The Hermite Polynomials are defined using the formula $$\mathscr{H}_n(x) = (-1)^n e^{x^2}\, \frac{d^n}{dx^n} \Bigg\{ e^{-x^2}\Bigg\}$$. The specific polynomial for n=5 is given by $$\mathscr{H}_5(x) = 32 x^5 - 160 x^3 + 120 x$$. The recursive relation for Hermite Polynomials is established as $$H_{n+1}(x) = 2\ x\ H_{n}(x) - 2\ n\ H_{n-1}(x)$$, with initial conditions $$H_{0}= 1$$ and $$H_{1} = 2\ x$$. This discussion provides a comprehensive overview of the definition and calculation methods for Hermite Polynomials.
PREREQUISITESMathematicians, physicists, and engineers interested in polynomial theory, quantum mechanics, and numerical methods will benefit from this discussion on Hermite Polynomials.
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
$\chi$ $\sigma$