Wielomiany Legendre
Równanie różniczkowe
\(\displaystyle{
\left(1-x^2\right)y'' - 2xy' + n\left(n+1\right)y=0\\
}\)
Wzór Rodriguesa
\(\displaystyle{
Q\left(x\right) = 1-x^2\\
L\left(x\right) = -2x \\
R'\left(x\right) = \frac{L\left(x\right)}{Q\left(x\right)}R\left(x\right)\\
R'\left(x\right) = \frac{-2x}{1-x^2}R\left(x\right)\\
\frac{R'\left(x\right)}{R\left(x\right)} = \frac{-2x}{1-x^2}\\
\ln{\left|R\left(x\right)\right|} = \ln{\left|1-x^2\right|} + C\\
R\left(x\right) = \left(1-x^2\right)\\
W\left(x\right) = \frac{\left(1-x^2\right)}{\left(1-x^2\right)} = 1\\
P_{n}\left(x\right) = \frac{1}{e_{n}}\cdot 1 \cdot\frac{\mbox{d}^n}{\mbox{d}x^n}\left(1\cdot\left(1-x^2\right)^n\right)\\
P_{n}\left(x\right) = \frac{1}{e_{n}}\frac{\mbox{d}^n}{\mbox{d}x^n}\left(\left(1-x^2\right)^n\right)\\
P_{n}\left(1\right) = 1\\
P_{0}\left(x\right) = \frac{1}{e_{0}}\cdot 1\\
\frac{1}{e_{0}} = 1\\
P_{1}\left(x\right) = \frac{1}{e_{1}}\cdot \left(-2x\right)\\
P_{1}\left(1\right) = \frac{1}{e_{1}}\cdot \left(-2\right)\\
\frac{1}{e_{1}} = -\frac{1}{2}\\
P_{2}\left(x\right) = \frac{1}{e_{2}}\left(-4x+12x^2\right)\\
P_{2}\left(1\right) = \frac{1}{e_{2}}\cdot 8\\
\frac{1}{e_{2}} = \frac{1}{8}\\
P_{3}\left(x\right) = \frac{1}{e_{3}}\left(72x-120x^3\right)\\
P_{3}\left(1\right) = \frac{1}{e_{3}}\cdot\left(-48\right)\\
\frac{1}{e_{3}} = \frac{-1}{48}\\
P_{n}\left(x\right) = \frac{\left(-1\right)^n}{2^n\cdot n!}\cdot\frac{\mbox{d}^n}{\mbox{d}x^n}\left(\left(1-x^2\right)^n\right)\\
P_{n}\left(x\right) = \frac{1}{2^n\cdot n!}\cdot\frac{\mbox{d}^n}{\mbox{d}x^n}\left(\left(x^2-1\right)^n\right)\\
}\)
# Równanie rekurencyjne
\(\displaystyle{
P_{n}\left(x\right) = \frac{1}{2^n\cdot n!}\cdot\frac{\mbox{d}^n}{\mbox{d}x^n}\left(\left(x^2-1\right)^n\right)\\
P_{n+1}\left(x\right) = \frac{1}{2^{n+1}\cdot \left(n+1\right)!}\cdot\frac{\mbox{d}^{n+1}}{\mbox{d}x^{n+1}}\left(\left(x^2-1\right)^{n+1}\right)\\
P_{n+1}\left(x\right) = \frac{1}{2^{n+1}\cdot \left(n+1\right)!}\cdot\frac{\mbox{d}^{n-1}}{\mbox{d}x^{n-1}}\left(\frac{\mbox{d}^2}{\mbox{d}x^2}\left(x^2-1\right)^{n+1}\right)\\
P_{n+1}\left(x\right) = \frac{1}{2^{n+1}\cdot \left(n+1\right)!}\cdot\frac{\mbox{d}^{n-1}}{\mbox{d}x^{n-1}}\left(\frac{\mbox{d}}{\mbox{d}x}\left(2\left(n+1\right)x\left(x^2-1\right)^{n}\right)\right)\\
P_{n+1}\left(x\right) = \frac{1}{2^{n}\cdot n!}\cdot\frac{\mbox{d}^{n-1}}{\mbox{d}x^{n-1}}\left(\frac{\mbox{d}}{\mbox{d}x}\left(x\left(x^2-1\right)^{n}\right)\right)\\
