"자코비 다항식"의 두 판 사이의 차이

2010년 5월 2일 (일) 09:26 판

이 항목의 스프링노트 원문주소

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

3항 점화식

미분방정식

자 코비 다항식은 다음을 만족시킨다
\((1-x^2)y'' + ( \beta-\alpha - (\alpha + \beta + 2)x )y'+ n(n+\alpha+\beta+1) y = 0\)

직교성

weight함수와 구간
\(w(x) = (1-x)^{\alpha} (1+x)^{\beta}\)
\([-1,1]\)
\(m\neq n\) 일 때
\(\int_{-1}^1 (1-x)^{\alpha} (1+x)^{\beta} P_m^{(\alpha,\beta)} (x)P_n^{(\alpha,\beta)} (x) \; dx= 0\)
\(m=n\) 일 때
\(\int_{-1}^1 (1-x)^{\alpha} (1+x)^{\beta} P_m^{(\alpha,\beta)} (x)P_n^{(\alpha,\beta)} (x) \; dx= \frac{2^{\alpha+\beta+1}}{2n+\alpha+\beta+1} \frac{\Gamma(n+\alpha+1)\Gamma(n+\beta+1)}{\Gamma(n+\alpha+\beta+1)n!} \delta_{nm}\)
예
\(\int_{-1}^1 (1-x)^{\alpha} (1+x)^{\beta} P_m^{(\alpha,\beta)} (x)P_n^{(\alpha,\beta)} (x) \; dx= \frac{2^{\alpha+\beta+1}}{2n+\alpha+\beta+1} \frac{\Gamma(n+\alpha+1)\Gamma(n+\beta+1)}{\Gamma(n+\alpha+\beta+1)n!} \delta_{nm}\)

목록

P_0(z)=1
P_1(z)=1/2 (a-b+(2+a+b) z)
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
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
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

재미있는 사실

Math Overflow http://mathoverflow.net/search?q=
네이버 지식인 http://kin.search.naver.com/search.naver?where=kin_qna&query=

역 사

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사전 형태의 자료

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네이버 오늘의과학
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Mathematical Moments from the AMS
BetterExplained

@@ 16번째 줄: / 16번째 줄: @@
 * [[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>
-*  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>  <br>
+*  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>
-<h5>직교성</h5>
-*  weight함수와 구간<br><math>w(x) = (1-x)^{\alpha} (1+x)^{\beta}</math><br><math>[-1,1]</math><br>
-* <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>
-* <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>
@@ 40번째 줄: / 34번째 줄: @@
 <h5>미분방정식</h5>
-*  자코비 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>
+*  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>
+<h5>직교성</h5>
+*  weight함수와 구간<br><math>w(x) = (1-x)^{\alpha} (1+x)^{\beta}</math><br><math>[-1,1]</math><br>
+* <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>
+* <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<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>
@@ 48번째 줄: / 50번째 줄: @@
 <h5>목록</h5>
-P_0(z)=1<br> P_1(z)=1/2 (\[Alpha]-\[Beta]+z (2+\[Alpha]+\[Beta]))<br> P_2(z)=1/2 (1+\[Alpha]) (2+\[Alpha])+1/2 (-1+z) (2+\[Alpha]) (3+\[Alpha]+\[Beta])+1/8 (-1+z)^2 (3+\[Alpha]+\[Beta]) (4+\[Alpha]+\[Beta])<br> P_3(z)=1/6 (1+\[Alpha]) (2+\[Alpha]) (3+\[Alpha])+1/4 (-1+z) (2+\[Alpha]) (3+\[Alpha]) (4+\[Alpha]+\[Beta])+1/8 (-1+z)^2 (3+\[Alpha]) (4+\[Alpha]+\[Beta]) (5+\[Alpha]+\[Beta])+1/48 (-1+z)^3 (4+\[Alpha]+\[Beta]) (5+\[Alpha]+\[Beta]) (6+\[Alpha]+\[Beta])<br> P_4(z)=1/24 (1+\[Alpha]) (2+\[Alpha]) (3+\[Alpha]) (4+\[Alpha])+1/12 (-1+z) (2+\[Alpha]) (3+\[Alpha]) (4+\[Alpha]) (5+\[Alpha]+\[Beta])+1/16 (-1+z)^2 (3+\[Alpha]) (4+\[Alpha]) (5+\[Alpha]+\[Beta]) (6+\[Alpha]+\[Beta])+1/48 (-1+z)^3 (4+\[Alpha]) (5+\[Alpha]+\[Beta]) (6+\[Alpha]+\[Beta]) (7+\[Alpha]+\[Beta])+1/384 (-1+z)^4 (5+\[Alpha]+\[Beta]) (6+\[Alpha]+\[Beta]) (7+\[Alpha]+\[Beta]) (8+\[Alpha]+\[Beta])
+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