Re: Differentiation
- From: Narcoleptic Insomniac <i_have_narcoleptic_insomnia@xxxxxxxxx>
- Date: Mon, 21 Aug 2006 16:50:36 EDT
On Aug 21, 2006 3:24 AM CT, William Elliot wrote:
From: Narcoleptic Insomniac
<i_have_narcoleptic_insomnia@xxxxxxxxx>
Newsgroups: sci.math
Subject: Re: Differentiation
Narcoleptic Insomniac fell asleep being kept awake
thinking
2(1 + 2x^2)e^(x^2)differentiate exp(x^2)?
e^(x^2) * 2x
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?
It looks as if 2^[n/2], where [] denotes the ceiling,
will always divide the polynomial. I suppose one could
prove this by using the fact that
(-1)^n (d^n e^(-x^2)) / dx^n = H_n(x) e^(-x^2)
...and proceed via induction.
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 of orthogonal
polynomials over the domain (-infty,infty) with
weighting function e^(-x^2) ,
I read that too and I understand the orthogonality, but I
haven't encountered the notion of a weighting function
yet.
Are they the polynomials with
(d^n e^(-x^2)) / dx^n = H_n(x) e^(-x^2) ?
It looks that they satisfy
(-1)^n (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) ?
^_^ Heh. Yes, we can define K_n(x) explicity as
K_n(x) = sum_{k = 0}^{n} T(n, k) x^k
...where the coefficients T(n, k) are given by
T(n, k) = n! 2^k / {k! ((n - k) / 2)!}
...if n - k == 0 (mod 2) and T(n, k) = 0 otherwise.
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? ;-)
Yes, I will prepare the drinks!
.
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