1 / 47

Numerical Mathematic

Numerical Mathematic. Wai Hung, Chan Thomas, Nguyen. Numerical Mathematic. Number Representation Round off errors Overflow and Underflow Classical Numerical Algorithms Linear Algebra. Number and their representation. ASCII – text characters Easy read and write of numbers Binary

nadine
Télécharger la présentation

Numerical Mathematic

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Numerical Mathematic • Wai Hung, Chan • Thomas, Nguyen

  2. Numerical Mathematic Number Representation Round off errors Overflow and Underflow Classical Numerical Algorithms Linear Algebra

  3. Number and their representation ASCII – text characters Easy read and write of numbers Binary Natural form of computer

  4. Binary Numbering System • The binary number system is similar to the decimal number system except that • All values are composed of 0’s and 1’s (instead of 0-9) • Each position in a number represents a power of 2 (instead of a power of 10) • Decimal: 729 – 7 in the 100’s position and 2 in the 10’s position and 9 in the 1’s position • Binary: 1101 – 1 in the 8’s position and 1 in the 4’s position and 0 in the 2’s position and 1 in the 1’s position

  5. Positive Integer Representations • Convert a binary value to its equivalent decimal value: • Examples: 011010102 0*27 + 1*26 + 1*25 + 0*24 + 1*23 + 0*22 + 1*21 + 0*20 = 64 + 32 + 8 + 2 = 10610

  6. Number and their representation (Cont) Binary number (base 2) • Things may become complicated • Numbers are finite (overflow) • Fraction and real numbers • Negative numbers • How do we represent negative number?

  7. Representing Negative Integers • Signed Magnitude • Add 1 bit to the number at the leading end, this will be the sign bit • Positive numbers sign bit = 0 • Negative numbers sign bit = 1 • Examples: • 34 = 00100010 • -34 = 10100010

  8. Representing Negative Integers (Cont) • Two’s Complement • Improvement over Signed Magnitude because it doesn’t have either of the problems • Representation is the same for positive numbers • For negative numbers, negate the number and then add 1 • Example: • 42  00101010 • -42  11010101 + 1 = 11010110 • Notice that the leading bit is still a sign bit • +3 = 00000011 -3 = 11111101 • +2 = 00000010 -2 = 11111110 • +1 = 00000001 -1 = 11111111

  9. Working with 2’s complement • Negate a number • Invert every single bit (0  1, 10) • Add 1 to the result • Example 0000 0010 =2 1111 1101 inverted 1111 1110 1 added =-2 0000 0001 inverted 0000 0010 1 added =2

  10. Round off errors • Any computer can only retain a finite number of significant digit to represent the results of an operation. When an result can not be represent exactly, a round off error introduced. • These are the errors the computer make in doing arithmetic. (For example, the error a computer or calculator makes in evaluating (1/3 + 1/7)). Even if we have a good formula to solve a problem it may not produce good answers when it evaluated on a computer.

  11. Round off errors (Cont) • Computers make small error when they do arithmetic. For example: 11 * (15/11) -15 = 0 (?)

  12. Overflow and Underflow • Overflow: If the result of a computation is larger than that allowed by the computer you have an overflow. • Underflow: In computing, a condition occurring when a machine calculation produces a non-zero result that is smaller than the smallest non-zero quantity that the machine's storage unit is capable of storing or representing.

  13. Example: 58 – 83 = -25 • Convert 58 = 0011 1010 83 = 0101 0011 • Get 2’s complement 0101 0011 Original number (83) 1010 1100 Flip the bits +1 Add 1 1010 1101 2’s complement • Add the two numbers: 0011 1010 =58 + 1010 1101 2’s complement of 83 1110 0111 Answer = -25 (No Overflow)

  14. Overflow (Cont) • The sum of two unsigned numbers can exceed any representation 0101 0011 = 83 + 0010 1111 = 47 1000 0010 = -126 (Overflow)

  15. Detecting Overflow • No overflow when adding a +ve and a -ve number. • No overflow when sign are the same for subtraction. • Overflow when adding two positive yields a negative • Or, adding two negative give a positive • Or, subtract a negative from a positive and get a negative • Or, subtract a positive from a negative and get a positive

