Replicator equation

In mathematical analysis, the Bohr–Mollerup theorem is named after the Danish mathematicians Harald Bohr and Johannes Mollerup, who proved it. The theorem characterizes the gamma function, defined for x > 0 by

$\mathrm{\Gamma }(x)=\underset{0}{\overset{\mathrm{\infty }}{\int }}{t}^{x-1}{e}^{-t}dt$

as the only function f on the interval x > 0 that simultaneously has the three properties

• $f(1)=1,$ and
• $f(x+1)=xf(x)\text{ for }x>0,$ and
• f is logarithmically convex.

An elegant treatment of this theorem is in Artin's book The Gamma Function, which has been reprinted by the AMS in a collection of Artin's writings.

The theorem was first published in a textbook on complex analysis, as Bohr and Mollerup thought it had already been proved.

Proof

Statement of the theorem

$\mathrm{\Gamma }(x)$ is the only function that satisfies $f(x+1)=xf(x)$ with $\log (f(x))$ convex and also with $f(1)=1$.

Proof

Let $\mathrm{\Gamma }(x)$ be a function with the assumed properties established above: $\mathrm{\Gamma }(x+1)=x\mathrm{\Gamma }(x)$ and $\log (\mathrm{\Gamma }(x))$ is convex, and $\mathrm{\Gamma }(1)=1$. From the fact that $\mathrm{\Gamma }(x+1)=x\mathrm{\Gamma }(x)$ we can establish

$\begin{array}{r}\mathrm{\Gamma }(x+n)=(x+n-1)(x+n-2)(x+n-3)\cdots (x+1)x\mathrm{\Gamma }(x)\end{array}$

The purpose of the stipulation that $\mathrm{\Gamma }(1)=1$ forces the $\mathrm{\Gamma }(x+1)=x\mathrm{\Gamma }(x)$ property to duplicate the factorials of the integers so we can conclude now that $\mathrm{\Gamma }(n)=(n-1)!$ if $n\in \mathbb{\mathbb{N}}$ and if $\mathrm{\Gamma }(x)$ exists at all. Because of our relation for $\mathrm{\Gamma }(x+n)$, if we can fully understand $\mathrm{\Gamma }(x)$ for $0<x\le 1$ then we understand $\mathrm{\Gamma }(x)$ for all values of $x$.

The slope of a line connecting two points $(x_1,f(x_1))$ and $(x_2,f(x_2))$, call it $\mathcal{ℳ}(x_1,x_2)$ is monotonically increasing for convex functions with $x_1<x_2$. Since we have stipulated $\log (\mathrm{\Gamma }(x))$ is convex we know

$\begin{array}{rl}\mathcal{ℳ}(n-1,n)& \le \mathcal{ℳ}(n,n+x)\le \mathcal{ℳ}(n,n+1)\mathrm{w}\mathrm{h}\mathrm{e}\mathrm{n}0<x\le 1\\ \frac{\log (\mathrm{\Gamma }(n))-\log (\mathrm{\Gamma }(n-1))}{n-(n-1)}& \le \frac{\log (\mathrm{\Gamma }(n))-\log (\mathrm{\Gamma }(n+x))}{n-(n+x)}\le \frac{\log (\mathrm{\Gamma }(n))-\log (\mathrm{\Gamma }(n+1))}{n-(n+1)}\\ \frac{\log ((n-1)!)-\log ((n-2)!)}{1}& \le \frac{\log (\mathrm{\Gamma }(n+x))-\log ((n-1)!)}{x}\le \frac{\log \left(n!\right)-\log ((n-1)!)}{1}\\ \log \left(\frac{(n-1)!}{(n-2)!}\right)& \le \frac{\log (\mathrm{\Gamma }(n+x))-\log ((n-1)!)}{x}\le \log \left(\frac{n!}{(n-1)!}\right)\\ \log (n-1)& \le \frac{\log (\mathrm{\Gamma }(n+x))-\log ((n-1)!)}{x}\le \log \left(n\right)\\ x\cdot \log (n-1)+\log ((n-1)!)& \le \log (\mathrm{\Gamma }(n+x))\le x\cdot \log \left(n\right)+\log ((n-1)!)\\ \log ((n-1{)}^x(n-1)!)& \le \log (\mathrm{\Gamma }(n+x))\le \log (n^x(n-1)!)\end{array}$