P_{n+1}\left(x\right) = \frac{1}{2^{n}\cdot n!}\cdot\frac{\mbox{d}^{n-1}}{\mbox{d}x^{n-1}}\left(\left(\left(x^2-1\right)^{n} + x\cdot 2nx\left(x^2-1\right)^{n-1}\right)\right)\\
P_{n+1}\left(x\right) = \frac{1}{2^{n}\cdot n!}\cdot\frac{\mbox{d}^{n-1}}{\mbox{d}x^{n-1}}\left(\left(\left(x^2-1\right)^{n} + 2nx^2\left(x^2-1\right)^{n-1}\right)\right)\\
P_{n+1}\left(x\right) = \frac{1}{2^{n}\cdot n!}\cdot\frac{\mbox{d}^{n-1}}{\mbox{d}x^{n-1}}\left(\left(\left(x^2-1\right)^{n} + 2n\left(\left(x^2-1\right)+1\right)\left(x^2-1\right)^{n-1}\right)\right)\\
P_{n+1}\left(x\right) = \frac{1}{2^{n}\cdot n!}\cdot\frac{\mbox{d}^{n-1}}{\mbox{d}x^{n-1}}\left(\left(\left(x^2-1\right)^{n} + 2n\left(x^2-1\right)^{n}+2n\left(x^2-1\right)^{n-1}\right)\right)\\
P_{n+1}\left(x\right) = \frac{1}{2^{n}\cdot n!}\cdot\frac{\mbox{d}^{n-1}}{\mbox{d}x^{n-1}}\left(\left( \left(2n+1\right)\left(x^2-1\right)^{n}+2n\left(x^2-1\right)^{n-1}\right)\right)\\
P_{n+1}\left(x\right) = \frac{2n+1}{2^{n}\cdot n!}\cdot\frac{\mbox{d}^{n-1}}{\mbox{d}x^{n-1}}{\left(\left(x^2-1\right)^{n}\right)}+\frac{1}{2^{n-1}\left(n-1\right)!}\cdot\frac{\mbox{d}^{n-1}}{\mbox{d}x^{n-1}}\left(\left(x^2-1\right)^{n-1}\right)\\
P_{n+1}\left(x\right) = \frac{2n+1}{2^{n}\cdot n!}\cdot\frac{\mbox{d}^{n-1}}{\mbox{d}x^{n-1}}{\left(\left(x^2-1\right)^{n}\right)}+P_{n-1}\left(x\right)\\
P_{n+1}\left(x\right) = \frac{1}{2^{n+1}\cdot \left(n+1\right)!}\cdot\frac{\mbox{d}^{n+1}}{\mbox{d}x^{n+1}}\left(\left(x^2-1\right)^{n+1}\right)\\
P_{n+1}\left(x\right) = \frac{1}{2^{n+1}\cdot \left(n+1\right)!}\cdot\frac{\mbox{d}^{n}}{\mbox{d}x^{n}}\left(\frac{\mbox{d}}{\mbox{d}x}\left(\left(x^2-1\right)^{n+1}\right)\right)\\
P_{n+1}\left(x\right) = \frac{1}{2^{n+1}\cdot \left(n+1\right)!}\cdot\frac{\mbox{d}^{n}}{\mbox{d}x^{n}}\left(\left(2\left(n+1\right)x\left(x^2-1\right)^{n}\right)\right)\\
P_{n+1}\left(x\right) = \frac{1}{2^{n}\cdot n!}\frac{\mbox{d}^n}{\mbox{d}x^n}\left(x\left(x^2-1\right)^{n}\right)\\
P_{n+1}\left(x\right) = \frac{1}{2^{n}\cdot n!}\left(x\frac{\mbox{d}^n}{\mbox{d}x^n}\left(\left(x^2-1\right)^{n}\right)+n\frac{\mbox{d}^{n-1}}{\mbox{d}x^{n-1}}\left(\left(x^2-1\right)^{n}\right)\right)\\
P_{n+1}\left(x\right) = x\cdot \frac{1}{2^{n}\cdot n!}\frac{\mbox{d}^n}{\mbox{d}x^n}\left(\left(x^2-1\right)^{n}\right)+\frac{n}{2^{n}\cdot n!}\frac{\mbox{d}^{n-1}}{\mbox{d}x^{n-1}}\left(\left(x^2-1\right)^{n}\right)\\
P_{n+1}\left(x\right) = xP_{n}\left(x\right) + \frac{n}{2^{n}\cdot n!}\frac{\mbox{d}^{n-1}}{\mbox{d}x^{n-1}}\left(\left(x^2-1\right)^{n}\right)\\
\begin{cases}P_{n+1}\left(x\right) = \frac{2n+1}{2^{n}\cdot n!}\cdot\frac{\mbox{d}^{n-1}}{\mbox{d}x^{n-1}}{\left(\left(x^2-1\right)^{n}\right)}+P_{n-1}\left(x\right)\\P_{n+1}\left(x\right) = xP_{n}\left(x\right) + \frac{n}{2^{n}\cdot n!}\frac{\mbox{d}^{n-1}}{\mbox{d}x^{n-1}}\left(\left(x^2-1\right)^{n}\right)\end{cases}\\
\begin{cases}nP_{n+1}\left(x\right) = \frac{n\left(2n+1\right)}{2^{n}\cdot n!}\cdot\frac{\mbox{d}^{n-1}}{\mbox{d}x^{n-1}}{\left(\left(x^2-1\right)^{n}\right)}+nP_{n-1}\left(x\right)\\\left(2n+1\right)P_{n+1}\left(x\right) = \left(2n+1\right)xP_{n}\left(x\right) + \frac{\left(2n+1\right)n}{2^{n}\cdot n!}\frac{\mbox{d}^{n-1}}{\mbox{d}x^{n-1}}\left(\left(x^2-1\right)^{n}\right)\end{cases}\\
\left(2n+1-n\right)P_{n+1}\left(x\right) = \left(2n+1\right)xP_{n}\left(x\right) - nP_{n-1}\left(x\right)\\
\left(n+1\right)P_{n+1}\left(x\right) = \left(2n+1\right)xP_{n}\left(x\right) - nP_{n-1}\left(x\right)\\
\begin{cases}\left(n+1\right)P_{n+1}\left(x\right) = \left(2n+1\right)xP_{n}\left(x\right) - nP_{n-1}\left(x\right) \qquad n > 0 \\P_{0}\left(x\right)=1\\P_{1}\left(x\right)=x\end{cases}\\
}\)
Przesuńmy indeksy w powyższej rekurencji
\(\displaystyle{
\begin{cases}\left(n+2\right)P_{n+2}\left(x\right) = \left(2n+3\right)xP_{n+1}\left(x\right) - \left(n+1\right)P_{n}\left(x\right) \qquad n > 0 \\P_{0}\left(x\right)=1\\P_{1}\left(x\right)=x\end{cases}\\
\begin{cases}P_{n+2}\left(x\right) = \frac{\left(2n+3\right)}{\left(n+2\right)}xP_{n+1}\left(x\right) - \frac{\left(n+1\right)}{\left(n+2\right)}P_{n}\left(x\right) \qquad n > 0 \\P_{0}\left(x\right)=1\\P_{1}\left(x\right)=x\end{cases}\\
}\)
Zdefiniujmy teraz zwykłą funkcję tworzącą
\(\displaystyle{
G\left(x,t\right) = \sum\limits_{n=0}^{\infty}{P_{n}\left(x\right)t^{n}}\\
}\)
Wstawmy funkcję tworzącą do równania rekurencyjnego
\(\displaystyle{
\sum\limits_{n=0}^{\infty}{P_{n+2}\left(x\right)t^{n}} = \sum\limits_{n=0}^{\infty}{\left(\frac{\left(2n+3\right)}{\left(n+2\right)}xP_{n+1}\left(x\right) - \frac{\left(n+1\right)}{\left(n+2\right)}P_{n}\left(x\right)\right)t^{n}}\\
\sum\limits_{n=0}^{\infty}{P_{n+2}\left(x\right)t^{n}} = x\left(\sum\limits_{n=0}^{\infty}\frac{\left(2n+3\right)}{\left(n+2\right)}P_{n+1}\left(x\right)t^{n}\right)-\left(\frac{\left(n+1\right)}{\left(n+2\right)}P_{n}\left(x\right)t^{n}\right)\\
