1
vote
1answer
32 views

A better approximation of $({n \over e})^n$ than by $ \Gamma(1+n-1/2)$ ? (Focus is “reverse” to the Stirling approximation)

In a formula in my self-study of a summation method based on the matrix of Eulerian numbers (which I thus call "Eulerian summation") I am considering terms like $$ \Big({n \over e}\Big)^n \cdot {1 ...
1
vote
0answers
9 views

Simplifcation of the Function Required

I want to simplify the following expression $$ \sum_{x=0}^{D+m} \frac {x!} {(n+1)_x} (\frac {\theta_{eff}} {\mu})^x {D+m \choose x} $$ The parameters $D, m, n, \theta_{eff}, \mu, $ are constants. ...
1
vote
4answers
92 views

Demonstrate that $\displaystyle \frac{(2n - 2)!!}{(2n - 3)!!} \simeq 1.7 \sqrt{n}$

As in the title, I know that $\displaystyle \frac{(2n - 2)!!}{(2n - 3)!!} = \frac{(2n - 2)(2n - 4)\cdots 4 \cdot 2}{(2n - 3)(2n - 5) \cdots 3 \cdot 1} \simeq 1.7 \sqrt{n}$ Could you give some hint ...
2
votes
1answer
46 views

Approximation of a factorial

In books I can find a lot of approximations of factorial, which doesn't require the a lot of multiplications, like for example Stirling's approximation. Very popular is also the Euler's solution which ...
1
vote
0answers
40 views

Approximating the probability that a range bounds a given number, with very large numbers

Let $m$ numbers be chosen uniformly from $0,\dots,n-1$ without replacement and then sorted in ascending order as $\ell_0,\dots,\ell_{m-1}$. Let there be $b,e,x$ such that $0 \le b \le e \le m$ and $0 ...
1
vote
1answer
62 views

Understanding an Approximation

I am reading the paper A Group-theoretic Approach to Fast Matrix Multiplication and there is an approximation in the paper I don't fully understand. In the proof of Theorem 3.3. it is stated that $$ ...
4
votes
2answers
100 views

Approximation of $ \frac{n!}{(n-2x)!}(n-1)^{-2x} $

I would like to find an approximation when $ n \rightarrow\infty$ of $ \frac{n!}{(n-2x)!}(n-1)^{-2x} $. Using Stirling formula, I obtain $$e^{\frac{-4x^2+x}{n}}. $$ The result doesn't seem right! ...
3
votes
0answers
40 views

Bound on Permutations [duplicate]

I am trying to prove the following inequality, $$n^{(l)} = n(n-1)\cdots(n-l+1) \geq \frac{n^l}{e}\quad\text{ for }\quad 2 \leq l \leq \sqrt{n}\;.$$ So my approach is to observe that $n^{(l)} = ...
1
vote
0answers
48 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 ...
2
votes
2answers
384 views

How many bits are in factorial?

I am interested in good integer approximation from below and from above for binary Log(N!). The question and the question provides only a general idea but not exact values. In other words I need ...
2
votes
0answers
141 views

Generating all positive integers from three operations

This question arose on sci.math, but since almost all the competent mathematicians there have migrated here, I thought I'd give the question a wider audience. Starting with an integer $t>2$, ...
3
votes
3answers
248 views

How to approximate $\sum_{k=1}^n k!$ using Stirling's formula?

How to find summation of the first $n$ factorials, $$1! + 2! + \cdots + n!$$ I know there's no direct formula, but how can it be estimated using Stirling's formula? Another question : Why can't ...
5
votes
3answers
2k views

Approximating log of factorial

I'm wondering if people had a recommendation for approximating $\log(n!)$. I've been using Stirlings formula, $ (n + \frac{1}{2})\log(n) - n + \frac{1}{2}\log(2\pi) $ but it is not so great for ...
21
votes
9answers
3k views

What is the purpose of Stirling's approximation to a factorial?

Stirling approximation to a factorial is $$ n! \sim \sqrt{2 \pi n} \left(\frac{n}{e}\right)^n. $$ I wonder what benefit can be got from it? From computational perspective (I admit I don't ...
9
votes
2answers
3k 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 $\log {m \choose n}$?
27
votes
1answer
633 views

A series problem by Knuth

I came across the following problem, known as Knuth's Series which originally was an American Mathematical Monthly problem. Prove that $$\sum_{n=1}^\infty ...
16
votes
5answers
1k views

How best to explain the $\sqrt{2\pi n}$ term in Stirling's?

I recently showed my Algorithms class how to bound $\ln n! = \sum \ln n$ by integrals, thereby obtaining the simple factorial approximation $$ e \left(\frac{n}{e}\right)^{n} \leq n! \leq ...