十讲ppt.ppt

1、Standard Problems of Numerical Linear Algebra,Linear systems of equations Solve Ax = b. Here A is a given n-by-n nonsingular real or complex matrix, b is a given column vector with n entries, and x is a column vector with n entries that we wish to compute. Least squares problems Compute the x that m

2、inimizes |Axb|2. Here A is m-by-n, b is m-by-1, x is n-by-1, and |y|2 is called the two-norm of the vector y.,Standard Problems of Numerical Linear Algebra,If m n so that we have more equations than unknowns, the system is called overdetermined. In this case we can not generally solve Ax = b exactly

3、. If m n, the system is called underdetermined, and we will have infinitely many solutions. Eigen value problems Given an n-by-n matrix A, find an n-by-1 nonzero vector x and a scalar l so that Ax = lx.,A set X is said to be endowed with a metric or a metric space structure or to be a metric space i

4、f a function d: X X R is exhibited satisfying the following conditions: a) d(x1 , x2) = 0 x1 = x2 , b) d(x1, x2) = d(x1, x2) (symmetry), c) d(x1, x3) d(x1, x2) + d(x2, x3) (the triangle inequality), where x1 , x2 , x3 are arbitrary elements of X.,【Definition 1】,设 X 是一个拓扑空间。称X是一个正规空间，如果X中的任何两个互不相交的闭子

5、集有互不相交的开邻域。,【定义2】,A normal space is a topological space X that satisfies every two disjoint closed subsets of X have disjoint open neighborhoods.,对正规空间X的任意两个互不相交的闭子集A、B，存在有一个连续映射 f : X0, 1，使得 当xA时，f(x) = 0，当xB时，f(x) = 1。,【乌里申引理】,【Urysohns Lemma】 For every two disjoint closed subsets A, B of a normal

6、 space X there exists a continuous mapping f : X0, 1 such that f(x) = 0 for xA and f(x) = 1 for xB.,【 Nested Intervals Theorem 】,What is the nested interval theorem? And please proof it in English.,For each n, let an , bn be a (nonempty) bounded interval of real numbers such that a1 , b1 a2 , b2 a3

7、, b3 and bn an 0. Then an , bn contains only one point.,Proof: Note that the sequences an and bn are respec-tively increasing and decreasing sequences, moreover both are bounded. Hence both converge, say ana and bnb .,【 Nested Intervals Theorem 】,For each n, let an , bn be a (nonempty) bounded inter

8、val of real numbers such that a1 , b1 a2 , b2 a3 , b3 and bn an 0. Then an , bn contains only one point.,Proof: Then ana and bbn for all nN. Since b a = lim (bn - an) = 0, a = b. Since anbn for all n we have aan , bn . Clearly if xa then x does not belong to an , bn.,【Definition 3】 Open and closed s

9、ets,A subset URn is said to be open if for every vector xU, a neighbourhood of x can be found, N(x, d) = yRn | | yx | d such that N(x, d)U. On the other hand, a set W is closed if and only if its complement in Rn is an open set.,【Definition 4】 Boundary of a set,A point x is said to be a boundary poi

10、nt of a subset URn if every neighbourhood of x contains at least one point of U and one point not belonging to U. The set of all boundary points of U, which is denoted by U, is called the boundary of U.,【Definition 5】 Interior and closure of a set,The interior of a set S is S - S . The closure of a

11、set S, denoted by , is the union of S and its boundary. An open set is equal to its interior. A closed set is equal to its closure.,【Proposition】,Every open subset V of R is a countable union of disjoint open intervals. Proof. For each xV , there is an open interval Ix with rational endpoints such t

12、hat xIxV. Then the collection IxxV is evidently countable and V =xV Ix .,Then (an , bn)nN is a disjoint collection of open intervals with union V.,Next, we prove it is always possible to have a disjoint collection. Since IxxV is a countable collection, we can enumerate the open intervals as I1 = (a1

13、, b1), I2 = (a2, b2), . For each nN, define,【Integrable Harmonic Functions on Rn】,Abstract A class of radial measure m on Rn is defined so that integrable harmonic functions f on Rn can be characterized as solutions of convolution equations f *m = f . In particular, it is proved that is harmonic if and only if n 9.,【Lagrange Interpolation on Conics and Cubics】,Abstract A bivariate polynomial interpolation problem for point lying on an al