It is evident from this last line that a function is being sandwiched between two expressions, a common analysis technique to prove various things such as the existence of a limit, or convergence. Now we recall that the function $\log ()$ and $e^{()}$ are both monotonically increasing. Therefore if we exponentiate each term of the inequality, we will preserve the inequalities. Continuing:

$\begin{array}{rl}(n-1{)}^x(n-1)!& \le \mathrm{\Gamma }(n+x)\le n^x(n-1)!\\ (n-1{)}^x(n-1)!& \le (x+n-1)(x+n-2)\cdots (x+1)x\mathrm{\Gamma }(x)\le n^x(n-1)!\\ \frac{(n-1{)}^x(n-1)!}{(x+n-1)(x+n-2)\cdots (x+1)x}\le \mathrm{\Gamma }(x)& \le \frac{n^x(n-1)!}{(x+n-1)(x+n-2)\cdots (x+1)x}\\ \frac{(n-1{)}^x(n-1)!}{(x+n-1)(x+n-2)\cdots (x+1)x}& \le \mathrm{\Gamma }(x)\le \frac{n^{x}n!}{(x+n)(x+n-1)\cdots (x+1)x}\left(\frac{n+x}{n}\right)\\ \end{array}$

The last line is a strong statement. In particular, it is true for all values of $n$. That is $\mathrm{\Gamma }(x)$ is less than the right hand side for any choice of $n$ and likewise, $\mathrm{\Gamma }(x)$ is greater than the left hand side for any other choice of $n$. Each single inequality stands alone and may be interpreted as an independent statement. Because of this fact, we are free to choose different values of $n$ for the RHS and the LHS. In particular, if we keep $n$ for the RHS and choose $n+1$ for the LHS and get:

$\begin{array}{rl}\frac{((n+1)-1{)}^x((n+1)-1)!}{(x+(n+1)-1)(x+(n+1)-2)\cdots (x+1)x}& \le \mathrm{\Gamma }(x)\le \frac{n^{x}n!}{(x+n)(x+n-1)\cdots (x+1)x}\left(\frac{n+x}{n}\right)\\ \frac{n^{x}n!}{(x+n)(x+n-1)\cdots (x+1)x}& \le \mathrm{\Gamma }(x)\le \frac{n^{x}n!}{(x+n)(x+n-1)\cdots (x+1)x}\left(\frac{n+x}{n}\right)\\ \end{array}$

Now let $n\to \mathrm{\infty }$. The limit drives $\frac{n+x}{n}\to 1$ so the left side of the last inequality is driven to equal the right side. $\frac{n^{x}n!}{(x+n)(x+n-1)\cdots (x+1)x}$ is sandwiched in between. This can only mean that $\underset{n\to \mathrm{\infty }}{\lim }\frac{n^{x}n!}{(x+n)(x+n-1)\cdots (x+1)x}$ is equal to $\mathrm{\Gamma }(x)$. In the context of this proof this means that $\underset{n\to \mathrm{\infty }}{\lim }\frac{n^{x}n!}{(x+n)(x+n-1)\cdots (x+1)x}$ has the three specified properties belonging to $\mathrm{\Gamma }(x)$. Also, the proof provides a specific expression for $\mathrm{\Gamma }(x)$. And the final critical part of the proof is to remember that the limit of a sequence is unique. This means that for any choice of $x\in (0,1]$ only one possible number $\mathrm{\Gamma }(x)$ can exist. Therefore there is no other function with all the properties assigned to $\mathrm{\Gamma }(x)$. the assumptions of this theorem to

The remaining loose end is the question of proving that $\mathrm{\Gamma }(x)$ makes sense for all $x$ where $\underset{n\to \mathrm{\infty }}{\lim }\frac{n^{x}n!}{(x+n)(x+n-1)\cdots (x+1)x}$ exists. The problem is that our first double inequality

