The discussion focuses on the definition and calculation of Hermite Polynomials, particularly through their recursive relations and explicit forms. Participants explore both the definition using derivatives and the recursive method for generating these polynomials.
Participants present similar definitions and recursive relations for Hermite Polynomials, but there is no explicit consensus on a single method or proof being superior. The discussion remains open to various approaches.
The discussion does not address potential limitations or assumptions in the definitions or proofs provided, nor does it explore the implications of the recursive relation in depth.
DreamWeaver said:Defining the Hermite Polynomials by:
$$\mathscr{H}_n(x) = (-1)^n e^{x^2}\, \frac{d^n}{dx^n} \Bigg\{ e^{-x^2}\Bigg\}$$Show that
$$\mathscr{H}_5(x) = 32 x^5-160 x^3+120 x$$
lfdahl said:\[H_5 = -e^{x^2}\cdot \frac{\mathrm{d^5} }{\mathrm{d} x^5}\left \{ e^{-x^2} \right \} =-e^{x^2}\cdot \frac{\mathrm{d^4} }{\mathrm{d} x^4}\left \{ -2xe^{-x^2} \right \}\\\\ =-e^{x^2}\cdot \frac{\mathrm{d^3} }{\mathrm{d} x^3}\left \{ (4x^2-2)e^{-x^2} \right \} =-e^{x^2}\cdot \frac{\mathrm{d^2} }{\mathrm{d} x^2}\left \{ (-8x^3+12x)e^{-x^2} \right \}\\\\ =-e^{x^2}\cdot \frac{\mathrm{d} }{\mathrm{d} x}\left \{ (16x^4-48x^2+12)e^{-x^2} \right \} =-e^{x^2}\cdot \left \{ (-32x^5+160x^3-120x)e^{-x^2} \right \}\\\\ =32x^5-160x^3+120x\]
Two excellent, elegant, and - even better - different proofs. Thanks for taking part, people! (Sun)chisigma said:[sp]The recursive relation for Hermite Polynomials is...
$\displaystyle H_{n+1}(x) = 2\ x\ H_{n}(x) - 2\ n\ H_{n-1} (x)\ (1)$
... and is...
$\displaystyle H_{0}= 1$
$\displaystyle H_{1} = 2\ x$
... so that...
$\displaystyle H_{2} = 4\ x^{2} - 2$
$\displaystyle H_{3} = 8\ x^{3} - 12\ x$
$\displaystyle H_{4} = 16\ x^{4} - 48\ x^{2} + 12$
$\displaystyle H_{5} = 32\ x^{5} - 160\ x^{3} + 120\ x$
[/sp]
Kind regards
$\chi$ $\sigma$