CS321 Introduction To Numerical Methods

Transcription

1 CS3 Introduction To Numerical Methods Fuhua (Frank) Cheng Department of Computer Science University of Kentucky Lexington KY

2 - - Table of Contents Errors and Number Representations 3 Error Types 3 Review of Taylor Series 3 3 Numbers on Computers 4 3 Representation of Numbers 4 3 Conversion 5 33 Floating-Point Number System 8 34 Arithmetic Operations 35 Round-off Errors 3 36 Control of Round-Off Errors 8 Numerical Linear Algebra System of Linear Equations Gaussian Elimination without Pivoting Gaussian Elimination with Partial Pivoting 5 3 Errors 8 4 Iterative Improvement 33 3 Eigenvalues and Eigenvectors 34 3 The Power Method 35 3 The QR Algorithm 38 4 Polynomial Interpolation 44 4 Lagrange Form 46 4 Newton Form 49 5 Solution of Nonlinear Equations 53 5 Solving a Single Nonlinear Equation 53 5 The Special Case of Polynomials 56 6 Numerical Integration 6 7 Piecewise Polynomial (Spline) Interpolation 64 7 Hermite Interpolation 64 7 Spline Interpolation 66 8 Numerical Differentiation 7 9 References 75 Index 76

3 - 3 - Error Types Errors and Number Representations Two major errors exist in numerical computations: truncation error and round-off error Truncation error is caused by the approximations used in the mathematical formula of the scheme For instance, when we use finite terms of the Taylor series to approximate a function, truncation error occurs Round-off error is caused by the finite-precision nature of a computer The number of digits that can be used to represent a number in a computer is finite In a lot of cases, we simply can not get a precise representation for a number in a computer Consequently, round-off error occurs To study these errors, one needs to know how Taylor series are used to derive numerical schemes and how numbers are stored and operated in a computer Review of Taylor Series A function f (x ) is said to be analytic at x a if f (x ) can be represented by a power series in powers of (x a ) within a radius of convergence, D > x a > The power series is given by f (x ) f (a ) + (x a )*f (a ) + (x a ) *f (a ) +! i (x a ) i *f (i ) (a ) () i! The Taylor series of a function about x is called the Maclaurin series The following are some familiar examples of Taylor series: e x x + x x +! 3! x 3 + x 5 + sinx x 3! 5! lnx (x ) i x + x 4 + cosx! 4! i + x + x + x 3 + x (x ) + i x i ( a, D ) () i! ( ) i x i + ( a, D ) (3) (i +)! ( ) i x i ( a, D ) (4) (i )! i (x ) 3 3 i x i ( a, D ) (5) ( ) i (x ) i ( a, D ) (6) i Usually, one uses finite terms in the Taylor series to approximate a function This causes truncation error For instance, in (), if we use the first (n +) terms to approximate f (x ), we get an error term E n + defined as follows:

4 - 4 - E n + i n + The following theorem is useful in getting an estimate for E n + (x a ) i *f (i ) (a ) (7) Theorem (Taylor s Theorem) If the function f (x ) possesses continuous derivatives of order,,, (n +) in an open interval I (c, d ), then for any a in I, where x is any value in I and f (x ) n i i! i! (x a ) i *f (i ) (a ) + R n + (8) (x a ) n + R n + *f (n +) (ξ) (9) (n +)! for some ξ between a and x R n + is called the remainder Theorem shows that estimating the truncation error (7) can be achieved by estimating (9) as far as one can get an upper bound on the value of f (n +) (ξ) For instance, in (), if we use the first 5 terms to approximate the value of e x at x, the truncation error E 5 would be e + () + () + () () 4 4 () 5 E 5 *e ξ for some ξ between and By taking an upper bound for e ξ in the above equation for E 5, say 3, one gets the following estimate for the truncation error E Numbers Used in Computers 3 Representation of Numbers The number system that we are most familiar with is the so called decimal number system A positive number in a decimal number system can always be expressed as x (a n a n a b b )

