Pythagoras0: 새 문서: ==five-term relation== * sequence defined recurrence : $1-x_{i}=x_{i-1}x_{i+1},\, x_0=x,\, x_2=y$ * solution : $x_0=x, x_1=1-xy, x_2=y, x_3=\frac{1-y}{1-xy}, x_4=\frac{1...$

2023-12-13T07:33:55Z

새 문서: ==five-term relation== * sequence defined recurrence :<math>1-x_{i}=x_{i-1}x_{i+1},\, x_0=x,\, x_2=y</math> * solution :<math>x_0=x, x_1=1-xy, x_2=y, x_3=\frac{1-y}{1-xy}, x_4=\frac{1...

새 문서

==five-term relation==
* sequence defined recurrence
:<math>1-x_{i}=x_{i-1}x_{i+1},\, x_0=x,\, x_2=y</math>
* solution
:<math>x_0=x, x_1=1-xy, x_2=y, x_3=\frac{1-y}{1-xy}, x_4=\frac{1-x}{1-xy}, x_5=x,x_6=1-xy,\dots</math>
* 5-periodic
[[파일:5항 관계식 (5-term relation)3.png]]
* if <math>0<x<1,\,0<y<1</math>, then <math>0<x_i<1</math> for all <math>i</math>
* if <math>x=a/(1+a),y = b/(1+b)</math>, then <math>x_0=\frac{a}{a+1},x_1\frac{a+b+1}{ab+a+b+1},\frac{b}{b+1},\frac{a+1}{a+b+1},\frac{b+1}{a+b+1},\dots</math>
<math>
\newcommand{\vol}{\operatorname{vol}}
</math>

==dilogarithm fuction==
* Define
:<math>\operatorname{Li}_ 2(z)= \sum_{n=1}^\infty {z^n \over n^2},\, |z|<1</math>
* extend domain
:<math>\operatorname{Li}_ 2(z) = -\int_0^z{{\log (1-t)}\over t} dt,\, z\in \mathbb C\backslash [1,\infty) </math>

===functional equations===
; five-term relation
:<math>\mbox{Li}_ 2(x)+\mbox{Li}_ 2(y)+\mbox{Li}_ 2 \left( \frac{1-x}{1-xy} \right)+\mbox{Li}_ 2(1-xy)+\mbox{Li}_ 2 \left( \frac{1-y}{1-xy} \right)=\text{elementary}</math>

Let us state this in terms of the Rogers dilogarithm (no worry about the branches)
:<math>L(x): =\operatorname{Li}_ 2(x)+\frac{1}{2}\log x\log (1-x)=-\frac{1}{2}\int_{0}^{x}\left(\frac{\log(y)}{1-y}+\frac{\log(1-y)}{y}\right)dy,\, x\in (0,1)</math>
with <math>L(0)=0,\, L(1)=\pi^2/6</math>
* <math>0< x,y< 1</math>
:<math>L(x)+L(1-xy)+L(y)+L\left(\frac{1-y}{1-xy}\right)+L\left(\frac{1-x}{1-xy} \right)=\frac{\pi^2}{2}</math>

;proof
Let us show that the partial derivatives of <math>F(x,y):=L(x)+L(1-xy)+L(y)+L\left(\frac{1-y}{1-xy}\right)+L\left(\frac{1-x}{1-xy} \right)</math> are 0.
Note
:<math>
\frac{d}{dx}L(h(x)) = \frac{1}{2}[\frac{-h'(x) \log (h(x))}{(1-h(x))}-\frac{h'(x) \log (1-h(x))}{h(x)}].
</math>

:<math>
\begin{aligned}
2F_x = &
\left(-\frac{\log (x)}{1-x}-\frac{\log (1-x)}{x}\right)+\left(\frac{\log (1-x y)}{x}+\frac{y \log (x y)}{1-x y}\right)+0 \\
& -\frac{(1-y)
\log \left(\frac{1-y}{1-x y}\right)+(1-x) y \log \left(\frac{(1-x) y}{1-x y}\right)}{(1-x) (1-x y)}
+\frac{(1-x) \log \left(\frac{1-x}{1-x y}\right)+x (1-y)
\log \left(\frac{x (1-y)}{1-x y}\right)}{(1-x) x (1-x y)} \\
& =\log (x)\left(\frac{-1}{1-x}+\frac{y}{1-xy}+\frac{(1-y)}{(1-x) (1-x y)} \right)+\dots \\
& =0
\end{aligned}
</math>
Do the same to show <math>F_y=0</math>.

<math>\exists</math> more intelligent way to control cancellations

Observe
:<math>\frac{d}{dx}L(h(x))=\frac{1}{2}[\log(h(x))\frac{d}{dx}\log (1-h(x))-\log(1-h(x))\frac{d}{dx}\log h(x)]</math>

For <math>f,g\in \mathbb{Q}(x,y)^{\times}</math>, define formally
:<math>
f\wedge g : = \frac{1}{2}[\log (f) d (\log (g))-\log (g) d (\log (f))]
</math>
where <math>df = f_x dx + f_y dy</math>.

Then
*<math>f\wedge g=-g \wedge f</math>
*<math>(f_1f_2)\wedge g=f_1\wedge g+f_2\wedge g</math>

For example,
:<math>dL(h(x,y))=h\wedge (1-h)</math>

So
:<math>
dF = F_x dx+F_y dy =x\wedge (1-x)+(1-x y)\wedge (x y)+y\wedge (1-y)+\frac{1-y}{1-x y}\wedge \left(\frac{y(1-x)}{1-x y}\right)+\frac{1-x}{1-x y}\wedge \left(\frac{x(1-y)}{1-xy}\right) =0
</math>

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Pythagoras0: 새 문서: ==five-term relation== * sequence defined recurrence :1-x_{i}=x_{i-1}x_{i+1},\, x_0=x,\, x_2=y * solution :x_0=x, x_1=1-xy, x_2=y, x_3=\frac{1-y}{1-xy}, x_4=\frac{1...

Pythagoras0: 새 문서: ==five-term relation== * sequence defined recurrence : $1-x_{i}=x_{i-1}x_{i+1},\, x_0=x,\, x_2=y$ * solution : $x_0=x, x_1=1-xy, x_2=y, x_3=\frac{1-y}{1-xy}, x_4=\frac{1...$