\sum\limits_{n=0}^{\infty}{P_{n+2}\left(x\right)t^{n}} = x\left(\sum\limits_{n=0}^{\infty}\frac{\left(2n+4-1\right)}{\left(n+2\right)}P_{n+1}\left(x\right)t^{n}\right)-\left(\frac{\left(n+2-1\right)}{\left(n+2\right)}P_{n}\left(x\right)t^{n}\right)\\
\sum\limits_{n=0}^{\infty}{P_{n+2}\left(x\right)t^{n}} = 2x\left(\sum\limits_{n=0}^{\infty}{P_{n+1}\left(x\right)t^{n}}\right) - x\left(\sum\limits_{n=0}^{\infty}{\frac{1}{n+2}P_{n+1}\left(x\right)t^{n}}\right)-\left(\sum\limits_{n=0}^{\infty}{P_{n}\left(x\right)t^{n}}\right)+\left(\sum\limits_{n=0}^{\infty}{\frac{1}{n+2}\cdot P_{n}\left(x\right)t^{n}}\right)\\
\frac{1}{t^2}\left(\sum\limits_{n=0}^{\infty}{P_{n+2}\left(x\right)t^{n+2}}\right) = \frac{2x}{t}\left(\sum\limits_{n=0}^{\infty}{P_{n+1}\left(x\right)t^{n+1}}\right) - \frac{x}{t}\left(\sum\limits_{n=0}^{\infty}{\frac{1}{n+2}P_{n+1}\left(x\right)t^{n+1}}\right)-\left(\sum\limits_{n=0}^{\infty}{P_{n}\left(x\right)t^{n}}\right)+\frac{1}{t^2}\left(\sum\limits_{n=0}^{\infty}{\frac{1}{n+2}\cdot P_{n}\left(x\right)t^{n+2}}\right)\\
\left(\sum\limits_{n=0}^{\infty}{P_{n+2}\left(x\right)t^{n+2}}\right) = 2xt\left(\sum\limits_{n=0}^{\infty}{P_{n+1}\left(x\right)t^{n+1}}\right) - xt\left(\sum\limits_{n=0}^{\infty}{\frac{1}{n+2}P_{n+1}\left(x\right)t^{n+1}}\right)-t^2\left(\sum\limits_{n=0}^{\infty}{P_{n}\left(x\right)t^{n}}\right)+\left(\sum\limits_{n=0}^{\infty}{\frac{1}{n+2}\cdot P_{n}\left(x\right)t^{n+2}}\right)\\
\left(\sum\limits_{n=2}^{\infty}{P_{n}\left(x\right)t^{n}}\right) = 2xt\left(\sum\limits_{n=1}^{\infty}{P_{n}\left(x\right)t^{n}}\right) - xt\left(\sum\limits_{n=1}^{\infty}{\frac{1}{n+1}P_{n}\left(x\right)t^{n}}\right)-t^2\left(\sum\limits_{n=0}^{\infty}{P_{n}\left(x\right)t^{n}}\right)+\left(\sum\limits_{n=0}^{\infty}{\frac{1}{n+2}\cdot P_{n}\left(x\right)t^{n+2}}\right)\\
\left(\sum\limits_{n=0}^{\infty}{P_{n}\left(x\right)t^{n}} - 1 - xt\right) = 2xt\left(\sum\limits_{n=0}^{\infty}{P_{n}\left(x\right)t^{n}} - 1\right) - xt\left(\sum\limits_{n=0}^{\infty}{\frac{1}{n+1}P_{n}\left(x\right)t^{n} - 1}\right)-t^2\left(\sum\limits_{n=0}^{\infty}{P_{n}\left(x\right)t^{n}}\right)+\left(\sum\limits_{n=0}^{\infty}{\frac{1}{n+2}\cdot P_{n}\left(x\right)t^{n+2}}\right)\\
