"자코비 다항식"의 두 판 사이의 차이
Pythagoras0 (토론 | 기여) |
Pythagoras0 (토론 | 기여) |
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* 직교다항식 | * 직교다항식 | ||
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| − | ==정의== | + | ===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 | + | * [[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 | + | * 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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| − | + | ==로드리게스 공식== | |
| + | * 다음이 성립한다 | ||
| + | $$ | ||
| + | (1-x)^{\alpha } (1+x)^{\beta } P_n^{(\alpha ,\beta )}(x)=\frac{(-1)^n}{2^n n!}\frac{d^n}{dx^n}\left[\left((1-x)^{\alpha +n} (1+x)^{\beta +n}\right)\right] \label{RF} | ||
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==미분방정식== | ==미분방정식== | ||
| − | * | + | * 자코비 다항식은 다음을 만족시킨다:<math>(1-x^2)y'' + ( \beta-\alpha - (\alpha + \beta + 2)x )y'+ n(n+\alpha+\beta+1) y = 0</math> |
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==직교성== | ==직교성== | ||
| + | * weight함수와 구간 | ||
| + | :<math>w(x) = (1-x)^{\alpha} (1+x)^{\beta}, x\in [-1,1]</math> | ||
| + | * 다음이 성립한다 | ||
| + | $$ | ||
| + | \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)} | ||
| + | $$ | ||
| + | ;(정리) | ||
| + | * $m,n\in \mathbb{Z}_{\geq 0}$에 대하여, | ||
| + | :<math>\int_{-1}^1 (1-x)^{\alpha} (1+x)^{\beta} P_m^{(\alpha,\beta)} (x)P_n^{(\alpha,\beta)} (x) \; dx= \frac{2^{\alpha+\beta+1}}{2n+\alpha+\beta+1} \frac{\Gamma(n+\alpha+1)\Gamma(n+\beta+1)}{\Gamma(n+\alpha+\beta+1)n!} \delta_{nm}</math> | ||
| + | ===예=== | ||
| + | * <math>\alpha=1/2,\beta=1/2,m=n=2</math>인 경우 | ||
| + | :<math>\int_{-1}^1 (1-x)^{\frac{1}{2}} (1+x)^{\frac{1}{2}} P_2^{(\frac{1}{2},\frac{1}{2})} (x)P_2^{(\frac{1}{2},\frac{1}{2})} (x) \; dx= \frac{4}{6} \frac{\Gamma(3+\frac{1}{2})\Gamma(3+\frac{1}{2})}{\Gamma(4)2!}=\frac{4(\frac{15\sqrt{\pi}}{8})^2}{12\cdot 3!}=\frac{25\pi}{128}</math> | ||
| + | ===증명=== | ||
| + | $P_m^{\alpha,\beta}$는 $m$차 다항식이므로, 다음과 같이 쓸 수 있다 | ||
| + | $$ | ||
| + | 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} | ||
| + | $$ | ||
| + | ■ | ||
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| − | + | ==테이블== | |
| + | $$ | ||
| + | \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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==메모== | ==메모== | ||
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==관련된 항목들== | ==관련된 항목들== | ||
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| − | + | ==매스매티카 파일 및 계산 리소스== | |
| + | * https://docs.google.com/file/d/0B8XXo8Tve1cxb0FqQ2ZaRG9oVVE/edit | ||
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| − | ==사전 | + | ==사전 형태의 자료== |
* http://ko.wikipedia.org/wiki/ | * http://ko.wikipedia.org/wiki/ | ||
| 90번째 줄: | 96번째 줄: | ||
* http://www.wolframalpha.com/input/?i= | * http://www.wolframalpha.com/input/?i= | ||
* [http://dlmf.nist.gov/ NIST Digital Library of Mathematical Functions] | * [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] | + | * [http://www.research.att.com/%7Enjas/sequences/index.html The On-Line Encyclopedia of Integer Sequences] |
** http://www.research.att.com/~njas/sequences/?q= | ** http://www.research.att.com/~njas/sequences/?q= | ||
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2014년 9월 25일 (목) 01:28 판
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
$$ \int_{-1}^1(1-x)^{\alpha} (1+x)^{\beta}\,dx=2^{\alpha+\beta+1}\frac{\Gamma(\alpha+1)\Gamma(\beta+1)}{\Gamma(\alpha+\beta+2)} $$
- (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
$$ \begin{array}{c|c} n & P_n^{(\alpha ,\beta )}(x) \\ \hline 0 & 1 \\ 1 & \frac{1}{2} (\alpha -\beta +z (\alpha +\beta +2)) \\ 2 & \frac{1}{2} (\alpha +1) (\alpha +2)+\frac{1}{8} (z-1)^2 (\alpha +\beta +3) (\alpha +\beta +4)+\frac{1}{2} (\alpha +2) (z-1) (\alpha +\beta +3) \\ 3 & \frac{1}{6} (\alpha +1) (\alpha +2) (\alpha +3)+\frac{1}{48} (z-1)^3 (\alpha +\beta +4) (\alpha +\beta +5) (\alpha +\beta +6)+\frac{1}{8} (\alpha +3) (z-1)^2 (\alpha +\beta +4) (\alpha +\beta +5)+\frac{1}{4} (\alpha +2) (\alpha +3) (z-1) (\alpha +\beta +4) \end{array} $$
메모
관련된 항목들
매스매티카 파일 및 계산 리소스