notes from math classes

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diff --git a/li/02_solving b/li/02_solving
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+Ax = b where A = mxn, x = 1xm, b = 1xn. x_i and b_i are defined to be in
+the ith row of their corresponding vectors.
+
+If C is an invertible mxm matrix, CAx = Cb is equivalent to Ax = b.
+(1)
+    If x_0 satisfies Ax_0 = b, CAx_0 = Cb is satisfied.
+(2)
+    If x_1 satisfies CAx_0 = Cb, C^{-1} CAx_0 = C^{-1} Cb \to Ax_0 = b
+    is satisfied.
+
+(I believe this proof requires associativity)
+
+    LU Decomposition/Factorization
+
+Goal: Factorize A into a lower triangular and upper triangular matrices
+L and U respectively. (These are very non-unique)
+
+A = LU. Conventionally, L is mxm, and U is mxn.
+Ax = b ---> LUx = b ---> Y = Ux, LY = b.
+
+If we assert that L is an invertible matrix (e.g. ones down the
+diagonal), this is trivial to solve
+  1    0  0           y1       b1
+( c21  1  0 ... ) * ( y2 ) = ( b2 )
+  c31 c32 1           y3       b3
+      ...             ...      ...
+Y has a unique solution by construction. ( ex: y1 = b1, c21*y1 + y2 = b2 ...)
+
+U is upper triangular, meaning that it is strictly zero below the
+diagonal started at the top-left. (However, we have no control over the
+values of the diagonal or the values above the diagonal)
+
+[Note] How do we ensure that L and U retain the properties of the
+original matrix's solution? (i.e. not just matrices of zero)
+
+Using a series of row operations (written as invertible matrices C_1,
+C_2, ..., C_k -- that is, adding a multiple to a lower row) will turn A
+into echelon (upper triangular), and C_1^{-1}C_2^{-1}*...*C_k^{-1} = L
+(lower triangular). U is not reduced-echelon, but it is echelon/upper
+triangular.
diff --git a/zhilova/02_set_theory b/zhilova/02_set_theory
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+Review of Basic Set Theory: 1.2
+
+Sets are represented by capital letters like A, B, C.
+
+Sets can be like {1, 2, 3, ...}, (0, 1), reals, R^k, etc.
+
+    Operations
+
+A \cup B - union of A and B, refers to a set such that x \in A \cup B
+iff x\in A OR x\in B.
+
+A \cap B - intersection, refers to a set such that x \in A \cup B iff
+x\in A AND x\in B.
+
+A \setminus B (written as \) - difference, x\in A \setminus B iff x\in A
+and x\not\in B
+
+B^c = \overbar{B} - complement, equivalent to \Omega \setminus B or
+C \setminus B, where C or \Omega is the sample space of the problem.
+
+Universe ex: all possible combinations of two coin flips.
+
+Countable union:
+
+ +\infty               n
+   U     A_i    or     U
+  i=1                 i=1
+
+Represents union of a list of indexed sets.
+
+Countable intersection is analagous.
+
+UNcountable intersection/union is a union or intersection over an
+uncountable set like the reals or (0, 1).
+
+Symmetric difference: (A \union B) \setminus (A \cap B). (think of XOR)
+
+    DeMorgan's Laws (and proofs)
+
+(A \cap B)^c = A^c \cup B^c
+
+Let x \in (A\cap B)^c \to x \not\in A \cap B.
+
+If x\in A, x\not\in B \to x\in B^c. If x\in B, x\not\in A \to x\in A^c.
+Therefore, x\in A^C \cup B^c.
+
+(The proof the other way is similarly trivial)
+
+(A \cup B)^C = A^C \cap B^c
+
+These laws can be generalized to infinite unions (although not through
+induction, which merely proves it for arbitrary large, finite sets).
+
+Prove: Let B be the union of all A^c where A \in C. x\in B iff x \in
+(infinite intersection of all A where A \in C)^c
+
+
+[Proof omitted at present]
+
+    Distributive laws
+
+C_1 \cap (C_2 \cup C_3) = (C_1 \cap C_2) \cup (C_1 \cap C_3)
+C_1 \cup (C_2 \cap C_3) = (C_1 \cup C_2) \cap (C_1 \cup C_3)
+
+    Sample space and probability
+
+Example sample space is n ordered Bernoulli trials [finite]
+                     or (0,1) or (0,1)^2 or R      [continuous]
+
+C = { w_1, w_2, ... } where w_i is an elementary event. ??MUST?? be
+countable.
+
+If A \subseteq C, A is an event (random event) in C.
+
+    Probability set function
+
+Assign to each w_j a number p(w_j) \in [0, 1].
+
+\sum_{j=1}^\infty p(w_j) = 1.
+p_j := p(w_j) also known as random weights
+
+P(A) = \sum_{j : w_j \in A} p(w_j) [ the sum can be generalized to integral]
+
+    Properties of P(A)
+
+P : subsets of C --> [0, 1]
+
+1) P(C) = 1
+2) \forall A \subseteq C, 0 \leq P(A) \leq 1
+3) If A_1, A_2 \subseteq C and A_1 \cap A_2 = \empty,
+    P(A_1 \cup A_2) = P(A_1) + P(A_2)
+4) \forall A_1, A_2 \subseteq C,
+    P(A_1\cup A_2) = P(A_1) + P(A_2) - P(A_1 \cap A_2)
+5) \forall A \subseteq C
+    P(A) + P(A^c) = 1.
+6) P(\empty) = 0
+7) P(A) is nondecreasing: If A_1 \subseteq A_2 \subseteq C,
+P(A_1) \leq P(A_2)
+
+sigma-algebra backed by C is a set which contains C, and is closed under
+countable union and complement w.r.t C. (these are not unique, e.g.
+{C, \empty} and P(C) are both valid sigma-algebras).