User:Catfive/Scratchpad

Partial Fraction Decomposition

If $n<m$ and

$p(x)$	$=x^{n}+a_{n-1}x^{n-1}+\cdots +a_{0}$
$q(x)$	$=x^{m}+b_{m-1}x^{m-1}+\cdots +b_{0}$
	$=(x-\alpha _{1})^{r_{1}}\cdot \ldots \cdot (x-\alpha _{k})^{r_{k}}(x^{2}+\beta _{1}x+\gamma _{1})^{s_{1}}\cdot \ldots \cdot (x^{2}+\beta _{j}x+\gamma _{j})^{s_{j}}$

then $p(x)/q(x)$ can be written in the form

$\left[{\frac {a_{1,1}}{x-\alpha _{1}}}+\cdots +{\frac {a_{1,r_{1}}}{(x-\alpha _{1})^{r_{1}}}}\right]+\cdots +\left[{\frac {a_{k,1}}{x-\alpha _{k}}}+\cdots +{\frac {a_{k,r_{k}}}{(x-\alpha _{k})^{r_{k}}}}\right]$

$+\left[{\frac {b_{1,1}x+c_{1,1}}{x^{2}+\beta _{1}x+\gamma _{1}}}+\cdots +{\frac {b_{1,s_{1}}x+c_{1,s_{1}}}{(x^{2}+\beta _{1}x+\gamma _{1})^{s_{1}}}}\right]+\cdots$

$+\left[{\frac {b_{j,1}x+c_{j,1}}{x^{2}+\beta _{j}x+\gamma _{j}}}+\cdots +{\frac {b_{j,s_{j}}x+c_{j,s_{j}}}{(x^{2}+\beta _{j}x+\gamma _{j})^{s_{j}}}}\right]$.

The Laplace Expansion for Determinants

(moved to Laplace expansion article)

Calculus on Manifolds

2-4 Suppose that f is differentiable at (0,0) and let λ = Df(0,0). Then

$(1)\quad \lim _{(h,k)\to 0}{\frac {\left||(h,k)|g\left({\frac {(h,k)}{|(h,k)|}}\right)-\lambda (h,k)\right|}{\left|(h,k)\right|}}=0$, or

$(2)\quad \lim _{(h,k)\to 0}\left|g\left({\frac {(h,k)}{|(h,k)|}}\right)-\lambda \left({\frac {(h,k)}{|(h,k)|}}\right)\right|=0$, so that in particular

$(3)\quad \lim _{h\to 0}\left|g(1,0)-\lambda (1,0)\right|=0$ so that λ(1,0) = 0, and similarly

$(4)\quad \lim _{k\to 0}\left|g(0,1)-\lambda (0,1)\right|=0$ so that λ(0,1) = 0, hence λ = 0 and

$(5)\quad \lim _{(h,k)\to 0}\left|g\left({\frac {(h,k)}{|(h,k)|}}\right)\right|=0$ or

$(6)\quad \lim _{x\to 0}\left|g\left({\frac {x}{|x|}}\right)\right|=0$, hence

$(7)\quad \lim _{t\to 0}\left|g\left({\frac {tx}{|t||x|}}\right)\right|=0$; but

$(8)\quad \left|g\left({\frac {tx}{|t||x|}}\right)\right|=\left|g\left({\frac {x}{|x|}}\right)\right|$

for all t ≠ 0, so

$(9)\quad \left|g\left({\frac {x}{|x|}}\right)\right|=0$

for all x ≠ 0.

Rectangles!

Definition. An open rectangle in Rⁿ is a set of the form $\{(x_{1},\ldots ,x_{n}):a_{i}<x_{i}<b_{i},i=1,\ldots ,n\}\,$, given real numbers $a_{i}\leq b_{i},i=1,\ldots ,n\,$. Notice that the empty set $\emptyset \,$ 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 $v(R)\,$ of a rectangle R defined as above is $\prod _{i=1}^{n}(b_{i}-a_{i})$. It follows that $v(\emptyset )=0\,$.

Definition. A partition ${\mathcal {P}}$ of a rectangle $R\,$ in Rⁿ is a finite collection of pairwise-disjoint open subrectangles of $R\,$ such that the union of their closures is the closure ${\overline {R}}$ of $R\,$.

Lemma. Suppose ${\mathcal {C}}$ is a finite collection of rectangles and ${\mathcal {H}}$ is a finite collection of hyperplanes parallel to an axis. Then ${\mathcal {H}}$ determines a unique partition of each rectangle in ${\mathcal {C}}$.

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 ${\mathcal {P}}$ is a partition of a rectangle R and the open rectangle A is a subset of R. Then the collection $\{p\cap A:p\in {\mathcal {P}}\}\,$ is a partition of A.

