integration by parts and differential operators
Stephen White
I think it's becoming apparent my differential operator theory is
not up to snuff.
A problem was assigned, to show that the Hermite polynomials are
orthogonal wrt a Gaussian weight measure. The professor defined them
by
h_n(x) = (2x - d/dx)^n 1.
This seems like an extremely peculiar way to define them, and the only
way I could make sense of it was to unravel it into a recurrence
relation
h_n(x) = 2xh_{n-1}(x) - (h_{n-1})'(x), h_0(x)=1.
After doing this, I was able to solve the problem, with extreme
difficulty, and with generous help from David McAnally. The whole
thing is best described as "a huge ugly mess".
But the professor's position is that the problem is transparently easy,
and this is supposedly by using integration by parts directly on (2x -
d/dx)^n 1. This doesn't make any sense to me. How does this work? I
have never formally studied differential operator theory and don't
understand how to generalize integration by parts to arbitrary
differential operators, short of just unravelling the thing like above.
I'd be very grateful if someone could explain how this works, not just
for the particular problem above but for general differential
operators.
S.W.
Robert Israel
Stephen White <mildre...@yahoo.com> wrote:
> I think it's becoming apparent my differential operator theory is
>not up to snuff.
>
> A problem was assigned, to show that the Hermite polynomials are
>orthogonal wrt a Gaussian weight measure. The professor defined them
>by
>h_n(x) = (2x - d/dx)^n 1.
>This seems like an extremely peculiar way to define them, and the only
>way I could make sense of it was to unravel it into a recurrence
>relation
>h_n(x) = 2xh_{n-1}(x) - (h_{n-1})'(x), h_0(x)=1.
>
>After doing this, I was able to solve the problem, with extreme
>difficulty, and with generous help from David McAnally. The whole
>thing is best described as "a huge ugly mess".
>
>But the professor's position is that the problem is transparently easy,
>and this is supposedly by using integration by parts directly on (2x -
>d/dx)^n 1. This doesn't make any sense to me. How does this work? I
>have never formally studied differential operator theory and don't
>understand how to generalize integration by parts to arbitrary
>differential operators, short of just unravelling the thing like above.
OK, you want to show that
int_{-infty}^infty h_n(x) h_m(x) exp(-x^2) dx = 0
if n <> m.
Now integration by parts says that (for suitable functions f and g)
int_{-infty}^infty f'(x) g(x) exp(-x^2) dx
= - int_{-infty}^infty f(x) (g(x) exp(-x^2))' dx
= - int_{-infty}^infty f(x) (-2x g(x) + g'(x)) exp(-x^2) dx
= int_{-infty}^infty f(x) T(g)(x) exp(-x^2) dx
where T is the operator 2 x - d/dx. Repeating this,
int_{-infty}^infty f(x) (T^n)(g)(x) exp(-x^2) dx
= int_{-infty}^infty f^(n)(x) g(x) exp(-x^2) dx
where f^(n) = (d/dx)^n f.
Note that h_n = T^n 1 is a polynomial of degree n. So if
n < m,
int_{-infty}^infty h_n(x) h_m(x) exp(-x^2) dx
= int_{-infty}^infty h_n(x) (T^m)(1)(x) exp(-x^2) dx
= int_{-infty}^infty (h_n)^(m)(x) exp(-x^2) dx = 0
because (h_n)^(m) = 0.
Robert Israel isr...@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada