Tagged Questions
1
vote
0answers
101 views
Using the gamma function as an upper and lower bound to the logarithm of a factorial function.
I am trying to find an upper and lower bound for the following function:
$$f(x) = \ln(\lfloor\frac{x}{b_1}\rfloor!) - \ln(\lfloor\frac{x}{b_2}\rfloor!) - \ln(\lfloor\frac{x}{b_3}\rfloor!)$$
where
...
0
votes
3answers
176 views
Maple Error on Asymptotic Analysis of $\ln(n)!$
In Maple, the command asmypt($f$,$x$) computes the asymptotic expansion of the function $f$ with respect to the variable $x$ (as $x \rightarrow \infty$).
The command
asympt(ln(n)!,n);
gives the ...
0
votes
1answer
83 views
How to compare big numbers that are outcome of different functions.
How is the best way to compare big numbers? They are result of two functions with different asymptotic growth. For example:
Googleplex which is $10^{{10}^{100}}$ to $1000!$
2
votes
2answers
682 views
Why is $\log(n!)$ $O(n\log n)$?
I thought that $\log(n!)$ would be $\Omega(n \log n )$, but I read somewhere that $\log(n!) = O(n\log n)$.
Why?
13
votes
2answers
327 views
On the Limit of Stirling's Approximation
I have recently proven the following curious identity: For real $x \geqslant 1$,
\begin{align}
\lfloor x \rfloor! = x^{\lfloor x \rfloor} e^{1-x} e^{\int_{1}^{x} \text{frac}(t)/t \ dt}
\end{align}
...
12
votes
6answers
563 views
A question on the Stirling approximation, and $\log(n!)$
In the analysis of an algorithm this statement has come up:$$\sum_{k = 1}^n\log(k) \in \Theta(n\log(n))$$ and I am having trouble justifying it. I wrote $$\sum_{k = 1}^n\log(k) = \log(n!), \ \ ...
20
votes
9answers
2k views
What is the purpose of Stirling's approximation to a factorial?
Stirling approximation to a factorial is
$$
n! \sim \sqrt{2 \pi n} \left(\frac{n}{e}\right)^n.
$$
I wonder what benefit can be got from it?
From computational perspective (I admit I don't ...
13
votes
6answers
2k views
Stirling's formula: proof?
Suppose we want to show that $$ n! \sim \sqrt{2 \pi} n^{n+(1/2)}e^{-n}$$
Instead we could show that $$\lim_{n \to \infty} \frac{n!}{n^{n+(1/2)}e^{-n}} = C$$ where $C$ is a constant. Maybe $C = ...
5
votes
2answers
365 views
Asymptotics of terms and errors in Stirling's Approximation
I have two related questions. Both are related to the asymptotics of Stirling's approximation, which is why I have included them in the same question. I will separate the questions if it is deemed ...
