Re: Gamma function question
- From: David C. Ullrich <ullrich@xxxxxxxxxxxxxxxx>
- Date: Sat, 23 Jun 2007 06:03:44 -0500
On Fri, 22 Jun 2007 14:38:57 -0700, mike3 <mike4ty4@xxxxxxxxx> wrote:
Hi.
Maybe the answer is real obvious and this is real stupid, but it's
been bugging me
recently. If the gamma function is supposed to be a generalization of
the factorial,
then why do we define
Gamma(z) = int_{t=0...inf} e^-t t^(z-1) dt = (z-1)!
instead of
"Gamma"(z) = int_{t=0...inf} e^-t t^z dt = Gamma(z+1) = z!
?
What is the point of the shift by 1, anyway?
A possible reason is this: dt/t is the Haar measure
on the group of positive reals with multiplication
for the group operation. If you don't know what
that means, what it amounts to is that integrals
of the form
int_0^infinity f(t) dt/t
transform very nicely under various changes of variables,
for example (writing int for int_0^infinity) if c > 0 and
a is real, a <> 0 then
int f(t) dt/t = int f(ct) dt/t = int f(1/t) dt/t
= |a| int f(t^a) dt/t
= int_{-infinity}^infinity f(e^t) dt,
etc. With dt instead of dt/t all those formulas look
more complicated.
I'm not saying that that's _the_ reason, but I think
of the definition as
Gamma(x) = int e^{-t} t^x dt/t
and it makes more sense, at least to me.
(Even if those transformations don't come up,
thinking about int f(t) dt/t instead of
int f(t) dt just seems "natural" from the
right point of view.)
If you know a little real analysis, in particular
what an "L^p space" is, then read on:
Maybe the reason it makes sense to me to think of
it this way is beacuse of various formulas that
come up in harmonic analysis. For example, in
one characterization of "Besov spaces" you see a
definition of the form
(*) something = [int (f(t)^p)/t^(ap+1) dt]^{1/p],
which is supposed to interpreted as
(**) sup_t f(t)/t^a
when p = infinity. When I look at the formula
written like that it makes no sense to me,
it's not clear why (*) should become (**)
for p = infinity, and I can't imagine how anyone
would keep it straight. Instead I write the formula as
(*') [int (f(t)/t^a)^p dt/t]^(1/p)
and it makes perfect sense - it's just an L^p
norm with respect to the measure dt/t (this
makes the formula much easier for me to remember
and it also makes it clear why it becomes
(**) for p = infinity.)
************************
David C. Ullrich
.
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- From: mike3
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