From: zeno@Athena.MIT.EDU
Received: by madman.MIT.EDU (5.61/4.7) id AA10099; Sat, 26 Oct 91 23:06:42 0400
MessageId: <9110270306.AA10099@madman.MIT.EDU>
To: mkgray@Athena.MIT.EDU
Subject: jacobians
Date: Sat, 26 Oct 91 23:06:41 EDT
Well, it's a bit hard over zephyr. See if this helps at all. Say
f : R^n > R^n is differentiable on some open set U in R^n.
For n = 1 this is just ordinary differentiability. For higher n
there are several equivalent ways of describing it, one of which is to
say that f is "approximable by linear maps" on the set U. So for
any point (vector) x in U, this means there is some linear operator
L : R^n > R^n which approximates f near x in this sense:
the map g(v) := f(x + v)  f(x), which is defined in R^n for all v
near 0 and takes values in R^n, is close to L :
g(v)  Lv f(x + v)  f(x)  Lv
lim  = lim  = 0 .
v>0 v v>0 v
v and the absolute values are in R^n, of course, so those limits mean
"as v goes to the zero vector in all possible ways". This L is just for
this particular point x, and describes exactly how f behaves "linearly" near
x. *Each* point x in U gives rise to its own fapproximating map L(x),
since f is assumed differentiable throughout U.
Now, any linear operator on R^n is completely described by its n x n matrix
w.r.t. the standard basis. It just so happens, as must have been shown in the
course, that the map L(x) which approximates f at x has matrix whose
(i,j)entry, for 1<=i<=n, 1<=j<=n, is exactly the number "Df_{ij}(x)", i.e.
df_i
(x) , where f_i : R^n > R is the ith coordinate function of f.
dx_j
(Part of the proof of this is also the fact that each f_i has all n par
tial derivatives defined at x). Since each such partial derivative is de
fined for all x in U, we can write (w.r.t. the standard basis)
_ _
 Df_{11}(x) . . . Df_{1n}(x) 
 . . . . . 
L(x) =  . . . . .  for all x in U.
 . . . . . 
 . . . . . 
_Df_{n1}(x) . . . Df_{nn}(x)_
To summarize, at each point x of U, f is "very much like its matrix
of partials at x". Okay, now that we have L as a matrix of n^2
different functions from R to R, I can mention the fact that f's
differentiability on U implies that all n^2 of these functions are
continuous on U. Hence, the function Jac_f : R^n > R definedby Jac_f(x) = det L(x) (L as defined above) is continuous. Therefore
the multiple integral
/

 Jac_f(x) dx

/
U
is welldefined  this is the jacobian of f on the region U. It
measures the "change of volume" that f accomplishes on U ; more details
about that can be discussed. What's important to realize is that this is
a welldefined number even if f is so complicated that we can't feasibly
compute it. But in the examples you will be seeing, f is in fact simple
enough that its partials Df_{ij}(x) can be actually written as "closed
form" functions on U, instead of being just these abstract continuous
functions from U into R. In this situation, it usually becomes easy
to compute the determinant as a "uniform" function over U, and thereby
calculate the jacobian.