$\begin{array}{r}\mathcal{ℳ}(n-1,n)\le \mathcal{ℳ}(n+x,n)\le \mathcal{ℳ}(n+1,n)\end{array}$

was constructed with the constraint $0<x\le 1$. If, say, $x>1$ then the fact that $\mathcal{ℳ}$ is monotonically increasing would make $\mathcal{ℳ}(n+1,n)<\mathcal{ℳ}(n+x,n)$, contradicting the inequality upon which the entire proof is constructed. But notice

$\begin{array}{rl}\mathrm{\Gamma }(x+1)& =\underset{n\to \mathrm{\infty }}{\lim }x\cdot \left(\frac{n^{x}n!}{(x+n)(x+n-1)\cdots (x+1)x}\right)\frac{n}{n+x+1}\\ \mathrm{\Gamma }(x)& =\left(\frac{1}{x}\right)\mathrm{\Gamma }(x+1)\end{array}$

which demonstrates how to bootstrap $\mathrm{\Gamma }(x)$ to all values of $x$ where the limit is defined.

References

• Other Sports Official Kull from Drumheller, has hobbies such as telescopes, property developers in singapore and crocheting. Identified some interesting places having spent 4 months at Saloum Delta.
  
  my web-site http://himerka.com/
• 
  
  I had like 17 domains hosted on single account, and never had any special troubles. If you are not happy with the service you will get your money back with in 45 days, that's guaranteed. But the Search Engine utility inside the Hostgator account furnished an instant score for my launched website. Fantastico is unable to install WordPress in a directory which already have any file i.e to install WordPress using Fantastico the destination directory must be empty and it should not have any previous installation files. When you share great information, others will take note. Once your hosting is purchased, you will need to setup your domain name to point to your hosting. Money Back: All accounts of Hostgator come with a 45 day money back guarantee. If you have any queries relating to where by and how to use Hostgator Discount Coupon, you can make contact with us at our site. If you are starting up a website or don't have too much website traffic coming your way, a shared plan is more than enough. Condition you want to take advantage of the worldwide web you prerequisite a HostGator web page, -1 of the most trusted and unfailing web suppliers on the world wide web today. Since, single server is shared by 700 to 800 websites, you cannot expect much speed.
  
  
  
  Hostgator tutorials on how to install Wordpress need not be complicated, especially when you will be dealing with a web hosting service that is friendly for novice webmasters and a blogging platform that is as intuitive as riding a bike. After that you can get Hostgator to host your domain and use the wordpress to do the blogging. Once you start site flipping, trust me you will not be able to stop. I cut my webmaster teeth on Control Panel many years ago, but since had left for other hosting companies with more commercial (cough, cough) interfaces. If you don't like it, you can chalk it up to experience and go on. First, find a good starter template design. When I signed up, I did a search for current "HostGator codes" on the web, which enabled me to receive a one-word entry for a discount. Your posts, comments, and pictures will all be imported into your new WordPress blog.
• Template:Planetmath reference
• Template:Planetmath reference
• 20 year-old Real Estate Agent Rusty from Saint-Paul, has hobbies and interests which includes monopoly, property developers in singapore and poker. Will soon undertake a contiki trip that may include going to the Lower Valley of the Omo.
  
  My blog: http://www.primaboinca.com/view_profile.php?userid=5889534
• 20 year-old Real Estate Agent Rusty from Saint-Paul, has hobbies and interests which includes monopoly, property developers in singapore and poker. Will soon undertake a contiki trip that may include going to the Lower Valley of the Omo.
  
  My blog: http://www.primaboinca.com/view_profile.php?userid=5889534
• 20 year-old Real Estate Agent Rusty from Saint-Paul, has hobbies and interests which includes monopoly, property developers in singapore and poker. Will soon undertake a contiki trip that may include going to the Lower Valley of the Omo.
  
  My blog: http://www.primaboinca.com/view_profile.php?userid=5889534 (Textbook in Complex Analysis)