Wielomiany Hermite wyprowadzenie wzoru

Własności wielomianów; pierwiastki, współczynniki. Dzielenie wielomianów. Wzory Viete'a. RÓWNANIA I NIERÓWNOŚCI wielomianowe (wyższych stopni). Rozkład na czynniki.
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Wielomiany Hermite wyprowadzenie wzoru

Post autor: Mariusz M »

Wyprowadźmy wzór na wykładniczą funkcję tworzącą wielomianów Hermite

Zacznijmy od wzoru rekurencyjnego na wielomiany Hermite

\(\displaystyle{ \begin{cases} H_{n}\left( x\right) = 1 \qquad n=0 \\ H_{n}\left( x\right) = 2x \qquad n=1 \\ H_{n+1}\left( x\right)=2x H_{n}\left( x\right)-2nH_{n-1}\left( x\right) \qquad n\ge 1 \end{cases} }\)

Przesuńmy teraz indeksy

\(\displaystyle{ \begin{cases} H_{n}\left( x\right) = 1 \qquad n=0 \\ H_{n}\left( x\right) = 2x \qquad n=1 \\ H_{n}\left( x\right)=2x H_{n-1}\left( x\right)-2\left( n-1\right) H_{n-2}\left( x\right) \qquad n\ge 2 \end{cases}}\)

Zdefiniujmy sobie wykładniczą funkcję tworzącą wielomianów Hermite


\(\displaystyle{ E\left( x,t\right) = \sum_{n=0}^{ \infty }{H_{n}\left( x\right) \cdot \frac{t^n}{n!} } }\)