  16. Overflow (Cont) General Overflow Condition

  17. Sources • http://wwwmaths.anu.edu.au/DoM/secondyear/MATH2501/lect-05-4.pdf • http://www.maths.uq.edu.au/~gac/math2200/mn_roff.pdf • http://lapwww.epfl.ch/courses/archord1/Computer%20Arithmetic.pdf • http://www.cse.psu.edu/~cg575/lectures/cse575-fpops.pdf

  18. CLASSICAL NUMERICAL ALGORITHM • TRAPEZOID RULE: •  (a to b) (x) dx  Tn = x/2 [(Xo) + 2 (X1) + 2(X2) +….+ 2 (Xn -1) + 2 (Xn) • where x = (b - a) /n and Xi = a + i x. • example : Use trapezoid rule with n=5 to approximate integral (1 to 2) (1/x) dx • with n=5, a =1, and b=2, we have x =(2-1)/5=0.2 • (1to 2) 1/x dx T5 = .2/2[(1)+ 2(1.2)+….. (2) • = .1[1/1 + 2/1.2 + 2/1.4+ 2/1.6 +..+1/2] •  0.695635

  19. MIDPOINT RULE: •  (a to b) (x) dx  Mn = x [(X1) + (X2) + • +….+ (Xn) where • x = (b - a) /n and Xi =1/2(Xi-1+Xi) = midpoint[Xi-1, Xi] • example : Use midpoint rule with n=5 to approximate integral (1 to 2) (1/x) dx • with n=5, a =1, and b=2, we have x =(2-1)/5=0.2 • (1to 2) 1/x dx = 1/5 [(1.1)+ 2(1.3)+….. (1.9) • = 0.2[1/1.1 + 2/1.3 + 2/1.4+ 2/1.5 +..+1/1.9] •  0.691908

  20. SIMPSON’S RULE •  (a to b) (x) dx  Sn = x/3 [(Xo) + 4 (X1) + 2(X2) + 4(X3) +2(Xn -2) + 4(Xn -1)+(Xn)] • where n is even and x = (b - a) / n • Example: use simpson’s rule to approximate • (1 to 2) (1/x) dx with n=10. • We have x =1/10=0.1 • (1 to 2) (1/x) dx  S10 = x/3 [(1)+4(1.1)+ 2(1.2)+ 4(1.3)+…..+ 2(1.8)+4(1.9)+ (2) • = 0.1/3 [1/1 + 4/1.1 + 2/1/2 + 4/1.3 + 2/1.4 + 4/1.5 + 2/1.6 + 4/1.7 + 2/1.8 + 4/1.9 + 1/2] •  0.0.693135

  21. LINEAR EQUATION 1. Introduction to linear equations • A linear equation in n unknowns x1; x2; … xn is an equation of the form: • a1x1 + a2x2 + + anxn = b; • where a1; a2; : : : ; an= b are given real numbers. • For example, with x and y instead of x1 and x2, the linear equation 2x + 3y = 6 describes the line passing through the points (3; 0) and (0; 2). • A system of m linear equations in n unknowns x1; x2; ; xn is a family of linear equations a11 x1 + a12 x2 + .... + a1n xn = b1 a21 x1 + a22 x2 + …. + a2n xn = b2 ... am1 x1 + am2 x2 + .. + amn xn = bm:

  22. INTRO(CONT) • Note that the above system can be written concisely as j=1 • Σ (i=1 to m) aij xj = bi; i = 1; 2; ; m: The matrix: a11 a12 a1n a21 a22 a2n ... am1 am2 amn • is called the coefficient matrix of the system, while the matrix a11 a12 a1n b1 a21 a22 a2n b2 … .. am1 am2 amn bm is called the augmented matrix of the system.

  23. EXAMPLE : Find a polynomial of the form y = a0+a1x+a2x^2+a3x^3which passes through the points (-3, -2), (-1, -2), (2, 1). • Solution. When x has the values -3;-1; 1; 2, then y takes corresponding values -2; 2; 5; 1 and we get four equations in the unknowns a0; a1; a2; a3. a0 -3a1 + 9a2-27a3 = -2 a0 - a1 + a2 -a3 = 2 a0 + a1 + a2 + a3 = 5 a0 + 2a1 + 4a2 + 8a3 = 1: • This system has the unique solution a0 = 93/20; a1 = 221/120; a2 = -23/20; a3 = -41/120. So the required polynomial is: y = 93/20 +221/20x –23/20x^2-41/20x^3

  24. Solving linear equations • DEFINITION : (Row echelon form) A matrix is in row echelon form if • (i) all zero rows (if any) are at the bottom of the matrix (ii) if two successive rows are non zero, the second row starts with more zeros than the first (moving from left to right). For example: the matrix is in row echelon form 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 The matrix is not in row echelon form 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 The zero matrix of any size is always in row echelon form.

