"자코비 다항식"의 두 판 사이의 차이

수학노트
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(같은 사용자의 중간 판 5개는 보이지 않습니다)
1번째 줄: 1번째 줄:
 
==개요==
 
==개요==
* $n\in \mathbb{Z}_{\geq 0}, \alpha, \beta$를 매개변수로 갖는 직교다항식 $P_{n}^{(\alpha\,\beta)}(x)$
+
* <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
 
* 다양한 직교다항식을 특수한 경우로 가짐
 
   
 
   
12번째 줄: 12번째 줄:
 
===특수한 경우===
 
===특수한 경우===
 
* [[게겐바워 다항식(ultraspherical polynomials)]]
 
* [[게겐바워 다항식(ultraspherical polynomials)]]
$$
+
:<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)
 
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>
 
* [[체비셰프 다항식]]
 
* [[체비셰프 다항식]]
$$
+
:<math>
 
T_n(x)=\frac{2^{2 n} (n!)^2}{(2 n)!}P_n^{\left(-\frac{1}{2},-\frac{1}{2}\right)}(x)
 
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)
 
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>
 
* [[르장드르 다항식]]
 
* [[르장드르 다항식]]
$$
+
:<math>
 
P_n(x)=P_n^{(0,0)}(x)
 
P_n(x)=P_n^{(0,0)}(x)
$$
+
</math>
 
* [[라게르 다항식]]
 
* [[라게르 다항식]]
$$
+
:<math>
 
L_n^{\alpha }(x)=\lim_{\beta \to \infty } \, P_n^{(\alpha ,\beta )}\left(1-\frac{2 x}{\beta }\right)
 
L_n^{\alpha }(x)=\lim_{\beta \to \infty } \, P_n^{(\alpha ,\beta )}\left(1-\frac{2 x}{\beta }\right)
$$
+
</math>
 
* [[에르미트 다항식(Hermite polynomials)]]
 
* [[에르미트 다항식(Hermite polynomials)]]
$$
+
:<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)
 
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==
 
==성질==
 
* 로드리게스 공식
 
* 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}
 
(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>
 
   
 
   
  
47번째 줄: 47번째 줄:
  
  
* 직교성, $m,n\in \mathbb{Z}_{\geq 0}$에 대하여,
+
* 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>  
 
:<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>  
  
63번째 줄: 63번째 줄:
 
;보조정리
 
;보조정리
 
다음이 성립한다
 
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)}
 
\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)
 
;(증명)
$t=(1-x)/2$로 치환하면,
+
<math>t=(1-x)/2</math>로 치환하면,
$$
+
:<math>
 
\begin{aligned}
 
\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 \\
 
\int_{-1}^1(1-x)^{\alpha} (1+x)^{\beta}\,dx=&\int_0^1 2^{\alpha+\beta+1}t^{\alpha}(1-t)^{\beta}\, dt \\
74번째 줄: 74번째 줄:
 
=&2^{\alpha+\beta+1}\frac{\Gamma(\alpha+1)\Gamma(\beta+1)}{\Gamma(\alpha+\beta+2)}
 
=&2^{\alpha+\beta+1}\frac{\Gamma(\alpha+1)\Gamma(\beta+1)}{\Gamma(\alpha+\beta+2)}
 
\end{aligned}
 
\end{aligned}
$$
+
</math>
여기서 $B(x,y)$는 [[오일러 베타적분(베타함수)]]
+
여기서 <math>B(x,y)</math>는 [[오일러 베타적분(베타함수)]]
 
 
  
  
 
;(정리)
 
;(정리)
* $m,n\in \mathbb{Z}_{\geq 0}$에 대하여,
+
* <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>  
 
