{1}{2}n\log n+cn\rfloor$ for some integer $n$ and real constant $c$, then how would one go about showing the following identity where $k$ is a fixed integer: $$\lim_{n\rightarrow ...
4
votes
2answers
375 views

computation of the sum

I am having trouble to compute the following sum: $$ \sum_{k=0}^n(n-2k)^p \frac{{n \choose k}{2m-n \choose m-k}}{{2m \choose m}} $$ Here $p\geq 2$. To simplify the question, we can even assume that ...
9
votes
2answers
4k views

Approximating the logarithm of the binomial coefficient

We know that by using Stirling approximation: $\log n! \approx n \log n$ So how to approximate $\log {m \choose n}$?
6
votes
4answers
4k views

How to simplify or calculate a formula with very big factorials

I'm facing a practical problem where I've calculated a formula that, with the help of some programming, can bring me to my final answer. However, the numbers involved are so big that it takes ages to ...