  25. DEFINITION: (Reduced row echelon form) A matrix is in reduced row echelon form if (i). it is in row echelon form(ii). the leading (leftmost nonzero) entry in each non zero row is 1, (iii). all other elements of the column in which the leading entry 1 occurs are zeros. • For example: • 1 0 • 0 1 • 0 1 2 0 0 2 • 0 0 0 1 0 3 0 0 0 0 1 4 0 0 0 0 0 0

  26. DEFINITION (Elementary row operations) There are three types of elementary row operations that can be performed on matrices: 1. Interchanging two rows: Ri <->Rj interchanges rows i and j. 2. Multiplying a row by a nonzero scalar: Ri ->t Ri multiplies row i by the nonzero scalar t. 3. Adding a multiple of one row to another row:Rj -> Rj + tRi adds t times row i to row j. • 1 2 0 1 2 0 • A = 2 1 1 R2-->R2 +2R3 4 -1 5 • 1 -1 -2 1 -1 2 • 1 2 0 2 4 0 • R2<-->R3 1 -1 2 R1-->2R1 1 -1 2 =B • 4 -1 5 4 -1 5

  27. The Gauss-Jordan algorithm • This is a process which starts with a given matrix Aand produces a matrix Bin reduced row echelon form, which is row equivalent to A. If A is the augmented matrix of a system of linear equations, then Bwill be a much simpler matrix than A • STEP 1.:Find the first nonzero column moving from left to right, (column c1) and select a non zero entry from this column. By interchanging rows, if necessary, ensure that the first entry in this column is nonzero. Multiply row 1 by the multiplicative inverse of a1c1 thereby converting a1c1 to 1. For each non zero element aic1 ; i > 1, (if any) in column c1, add -aic1,time row 1 to row I, thereby, we can find all element in column c1 is apart from the first zero. • STEP 2. : If the matrix obtained at Step 1 has its 2nd; : : : ;mth rows all zero, the matrix is in reduced row echelon form. Otherwise suppose that the rst column which has a non zero element in the rows below the rst is column c2. Then c1 < c2. By interchanging rows below the first, if necessary, ensure that a2c2 is non{zero. Then convert a2c2 to 1 and by adding suitable multiples of row 2 to the remaning rows, where necessary, ensure that all remaining elements in column c2 are zero.

  28. EXAMPLE • 0 0 4 0 2 2 -2 5 1 1 -1 5/2 • 2 2 -2 5 0 0 4 0 0 0 4 0 • 5 5 -1 5 R1 <->R2 5 5 -1 5 R1->-1/2 R1 5 5 -1 5 • 1 1 -1 5/2 1 1 -1 5/2 • R3 ->R3-5R1 0 0 4 2 R2->1/4R2 0 0 1 0 • 0 0 4 -15/2 0 0 4 -15/2 • R1->R1 + R2 1 1 0 5/2 1 1 0 5/2 • R3->R3- 4R2 0 0 1 0 R3->-2/15R3 0 0 1 0 • 0 0 0 -15/2 0 0 0 1 • 1 1 0 0 • R1 -> R1- 5/2R3 0 0 1 0 • 0 0 0 1 • The last matrix is in reduced row echelon form.

  29. MATRICES • Matrix arithmetic • Matrices will usually be denoted by capital letters and the equation A = [aij ] means that the element in the ith row and jth column of the matrix A equals aij . It is also occasionally convenient to write aij = (A)ij . For the present all matrices will have rational entries, unless otherwise stated. • EXAMPLE 2.1.1 The formula aij = 1/(i + j) for 1<=i<=3; 1<= j<=4, defines a 3x4 matrix A = [aij ], namely • 1/2 1/3 1/4 1/5 • A = 1/3 1/4 1/5 1/6 • 1/4 1/5 1/6 1/7

  30. DEFINITION:(Equality of matrices) Matrices A and B are said to be equal if A and B have the same size and corresponding elements are equal; that is A and B є Mmxn(F) and A = [aij ]; B = [bij ], with aij = bij for 1<= i <= m; 1<= j<= n. • DEFINITION:(Addition of matrices) Let A = [aij] and B =[bij] be of the same size. Then A + B is the matrix obtained by adding corresponding elements of A and B; that is A + B = [aij] + [bij] = [aij+ bij]. • DEFINITION :(Scalar multiple of a matrix) Let A = [aij ] and t є F (that is t is a Scalar). Then tA is the matrix obtained by multiplying all elements of A by t; that is tA = t[aij ] = [taij ].