\sum\limits_{n=0}^{\infty}{P_{n}\left(x\right)t^{n}} = 1 + xt - 2xt + xt + 2xt\left(\sum\limits_{n=0}^{\infty}{P_{n}\left(x\right)t^{n}}\right) - xt\left(\sum\limits_{n=0}^{\infty}{\frac{1}{n+1}P_{n}\left(x\right)t^{n}}\right)-t^2\left(\sum\limits_{n=0}^{\infty}{P_{n}\left(x\right)t^{n}}\right)+\left(\sum\limits_{n=0}^{\infty}{\frac{1}{n+2}\cdot P_{n}\left(x\right)t^{n+2}}\right)\\
\sum\limits_{n=0}^{\infty}{P_{n}\left(x\right)t^{n}} = 1 + 2xt\left(\sum\limits_{n=0}^{\infty}{P_{n}\left(x\right)t^{n}}\right) - x\left(\sum\limits_{n=0}^{\infty}{\frac{1}{n+1}P_{n}\left(x\right)t^{n+1}}\right)-t^2\left(\sum\limits_{n=0}^{\infty}{P_{n}\left(x\right)t^{n}}\right)+\left(\sum\limits_{n=0}^{\infty}{\frac{1}{n+2}\cdot P_{n}\left(x\right)t^{n+2}}\right)\\
\left(1-2xt+t^2\right)\left(\sum\limits_{n=0}^{\infty}{P_{n}\left(x\right)t^{n}}\right) = 1- x\left(\sum\limits_{n=0}^{\infty}{\frac{1}{n+1}P_{n}\left(x\right)t^{n+1}}\right)+\left(\sum\limits_{n=0}^{\infty}{\frac{1}{n+2}\cdot P_{n}\left(x\right)t^{n+2}}\right)\\
\frac{\mbox{d}}{\mbox{d}t}\left(\left(1-2xt+t^2\right)\left(\sum\limits_{n=0}^{\infty}{P_{n}\left(x\right)t^{n}}\right)\right) = \frac{\mbox{d}}{\mbox{d}t}\left(1- x\left(\sum\limits_{n=0}^{\infty}{\frac{1}{n+1}P_{n}\left(x\right)t^{n+1}}\right)+\left(\sum\limits_{n=0}^{\infty}{\frac{1}{n+2}\cdot P_{n}\left(x\right)t^{n+2}}\right)\right)\\
\frac{\mbox{d}}{\mbox{d}t}\left(\left(1-2xt+t^2\right)\left(\sum\limits_{n=0}^{\infty}{P_{n}\left(x\right)t^{n}}\right)\right) = \frac{\mbox{d}}{\mbox{d}t}\left(1\right) - x\frac{\mbox{d}}{\mbox{d}t}\left(\sum\limits_{n=0}^{\infty}{\frac{1}{n+1}P_{n}\left(x\right)t^{n+1}}\right) + \frac{\mbox{d}}{\mbox{d}t}\left(\sum\limits_{n=0}^{\infty}{\frac{1}{n+2}\cdot P_{n}\left(x\right)t^{n+2}}\right)\\
\left(2t-2x\right)\left(\sum\limits_{n=0}^{\infty}{P_{n}\left(x\right)t^{n}}\right)+\left(1-2xt+t^2\right)\frac{\mbox{d}}{\mbox{d}t}\left(\sum\limits_{n=0}^{\infty}{P_{n}\left(x\right)t^{n}}\right) = -x\left(\sum\limits_{n=0}^{\infty}{P_{n}\left(x\right)t^{n}}\right)+t\left(\sum\limits_{n=0}^{\infty}{P_{n}\left(x\right)t^{n}}\right)\\
-2\left(x - t\right)\left(\sum\limits_{n=0}^{\infty}{P_{n}\left(x\right)t^{n}}\right)+\left(1-2xt+t^2\right)\frac{\mbox{d}}{\mbox{d}t}\left(\sum\limits_{n=0}^{\infty}{P_{n}\left(x\right)t^{n}}\right) = -\left(x-t\right)\left(\sum\limits_{n=0}^{\infty}{P_{n}\left(x\right)t^{n}}\right)\\
\left(1-2xt+t^2\right)\frac{\mbox{d}}{\mbox{d}t}\left(\sum\limits_{n=0}^{\infty}{P_{n}\left(x\right)t^{n}}\right)=\left(x-t\right)\left(\sum\limits_{n=0}^{\infty}{P_{n}\left(x\right)t^{n}}\right)\\
