数值分析(英文版)

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Chapter 0 Foundation of Algebra and Calculus

§0 .1 Foundation of Linear Algebra1, n-dimensional vector x=(x1,x2,…,xn), y=(y1,y2,…,yn). operations: x+y, x-y, cx. inner product of two vectors (x,y) length of a vector |x| a vector space V, a subspace of V linear independent of vectors

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2, Matrix

a11 a21 A am1

a12 a22 am 2

a1n a2 n amn

m rows, n columns operations: A+B, A-B, cA, AB, AT |A| (determinant of square matrix ) singular or nonsingular matrix A-1 inverse of a square matrix How to decide whether A is invertible or not? How to calculate A-1 ?2013-12-9 2

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(Continue) some special matrices (1) unit matrix (2) diagonal matrix (3) symmetric matrix (4) orthogonal matrix (5) triangular matrix (upper or lower) (6) elementary matrix 3, Eigenvalue and Eigenvector of matrix

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3, Eigenvalue and Eigenvector of matrix Eigenpolynomial of matrix Eigenvalue of matrix Eigenvector of matrix similar matrix diagonalization of matrix 4, Quadratic form quadratic form canonical form of matrix positive (negative) definite matrix half positive definite matrix2013-12-9 4

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Exercise 1, x=(2,1,4)T, y=(1,1,1) T, please show x+2y, 3x-4y, |x-y|, (2y,x). 2, 2 1 1 2 1 3 2 1 A 0 3 2 , B 0 4 ,C . 4 1 0 4 1 0 1 1

Please give AT, 2A, |A|, A-1, Rank(A), BC,AB-B.

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3, Please describe Cramer Rule about the linear system of equations. 4, Solve the following linear system of equations x1 2 x2 x3 3 4 x1 x2 3 x3 5 x 2 x 2 x 1 2 3 1and

2 x1 x2 x3 3 4 x1 x2 3 x3 5 x x 1 2 3

5,

1 3 4 3 2 4 2 1 A , B 0 1 3 , C 2 6 2 . 1 1 0 0 2 4 2 3 6

Please show all eigenvalues and eigenvectors of 2013-12-9 them.

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6,

1 3 0 A 3 6 1 . 0 1 2

Please change A into canonical form, and decide whether A is positive definite or not. 7, 1 a 0 A a 1 0 . 0 0 1

Please show the condition under which A is positive definite.2013-12-9 7

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§0.2 Foundation of Calculus1, derivative and differentiation derivative of elementary functions (sinx)’, (cosx)’,(xa)’,(lnx)’,(ax)’ derivative of composite function (sin2x+a2x)’ (tan(1+x2))’ (f(g(x)))’ high order derivative of functions the Taylor’s expansion of a function f(x)f ''( x*) f ( n ) ( x*) f ( x) f ( x*) f '( x*)( x x*) ( x x*) 2 ( x x*) n Rn ( x) 2! n!

show the Taylor’s expansion the following functions about the point x=0: sinx, 1/(1+x),ex2013-12-9 8

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§1 .2 Foundation of Calculus1, multivariables function and its partial derivative f ( x, y )

f ( x, y ) z=f(x,y), we can define x , y show the partial derivatives of the following functions. z=x2+y2, z=x2y2+sinxy, z=1/(x+y) The Taylor’s expansion of two variables function f ( x*, y*) f ( x*, y*) f ( x, y ) f ( x*, y*) ( x x*) ( y y*) R( x, y ) x y2013-12-9 9

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What is a Taylor series?Some examples of Taylor series which you must have seenx2 x4 x6 cos(x) 1 2! 4! 6!

x3 x5 x7 sin(x) x 3! 5! 7!x2 x3 e 1 x 2! 3!x

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General Taylor SeriesThe general form of the Taylor series is given by f x 2 f x 3 f x h f x f x h h h 2! 3!

provided that all derivatives of f(x) are continuous and exist in the interval [x,x+h]What does this mean in plain English?As Archimedes would have said, “Give me the value of the function at a single point, and the value of all (first, second, and so on) its derivatives at that single point, and I can give you the value of the function at any other point”2013-12-9 11

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Example—Taylor SeriesFind the value of f 6 given that f 4 125, f 4 74, f 4 30, f 4 6 and all other higher order derivatives of f x at x 4 are zero. Solution:h2 h3 f x h f x f x h f x f x 2! 3! x 4 h 6 4 2

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Example (cont.)Solution: (cont.) Since the higher order derivatives are zero,22 23 f 4 2 f 4 f 4 2 f 4 f 4 2! 3! 2 2 23 f 6 125 74 2 30 6 2! 3! 125 148 60 8

341

Note that to find f 6 exactly, we only need the value of the function and all its derivatives at some other point, x 4 in this case2013-12-9 13

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Derivation for Maclaurin Series for exDerive the Maclaurin seriesx2 x3 e 1 x 2! 3!x

The Maclaurin series is simply the Taylor series about the point x=0h2 h3 h4 h5 f x h f x f x h f x f x f x f x 2! 3! 4 5 h2 h3 h4 h5 f 0 h f 0 f 0 h f 0 f 0 f 0 f 0 2! 3! 4 5

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Derivation (cont.)Sincef ( x) e x , f ( x) e x , f ( x) e x , ... , f n ( x) e x

and

f n (0) e 0 1

the Maclaurin series is then(e 0 ) 2 (e 0 ) 3 f ( h ) (e ) (e ) h h h ... 2! 3! 1 1 1 h h 2 h3 ... 2! 3!0 0

So,x 2 x3 f ( x) 1 x ... 2! 3!2013-12-9 15

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Error in Taylor SeriesThe Taylor polynomial of order n of a function f(x) wi

th (n+1) continuous derivatives in the domain [x,x+h] is given byh2 hn n f x h f x f x h f ' ' x f x Rn x 2! n!

where the remainder is given by x h n 1 f n 1 c R x n

(n 1)!

wherex c x h

that is, c is some point in the domain [x,x+h]2013-12-9 16

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Example—error in Taylor seriesThe Taylor series forx

e x at point x 0 is given by

x 2 x3 x 4 x5 e 1 x 2! 3! 4! 5!

It can be seen that as the number of terms used increases, the error bound decreases and hence a better estimate of the function can be found. How many terms would it require to get an approximation of e1 within a magnitude of true error of less than 10-6.

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Example—(cont.)Solution: Using n 1 terms of Taylor series gives error x h n 1 f n 1 c x 0, h 1, f ( x) e x Rn x n 1 ! bound of

0 1 n 1 f n 1 c Rn 0 n 1 ! 1 n 1 e c n 1 ! Sincex c x h 0 c 0 1 0 c 11 e Rn 0 (n 1)! (n 1)!18

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Example—(cont.)Solution: (cont.) So if we want to find out how many terms it would 1 require to get an approximation of e within a 10 6 , magnitude of true error of less thane 10 6 ( n 1)!

(n 1)! 10 6 e(n 1)! 10 6 3

n 9

So 9 terms or more are needed to get a true error less than 10 62013-12-9 19

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