숫자 5
" I like explicit, hands-on formulas. To me they have a beauty of their own. They can be deep or not. As an example, imagine you have a series of numbers such that if you add 1 to any number you will get the product of its left and right neighbors. Then this series will repeat itself at every fifth step! For instance, if you start with 3, 4 then the sequence continues: 3, 4, 5/3, 2/3, 1, 3, 4, 5/3, etc. The difference between a mathematician and a nonmathematician is not just being able to discover something like this, but to care about it and to be curious about why it's true, what it means, and what other things in mathematics it might be connected with. In this particular case, the statement itself turns out to be connected with a myriad of deep topics in advanced mathematics: hyperbolic geometry, algebraic K-theory, the Schrodinger equation of quantum mechanics, and certain models of quantum field theory. I find this kind of connection between very elementary and very deep mathematics overwhelmingly beautiful." Don Zagier (Mathematicians: An Outer View of the Inner World)
\(x_{i}+1=x_{i-1}x_{i+1}\), \(x_0=a\), \(x_1=b\)로 정의된 점화식이 있다고 하자.
계산을 해보면, 다음과 같은 수열을 얻게 된다.
\(a,b,\frac{1+b}{a},\frac{1+a+b}{a b},\frac{1+a}{b},a,b,\frac{1+b}{a},\frac{1+a+b}{a b},\frac{1+a}{b}\cdots\)
주기가 5인 수열이 됨을 확인할 수 있다.
\(a=3,b=4\) 인 경우라면, 위에서처럼 3,4,5/3,2/3,1,3,4,5/3 … 을 얻게 된다.
아무런 맥락없이 본다면, 매우 단순한 사실이지만 실제로 나는 요즘 이 주기가 5인 수열과 씨름을 하는 중이다.
\(\int_{0}^{\frac{\sqrt{5}-1}{2}}\frac{\log(1-t)}{t}+\frac{\log(t)}{1-t}dt=-\frac{\pi^2}{5}\)
\(L(x)=-\frac{1}{2}\int_{0}^{x}\frac{\log(1-y)}{y}+\frac{\log(y)}{1-y}dy\) 로 정의된다.
\(x\in [0,1]\) 에서의 그래프는 다음과 같이 생겼다.
[/pages/4855791/attachments/3056365 Roger_dilogarithm.jpg]
이 함수가 만족시키는 가장 중요한 성질로 5항 관계식 (5-term relation) 이라는 것이 있는데, 다음과 같다.
\(0\leq x,y\leq 1\) 일 때, \(L(x)+L(1-xy)+L(y)+L(\frac{1-y}{1-xy})+L\Left( \frac{1-x}{1-xy} )\right)=\frac{\pi^2}{2}\)
여기서 \(x=y=\frac{1}{2} \left(\sqrt{5}-1\right)\) 로 두면, \(x=1-x y=y=\frac{1-y}{1-x y}=\frac{1-x}{1-x y}=\frac{1}{2} \left(\sqrt{5}-1\right)\) 가 된다.
\(L(\frac{-1+\sqrt{5}}{2})=\frac{\pi^2}{10}\) 를 얻을 수 있다.
이러한 적분을 이용하면,
\(G(q) = \sum_{n=0}^\infty \frac {q^{n^2}} {(q;q)_n} = \frac {1}{(q;q^5)_\infty (q^4; q^5)_\infty} =1+ q +q^2 +q^3 +2q^4+2q^5 +3q^6+\cdots\)
\(H(q) =\sum_{n=0}^\infty \frac {q^{n^2+n}} {(q;q)_n} = \frac {1}{(q^2;q^5)_\infty (q^3; q^5)_\infty} =1+q^2 +q^3 +q^4+q^5 +2q^6+\cdots\)
\(q=e^{-t}\) 으로 두면 \(t\sim 0\) 일 때,
\(H(q)=\sum_{n=0}^\infty \frac {q^{n^2+n}} {(q;q)_n} \sim \sqrt\frac{2}{5+\sqrt{5}}\exp(\frac{\pi^2}{15t}+\frac{11t}{60})+o(1)\)
\(G(q)=\sum_{n=0}^\infty \frac {q^{n^2}} {(q;q)_n} \sim \sqrt\frac{2}{5-\sqrt{5}}\exp(\frac{\pi^2}{15t}-\frac{t}{60})+o(1)\)
와 같은 식을 보이는데 쓸 수 있다.
\(\cfrac{1}{1 + \cfrac{e^{-2\pi}}{1 + \cfrac{e^{-4\pi}}{1+\dots}}} = \left({\sqrt{5+\sqrt{5}\over 2}-{\sqrt{5}+1\over 2}}\right)e^{2\pi/5} = e^{2\pi/5}\left({\sqrt{\varphi\sqrt{5}}-\varphi}\right) = 0.9981360\dots\)
Andrei Zelevinsky,What is... a Cluster Algebra,Notices of the AMS,54 (2007),no .11,494-1495. http://www.ams.org/notices/200711/tx071101494p.pdf