Re: Differentiation

From: Narcoleptic Insomniac <i_have_narcoleptic_insomnia@xxxxxxxxx>
Newsgroups: sci.math
Subject: Re: Differentiation

Narcoleptic Insomniac fell asleep being kept awake thinking

differentiate exp(x^2)?

e^(x^2) * 2x

2(1 + 2x^2)e^(x^2)
4x(3 + 2x^2)e^(x^2)

4(3 + 12x^2 + 4x^4)e^(x^2)
8x(15 + 20x^2 + 4x^4)e^(x^2)
8(15 + 90x^2 + 60x^4 + 8x^6)e^(x^2)
16x(105 + 210x^2 + 84x^4 + 8x^6)e^(x^2)
16(105 + 840x^2 + 840x^4 + 224x^6 + 16x^8)e^(x^2)
32x(945 + 2520x^2 + 1512x^4 + 288x^6 + 16x^8)e^(x^2)
32(945 + 9450x^2 + 12600x^4 + 5040x^6 + 720x^8 + 32x^10)e^(x^2)

(d^n e^(x^2)) / dx^n = ?

It's not too hard to see that the n'th derivative is
an n degree polynomial of the form

e^(x^2) * (a_0 + a_1 x + ... + a_n x^n)

...where a_n = 2^n. Moreover, if n is even then a_k = 0 for
all odd k; similarly if n is odd then a_k = 0 for all even k.

I'm not positive but it also looks like a_0 = 0 if n is odd and

a_0 = 2^(n/2) * ... * (n - 3) * (n - 1)

...if n is even.

I tried to ignore the e^(x^2) factor and look for more
patterns by factorizing the coefficients.

Does 2^[n/2] always factors out of the polynomial?

Long story short, it looks like

[d^n e^(x^2)] / dx^n = e^(x^2) * H_n(x)

...where H_n(x) is an n'th degree Hermite polynomial

http://mathworld.wolfram.com/HermitePolynomial.html .

I totally ignored the fact that these Hermite polynomials
are alternating and these derivatives are not. So that
formula for the n'th derivative isn't correct, but they
must be closely related.

The Hermite polynomials H_n(x) are set of orthogonal polynomials
over the domain (-infty,infty) with weighting function e^(-x^2) ,

Are they the polynomials with
(d^n e^(-x^2)) / dx^n = H_n(x) e^(-x^2) ?

Kyle Czarnecki

Similary the Kyle polynomials
(d^n e^(x^2)) / dx^n = K_n(x) e^(x^2) ?

and the hyper-Kyle and hyper-Hermite polynomials
(d^n e^(x^k)) / dx^n = HK^k_n(x) e^(x^k)
(d^n e^(-x^k)) / dx^n = HH^k_n,k(x) e^(-x^k)

Should I prepare a large dose of confetti? ;-)

----
.

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