  31. DEFINITION 2.1.4 (Additive inverse of a matrix) Let A = [aij ] .Then ¡A is the matrix obtained by replacing the elements of A by their additive inverses; that is A = -[aij ] = [-aij ]: • DEFINITION 2.1.5 (Subtraction of matrices) Matrix subtraction is defined for two matrices A = [aij ] and B = [bij ] of the same size, in the usual way; that is A - B = [aij ] -[bij ] = [aij -bij ]: • DEFINITION 2.1.6 (The zero matrix) For each m; n the matrix inMmxn(F), all of whose elements are zero, is called the zero matrix (of size mxn) and is denoted by the symbol 0.

  32. The matrix operations of addition, scalar multiplication, additive inverse and subtraction satisfy the usual laws of arithmetic. • 1. (A + B) + C = A + (B + C); • 2. A + B = B + A; • 3. 0 + A = A; • 4. A + (-A) = 0; • 5. (s + t)A = sA + tA, (s x t)A = sA xtA; • 6. t(A + B) = tA + tB, t(A +B) = tA +tB; • 7. s(tA) = (st)A; • 8. 1A = A, 0A = 0, (-1)A = -A; • 9. tA = 0 => t = 0 or A = 0.

  33. DEFINITION: (Matrix product) Let A = [aij ] be a matrix of size m x n and B = [bjk] be a matrix of size n x p; (that is the number of columns of A equals the number of rows of B). Then AB is the m x p matrix C = [cik] whose (i, k)th element is defined by the formula: • cik = Σ(j=1 to n) aijbjk = ai1*b1k + .. + ain*bnk. • Example: • 1 2 5 6 = 1*5 +2*7 1*6 +2*8 = 19 22 • 3 4 7 8 385 +4*7 3*6 +4*8 43 50 • 5 6 1 2 23 34 1 2 5 6 • 7 8 3 4 = 41 46 = 3 4 7 8

  34. Matrix product (cont) • Matrix multiplication obeys many of the familiar laws of arithmetic apart from the commutative law. • 1. (AB)C = A(BC) if A; B; C are mxn; nxp; pxq, respectively; • 2. t(AB) = (tA)B = A(tB), A(-B) = (-A)B = -(AB); • 3. (A + B)C = AC + BC if A and B are mxn and C is nxp; • 4. D(A + B) = DA + DB if A and B are mxn and D is pxm.

  35. Example: • Let A, B, C, D be matrices defined by • A = 3 0 B = 1 5 2 C = -3 -1 • -1 2 -1 1 0 2 1 • 1 1 -4 1 3 4 3 • C = 4 -1 • 2 0 • Which of the following matrices are defined? • A + B; A + C; AB; BA; CD; DC; D^2:

  36. THEOREM (Cramer's rule for 2 equations in 2 unknowns) • The system: ax + by = e • cx + dy = f a b • has a unique solution if Δ = c d = 0 namely • X= Δ1/Δ, Y= Δ2/Δ where Δ1 = e b , Δ2 = a e • f d c f EXAMPLE : the system 7x + 8y = 100 • 2x - 9y = 10 • Δ= 7 8 Δ1= 100 8 Δ2= 7 100 • 2 -9 = -79 10 9 =-980 2 10 =-130 So the system has unique solution • X = 980 / 79 , Y = 130 / 79

  37. DETERMINANTS • DEFINITION: If A = a11 a12 • a21 a22 ¸ we define the determinant of A, (also denoted by det A,) to be the scalar det A = a11*a22 - a12*a21: • DEFINITION: (Minor) Let Mij(A) (or simply Mij if there is no ambiguity) denote the determinant of the (n -1) and (n - 1) sub matrix of A formed by deleting the ith row and jth column of A. (Mij(A) is called the (i, j) minor of A.) • The determinant function has been defined for matrices of size (n-1)x(n-1). Then det A is defined by the so called first-row Laplace expansion.

