MHB How are the Hermite Polynomials Defined and Calculated?

[DreamWeaver]

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]

Spoiler

\[H_5 = -e^{x^2}\cdot \frac{\mathrm{d^5} }{\mathrm{d} x^5}\left \{ e^{-x^2} \right \} =-e^{x^2}\cdot \frac{\mathrm{d^4} }{\mathrm{d} x^4}\left \{ -2xe^{-x^2} \right \}\\\\ =-e^{x^2}\cdot \frac{\mathrm{d^3} }{\mathrm{d} x^3}\left \{ (4x^2-2)e^{-x^2} \right \} =-e^{x^2}\cdot \frac{\mathrm{d^2} }{\mathrm{d} x^2}\left \{ (-8x^3+12x)e^{-x^2} \right \}\\\\ =-e^{x^2}\cdot \frac{\mathrm{d} }{\mathrm{d} x}\left \{ (16x^4-48x^2+12)e^{-x^2} \right \} =-e^{x^2}\cdot \left \{ (-32x^5+160x^3-120x)e^{-x^2} \right \}\\\\ =32x^5-160x^3+120x\]

 

[chisigma]

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

$$\mathscr{H}_5(x) = 32 x^5-160 x^3+120 x$$

[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

$\chi$ $\sigma$

 

[DreamWeaver]

Spoiler

\[H_5 = -e^{x^2}\cdot \frac{\mathrm{d^5} }{\mathrm{d} x^5}\left \{ e^{-x^2} \right \} =-e^{x^2}\cdot \frac{\mathrm{d^4} }{\mathrm{d} x^4}\left \{ -2xe^{-x^2} \right \}\\\\ =-e^{x^2}\cdot \frac{\mathrm{d^3} }{\mathrm{d} x^3}\left \{ (4x^2-2)e^{-x^2} \right \} =-e^{x^2}\cdot \frac{\mathrm{d^2} }{\mathrm{d} x^2}\left \{ (-8x^3+12x)e^{-x^2} \right \}\\\\ =-e^{x^2}\cdot \frac{\mathrm{d} }{\mathrm{d} x}\left \{ (16x^4-48x^2+12)e^{-x^2} \right \} =-e^{x^2}\cdot \left \{ (-32x^5+160x^3-120x)e^{-x^2} \right \}\\\\ =32x^5-160x^3+120x\]

[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

$\chi$ $\sigma$

Two excellent, elegant, and - even better - different proofs. Thanks for taking part, people! (Sun)

Gethin