proving part (1).
The vectors
and
are the columns of the matrix
and so, from elementary linear algebra, we see that they are linearly independent if and only if the
determinant of
is nonzero.
Hence, part (2) follows from part (1).Similarly, part (3) is a consequence of part (1).
It may come as no surprise that the contour integral
of a function
around the boundary of a geometric set
is not
necessarily 0 if the function
is not differentiable at each point in the interior of
However, it is exactly these kinds of contour integrals that will occupy our attention in the rest of this chapter,
and we shouldn't jump to any conclusions.
Let
be a point in
and let
be the geometric set that is a closed disk
Let
be the parameterization of the boundary
of
given by
for
For each integer
define
- Show that
for all
- Show that
There is a remarkable result about contour integrals of certain functions that aren't
differentiable everywhere within a geometric set, and it is what has been called theFundamental Theorem of Analysis, or Cauchy's Theorem.
This theorem has many general statements, but we present one here that is quite broad and certainlyadequate for our purposes.
Cauchy's theorem, fundamental theorem of analysis
Let
be a piecewise smooth geometric set whose boundary
has finite length, and
let
be a piecewise smooth geometric set, whose boundary
also is of finite length.
Suppose
is continuous on
i.e., at
every point
that is in
but not in
and assume that
is differentiable on
i.e., at every point
in
but not in
(We think of these sets as being the points “between” the boundary curves of these geometric sets.)
Then the two contour integrals
and
are equal.
Let the geometric set
be determined by the interval
and the
two bounding functions
and
and let the geometric set
be determined by the subinterval
of
and the two bounding functions
and
Because
we know that
and
for all
We define four geometric sets
as follows:
-
is determined by the interval
and the two bounding functions
and
restricted to that interval.
-
is determined by the interval
and the two bounding functions
and
restricted to that interval.
-
is determined by the interval
and the two bounding functions
and
restricted to that interval.
-
is determined by the interval
and the two bounding
functions
and
restricted to that interval.
Observe that the five sets
constitute a partition of the geometric set
The corollary to
[link] applies to each of the four geometric sets
Hence, the contour integral of
around each of the four boundaries of these geometric sets is 0.
So, by
[link] ,
as desired.
- Draw a picture of the five geometric sets in the proof above and
justify the claim that the sum of the four contour integrals around the geometric sets
is the integral around
minus the integral around
- Let
be pairwise disjoint, piecewise smooth geometric sets, each
having a boundary of finite length, and each containedin a piecewise smooth geometric set
whose boundary also has finite length.
Prove that the
's are some of the elements of a partition
of
each of which is piecewise smooth and has a boundary of finite length.
Show that, by reindexing,
can be chosen to be the first
elements of the partition
HINT: Just carefully adjust the proof of
[link] .
- Suppose
is a piecewise smooth geometric set whose boundary has finite length,
and let
be a partition of
for which each
is piecewise smooth
and has a boundary
of finite length.
Suppose
is continuous on each of the boundaries
of the
's as well as the boundary
of
and assume that
is continuous on each of the
's, for
and differentiable at each point of their interiors.
Prove that
- Prove the following generalization of the Cauchy Theorem:
Let
be pairwise disjoint, piecewise smooth geometric sets whose boundaries have finite length, all contained
in the interior of a piecewise smooth geometric set
whose boundary also has finite length.
Suppose
is continuous at each point of
that is not in the interior of any of the
's,
and that
is differentiable at each point of
that is not an element of any of the
's.
Prove that