"자코비 다항식"의 두 판 사이의 차이

수학노트
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잔글 (찾아 바꾸기 – “<h5>” 문자열을 “==” 문자열로)
 
(같은 사용자의 중간 판 31개는 보이지 않습니다)
1번째 줄: 1번째 줄:
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<h5 style="margin: 0px; line-height: 3.428em; color: rgb(34, 61, 103); font-family: 'malgun gothic',dotum,gulim,sans-serif; font-size: 1.166em; background-position: 0px 100%;">이 항목의 스프링노트 원문주소</h5>
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==개요==
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* <math>n\in \mathbb{Z}_{\geq 0}, \alpha, \beta</math>를 매개변수로 갖는 직교다항식 <math>P_{n}^{(\alpha\,\beta)}(x)</math>
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* 다양한 직교다항식을 특수한 경우로 가짐
 +
  
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* [[자코비 다항식]]
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===정의===
  
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* [[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>
  
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<h5 style="margin: 0px; line-height: 3.428em; color: rgb(34, 61, 103); font-family: 'malgun gothic',dotum,gulim,sans-serif; font-size: 1.166em; background-position: 0px 100%;">개요</h5>
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===특수한 경우===
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* [[게겐바워 다항식(ultraspherical polynomials)]]
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:<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)
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</math>
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* [[체비셰프 다항식]]
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:<math>
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T_n(x)=\frac{2^{2 n} (n!)^2}{(2 n)!}P_n^{\left(-\frac{1}{2},-\frac{1}{2}\right)}(x)
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</math>
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:<math>
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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)
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</math>
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* [[르장드르 다항식]]
 +
:<math>
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P_n(x)=P_n^{(0,0)}(x)
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</math>
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* [[라게르 다항식]]
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:<math>
 +
L_n^{\alpha }(x)=\lim_{\beta \to \infty } \, P_n^{(\alpha ,\beta )}\left(1-\frac{2 x}{\beta }\right)
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</math>
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* [[에르미트 다항식(Hermite polynomials)]]
 +
:<math>
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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)
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</math>
  
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* 직교다항식
 
  
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==성질==
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* 로드리게스 공식
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:<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}
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</math>
 +
  
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* 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>
  
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==정의</h5>
 
  
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* [[초기하급수(Hypergeometric series)]]를 통해 정의된다<br><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><br>
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* 직교성, <math>m,n\in \mathbb{Z}_{\geq 0}</math>에 대하여,
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*  다항식표현<br><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><br>
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:<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>  
  
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===예===
 +
* <math>\alpha=1/2,\beta=1/2,m=n=2</math>인 경우
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:<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>
  
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==3항 점화식</h5>
 
  
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==직교성의 증명==
  
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* 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)
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<math>t=(1-x)/2</math>로 치환하면,
 +
:<math>
 +
\begin{aligned}
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\int_{-1}^1(1-x)^{\alpha} (1+x)^{\beta}\,dx=&\int_0^1 2^{\alpha+\beta+1}t^{\alpha}(1-t)^{\beta}\, dt \\
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=&2^{\alpha+\beta+1}B(\alpha+1,\beta+1)\\
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=&2^{\alpha+\beta+1}\frac{\Gamma(\alpha+1)\Gamma(\beta+1)}{\Gamma(\alpha+\beta+2)}
 +
\end{aligned}
 +
</math>
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여기서 <math>B(x,y)</math>는 [[오일러 베타적분(베타함수)]]
 +
■
  
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==미분방정식</h5>
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;(정리)
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* <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>  
  
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*  자 코비 다항식은 다음을 만족시킨다<br><math>(1-x^2)y'' + ( \beta-\alpha - (\alpha + \beta + 2)x )y'+ n(n+\alpha+\beta+1) y = 0</math><br>
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;증명
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<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>
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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)}.
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</math>
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직교성은 \ref{RF}과 [[부분적분]]을 이용하여 증명할 수 있다. <math>m\leq n</math>이라 가정하자.
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:<math>
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\begin{aligned}
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\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\\
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=&\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\\
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=&\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!}
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\end{aligned}
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</math>
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■
  
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==테이블==
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:<math>
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\begin{array}{c|c}
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n & P_n^{(\alpha ,\beta )}(x) \\
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\hline
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0 & 1 \\
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1 & \frac{1}{2} (\alpha -\beta +z (\alpha +\beta +2)) \\
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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) \\
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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)
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\end{array}
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</math>
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==직교성</h5>
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==관련된 항목들==
 +
* [[오일러 베타적분(베타함수)]]
  
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*  weight함수와 구간<br><math>w(x) = (1-x)^{\alpha} (1+x)^{\beta}</math><br><math>[-1,1]</math><br>
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==매스매티카 파일 및 계산 리소스==
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* <math>m\neq n</math> 일 때<br><math>\int_{-1}^1 (1-x)^{\alpha} (1+x)^{\beta}  P_m^{(\alpha,\beta)} (x)P_n^{(\alpha,\beta)} (x) \; dx= 0</math><br>
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* https://docs.google.com/file/d/0B8XXo8Tve1cxb0FqQ2ZaRG9oVVE/edit
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* <math>m=n</math> 일 때<br><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><br> 예 <math>\alpha=2,\beta=2,m=n=2</math><br><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><br>
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==사전 형태의 자료==
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==목록</h5>
 
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*  매쓰매티카 코드<br>
 
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*# Do[Print["P_",i,"(z)=",JacobiP[i,a,b,z]],{i,0,4}]
 
