User:Catfive/Scratchpad

Partial Fraction Decomposition

If $n<m$ and

$p(x)$	$=x^n+a_{n-1}{x}^{n-1}+\cdots +a_0$
$q(x)$	$=x^m+b_{m-1}{x}^{m-1}+\cdots +b_0$
	$=(x-{\alpha }_1{)}^{r_1}\cdot \dots \cdot (x-{\alpha }_k{)}^{r_k}(x^2+{\beta }_{1}x+{\gamma }_1{)}^{s_1}\cdot \dots \cdot (x^2+{\beta }_{j}x+{\gamma }_j{)}^{s_j}$

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

$[\frac{a_{1,1}}{x-{\alpha }_1}+\cdots +\frac{a_{1,r_1}}{(x-{\alpha }_1{)}^{r_1}}]+\cdots +[\frac{a_{k,1}}{x-{\alpha }_k}+\cdots +\frac{a_{k,r_k}}{(x-{\alpha }_k{)}^{r_k}}]$

$+[\frac{b_{1,1}x+c_{1,1}}{x^2+{\beta }_{1}x+{\gamma }_1}+\cdots +\frac{b_{1,s_1}x+c_{1,s_1}}{(x^2+{\beta }_{1}x+{\gamma }_1{)}^{s_1}}]+\cdots$

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

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

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

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

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

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

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

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

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

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

for all t ≠ 0, so

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

for all x ≠ 0.

Rectangles!

Definition. An open rectangle in Rⁿ is a set of the form $\{(x_1,\dots ,x_n):a_i<x_i<b_i,i=1,\dots ,n\}$, given real numbers $a_i\le b_i,i=1,\dots ,n$. Notice that the empty set $\mathrm{\varnothing }$ 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 $v(\mathrm{\varnothing })=0$.

Definition. A partition ${𝒫}$ 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 $\bar{R}$ of $R$.

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

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

Proof: We must show that $D={\cup }_{p\in {𝒫}}\overline{p\cap A}=\bar{A}$. First we show that $\bar{A}\subset D$, which will follow from $A\subset D$ since then $\bar{A}\subset \bar{D}=D$. Suppose then that x is in A. Then x is in R and hence is in $\bar{p}$ for some p in ${𝒫}$. 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, $D\subset \bar{A}$. Thus, $D=\bar{A}$.

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

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

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

Theorem. Suppose $I_1,\dots ,I_r$ are pairwise-disjoint open rectangles, and $J_1,\dots ,J_s$ are open rectangles such that $\underset{i=1}{\overset{r}{\cup }}{I}_i\subset \underset{j=1}{\overset{s}{\cup }}{J}_j$. Then $v(I_1)+\cdots +v(I_r)\le v(J_1)+\cdots +v(J_s)$.

Proof: Let $R$ be a rectangle containing the $J_j$. Then by the second lemma there exists, for each j, a partition ${{ℛ}}_j$ of $R$ containing $J_j$. Let ${𝒫}$ be the common refinement of these partitions. Since ${𝒫}$ refines ${{ℛ}}_j$, for each p in ${𝒫}$ there exists an element of ${{ℛ}}_j$ 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 ${𝒫}$. Moreover, by the third lemma the elements of ${𝒫}$ contained in $J_j$ form a partition ${{𝒬}}_j$ of $J_j$, and

$\int {\chi }_J=\sum _{q\in {\cup }_j{{𝒬}}_j}v(q)\le \sum _{j=1}^s\sum _{q\in {{𝒬}}_j}v(q)=\sum _{j=1}^{s}v(J_j)$.

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 ${\chi }_I\le {\chi }_J$, it is also true that $\int {\chi }_I\le \int {\chi }_J$; that is,

$\sum _{i=1}^{r}v(I_i)=\int {\chi }_I\le \int {\chi }_J\le \sum _{j=1}^{s}v(J_j)$.

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 ${\chi }_J\le \sum _{j=1}^s{\chi }_{J_j}$, and since ${\chi }_I\le {\chi }_J$, it is also true that $\int {\chi }_I\le \int {\chi }_J$. Thus

$\sum _{i=1}^{r}v(I_i)=\sum _{i=1}^r\int {\chi }_{I_i}=\int \sum _{i=1}^r{\chi }_{I_i}=\int {\chi }_I\le \int {\chi }_J\le \int \sum _{j=1}^s{\chi }_{J_j}=\sum _{j=1}^s\int {\chi }_{J_j}=\sum _{j=1}^{s}v(J_j)$.