Wstawmy wykładniczą funkcję tworzącą do równania rekurencyjnego


\(\displaystyle{ \sum_{n=2}^{ \infty }H_{n}\left( x\right) \cdot \frac{t^n}{n!}= \sum_{n=2}^{ \infty }{\left( 2xH_{n-1}\left( x\right)-2\left(n-1 \right)H_{n-2}\left( x\right) \right) \cdot \frac{t^n}{n!} } \\
\sum_{n=2}^{ \infty }H_{n}\left( x\right) \cdot \frac{t^n}{n!}=\left( \sum_{n=2}^{ \infty }2xH_{n-1}\left( x\right) \cdot \frac{t^n}{n!} \right) +\left( \sum_{n=2}^{ \infty }{\left( -2\left(n-1 \right)H_{n-2}\left( x\right) \cdot \frac{t^n}{n!} \right) } \right) \\
\sum_{n=2}^{ \infty }H_{n}\left( x\right) \cdot \frac{t^n}{n!}=2x\left( \sum_{n=2}^{ \infty }H_{n-1}\left(x \right) \cdot \frac{t^n}{n!} \right)-2\left( \sum_{n=2}^{ \infty }{\left( n-1\right)H_{n-2}\left( x\right) \cdot \frac{t^n}{n!} } \right) \\
\sum_{n=2}^{ \infty }H_{n}\left( x\right) \cdot \frac{t^n}{n!}=2x\left( \sum_{n=1}^{ \infty }H_{n}\left( x\right) \cdot \frac{t^{n+1}}{\left( n+1\right)! } \right) -2\left( \sum_{n=0}^{ \infty } \left( n+1\right)H_{n}\left( x\right) \cdot \frac{t^{n+2}}{\left( n+2\right)! } \right) \\
\sum_{n=2}^{ \infty }H_{n}\left( x\right) \cdot \frac{t^n}{n!}=2x\left( \sum_{n=1}^{ \infty }H_{n}\left( x\right) \cdot \frac{t^{n+1}}{\left( n+1\right)! } \right) -2\left( \sum_{n=0}^{ \infty }\left( n+2\right)H_{n}\left( x\right) \cdot \frac{t^{n+2}}{\left( n+2\right)! } - \sum_{n=0}^{ \infty }{H_{n}\left( x\right) \cdot \frac{t^{n+2}}{\left( n+2\right)! } } \right) \\
\sum_{n=2}^{ \infty }H_{n}\left( x\right) \cdot \frac{t^n}{n!}=2x\left( \sum_{n=1}^{ \infty }H_{n}\left( x\right) \cdot \frac{t^{n+1}}{\left( n+1\right)! } \right) - 2t\left( \sum_{n=0}^{ \infty}H_{n}\left( x\right) \cdot \frac{t^{n+1}}{\left( n+1\right) !}\right)+2\left( \sum_{n=0}^{ \infty}H_{n}\left( x\right) \cdot \frac{t^{n+2}}{\left( n+2\right) !}\right) \\
\sum_{n=0}^{ \infty }{H_{n}\left( x\right) \cdot \frac{t^n}{n!} } - 1 - 2xt = 2x\left( \sum_{n=0}^{ \infty } {H_{n}\left( x\right) \cdot \frac{t^{n+1}}{\left( n+1\right) !} }-t\right)+2\left( \sum_{n=0}^{ \infty}H_{n}\left( x\right) \cdot \frac{t^{n+2}}{\left( n+2\right) !}\right) -2t\left( \sum_{n=0}^{ \infty}H_{n}\left( x\right) \cdot \frac{t^{n+1}}{\left( n+1\right) !}\right) \\
\sum_{n=0}^{ \infty }{H_{n}\left( x\right) \cdot \frac{t^n}{n!} } - 1 - 2xt = -2xt + 2x\left( \sum_{n=0}^{ \infty } {H_{n}\left( x\right) \cdot \frac{t^{n+1}}{\left( n+1\right) !} }\right) + 2\left( \sum_{n=0}^{ \infty}H_{n}\left( x\right) \cdot \frac{t^{n+2}}{\left( n+2\right) !}\right) -2t\left( \sum_{n=0}^{ \infty}H_{n}\left( x\right) \cdot \frac{t^{n+1}}{\left( n+1\right) !}\right) \\
\sum_{n=0}^{ \infty }{H_{n}\left( x\right) \cdot \frac{t^n}{n!} } - 1 = \left( 2x-2t\right) \left( \sum_{n=0}^{ \infty } {H_{n}\left( x\right) \cdot \frac{t^{n+1}}{\left( n+1\right) !} }\right) + 2\left( \sum_{n=0}^{ \infty}H_{n}\left( x\right) \cdot \frac{t^{n+2}}{\left( n+2\right) !}\right)\\
\sum_{n=0}^{ \infty }{H_{n}\left( x\right) \cdot \frac{t^n}{n!} }+\left( 2t-2x\right)\left( \sum_{n=0}^{ \infty } {H_{n}\left( x\right) \cdot \frac{t^{n+1}}{\left( n+1\right) !} }\right) - 2\left( \sum_{n=0}^{ \infty}H_{n}\left( x\right) \cdot \frac{t^{n+2}}{\left( n+2\right) !}\right) = 1\\
}\)


Niech \(\displaystyle{ y\left( t\right) = \sum_{n=0}^{ \infty }H_{n}\left( x\right) \cdot \frac{t^{n+2}}{\left( n+2\right)! } }\)

Mamy wówczas

\(\displaystyle{
y\left( t\right) = \sum_{n=0}^{ \infty }H_{n}\left( x\right) \cdot \frac{t^{n+2}}{\left( n+2\right)! }\\
y'\left( t\right) = \sum_{n=0}^{ \infty }\left( n+2\right) H_{n}\left( x\right) \cdot \frac{t^{n+1}}{\left( n+2\right)! }\\
y'\left( t\right) = \sum_{n=0}^{ \infty }\left( n+2\right) H_{n}\left( x\right) \cdot \frac{t^{n+1}}{\left( n+2\right) \left( n+1\right)! }\\
y'\left( t\right) = \sum_{n=0}^{ \infty } H_{n}\left( x\right) \cdot \frac{t^{n+1}}{ \left( n+1\right)! }\\
y''\left( t\right) = \sum_{n=0}^{ \infty } \left( n+1\right) H_{n}\left( x\right) \cdot \frac{t^{n}}{ \left( n+1\right)! }\\
y''\left( t\right) = \sum_{n=0}^{ \infty } \left( n+1\right) H_{n}\left( x\right) \cdot \frac{t^{n}}{ \left( n+1\right)n! }\\
y''\left( t\right) = \sum_{n=0}^{ \infty } H_{n}\left( x\right) \cdot \frac{t^{n}}{ n! }\\
}\)



\(\displaystyle{ y''\left( t\right)+\left( 2t-2x\right)y'\left( t\right) -2y\left( t\right) = 1 \\
y''\left( t\right)+2\left( t - x\right)y'\left( t\right) -2y\left( t\right) = 1\\
}\)


Otrzymujemy zatem następujące równanie różniczkowe z warunkami początkowymi
\(\displaystyle{
\begin{cases} y''\left( t\right)+2\left( t - x\right)y'\left( t\right) -2y\left( t\right) = 1
\\ y\left( 0\right)=0\\y'\left( 0\right)=0 \end{cases} \\
}\)


Gdy się uważnie przyjrzymy to z łatwością zauważymy całkę szczególną równania jednorodnego
Pozwoli to nam obniżyć rząd równania

Otóż \(\displaystyle{ y_{1}\left( t\right) = t - x }\)
jest całką szczególną równania jednorodnego

Sprawdźmy to
\(\displaystyle{
0+2\left( t-x\right) \cdot 1 - 2 \cdot \left( t-x\right)\\
0+2t-2x-2t+2x \\
=0\\
}\)


Podstawienie \(\displaystyle{ y=\left( t - x\right) \int{u\left( t\right)\mbox{d}t } }\)
obniży rząd równania

\(\displaystyle{
y\left( t\right) =\left( t - x\right) \int{u\left( t\right)\mbox{d}t }\\
y'\left( t\right) = \int{u\left( t\right)\mbox{d}t } + \left( t - x\right)u\left( t\right)\\
y''\left( t\right) = u\left( t\right) + u\left( t\right) + \left( t - x\right)u'\left( t\right)\\
y''\left( t\right) = 2u\left( t\right) + \left( t - x\right)u'\left( t\right)\\
}\)