5 - 5 - a n * n + a n * n + a * + b * + b * + () n i a i * i + i b i * i where a i and b i are integers between and 9 a n * n + a n * n + a * is called the integer part of x and b * + b * + is called the fractional part of x The fractional part could contain infinitely many terms is called the base of the decimal number system A negative decimal number is a positive decimal number with a negative sign in front of it However, it is not necessary to use as a base Any positive integer different from can be used as a base Actually, sometimes it is necessary to use positive integers different from as bases An obvious example is the base number system used in computers In general, if β > is chosen as a base, then a number x in the base β number system is expressed as x (a n a n a b b )β a n * β n + a n * β n + a * β + b * β + b * β + () n i a i * β i + i b i * β i where a i and b i are integers between and β Popular bases include, 8, and 6 (see Table ) When β is bigger than, it is usually convenient to use non-integer symbols to represent those a i and b i which are bigger than 9 For instance, when β 6, we use A, B, C, D, E and F to represent,,, 3, 4 and 5, respectively (see Table ) Examples of numbers in different number systems are shown below (57), (574), (), () (75) 8, (57) 8, (FC ) 6, (A 9C 75E ) 6 Table Popular number systems Base Number system Symbols binary, 8 octal,,,7 decimal,,,9 6 hexadecimal,,,9,a,b,c,d,e,f 3 Conversion Converting a number from one representation to another is necessary when the number is to be processed in a representation different from its input form The conversion is done for its integer part and fractional part separately

6 - 6 - To convert the integer part of a decimal number to a base β ( ) representation, simply divide the integer part repeatedly by β and take the remainders The repeated division stops when the quotient becomes zero For example, to convert (37) to a binary integer, we divide (37) by repeatedly until the quotient becomes zero The least significant digit in the binary representation is the first remainder, the second least significant digit is the second remainder,, etc Example (37) (??? ) Hence (37) () The above process follows from the observation that if (37) is expressed in the following binary form then by dividing both sides by, one gets (37) a n * n + a n * n + + a * + a * (37) an * n + a n * n + + a * + a q (a n * n + a n * n + + a * ) is an integer and a / is a fraction Therefore, q has to be the quotient and a has to be the remainder Similarly, by dividing q by again, one gets (a n * n + a n * n 3 + a * ) as the quotient and a as the quotient, etc To convert a base β ( ) integer to a decimal integer, simply evaluate its base β representation in the decimal number system Example (53) 8 (??? ) (53) 8 (5* 8 + 3* 8 + ) (346) In general, if (a n a n a a is given, its decimal equivalent can be evaluated as follows:

7 - 7 - t : a n ; for i : n downto do t : t* β + a i ; To convert the fractional part of a decimal number to a base β ( ) representation, simply multiply the fractional part repeatedly by β and take the integer parts This process stops when the fractional part becomes zero However, the fractional part might never become zero In that case, the fractional part of the base β representation has infinitely many terms Example (75) (??? ) 75 * 5 5 * Hence (75) () Example () (??? ) * * 4 4 * 8 8 * 6 6 * * 4 4 * 8 8 * 6 6 * Hence () ( ) (Note that the process repeats itself after the 5th step) To convert the fractional part of a base β ( ) representation to a decimal representation, one can use a similar approach, ie repeatedly multiply the fractional part by and take the integer parts The multiplication has to be performed in the base β number system ( has to be converted to a base β representation) The integer part of each product is converted to a decimal integer first before being used in the decimal form Example () (??? ) * * Hence () (75)

8 - 8 - If the fractional part f in a base β representation has finite terms, say f (b b b n, then it can also be converted to a decimal form using a nested loop as follows s : b n /β ; for i : n downto do s : (b i + s )/β ; Example () (??? ) s 4 5 s 3 ( + 5)/ 5 s ( + 5)/ 65 s ( + 65)/ 85 Hence () (85)

