자코비 다항식

수학노트
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개요

  • <math>n\in \mathbb{Z}_{\geq 0}, \alpha, \beta</math>를 매개변수로 갖는 직교다항식 <math>P_{n}^{(\alpha\,\beta)}(x)</math>
  • 다양한 직교다항식을 특수한 경우로 가짐


정의

  • 초기하급수(Hypergeometric series)를 통해 정의된다:<math>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)</math>
  • 다항식표현:<math>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</math>


특수한 경우

<math>

C_n^{(\lambda )}(x)=\frac{(2 \lambda)_{n}}{\left(\lambda +\frac{1}{2}\right)_n}P_n^{\left(\lambda -\frac{1}{2},\lambda -\frac{1}{2}\right)}(x) </math>

<math>

T_n(x)=\frac{2^{2 n} (n!)^2}{(2 n)!}P_n^{\left(-\frac{1}{2},-\frac{1}{2}\right)}(x) </math>

<math>

U_n(x)=\frac{2^{2 n+1} ((n+1)!)^2 }{(2 n+2)!}P_n^{\left(\frac{1}{2},\frac{1}{2}\right)}(x) </math>

<math>

P_n(x)=P_n^{(0,0)}(x) </math>

<math>

L_n^{\alpha }(x)=\lim_{\beta \to \infty } \, P_n^{(\alpha ,\beta )}\left(1-\frac{2 x}{\beta }\right) </math>

<math>

H_n(x)=\lim_{\alpha \to \infty } \, \frac{\left(2^n n!\right) }{\alpha ^{n/2}}P_n^{(\alpha ,\alpha )}\left(\frac{x}{\sqrt{\alpha }}\right) </math>


성질

  • 로드리게스 공식
<math>

(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} </math>


  • 자코비 다항식은 다음의 미분방정식 만족시킨다
<math>(1-x^2)y + ( \beta-\alpha - (\alpha + \beta + 2)x )y'+ n(n+\alpha+\beta+1) y = 0</math>


  • 직교성, <math>m,n\in \mathbb{Z}_{\geq 0}</math>에 대하여,
<math>\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}</math>


예

  • <math>\alpha=1/2,\beta=1/2,m=n=2</math>인 경우
<math>\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}</math>


직교성의 증명

  • weight함수와 구간
<math>w(x) = (1-x)^{\alpha} (1+x)^{\beta}, x\in [-1,1]</math>
보조정리

다음이 성립한다

<math>

\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)} </math>

(증명)

<math>t=(1-x)/2</math>로 치환하면,

<math>

\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} </math> 여기서 <math>B(x,y)</math>는 오일러 베타적분(베타함수) ■


(정리)
  • <math>m,n\in \mathbb{Z}_{\geq 0}</math>에 대하여,
<math>\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}</math>
증명

<math>P_m^{\alpha,\beta}</math>는 <math>m</math>차 다항식이므로, 적당한 상수 <math>c_{mk}, k=0,1,\cdots, m</math>에 대하여 다음과 같이 쓸 수 있다

<math>

P_m^{(\alpha,\beta)} (x)=\sum_{k=0}^m c_{mk}x^k,\,c_{mm}=\frac{\Gamma (2 m+\alpha +\beta +1)}{2^{m} m! \Gamma (m+\alpha +\beta +1)}. </math> 직교성은 \ref{RF}과 부분적분을 이용하여 증명할 수 있다. <math>m\leq n</math>이라 가정하자.

<math>

\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}\delta_{nk}\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} </math> ■

테이블

<math>

\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} </math>


관련된 항목들

매스매티카 파일 및 계산 리소스


사전 형태의 자료


리뷰, 에세이, 강의노트

관련논문

  • Oleg Szehr, Rachid Zarouf, On the asymptotic behavior of jacobi polynomials with varying parameters, arXiv:1605.02509 [math.CA], May 09 2016, http://arxiv.org/abs/1605.02509

메타데이터

위키데이터

Spacy 패턴 목록

  • [{'LOWER': 'jacobi'}, {'LEMMA': 'polynomial'}]
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