\frac{\mbox{d}G\left(t\right)}{\mbox{d}t} = \frac{x-t}{1-2xt+t^2}\cdot G\left(t\right)\\
\frac{\mbox{d}G\left(t\right)}{G\left(t\right)} = \frac{x-t}{1-2xt+t^2}\mbox{d}t\\
\frac{\mbox{d}G\left(t\right)}{G\left(t\right)} = -\frac{1}{2}\frac{-2x+2t}{1-2xt+t^2}\mbox{d}t\\
\ln{\left|G\left(t\right)\right|} = -\frac{1}{2}\ln{\left|1-2xt+t^2\right|}+C\\
G\left(x,t\right) = \frac{C\left(x\right)}{\sqrt{1-2xt+t^2}}\\
G\left(x,0\right) = 1 = \frac{C\left(x\right)}{1}\\
G\left(x,t\right) = \frac{1}{\sqrt{1-2xt+t^2}}\\
}\)
Rozwińmy funkcję tworzącą w szereg korzystając z dwumianu Newtona
\(\displaystyle{
\frac{1}{\sqrt{1-2xt+t^2}} = \left(1+t\left(t-2x\right)\right)^{-\frac{1}{2}}\\
\frac{1}{\sqrt{1-2xt+t^2}} = \sum\limits_{n=0}^{\infty}{{-\frac{1}{2} \choose n}t^{n}\left(t-2x\right)^{n}}\\
\frac{1}{\sqrt{1-2xt+t^2}} = \sum\limits_{n=0}^{\infty}{{-\frac{1}{2} \choose n}t^{n}\left(\sum\limits_{k=0}^{n}{ n \choose k}t^{k}\left(-1\right)^{n-k}\left(2x\right)^{n-k}\right)}\\
\frac{1}{\sqrt{1-2xt+t^2}} = \sum\limits_{n=0}^{\infty}{\sum\limits_{k=0}^{n}{{-\frac{1}{2} \choose n} \cdot{ n \choose k}\left(-1\right)^{n-k}\left(2x\right)^{n-k} t^{n+k}}}\\
}\)
Rozpiszmy teraz ten symbol Newtona aby pozbyć się tego ułamkowego argumentu
\(\displaystyle{
{\frac{-1}{2} \choose n} = \frac{-\frac{1}{2}\cdot\left(-\frac{1}{2}\right)\cdot _\cdots\cdot\left(-\frac{1}{2} - \left(n-1\right)\right)}{n!}\\
{\frac{-1}{2} \choose n}=\frac{\left(-1\right)^{n}}{2^{n}}\frac{1\cdot 3\cdot 5 \cdot _\cdots\cdot\left(2n-1\right)}{n!}\\
{\frac{-1}{2} \choose n}=\frac{\left(-1\right)^{n}}{2^{n}}\frac{1\cdot 3\cdot 5 \cdot _\cdots\cdot\left(2n-1\right)\cdot 2\cdot4\cdot 6 \cdot _\cdots \cdot 2n}{n!\cdot 2\cdot4\cdot 6 \cdot _\cdots \cdot 2n}\\
{\frac{-1}{2} \choose n}=\frac{\left(-1\right)^{n}}{2^{n}}\cdot\frac{\left(2n\right)!}{2^{n}n!\cdot n!}\\
{\frac{-1}{2} \choose n}=\frac{\left(-1\right)^{n}}{2^{2n}}\cdot\frac{\left(2n\right)!}{n!\cdot n!}\\
{\frac{-1}{2} \choose n}=\frac{\left(-1\right)^{n}}{2^{2n}}\cdot{2n \choose n}\\
}\)
Mamy zatem następujący wzór
\(\displaystyle{
\frac{1}{\sqrt{1-2xt+t^2}} = \sum\limits_{n=0}^{\infty}{\sum\limits_{k=0}^{n}{{-\frac{1}{2} \choose n} \cdot{ n \choose k}\left(-1\right)^{n-k}\left(2x\right)^{n-k} t^{n+k}}}\\
\frac{1}{\sqrt{1-2xt+t^2}} = \sum\limits_{n=0}^{\infty}{\sum\limits_{k=0}^{n}{{2n \choose n} \cdot{ n \choose k}\left(-1\right)^{n-k}\frac{\left(-1\right)^n}{2^{2n}}\left(2x\right)^{n-k} t^{n+k}}}\\