\(\displaystyle{
\left(2u\left( t\right) + \left( t - x\right)u'\left( t\right) \right) +2\left( t-x\right)\left(\int{u\left( t\right)\mbox{d}t } + \left( t - x\right)u\left( t\right) \right) - 2\left( t - x\right) \int{u\left( t\right)\mbox{d}t } = 1 \\
2u\left( t\right) + \left( t - x\right)u'\left( t\right) + 2\left( t-x\right)\int{u\left( t\right)\mbox{d}t }+2\left( t-x\right)^2u\left( t\right)-2\left( t - x\right) \int{u\left( t\right)\mbox{d}t } = 1\\
\left( t-x\right)u'\left( t\right) +2\left( t^2-2xt+x^2+1\right)u\left( t\right) = 1\\
\left( t-x\right)u'\left( t\right) +2\left( t^2-2xt+x^2+1\right)u\left( t\right) = 0\\
\left( t-x\right)u'\left( t\right) = -2\left( t^2-2xt+x^2+1\right)u\left( t\right)\\
u'\left( t\right) = -2 \frac{t^2-2xt+x^2+1}{t-x} \cdot u\left( t\right)\\
\frac{u'\left( t\right)}{u\left( t\right)} = -2 \cdot \frac{\left( t-1\right)^2 + 1 }{t - x}\\
\frac{u'\left( t\right)}{u\left( t\right)} = -2\left( t - x\right) - \frac{2}{t-x}\\
\ln{\left| u\left( t\right) \right| } = \int{\left(-2\left( t-x\right) - \frac{2}{t-x}\right)\mbox{d}x } \\
\ln{\left| u\left( t\right) \right| } = -\left( t-x\right)^2-2\ln{\left| t-x\right| }+C_{1}\\
u\left( t\right) = C \cdot \frac{1}{\left( t-x\right)^2 } \cdot e^{-\left( t-x\right)^2 }\\
u\left( t\right) = C\left( t\right) \cdot \frac{1}{\left( t-x\right)^2 } \cdot e^{-\left( t-x\right)^2}\\
u'\left( t\right) = C'\left( t\right) \cdot \frac{1}{\left( t-x\right)^2 } \cdot e^{-\left( t-x\right)^2}+C\left( t\right) \left( \frac{-2}{\left( t-x\right)^3 }e^{-\left( t-x\right)^2 }+\frac{1}{\left( t-x\right)^2 } \cdot \left( -2\left( t-x\right) \right) \cdot e^{-\left( t-x\right)^2 } \right)\\
u'\left( t\right) = C'\left( t\right) \cdot \frac{1}{\left( t-x\right)^2 } \cdot e^{-\left( t-x\right)^2}+C\left( t\right)\left( -\frac{2}{\left( t-x\right)^3 }e^{-\left( t-x\right)^2 }-\frac{2}{t-x}e^{-\left( t-x\right)^2 }\right) \\
u'\left( t\right) = C'\left( t\right) \cdot \frac{1}{\left( t-x\right)^2 } \cdot e^{-\left( t-x\right)^2}-2C\left( t\right)\left( \frac{1}{\left( t-x\right)^3 }+\frac{1}{t-x} \right) e^{-\left( t-x\right)^2 }\\
u'\left( t\right) = C'\left( t\right) \cdot \frac{1}{\left( t-x\right)^2 } \cdot e^{-\left( t-x\right)^2}-2C\left( t\right) \cdot \frac{t^2-2xt+x^2+1}{\left( t-x\right)^3 } \cdot e^{-\left( t-x\right)^2}\\
\left( t-x\right) \left( C'\left( t\right) \cdot \frac{1}{\left( t-x\right)^2 } \cdot e^{-\left( t-x\right)^2}-2C\left( t\right) \cdot \frac{t^2-2xt+x^2+1}{\left( t-x\right)^3 } \cdot e^{-\left( t-x\right)^2}\right) + 2\left( t^2-2xt+x^2+1\right) C\left( t\right) \frac{1}{\left( t-x\right)^2 } \cdot e^{-\left( t-x\right)^2 }=1\\
C'\left( t\right) \cdot \frac{1}{\left( t-x\right) } \cdot e^{-\left( t-x\right)^2}-2C\left( t\right) \cdot \frac{t^2-2xt+x^2+1}{\left( t-x\right)^2 } \cdot e^{-\left( t-x\right)^2}+ 2\left( t^2-2xt+x^2+1\right) C\left( t\right) \frac{1}{\left( t-x\right)^2 } \cdot e^{-\left( t-x\right)^2 }=1\\
C'\left( t\right) = \left( t-x\right)e^{\left( t-x\right)^2}\\
C\left( t\right) = \frac{1}{2}e^{\left( t-x\right)^2}+C_{1}\\
u\left( t\right) = \left(\frac{1}{2}e^{\left( t-x\right)^2}+C_{1} \right) \cdot \frac{1}{\left( t-x\right)^2 } \cdot e^{-\left( t-x\right)^2}\\
u\left( t\right) = \frac{1}{2} \cdot \frac{1}{\left( t-x\right)^2} + C_{1} \cdot \frac{1}{\left( t-x\right)^2 } \cdot e^{-\left( t-x\right)^2}\\
}\)


\(\displaystyle{ y''\left( t\right)=2u\left( t\right)+\left( t-x\right)u'\left( t\right) \\
y''\left( 0\right) = 1\\
u'\left( t\right) = -\frac{1}{\left( t-x\right)^3 }+C_{1}\left( -\frac{2}{\left( t-x\right)^3 }\cdot e^{-\left( t-x\right)^2}+\frac{1}{\left( t-x\right)^2 } \cdot \left( -2\left( t-x\right) \right)e^{-\left( t-x\right)^2 } \right)\\
u'\left( t\right) = -\frac{1}{\left( t-x\right)^3 } -2C_{1}\left( \frac{1}{\left( t-x\right)^3 }+ \frac{1}{t-x} \right)e^{-\left( t-x\right)^2 }\\
u'\left( t\right) = -\frac{1}{\left( t-x\right)^3 } -2C_{1} \cdot \frac{t^2-2xt+x^2+1}{\left( t-x\right)^3 } \cdot e^{-\left( t-x\right)^2 }\\
y''\left( t\right) = 2\left(\frac{1}{2} \cdot \frac{1}{\left( t-x\right)^2} + C_{1} \cdot \frac{1}{\left( t-x\right)^2 } \cdot e^{-\left( t-x\right)^2} \right)+\left( t-x\right) \left(-\frac{1}{\left( t-x\right)^3 } -2C_{1} \cdot \frac{t^2-2xt+x^2+1}{\left( t-x\right)^3 } \cdot e^{-\left( t-x\right)^2 } \right) \\
y''\left( t\right) = \frac{1}{\left( t-x\right)^2} + 2C_{1} \cdot \frac{1}{\left( t-x\right)^2 } \cdot e^{-\left( t-x\right)^2} - \frac{1}{\left( t-x\right)^2 } -2C_{1} \cdot \frac{t^2-2xt+x^2+1}{\left( t-x\right)^2 } \cdot e^{-\left( t-x\right)^2 }\\
y''\left( t\right) = 2C_{1} \cdot \frac{1}{\left( t-x\right)^2 } \cdot e^{-\left( t-x\right)^2} - 2C_{1}\left( 1+ \frac{1}{\left( t-x\right)^2 } \right)e^{-\left( t-x\right)^2 } \\
y''\left( t\right) = 2C_{1} \cdot \frac{1}{\left( t-x\right)^2 } \cdot e^{-\left( t-x\right)^2} - 2C_{1}e^{-\left( t-x\right)^2 } - 2C_{1} \cdot \frac{1}{\left( t-x\right)^2 } \cdot e^{-\left( t-x\right)^2 }\\
y''\left( t\right) = - 2C_{1}e^{-\left( t-x\right)^2 }\\
}\)


