Partial Fraction Decomposition
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If
and
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|
|
|
|
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then
can be written in the form
.
The Laplace Expansion for Determinants
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(moved to Laplace expansion article)
Calculus on Manifolds
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2-4 Suppose that f is differentiable at (0,0) and let λ = Df(0,0). Then
, or
, so that in particular
so that λ(1,0) = 0, and similarly
so that λ(0,1) = 0, hence λ = 0 and
or
, hence
; but
![{\displaystyle (8)\quad \left|g\left({\frac {tx}{|t||x|}}\right)\right|=\left|g\left({\frac {x}{|x|}}\right)\right|}](https://wikimedia.riteme.site/api/rest_v1/media/math/render/svg/669264ec34717a02c0c17cd469709ba4dd4bacc3)
for all t ≠ 0, so
![{\displaystyle (9)\quad \left|g\left({\frac {x}{|x|}}\right)\right|=0}](https://wikimedia.riteme.site/api/rest_v1/media/math/render/svg/857f08d9126f04a20142f5a530729e75245af9ef)
for all x ≠ 0.
Definition. An open rectangle in Rn is a set of the form
, given real numbers
. Notice that the empty set
is an open rectangle. A closed rectangle is defined similarly by replacing < with ≤. The term rectangle means either an open or a closed rectangle.
Definition. The elementary volume
of a rectangle R defined as above is
. It follows that
.
Definition. A partition
of a rectangle
in Rn is a finite collection of pairwise-disjoint open subrectangles of
such that the union of their closures is the closure
of
.
Lemma. Suppose
is a finite collection of rectangles and
is a finite collection of hyperplanes parallel to an axis. Then
determines a unique partition of each rectangle in
.
Proof: It suffices to prove the lemma in the case of one rectangle and one
hyperplane, and in this case the statement is obvious.
Lemma. Suppose A and B are rectangles and B is a subset of A. Then there exists a partition of A that contains B.
Proof: Let H be the collection of the 2n hyperplanes coinciding with the
boundary planes of B. By the first lemma H determines a unique partition
of A, and by construction B is an element of the partition.
Lemma. Suppose
is a partition of a rectangle R and the open rectangle A is a subset of R. Then the collection
is a partition of A.
Proof: We must show that
. First we show that
, which will follow from
since then
. Suppose then that x is in A. Then x is in R and hence is in
for some p in
. If x is in p then it's also in
, hence in D. If x is on the boundary of p and N is any neighbourhood of x, then
is a neighbourhood of x and so contains a point of p. Hence every neighbourhood of x contains a point of
, i.e. x is in
and so in D.
On the other hand, if y is in D, y is in
for some p so it can't be exterior to A. In other words,
. Thus,
.
Definition. A partition
is a refinement of a partition
if every member of
is contained in a member of
.
Definition. The common refinement of the partitions
and
of a rectangle is the collection
.
Proposition. The common refinement of the partitions
and
of a rectangle R is a partition of R.
Proof: Clearly the members of the common refinement are pairwise-disjoint open subrectangles of R. If x is in the closure of R then x is in
for some p in
. By the previous lemma,
induces a partition on p, i.e. there is some p' in
such that
contains x.
Theorem. Suppose
are pairwise-disjoint
open rectangles, and
are open rectangles such
that
. Then
.
Proof: Let
be a rectangle containing the
. Then by the second lemma there exists, for each j, a partition
of
containing
. Let
be the common refinement of these partitions. Since
refines
, for each p in
there exists an element of
containing p; hence p is either a subset of
or disjoint from it. Thus
, the characteristic function of
, is a well-defined step function on
. Moreover, by the third lemma the elements of
contained in
form a partition
of
, and
.
Now if
is the characteristic function of
then
because the
are pairwise-disjoint. Hence
. And since
, it is also true that
; that is,
.
Second proof: Let
and
be the characteristic functions of
and
. Then
and
, and since
, it is also true that
. Thus
.