User:Catfive/Scratchpad

Partial Fraction Decomposition
If $$n < m$$ and

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

$$ \left [ \frac{a_{1,1}}{x - \alpha_1} + \cdots + \frac{a_{1,r_1}}{(x - \alpha_1)^{r_1}}\right ] + \cdots + \left [ \frac{a_{k,1}}{x - \alpha_k} + \cdots + \frac{a_{k,r_k}}{(x - \alpha_k)^{r_k}} \right ] $$

$$ + \left [ \frac{b_{1,1}x + c_{1,1}}{x^2 + \beta_1x + \gamma_1} + \cdots + \frac{b_{1,s_1}x + c_{1,s_1}}{(x^2 + \beta_1x + \gamma_1)^{s_1}} \right ] + \cdots $$

$$ + \left [ \frac{b_{j,1}x + c_{j,1}}{x^2 + \beta_jx + \gamma_j} + \cdots + \frac{b_{j,s_j}x + c_{j,s_j}}{(x^2 + \beta_jx + \gamma_j)^{s_j}} \right ] $$.

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 &lambda; = 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 &lambda;(1,0) = 0, and similarly
 * $$(4) \quad \lim_{k\to 0} \left|g(0,1) - \lambda(0,1)\right| = 0$$ so that &lambda;(0,1) = 0, hence &lambda; = 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 &ne; 0, so
 * $$(9) \quad \left|g\left(\frac{x}{|x|}\right)\right| = 0$$

for all x &ne; 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 $$\empty\,$$ is an open rectangle. A closed rectangle is defined similarly by replacing &lt; with &le;. 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 $$v(\empty) = 0\,$$.

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
 * $$\sum_{i=1}^r v(I_i) = \sum_{i=1}^r \int \chi_{I_i} = \int \sum_{i=1}^r \chi_{I_i} = \int \chi_I \leq \int \chi_J \leq \int \sum_{j=1}^s \chi_{J_j} = \sum_{j=1}^s \int \chi_{J_j} = \sum_{j=1}^s v(J_j)$$.