Wyznaczmy teraz stałą z warunku \(\displaystyle{ y''\left( t\right) = 1 }\)

\(\displaystyle{
y''\left( t\right) = - 2C_{1}e^{-\left( t-x\right)^2 }\\
-2C_{1}e^{-x^2}=1\\
C_{1}=-\frac{1}{2}e^{x^2}\\
y''\left( t\right) =-2 \cdot \left( -\frac{1}{2}e^{x^2}\right)e^{-\left( t - x\right)^2 } \\
y''\left( t\right) =e^{x^2}e^{-t^2+2tx-x^2}\\
y''\left( t\right) = e^{x^2-t^2+2tx-x^2}\\
y''\left( t\right) = e^{2xt-t^2}\\
}\)


Zatem wykładnicza funkcja tworząca wielomianów Hermite to

\(\displaystyle{ E\left( x,t\right) = e^{2xt-t^2} }\)

Teraz mamy dwie możliwości albo liczymy n. pochodną i obliczamy jej wartość w zerze albo korzystamy z iloczynu Cauchyego szeregów
Skorzystanie ze wzoru Leibniza na pochodną iloczynu może być kłopotliwe bo obliczenie n. pochodnej czynników jest dość trudne
przynajmniej dla mnie
Skorzystanie z iloczynu Cauchego na pierwszy rzut oka może być kłopotliwe ale gdy jeden z szeregów rozłożymy na
sumę szeregów o wyrazach parzystych i nieparzystych to może się udać

\(\displaystyle{ \left( \sum_{n=0}^{ \infty }a_{n}x^n \right)\left( \sum_{k=0}^{ \infty }b_{k}x^k \right) = \sum_{n=0}^{ \infty }\left( \sum_{k=0}^{n}a_{k}b_{n-k} \right) x^{n} }\)

\(\displaystyle{ e^{2xt-t^2}=e^{2xt}e^{-t^2}\\
\left( \sum_{n=0}^{ \infty }\frac{\left( 2x\right)^n }{n!}t^n \right) \cdot \left( \sum_{k=0}^{ \infty } \frac{\left( -1\right)^k }{k!}t^{2k} \right) \\
\left( \sum_{n=0}^{ \infty }\frac{\left( 2x\right)^{2n} }{\left( 2n \right) !}t^{2n} + \sum_{m=0}^{ \infty } \frac{\left( 2x\right)^{2m+1} }{\left( 2m+1\right) !}t^{2m+1} \right) \cdot \left( \sum_{k=0}^{ \infty } \frac{\left( -1\right)^k }{k!}t^{2k} \right) \\
\left( \sum_{n=0}^{ \infty }\frac{\left( 2x\right)^{2n} }{\left( 2n \right) !}t^{2n}\right)\left(\sum_{k=0}^{ \infty } \frac{\left( -1\right)^k }{k!}t^{2k} \right) +\left( \sum_{m=0}^{ \infty } \frac{\left( 2x\right)^{2m+1} }{\left( 2m+1\right) !}t^{2m+1} \right) \left(\sum_{k=0}^{ \infty } \frac{\left( -1\right)^k }{k!}t^{2k} \right) \\
\left( \sum_{n=0}^{ \infty }\frac{\left( 2x\right)^{2n} }{\left( 2n \right) !}t^{2n}\right)\left(\sum_{k=0}^{ \infty } \frac{\left( -1\right)^k }{k!}t^{2k} \right) +t\left( \sum_{m=0}^{ \infty } \frac{\left( 2x\right)^{2m+1} }{\left( 2m+1\right) !}t^{2m} \right) \left(\sum_{k=0}^{ \infty } \frac{\left( -1\right)^k }{k!}t^{2k} \right) \\
}\)