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P_0(z)=1<br> P_1(z)=1/2 (a-b+(2+a+b) z)<br> P_2(z)=1/2 (1+a) (2+a)+1/2 (2+a) (3+a+b) (-1+z)+1/8 (3+a+b) (4+a+b) (-1+z)^2<br> P_3(z)=1/6 (1+a) (2+a) (3+a)+1/4 (2+a) (3+a) (4+a+b) (-1+z)+1/8 (3+a) (4+a+b) (5+a+b) (-1+z)^2+1/48 (4+a+b) (5+a+b) (6+a+b) (-1+z)^3<br> P_4(z)=1/24 (1+a) (2+a) (3+a) (4+a)+1/12 (2+a) (3+a) (4+a) (5+a+b) (-1+z)+1/16 (3+a) (4+a) (5+a+b) (6+a+b) (-1+z)^2+1/48 (4+a) (5+a+b) (6+a+b) (7+a+b) (-1+z)^3+1/384 (5+a+b) (6+a+b) (7+a+b) (8+a+b) (-1+z)^4
 
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==재미있는 사실</h5>
 
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* Math Overflow http://mathoverflow.net/search?q=
 
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* 네이버 지식인 http://kin.search.naver.com/search.naver?where=kin_qna&query=
 
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==역사</h5>
 
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* http://www.google.com/search?hl=en&tbs=tl:1&q=
 
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* [[수학사연표 (역사)|수학사연표]]
 
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*  
 
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==메모</h5>
 
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==관련된 항목들</h5>
 
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<h5 style="margin: 0px; line-height: 3.428em; color: rgb(34, 61, 103); font-family: 'malgun gothic',dotum,gulim,sans-serif; font-size: 1.166em; background-position: 0px 100%;">수학용어번역</h5>
 
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* 단어사전 http://www.google.com/dictionary?langpair=en|ko&q=
 
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* 발음사전 http://www.forvo.com/search/
 
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* [http://mathnet.kaist.ac.kr/mathnet/math_list.php?mode=list&ftype=&fstr= 대한수학회 수학 학술 용어집]<br>
 
−
** http://mathnet.kaist.ac.kr/mathnet/math_list.php?mode=list&ftype=eng_term&fstr=
 
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* [http://www.nktech.net/science/term/term_l.jsp?l_mode=cate&s_code_cd=MA 남·북한수학용어비교]
 
−
* [http://kms.or.kr/home/kor/board/bulletin_list_subject.asp?bulletinid=%7BD6048897-56F9-43D7-8BB6-50B362D1243A%7D&boardname=%BC%F6%C7%D0%BF%EB%BE%EE%C5%E4%B7%D0%B9%E6&globalmenu=7&localmenu=4 대한수학회 수학용어한글화 게시판]
 
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==사전 형태의 자료</h5>
 
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* http://ko.wikipedia.org/wiki/
 
 
* http://en.wikipedia.org/wiki/Jacobi_polynomials
 
* http://en.wikipedia.org/wiki/Jacobi_polynomials
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* http://www.wolframalpha.com/input/?i=
+
* [http://dlmf.nist.gov/18 NIST Digital Library of Mathematical Functions Chapter 18 Orthogonal Polynomials]
−
* [http://dlmf.nist.gov/ NIST Digital Library of Mathematical Functions]
 
−
* [http://www.research.att.com/%7Enjas/sequences/index.html The On-Line Encyclopedia of Integer Sequences]<br>
 
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** http://www.research.att.com/~njas/sequences/?q=
 
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==관련논문</h5>
 
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* http://www.jstor.org/action/doBasicSearch?Query=
 
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* http://www.ams.org/mathscinet
 
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* http://dx.doi.org/
 
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==관련도서</h5>
 
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*  도서내검색<br>
 
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** http://books.google.com/books?q=
 
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** http://book.daum.net/search/contentSearch.do?query=
 
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*  도 서검색<br>
 
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** http://books.google.com/books?q=
 
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** http://book.daum.net/search/mainSearch.do?query=
 
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** http://book.daum.net/search/mainSearch.do?query=
 
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==관 련기사</h5>
 
  
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*  네이버 뉴스 검색 (키워드 수정)<br>
 
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** http://news.search.naver.com/search.naver?where=news&x=0&y=0&sm=tab_hty&query=
 
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** http://news.search.naver.com/search.naver?where=news&x=0&y=0&sm=tab_hty&query=
 
−
** http://news.search.naver.com/search.naver?where=news&x=0&y=0&sm=tab_hty&query=
 
  
−
 
+
==리뷰, 에세이, 강의노트==
 +
* S. Ole Warnaar, [http://www.maths.uq.edu.au/%7Euqowarna/talks/Wien.pdf Beta Integrals]
 +
[[분류:특수함수]]
  
−
 
+
== 관련논문 ==
  
−
==블 로그</h5>
+
* 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
  
−
*  구글 블로그 검색<br>
+
==메타데이터==
−
** http://blogsearch.google.com/blogsearch?q=
+
===위키데이터===
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* [http://navercast.naver.com/science/list 네이버 오늘의과학]
+
* ID : [https://www.wikidata.org/wiki/Q371631 Q371631]
−
* [http://math.dongascience.com/ 수학동아]
+
===Spacy 패턴 목록===
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* [http://www.ams.org/mathmoments/ Mathematical Moments from the AMS]
+
* [{'LOWER': 'jacobi'}, {'LEMMA': 'polynomial'}]
−
* [http://betterexplained.com/ BetterExplained]
 

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,

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


예

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

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


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

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


관련된 항목들

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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'}]
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