  38. detA = a11M11(A) - a12 M12(A) + ……. + • (-1)^1+n M1n(A) • = Σ(j=1 to n) (-1)^1+j a1j M1j(A): • For example: if A = [aij ] is a 3 x 3 matrix, the Laplace expansion gives: • detA = a11 M11(A) - a12 M12(A) + a13 M13(A) • = a11(a22a33 - a23a32) - a12(a21a33- a23a31) + a13(a21a32 - a22a31) • = a11a22a33 - a11a23a32 - a12a21a33 + a12a23a31 + a13a21a32 - a13a22a31

  39. THEOREM: Let A = [aij ], where aij = 0 if i < j. Then detA = a11*a22*…*ann, an important special case is when A is a diagonal matrix. • a11 0 0 0 ... 0 • a21 a22 …...….0 • det A = … a33 0 • an1 an2 ……ann • det A = a11 (a22………ann) • 1 0 0 0 • det A = 3 3 0 0 = 18 • 4 3 3 0 • 1 3 4 2

  40. THEOREM: detA = Σ (j=1 to n) (-1)^i+j aij Mij(A)for i= 1,.,n(the so-called i-th row expansion) and detA = Σ(i=1to n)(-1)^i+j aij Mij(A)for j= 1,...,n(the so-called j-th column expansion). • The expression (-1)i+j obeys the chess board pattern of signs: • + - + - ……. • - + - +……. • + - + - …… • . • .

  41. DEFINITION: (Cofactor) The (i, j) cofactor of A, denoted by Cij(A) (or Cij if there is no ambiguity) is defined by Cij(A) = (-1)^i+j Mij(A). • DEFINITION: (Adjoint) If A = [aij ] is an nxn matrix, the ad-joint of A, denoted by adjA, is the transpose of the matrix of cofactors. • C11 C21 ….. Cn1 • adj A = C12 C22 …. Cn2 • .. ….. • C1n C2n …..Cnn • THEOREM: Let A be an n x n matrix. Then • A(adjA) = (adjA)A.

  42. If detA = 0, then A is nonsingular and • A^-1 = (1/ detA) adj A. • Example: 1 2 3 • det A = 4 5 6 = -3 = 0 • 7 8 9 • C11 C21 C31 • A^-1 = - 1/3 C12 C22 C32 • C13 C23 C33 • -3 6 -3 • A^-1 = -1/3 12 -15 6 • -8 8 -3

  43. EIGENVALUES AND EIGENVECTORS • Motivation: • We motivate the chapter on eigenvalues by discussing the equation: ax^2 + 2hxy + by^2 = c • where not all of a; h; b are zero. The expression ax2 + 2hxy + by2 is called a quadratic form in x and y and we have the identity • ax^2 + 2hxy + by^2 = x y a h x = X^t AX • h b y • where X = x a h • y and A = h b . A is called the matrix of the quadratic form.

  44. A has characteristic equation: • λ² - (a+b) λ + ab - h² = 0 • this called eigenvalue equation of the matrix A. • DEFINITION (Eigenvalue, eigenvector) • Let A be a complex square matrix. Then if ¸λ is a complex number and X a non-zero complex column vector satisfying AX = λX, we call X an • eigenvector of A, while ¸ λ is called an eigenvalue of A. • if ¸ λ is an eigenvalue of an nxn matrix A, with • corresponding eigenvector X, then (A -λIn)X = 0, with X = 0, so det (A - λIn) = 0 and there are at most n distinct eigenvalues of A.

  45. EXAMPLE: Find the eigenvalues and eigen- • vectors of A = 2 1 • 1 2 • Solution. The characteristic equation of A is • λ² - 4 λ + 3 = 0, or (λ -1)(λ -3) = 0 • Hence λ = 1 or 3. The eigenvector equation (A - λ In)X = 0 reduces to 2-λ 1 x 0 • 1 2- λ y = 0 • or (2 - λ)x + y = 0 • x+ (2 - λ)y= 0: • Taking λ =1 give x + y = 0 • x + y = 0

  46. Which has solution x = -y, y arbitrary.Consequently the eigenvectors corre-sponding • to λ = 1 are the vectors -y with y = 0 • y • Taking λ = 3 give -x + y = 0 • x + y = 0 • which has solution x = y, yarbitrary. Consequently the eigenvectors corre-sponding to • λ = 3 are the vectors y with y = 0 • y

  47. Reference sources • www.maths.uq.edu.au/~krm/ela.html

More Related