\(\displaystyle{
\sum_{n=0}^{ \infty }{H_{2n}\left( x\right) \cdot \frac{t^{2n}}{\left( 2n\right) !} } = \sum_{n=0}^{ \infty }\left( \sum_{k=0}^{n} \frac{\left( -1\right)^k }{k!} \cdot \frac{\left( 2x\right)^{2n-2k} }{\left( 2n-2k\right)! } \right) t^{2n} \\
\frac{H_{2n}\left( x\right)}{\left( 2n\right)! } = \sum_{k=0}^{n} \frac{\left( -1\right)^k }{k!} \cdot \frac{\left( 2x\right)^{2n-2k} }{\left( 2n-2k\right)! }\\
H_{2n}\left( x\right) = \sum_{k=0}^{n} \frac{\left( -1\right)^k \left( 2n\right)! }{k!} \cdot \frac{\left( 2x\right)^{2n-2k} }{\left( 2n-2k\right)! }\\
\sum_{n=0}^{ \infty }{H_{2n+1}\left( x\right) \cdot \frac{t^{2n+1}}{\left(2n+1\right)!} } = t\left( \sum_{n=0}^{ \infty }\left( \sum_{k=0}^{n} \frac{\left( -1\right)^k }{k!} \cdot \frac{\left( 2x\right)^{2n-2k+1} }{\left( 2n-2k+1\right) } \right) t^{2n} \right) \\
\sum_{n=0}^{ \infty }{H_{2n+1}\left( x\right) \cdot \frac{t^{2n+1}}{\left(2n+1\right)!} } = \sum_{n=0}^{ \infty }\left( \sum_{k=0}^{n} \frac{\left( -1\right)^k }{k!} \cdot \frac{\left( 2x\right)^{2n-2k+1} }{\left( 2n-2k+1\right) } \right) t^{2n+1}\\
\frac{H_{2n+1}\left( x\right)}{\left(2n+1\right)!} = \sum_{k=0}^{n} \frac{\left( -1\right)^k }{k!} \cdot \frac{\left( 2x\right)^{2n-2k+1} }{\left( 2n-2k+1\right) } \\
H_{2n+1}\left( x\right) = \sum_{k=0}^{n} \frac{\left( -1\right)^k\left( 2n+1\right)! }{k!} \cdot \frac{\left( 2x\right)^{2n-2k+1} }{\left( 2n-2k+1\right) } \\
}\)


\(\displaystyle{
\begin{cases} H_{2n}\left( x\right) = \sum_{k=0}^{n} \frac{\left( -1\right)^k \left( 2n\right)! }{k!} \cdot \frac{\left( 2x\right)^{2n-2k} }{\left( 2n-2k\right)! } \\ H_{2n+1}\left( x\right) = \sum_{k=0}^{n} \frac{\left( -1\right)^k\left( 2n+1\right)! }{k!} \cdot \frac{\left( 2x\right)^{2n-2k+1} }{\left( 2n-2k+1\right) } \end{cases} \\
}\)


Te dwa przypadki można połączyć w jeden

\(\displaystyle{
H_{n}\left( x\right) = \sum_{k=0}^{\lfloor \frac{n}{2} \rfloor}{ \frac{\left( -1\right)^k \cdot n! \cdot \left( 2x\right)^{n-2k} }{k!\left( n-2k\right)! } } }\)


Można by jeszcze wyprowadzić wzoru Rodriguesa
Pokazać że te wielomiany są ortogonalne
ODPOWIEDZ