:<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
 
;증명
$P_m^{\alpha,\beta}$$m$차 다항식이므로, 적당한 상수 $c_{mk}, k=0,1,\cdots, m$에 대하여 다음과 같이 쓸 수 있다
+
<math>P_m^{\alpha,\beta}</math><math>m</math>차 다항식이므로, 적당한 상수 <math>c_{mk}, k=0,1,\cdots, m</math>에 대하여 다음과 같이 쓸 수 있다
$$
+
:<math>
P_m^{(\alpha,\beta)} (x)=\sum_{k=0}^m c_{mk}x^k,\,c_{mm}=\frac{2^{-m} \Gamma (2 m+\alpha +\beta +1)}{m! \Gamma (m+\alpha +\beta +1)}.
+
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)}.
$$
+
</math>
직교성은 \ref{RF}과 [[부분적분]]을 이용하여 증명할 수 있다. $m\leq n$이라 가정하자.
+
직교성은 \ref{RF}과 [[부분적분]]을 이용하여 증명할 수 있다. <math>m\leq n</math>이라 가정하자.
$$
+
:<math>
 
\begin{aligned}
 
\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\\
 
\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\\
95번째 줄: 95번째 줄:
 
=&\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!}
 
=&\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}
 
\end{aligned}
$$
+
</math>
 
 
  
 
==테이블==
 
==테이블==
$$
+
:<math>
 
\begin{array}{c|c}
 
\begin{array}{c|c}
 
  n & P_n^{(\alpha ,\beta )}(x) \\
 
  n & P_n^{(\alpha ,\beta )}(x) \\
108번째 줄: 108번째 줄:
 
  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)
 
  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}
 
\end{array}
$$
+
</math>
 
   
 
   
  
126번째 줄: 126번째 줄:
 
==리뷰, 에세이, 강의노트==
 
==리뷰, 에세이, 강의노트==
 
* S. Ole Warnaar, [http://www.maths.uq.edu.au/%7Euqowarna/talks/Wien.pdf Beta Integrals]
 
* S. Ole Warnaar, [http://www.maths.uq.edu.au/%7Euqowarna/talks/Wien.pdf Beta Integrals]
 +
[[분류:특수함수]]
 +
 +
== 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===
 +
* ID :  [https://www.wikidata.org/wiki/Q371631 Q371631]
 +
===Spacy 패턴 목록===
 +
* [{'LOWER': 'jacobi'}, {'LEMMA': 'polynomial'}]

2021년 2월 17일 (수) 04:57 기준 최신판

개요

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


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\[P_n^{(\alpha,\beta)} (z) = \frac{\Gamma (\alpha+n+1)}{n!\Gamma (\alpha+\beta+n+1)} \sum_{m=0}^n {n\choose m} \frac{\Gamma (\alpha + \beta + n + m + 1)}{\Gamma (\alpha + m + 1)} \left(\frac{z-1}{2}\right)^m\]


특수한 경우

\[ C_n^{(\lambda )}(x)=\frac{(2 \lambda)_{n}}{\left(\lambda +\frac{1}{2}\right)_n}P_n^{\left(\lambda -\frac{1}{2},\lambda -\frac{1}{2}\right)}(x) \]

\[ T_n(x)=\frac{2^{2 n} (n!)^2}{(2 n)!}P_n^{\left(-\frac{1}{2},-\frac{1}{2}\right)}(x) \] \[ U_n(x)=\frac{2^{2 n+1} ((n+1)!)^2 }{(2 n+2)!}P_n^{\left(\frac{1}{2},\frac{1}{2}\right)}(x) \]

\[ P_n(x)=P_n^{(0,0)}(x) \]

\[ L_n^{\alpha }(x)=\lim_{\beta \to \infty } \, P_n^{(\alpha ,\beta )}\left(1-\frac{2 x}{\beta }\right) \]

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


성질

  • 로드리게스 공식

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


  • 자코비 다항식은 다음의 미분방정식 만족시킨다

\[(1-x^2)y'' + ( \beta-\alpha - (\alpha + \beta + 2)x )y'+ n(n+\alpha+\beta+1) y = 0\]


  • 직교성, \(m,n\in \mathbb{Z}_{\geq 0}\)에 대하여,

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


  • \(\alpha=1/2,\beta=1/2,m=n=2\)인 경우

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


직교성의 증명

  • weight함수와 구간

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

보조정리

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,

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

증명

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

테이블

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


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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'}]