\frac{1}{\sqrt{1-2xt+t^2}} = \sum\limits_{n=0}^{\infty}{\sum\limits_{k=0}^{n}{{2n \choose n} \cdot{ n \choose k}\frac{\left(-1\right)^{k}}{2^{2n}}\left(2x\right)^{n-k} t^{n+k}}}\\
}\)
ale my potrzebyjemy potęg t zależnych tylko od n
\(\displaystyle{
\frac{1}{\sqrt{1-2xt+t^2}} = \sum\limits_{n=0}^{\infty}{\sum\limits_{k=0}^{n}{{2n \choose n} \cdot{ n \choose k}\frac{\left(-1\right)^{k}}{2^{2n}}\left(2x\right)^{n-k} t^{n+k}}}\\
\frac{1}{\sqrt{1-2xt+t^2}} = \sum\limits_{m=0}^{\infty}{\sum\limits_{k=0}^{\lfloor\frac{m}{2}\rfloor}{{2m-2k \choose m-k} \cdot{ m-k \choose k}\frac{\left(-1\right)^{k}}{2^{2m-2k}}\left(2x\right)^{m-2k} t^{m}}}\\
\frac{1}{\sqrt{1-2xt+t^2}} = \sum\limits_{m=0}^{\infty}{\sum\limits_{k=0}^{\lfloor\frac{m}{2}\rfloor}{{2m-2k \choose m-k} \cdot{ m-k \choose k}\frac{\left(-1\right)^{k}}{2^{2m-2k}}\cdot \frac{2^{m}}{2^{2k}} x^{m-2k} t^{m}}}\\
\frac{1}{\sqrt{1-2xt+t^2}} = \sum\limits_{m=0}^{\infty}{\sum\limits_{k=0}^{\lfloor\frac{m}{2}\rfloor}{{2m-2k \choose m-k} \cdot{ m-k \choose k}\frac{\left(-1\right)^{k}}{2^{m}}x^{m-2k} t^{m}}}\\
\frac{1}{\sqrt{1-2xt+t^2}} = \sum\limits_{m=0}^{\infty}{\sum\limits_{k=0}^{\lfloor\frac{m}{2}\rfloor}{\frac{\left(2m-2k\right)!}{\left(m-k\right)!\cdot\left(m-k\right)!} \cdot\frac{\left(m-k\right)!\cdot m!}{k!\left(m-2k\right)!\cdot m!}\cdot\frac{\left(-1\right)^{k}}{2^{m}}x^{m-2k} t^{m}}}\\
\frac{1}{\sqrt{1-2xt+t^2}} = \sum\limits_{m=0}^{\infty}{\sum\limits_{k=0}^{\lfloor\frac{m}{2}\rfloor}{2m-2k \choose m}\cdot {m \choose k}\cdot\frac{\left(-1\right)^{k}}{2^{m}}x^{m-2k} t^{m}}\\
P_{m}\left(x\right) = \sum\limits_{k=0}^{\lfloor\frac{m}{2}\rfloor}{2m-2k \choose m}\cdot {m \choose k}\cdot\frac{\left(-1\right)^{k}}{2^{m}}x^{m-2k} \\
}\)
\(\displaystyle{
\int\limits_{-1}^{1}{P_{m}\left(x\right)P_{n}\left(x\right)\mbox{d}x} = \frac{2}{2n+1}\delta_{m,n}\\
}\)
Tylko jak uzasadnić przejście
\(\displaystyle{
\frac{1}{\sqrt{1-2xt+t^2}} = \sum\limits_{n=0}^{\infty}{\sum\limits_{k=0}^{n}{{2n \choose n} \cdot{ n \choose k}\frac{\left(-1\right)^{k}}{2^{2n}}\left(2x\right)^{n-k} t^{n+k}}}\\
\frac{1}{\sqrt{1-2xt+t^2}} = \sum\limits_{m=0}^{\infty}{\sum\limits_{k=0}^{\lfloor\frac{m}{2}\rfloor}{{2m-2k \choose m-k} \cdot{ m-k \choose k}\frac{\left(-1\right)^{k}}{2^{2m-2k}}\left(2x\right)^{m-2k} t^{m}}}\\
}\)
Wielomiany Legendre można wykorzystać do numerycznego całkowania funkcji za pomocą kwadratur Gaussa