자코비 다항식

개요

• 직교다항식

정의

• 초기하급수(Hypergeometric series)를 통해 정의된다\[P_n^{(\alpha,\beta)}(z)=\frac{(\alpha+1)_n}{n!} \,_2F_1\left(-n,1+\alpha+\beta+n;\alpha+1;\frac{1-z}{2}\right)\]
• 다항식표현\[P_n^{(\alpha,\beta)} (z) = \frac{\Gamma (\alpha+n+1)}{n!\Gamma (\alpha+\beta+n+1)} \sum_{m=0}^n {n\choose m} \frac{\Gamma (\alpha + \beta + n + m + 1)}{\Gamma (\alpha + m + 1)} \left(\frac{z-1}{2}\right)^m\]

로드리게스 공식

• 다음이 성립한다

$$ (1-x)^{\alpha } (1+x)^{\beta } P_n^{(\alpha ,\beta )}(x)=\frac{(-1)^n}{2^n n!}\frac{d^n}{dx^n}\left[\left((1-x)^{\alpha +n} (1+x)^{\beta +n}\right)\right] \label{RF} $$

미분방정식

• 자코비 다항식은 다음을 만족시킨다\[(1-x^2)y'' + ( \beta-\alpha - (\alpha + \beta + 2)x )y'+ n(n+\alpha+\beta+1) y = 0\]

직교성

• weight함수와 구간

\[w(x) = (1-x)^{\alpha} (1+x)^{\beta}, x\in [-1,1]\]

• 다음이 성립한다

$$ \int_{-1}^1(1-x)^{\alpha} (1+x)^{\beta}\,dx=2^{\alpha+\beta+1}\frac{\Gamma(\alpha+1)\Gamma(\beta+1)}{\Gamma(\alpha+\beta+2)} $$

(증명)

$t=(1-x)/2$로 치환하면, $$ \begin{aligned} \int_{-1}^1(1-x)^{\alpha} (1+x)^{\beta}\,dx=&\int_0^1 2^{\alpha+\beta+1}t^{\alpha}(1-t)^{\beta}\, dt \\ =&2^{\alpha+\beta+1}B(\alpha+1,\beta+1)\\ =&2^{\alpha+\beta+1}\frac{\Gamma(\alpha+1)\Gamma(\beta+1)}{\Gamma(\alpha+\beta+2)} \end{aligned} $$ 여기서 $B(x,y)$는 오일러 베타적분(베타함수) ■

(정리)

• $m,n\in \mathbb{Z}_{\geq 0}$에 대하여,

\[\int_{-1}^1 (1-x)^{\alpha} (1+x)^{\beta} P_m^{(\alpha,\beta)} (x)P_n^{(\alpha,\beta)} (x) \; dx= \frac{2^{\alpha+\beta+1}}{2n+\alpha+\beta+1} \frac{\Gamma(n+\alpha+1)\Gamma(n+\beta+1)}{\Gamma(n+\alpha+\beta+1)n!} \delta_{nm}\]

예

• \(\alpha=1/2,\beta=1/2,m=n=2\)인 경우

\[\int_{-1}^1 (1-x)^{\frac{1}{2}} (1+x)^{\frac{1}{2}} P_2^{(\frac{1}{2},\frac{1}{2})} (x)P_2^{(\frac{1}{2},\frac{1}{2})} (x) \; dx= \frac{4}{6} \frac{\Gamma(3+\frac{1}{2})\Gamma(3+\frac{1}{2})}{\Gamma(4)2!}=\frac{4(\frac{15\sqrt{\pi}}{8})^2}{12\cdot 3!}=\frac{25\pi}{128}\]

증명

$P_m^{\alpha,\beta}$는 $m$차 다항식이므로, 다음과 같이 쓸 수 있다 $$ P_m^{(\alpha,\beta)} (x)=\sum_{k=0}^m c_{mk}x^k $$ 이 때, $c_{mm}=$ 직교성은 \ref{RF}과 부분적분을 이용하여 증명할 수 있다. $m\leq n$이라 가정하자. $$ \begin{aligned} \int_{-1}^1 (1-x)^{\alpha} (1+x)^{\beta} P_m^{(\alpha,\beta)} (x)P_n^{(\alpha,\beta)} (x) \, dx=&\sum_{k=0}^m c_{mk}\frac{(-1)^n}{2^nn!}\int_{-1}^1x^k\frac{d^n}{dx^n}\left[(1-x)^{\alpha+n} (1+x)^{\beta+n}\right]\,dx\\ =&\sum_{k=0}^m\frac{ c_{mk}}{2^n}\int_{-1}^1\left[(1-x)^{\alpha+n} (1+x)^{\beta+n}\right]\,dx\\ =&\delta_{nm}\frac{2^{\alpha+\beta+1}}{2n+\alpha+\beta+1} \frac{\Gamma(n+\alpha+1)\Gamma(n+\beta+1)}{\Gamma(n+\alpha+\beta+1)n!} \end{aligned} $$ ■

테이블

$$ \begin{array}{c|c} n & P_n^{(\alpha ,\beta )}(x) \\ \hline 0 & 1 \\ 1 & \frac{1}{2} (\alpha -\beta +z (\alpha +\beta +2)) \\ 2 & \frac{1}{2} (\alpha +1) (\alpha +2)+\frac{1}{8} (z-1)^2 (\alpha +\beta +3) (\alpha +\beta +4)+\frac{1}{2} (\alpha +2) (z-1) (\alpha +\beta +3) \\ 3 & \frac{1}{6} (\alpha +1) (\alpha +2) (\alpha +3)+\frac{1}{48} (z-1)^3 (\alpha +\beta +4) (\alpha +\beta +5) (\alpha +\beta +6)+\frac{1}{8} (\alpha +3) (z-1)^2 (\alpha +\beta +4) (\alpha +\beta +5)+\frac{1}{4} (\alpha +2) (\alpha +3) (z-1) (\alpha +\beta +4) \end{array} $$

매스매티카 파일 및 계산 리소스

• https://docs.google.com/file/d/0B8XXo8Tve1cxb0FqQ2ZaRG9oVVE/edit

사전 형태의 자료

• http://en.wikipedia.org/wiki/Jacobi_polynomials
• NIST Digital Library of Mathematical Functions Chapter 18 Orthogonal Polynomials

리뷰, 에세이, 강의노트

• S. Ole Warnaar, Beta Integrals