9 Floating-Point Number System Numbers used in scientific computation are usually assigned a unit of memory called a word or a double word Each word consists of fixed number of digits For example, a word in a typical PC consists of 3 binary digits (bits) and a double word consists of 64 binary digits Numbers are usually stored in these digits in two modes: fixed-point (or, integer) number mode, and floating-point (or, real) number mode We will focus on the floating-point number mode only This is the representation scheme used by most computers (because it generates smaller relative errors, as will be seen later) A floating-point number system is characterized by 4 parameters (integers): β number base t number of digits L, U lower and upper bounds of the exponent and is denoted by F (β, t, L, U ) Each number in F (β, t, L, U ) is called a floating-point number and is expressed as ± (d d d t * β e with L e U and d i β but d β e is called the exponent and (± d d d t is called the mantissa The first digit of the mantissa is always non-zero, with one exception, is also a number of the floating-point number system but is expressed as +( )* β L with t zeros in the mantissa F (β, t, L, U ) contains (U L +)(β )β t + floating-point numbers The numbers are not uniformly spaced Since a floating-point number system contains only finite floating-point numbers, not every decimal number can be precisely represented by a floating-point number Given a decimal number x, if it can not be precisely represented by a floating-point number, a floating-point number closest to x is selected to represent x in the floating-point number system The process of replacing x by its nearest floating-point number is called rounding This floating-point number is called the floating-point representation of x in F (β, t, L, U ) and is denoted by f l (x ) The difference between x and f l (x ) is called the round-off error round-off error x f l (x ) The round-off error is an absolute error The relative round-off error is defined by relative round-off error x f l (x ) / x

10 - - Example F (,,, ) contains 3 floating-point numbers ± () * ± 3/ ± () * ± ± () * ± 3/4 ± () * ± / ± () * ± 3/8 ± () * ± /4 ± () * 3/8 3/8 3/ 3/4 / /4 /4 / 3/4 3/ Example If x (4) then f l (x ) in F (,,, ) is f l (x ) +() * /4 round-off error x f l (x ) relative round-off error x f l (x ) / x 466 5% Example If x (86554) then f l (x ) in F (, 4,, 3) is f l (x ) +(8655) * 3 round-off error x f l (x ) 4 relative round-off error x f l (x ) / x 464 5% Usually, floating-point number systems produce (relatively) large round-off error but smaller relative round-off error Smaller relative round-off errors are preferred than smaller round-off errors because the smaller the relative round-off error, the more the correct digits in f l (x ) The relationship between the number of correct digits in f l (x ) and the relative round-off error is stated in the following theorem Theorem Let f l (x ) be the floating-point representation of x in the floating-point number system F (β, t, L, U ) If there are m ( t ) zeros between the decimal point and the first non-zero digit in the relative round-off error of f l (x ), ie, x f l (x ) / x (r r with r i ; i,, m, but r m + then the number of correct digits in f l (x ) (starting from the left-most digit) is at least m Proof Let x (d d * β e and f l (x ) (g g g t * β e If the first n digits of f l (x ) are correct, ie, d i g i, i,, n, then

11 - - x f l (x ) ( h h * β e with n zeros between the decimal point and h Hence, Since x f l (x ) x (h h * β e n (d d * β e (h h * β n (d d ( (d d ( it follows that (h h * β n x f l (x ) (h h * β n x or ( h h x x f l (x ) ( h h with n zeros between the decimal point and h in the left term and n zeros between the decimal point and h in the right term That is, if the number of correct digits in f l (x ) is n then the number of zeros between the decimal points and the first non-zero digit in the relative round-off error is either n or n Therefore, if the number of zeros between the decimal point and the first non-zero digit in the relative round-off error is m then the correct digits in f l (x ) is either m or m + 34 Arithmetic Operations We assume a floating-point number system F (β, t, L, U ) is used in our computer Note that the way a floating-point number is stored in a word defines a unique floating-point number system For example, a floating-point number in a PC is stored in a word (single precision) as follows 8 3 sign bit exponent mantissa (8 bits) (3 bits) Therefore, the floating-point number system F (, 3, 8, 7) is used in a PC If a double word is used, the second word is usually used as a supplementary space for the mantissa Some computers use biased exponent For example, 7 bits can represent unsigned integers in the range 7 ( exponents 64 63); the number stored is the true exponent plus 64

