User:Catfive/Scratchpad

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[edit] 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_1x + \gamma_1)^{s_1} \cdot \ldots \cdot (x^2 + \beta_jx + \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_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 ] .


[edit] The Laplace Expansion for Determinants

(moved to Laplace expansion article)

[edit] 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.


[edit] 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 < 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(\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).