2. Review of Calculus Notation. C(X) all functions continuous on the set X. C[a, b] all functions continuous on the interval [a, b].

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1 CHAPTER Mathematical Preliminaries and Error Analysis. Review of Calculus Notation. C(X) all functions continuous on the set X. C[a, b] all functions continuous on the interval [a, b]. C n(x) all functions that have n derivatives continuous on X. C (X) all functions that have derivatives of all orders on X. Theorem (Mean Value Theorem). If f C[a, b] and f is di erentiable on (a, b), then a number c in (a, b) exists with f (b) f (a) f 0(c) b a or f (b) f (a) + f 0(c)(b a).

2 . MATHEMATICAL PRELIMINARIES AND ERROR ANALYSIS Theorem (Extreme Value Theorem). If f C[a, b], then c, c [a, b] exist with f (c) f (x) f (c) for all x [a, b]. In addition, if f is di erentiable on (a, b), then c and c occur either at the endoints of [a, b] or where f 0 is zero. Theorem (Intermediate Value Theorem). If f C[a, b] and K is any number between f (a) and f (b), then there exists c (a, b) such that f (c) K. Corollary (Intermediate Value Theorem). If f C[a, b] and f (a)f (b) < 0, then there exists c (a, b) such that f (c) 0. Note. This corollary is often used in aroximating roots of equations.

3 . REVIEW OF CALCULUS 3 Theorem (Taylor s Theorem). Let f C n [a, b] with f (n+) existing on [a, b] and x 0 [a, b]. For every x [a, b], there exists a number (x) between x 0 and x with where k0 f(x) P n (x) + R n (x) P n (x) f(x 0 ) + f 0 (x 0 )(x x 0 ) + f 00 (x 0 ) (x x 0 ) + + f (n) (x 0 ) (x x 0 ) n! n! nx f (k) (x 0 ) (x x 0 ) k k! and Note. R n (x) f (n+) ( (x)) (x x 0 ) n+. (n + )! () f is the sum of a Taylor olynomial of order n (also called a Maclaurin olynomial if x 0 0) and a remainder term (or truncation error). () The case n 0 is just the Mean Value Theorem (MVT). f(x) P 0 (x) + R 0 (x) f(x 0 ) {z } P 0 (x) + f 0 ( (x))(x x 0 ) {z } R o (x) (3) If f is a olynomial of degree r, then for n r, f(x) P n (x). Male. See taylor olynomials.mw and/or taylor olynomials.df

4 4. MATHEMATICAL PRELIMINARIES AND ERROR ANALYSIS Examle. Let f(x) x cos(x) (x ), x 0 0. (a)find P 3 (x) and P 4 (x) and use them to aroximate f(0.4). f 0 (x) cos(x) 4x sin(x) (x ) f 00 (x) 4 sin(x) 4 sin(x) 8x cos(x) 8sin(x) 8x cos(x) f 000 (x) 6 cos(x) 8 cos(x) + 6x sin(x) 4 cos(x) + 6x sin(x) f (4) (x) 48 sin(x) + 6 sin(x) + 3x cos(x) 64 sin(x) + 3x cos(x) f (5) (x) 8 cos(x) + 3 cos(x) 60 cos(x) 64x sin(x) 64x sin(x) f(0) 4; f 0 (0) 6; f 00 (0) ; f 000 (0) 4; f (4) (0) 0 P 3 (x) 4 + 6x + x x 3 4x 3 x + 6x 4 Since f (4) (0) 0, f(0.4) P 3 (0.4).06 P 4 (x) P 3 (x) 4x 3 x + 6x 4.

5 . REVIEW OF CALCULUS 5 (b) Use the error formula in Taylor s Theorem to find an uer bound for the error f(0.4) P 3 (0.4) and f(0.4) P 4 (0.4). By grahing, for 0 ale x ale 0.4, f (4) (x) ale 55. and f (5) (x) ale 60. Then R 3 (.4) f(0.4) P 3 (0.4) f (4) ( (x)) ale 55. 4! 4 (.4)4 ale by rounding u. Similarly, R 4 (.4) f(0.4) P 4 (0.4) f (5) ( (x)) ale 60 5! 0 (.4)5 ale.0366, a better bound than for R 3 (.4). The actual absolute error is f(0.4) P 3 (0.4) (.8 cos(.8).56) (.06) Notice that in this roblem we are dealing with error at a oint.