12 - - Given a number x, the process of finding its floating-point representation in F (β, t, L, U ) is performed as follows The number is normalized, ie, finding the following representation, first x (a a a t a t + * β e ; a We shall assume that L e U The fraction is then replaced with a t -digit fraction by one of the following methods chopping : digits beyond a t are dropped rounding : taking the first t digits of (a a at at + + β/) If e < L or e > U then we say an underflow or overflow has occurred An overflow is a fatal error Generally, when an overflow occurs, the execution of the program stops If an underflow occurs, usually the computer simply sets x to zero without interrupting the program Example If F (β, t, L, U ) F (, 4, 4, 3) and x (86557) then f l (x ) (8655) * by chopping and Fl (x ) (8656) * by rounding Example If F (β, t, L, U ) F (, 4, 4,3) and x () then f l (x ) () * by chopping and f l (x ) () * by rounding Four arithmetic operations: addition (+), subtraction ( ), multiplication (* ), and division (/) are used in F (β, t, L, U ) The operations are usually performed in double-length work area (registers two words long) Addition/subtraction of two floating-point numbers is performed as follows: The smaller-in-magnitude is adjusted (shifted) first so that the exponent is the same as the larger one This process is called "unnormalization" The operation is performed, the result normalized if necessary and then chopped or rounded to the correct length Multiplication/division of two floating-point numbers is performed as follows: Multiply/divide the fractions, and add/subtract the exponents The fraction is normalized if necessary (the fractional part will usually be larger than can be stored in a word (or, double word)) and then chopped or rounded to the correct number of digits

13 - 3 - Example If F (β, t, L, U ) F (, 4, 4, 3), a () and b () * *, then f l (a +b )? first unnormalize b to get b () * then perform the addition * * * Hence, f l (a +b ) () * for both chopping and rounding Example For the same floating-point number system if a () * and b () * then f l (a +b )? No unnormalization is required Simply perform the addition * + * * Normalize to get () * Hence, f l (a +b ) () * by both chopping and rounding Example For a () * and b () * in F (, 4, 4, 3), f l (a*b )??? a b + * s complement of - Working register (6 bits) No normalization is required The result is then chopped or rounded to the correct number of digits Hence, f l (a*b ) () * by chopping f l (a*b ) () * by rounding 35 Round-off Errors An important fact about relative round-off error is: if F (β, t, L, U ) is the floating-point number system used in the computer then the relative round-off error is always less than or equal

14 - 4 - Example For the same floating-point number system, if a () and b () * * then f l (a /b )? a b / s complement of - Working register (6 bits) No normalization is required The result is then chopped or rounded to the correct number of digits Hence, f l (a /b ) () * 3 by chopping f l (a /b ) () * 3 by rounding to a constant ε defined as follows x f l (x t, ε β t x /, for chopping for rounding () The proof of () can be sketched as follows Let x (x x x t x t + * β r If chopping is used in the rounding process, we have f l (x ) (x x x t * β r Hence, x f l (x ) ( xt + xt + )β * βr β t * β r β r t Consequently, x f l (x ) x β r t β t β β r If rounding is used in the rounding process, define a and b as follows: a (x x x t * β r b (x x x t * β r + β r t It is easy to see that if x t + < β/ then f l (x ) a and x lies in the left half of the interval [a, b ] If x t + β/ then f l (x ) b and x lies in the right half of the interval [a, b ] Hence, in either case, x f l (x ) a b β r t