Proof: We must show that $D=\cup _{p\in {\mathcal {P}}}\,{\overline {p\cap A}}={\overline {A}}$. First we show that ${\overline {A}}\subset D\,$, which will follow from $A\subset D\,$ since then ${\overline {A}}\subset {\overline {D}}=D\,$. Suppose then that x is in A. Then x is in R and hence is in ${\overline {p}}$ for some p in ${\mathcal {P}}$. If x is in p then it's also in $p\cap A\,$, hence in D. If x is on the boundary of p and N is any neighbourhood of x, then $N\cap A\,$ is a neighbourhood of x and so contains a point of p. Hence every neighbourhood of x contains a point of $p\cap A\,$, i.e. x is in ${\overline {p\cap A}}$ and so in D.

On the other hand, if y is in D, y is in ${\overline {p\cap A}}$ for some p so it can't be exterior to A. In other words, $D\subset {\overline {A}}$. Thus, $D={\overline {A}}$.

Definition. A partition ${\mathcal {P}}'$ is a refinement of a partition ${\mathcal {P}}$ if every member of ${\mathcal {P}}'$ is contained in a member of ${\mathcal {P}}$.

Definition. The common refinement of the partitions ${\mathcal {P}}$ and ${\mathcal {P}}'$ of a rectangle is the collection $\{p\cap p':p\in {\mathcal {P}},p'\in {\mathcal {P}}'\}$.

Proposition. The common refinement of the partitions ${\mathcal {P}}$ and ${\mathcal {P}}'$ 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 ${\overline {p}}$ for some p in ${\mathcal {P}}$. By the previous lemma, ${\mathcal {P}}'$ induces a partition on p, i.e. there is some p' in ${\mathcal {P}}'$ such that ${\overline {p\cap p'}}$ contains x.

Theorem. Suppose $I_{1},\ldots ,I_{r}\,$ are pairwise-disjoint open rectangles, and $J_{1},\ldots ,J_{s}\,$ are open rectangles such that $\cup _{i=1}^{r}I_{i}\subset \cup _{j=1}^{s}J_{j}\,$. Then $v(I_{1})+\cdots +v(I_{r})\leq v(J_{1})+\cdots +v(J_{s})$.

Proof: Let $R\,$ be a rectangle containing the $J_{j}\,$. Then by the second lemma there exists, for each j, a partition ${\mathcal {R}}_{j}\,$ of $R\,$ containing $J_{j}\,$. Let ${\mathcal {P}}$ be the common refinement of these partitions. Since ${\mathcal {P}}$ refines ${\mathcal {R}}_{j}\,$, for each p in ${\mathcal {P}}$ there exists an element of ${\mathcal {R}}_{j}\,$ containing p; hence p is either a subset of $J_{j}\,$ or disjoint from it. Thus $\chi _{J}\,$, the characteristic function of $\cup _{j}J_{j}\,$, is a well-defined step function on ${\mathcal {P}}$. Moreover, by the third lemma the elements of ${\mathcal {P}}$ contained in $J_{j}\,$ form a partition ${\mathcal {Q}}_{j}\,$ of $J_{j}\,$, and

$\int \chi _{J}=\sum _{q\in \cup _{j}{\mathcal {Q}}_{j}}v(q)\leq \sum _{j=1}^{s}\sum _{q\in {\mathcal {Q}}_{j}}v(q)=\sum _{j=1}^{s}v(J_{j})$.

Now if $\chi _{I}\,$ is the characteristic function of $\cup _{i}I_{i}\,$ then $\chi _{I}=\sum _{i}\chi _{I_{i}}\,$ because the $I_{i}\,$ are pairwise-disjoint. Hence $\int \chi _{I}=\sum _{i}\int \chi _{I_{i}}=\sum _{i=1}^{r}v(I_{i})$. And since $\chi _{I}\leq \chi _{J}$, it is also true that $\int \chi _{I}\leq \int \chi _{J}$; that is,

$\sum _{i=1}^{r}v(I_{i})=\int \chi _{I}\leq \int \chi _{J}\leq \sum _{j=1}^{s}v(J_{j})$.

Second proof: Let $\chi _{I}\,$ and $\chi _{J}\,$ be the characteristic functions of $\cup _{i}I_{i}\,$ and $\cup _{j}J_{j}\,$. Then $\chi _{I}=\sum _{i=1}^{r}\chi _{I_{i}}$ and $\chi _{J}\leq \sum _{j=1}^{s}\chi _{J_{j}}$, and since $\chi _{I}\leq \chi _{J}$, it is also true that $\int \chi _{I}\leq \int \chi _{J}$. Thus

$\sum _{i=1}^{r}v(I_{i})=\sum _{i=1}^{r}\int \chi _{I_{i}}=\int \sum _{i=1}^{r}\chi _{I_{i}}=\int \chi _{I}\leq \int \chi _{J}\leq \int \sum _{j=1}^{s}\chi _{J_{j}}=\sum _{j=1}^{s}\int \chi _{J_{j}}=\sum _{j=1}^{s}v(J_{j})$.