15 - 5 - But then x f l (x ) x β r t β t β β r Example For x 467 in F (, 4, 5, 5) If chopping is used, we have f l (x ) 46* and ε 3 x f l (x ) And indeed, 5648 x If rounding is used, we have f l (x ) 47* and ε 3 / x f l (x ) And indeed, x ε is usually referred to as machine precision, machine unit, or machine epsilon It can also be defined as the smallest, positive floating-point number x such that f l ( + x ) > (3) It is not difficult to prove that these two definitions of ε are equivalent In the following two examples we will show that the ε defined by () for the given floating-point number system is indeed the smallest, positive floating-point number that satisfies (3) Example For F (, 4, 5, 5) with chopping ε 3 and f l (+ε) f l () > If x is a positive number smaller than ε then x t for some t i 9, i,, t But then f l (+x ) f l (t t ) f l (t * t * ) Therefore, ε is the smallest positive number such that (3) is true δ f l (x ) x x ε plays an important role in estimating the error in floating-point arithmetic operations Note that if we set (4) then the floating-point representation of a given non-zero number x can be expressed in the following form

16 - 6 - Example For F (, 4, 5, 5) with rounding ε 3 and f l ( + ε) f l (5) > If x is a positive number smaller than ε then x t t for some t 4 and t i 9, i, 3, But then f l (+x ) f l (t t ) f l (t * t * ) (due to that t 4) Therefore, ε is the smallest positive number such that (3) is true f l (x ) x ( + δ) where δ ε (5) ie, the floating-point representation of x is actually a slight perturbation of x (if x is already a floating-point number then f l (x ) x ) x could be an expression If x a Θ b denotes the exact result of any of the four arithmetic operations (+,, *, /) on two floating-point numbers a and b, then following (5), we have where f l (a Θb ) (a Θ b )( + δ) δ f l (a Θb ) (a Θ b ) and δ ε a Θ b This shows that one can estimate the relative error generated in arithmetic operations using backward error analysis For example, given three floating-point numbers x, y, and z in the floating-point number system F (, 4, 64, 63), to estimate the relative round-off error in computing z* (x + y ), one compute (x + y ) first and produce f l (x + y ), then compute z*f l (x + y ) to produce f l (z*f l (x + y )) Note that 3, chopping and f l (x + y ) (x + y )( + δ ) for some δ where δ 4, f l (z* (x + y )) f l (z*f l (x + y )) (z*f l (x + y ))( + δ ) rounding for some δ which satisfies the same upper bounds as δ Consequently, f l (z* (x + y )) z (x + y )( + δ )( + δ ) z (x + y )( + δ + δ + δ δ ) Hence, f l (z (x + y )) z (x + y ) δ + δ + δ δ + 46, chopping z (x + y ) , rounding

17 - 7 - The second terms ( 46 and 48 ) on the right hand side of the above inequality can be ignored since their values are small compared with the first terms While holding in real number arithmetic, the following axioms do not hold in floating-point arithmetic () a + (b + c )) ((a + b ) + c ) () a * (b * c )) ((a * b ) * c ) (3) If a * b a * c and a then b c (4) a * (b + c ) a * b + a * c (5) a * (b / a ) b Examples which do not hold for the first three cases are shown below Examples which do not hold for the last two cases can be found in Sterbenz s book [4] Example For F (6, 6, 64, 63) with chopping, if a ( ) 6 * 6 63 b (55637) 6 * 6 63 and c (AD 9) 6 * 6 63 then f l (a + (b + c )) f l (a + f l (b + c )) (CF 558) 6 * 6 63 and f l ((a + b ) + c ) f l (f l (a + b ) + c ) (CCCCCC ) 6 * 6 63 Or, if a 6 6, b, and c then f l (a + (b + c )) 6 6 and f l ((a + b ) + c ) In either case, f l (a + (b + c )) f l ((a + b ) + c ) Example For F (, 6, 64, 63) with chopping, if a b + 5 and c * 5 then f l ((a * b ) * c ) f l (f l (a * b ) * c ) (999999) * and f l (a * (b * c )) f l (a * f l (b * c )) (999998) * Hence, f l ((a * b ) * c ) f l (a * (b * c )) Nevertheless, the following axioms hold in floating-point arithmetic () a + b b + a () a * b b * a (3) a + + a a (4) a * * a

18 - 8 - Example For F (, 6, 64, 63) with chopping, if a, b 9, and c then f l (a * b ) (8) * f l (a * c ) But obviously b c (5) a + ( a ) ( a ) + a 36 Control of Round-Off Errors Round-off error occurs in many occasions, such as when adding (subtracting) a very small number to (from) a very large number The most serious round-off error, however, occurs when a number is subtracted from another one of almost the same size In such a case, called catastrophic cancellation, one loses many significant digits in the result and, consequently, would get very large round-off error if the result is then used in other arithmetic operations Minimizing the effect of round-off errors is a major concern in handling floating-point arithmetic operations It can be achieved through improving the hardware and software features of the computers and through careful programming () Hardware features: use extra digits (bits) inside the arithmetic unit to get more accurate results () Software features: to design compilers so that addition is done before chopping or rounding of the result of a multiplication within a floating-point operation (flop) (3) Careful programming: use stable algorithms in floating-point arithmetic operations An algorithm is said to be stable if errors introduced into the calculation grow not too fast (means not at an exponential rate) as the computation proceeds The moral here is to avoid situations in which cancellation (subtraction), especially catastrophic cancellation, occurs Techniques that have been used include: grouping [Nakamura], changing variables, using Taylor expansions, and rewriting the equation If a subtraction is inevitable, then use double precision arithmetic The hardware and software features are machine-dependent The third approach, however, is under the complete control of us In the following, we will show examples that use the techniques listed in (3) to reduce round-off error In the case of solving an quadratic equation ax + bx + c for x, we know that x can be computed using the following formulas: x x b + b 4ac ; (6) a b b (7) a

19 - 9 - However, if the magnitude of b is much bigger than that of 4ac then either x or x will be poorly determined since, in this case, the magnitude of b would be very close to that of b 4ac and, consequently, one would get a catastrophic cancellation either in x or x For instance, if one is to solve the following quadratic equation in the floating-point number system F (, 5, 5, 5) with rounded arithmetic, x + x + since b 345 and b 4ac 34, it follows that If (6) is used to calculate x, one gets b 4ac 9 x + 9 The true value of x is 98 This results in a relative round-off error of about 9% Note that in the numerator of the above computation process of x, most of the significant digits cancel, only the least significant one is left to determine the result The other solution x does not have this problem Note that x 9 and the exact solution is 999 Hence, x is almost correct A better approach is to rationalize the numerator of x if b is positive and rationalize the numerator of x if b is negative, and use the new formulas to calculate x and x, as follows: b > x c ; x b + b 4ac b b 4ac (8) a (9) b x b + b 4ac ; x a c b + b 4ac For instance, in the above example, if x is computed using (8), we get () x The computed value is correct to five places now, with a relative round-off error of less than 8%

20 - - Given an x, one usually evaluates e x by adding up terms in () until some condition is satisfied, such as when the size of a term is smaller than a given tolerance, or simply by adding up certain number of terms in () Now consider the problem of evaluating e 55 in the floating-point number system F (, 5, 5, 5) with rounded arithmetic If we use the sum of the first 5 terms in () to approximate e x, then by adding up the first 5 terms in () for x 55, we get 6363 for e 55 But actually e The reason for this is that in the process of calculating the sum, errors that occur are emphasized by cancellation, see the computation process shown below We may use more digits However, this is more costly n x n /n! x n /n! A better approach is to use e x + ( x ) + ( x ) +! () ( x ) 3 + 3! to evaluate e x if x is negative Note that if x is negative then there is no cancellation involved when () is used to evaluate e x Indeed, with (), we get 4865 for e 55, a result that is correct to four places, in only terms Other examples are shown below Example Evaluate f (x ) x sinx when x sinx is close to x when x is close to zero To avoid cancellation, one should use (3) to replace sinx and use the new expression to evaluate f (x )

21 - - Example Evaluate f (x ) e x e x when x To avoid cancellation, one should use the following alternative when x f (x ) e x (e 3x (3x ) ) (3x + + ) e x! Example Evaluate f (x ) cos x sin x when x 4 π To avoid cancellation when x is close to π/4, one may use the alternative π f (x ) cos( ) 4 Example Evaluate f (x ) lnx when x e To avoid cancellation, use f (x ) lnx lne ln(x /e )