6 6. MATHEMATICAL PRELIMINARIES AND ERROR ANALYSIS Examle. Aroximate sin(4 ) with error less than or equal to 0 6. x Use Taylor olynomial with x 0 4, so x x To find n, with f(x) sin(x) and 7 30 ale (x) ale 4, 7 R n f (n+) ( (x)) 7 30 (n + )! 30 4 n+ ale (n + )! 60 n n+ (n + )!. 7 Then R and 30 3! 7 R < 0 6, so take n ! f(x) sin(x), f 0 (x) cos(x), f 00 (x) sin(x), f 000 (x) cos(x), so f 4, f 0 4, f 00 4, f Thus 7 7 sin(4 ) sin P f + f (x 0 x 0 ) + f 00 ( 4 ) (x x 0 ) + f 000 ( 4 ) (x x 0 ) 3 4 4! 3! From the TI-89, sin , 30 7 so sin(4 ) P Male. See taylor s error.mw and/or taylor s error.df

7 3. ROUND-OFF ERROR AND COMPUTER ARITHMETIC 7 3. Round-o Error and Comuter Arithmetic Male. See comuternumbers.mw and/or comuternumbers.df There are roblems in using a finite subset of the reals (rational numbers, actually) to reresent all reals. Suose each machine number can be reresented as a k-digit decimal machine number in floating-oint form as ±0.d d d k 0 n, ale d ale 9, 0 ale d i ale 9, i,..., k. Suose y 0.d d d k d k+ d k+ 0 n is in the numerical range of the machine. Two methods for getting the floating-oint form of k digits fl(y): () fl(y) 0.d d d k 0 n is choing. () rounding: add 5 0 n (k+) and then cho to get 0. k 0 m. We refer to both cases as round-o error. Examle (of rounding). Suose we wish to round and with k 4 and n 3, so 5 0 n (k+) Now cho:

8 8. MATHEMATICAL PRELIMINARIES AND ERROR ANALYSIS Measuring aroximation errors: Suose? aroximates : absolute error? relative error? Problem (.0, #a).,? 7. absolute error relative error Since 4 is the largest nonnegative value of t for which relative error < 5 0 t we say? aroximates to 4 significant digits. An aroximation to floating-oint arithmetic: x y fl(fl(x) + fl(y)) x y fl(fl(x) fl(y)) x y fl(fl(x) fl(y)) x y fl(fl(x) fl(y)) In Male, Digits:t; causes all arithmetic to be rounded to t digits (0 is the default), and, for examle, x y corresonds to evalf(evalf(x)+evalf(y)).

9 3. ROUND-OFF ERROR AND COMPUTER ARITHMETIC 9 There are roblems to be noted with subtraction: Examle. Use 4 digit choing fl( ) 3.4 relative error fl( 4 7 ) 3.4 relative error.73 0 Both reresentations have 4 significant digits, but 7 fl( ) 0.00 with relative error ( 7 ) 0.00 ( 7 ).09 0, yielding a result with only significant digit. Thus subtracting nearly equal numbers cancels significant digits, and these cannot be recovered. We get small absolute error, but large relative error.

10 0. MATHEMATICAL PRELIMINARIES AND ERROR ANALYSIS 4. Errors in Scientific Comutation Sequencing of oerations can make a di erence: Problem (.7, #b). Use four digit rounding to find the roots of 4 x 6 0. Note that the exact roots round to and x + 3 The quadratic formula for ax + bx + c 0 gives as solutions. x b + b 4ac a and x b b 4ac a a.3333, b 30.75, c.667, b 945.6, ac.0556, 4ac.4 b 4ac 945.8, b 4ac 30.75, b + b 4ac 0 b b 4ac 6.50, and a.6666, so x and x , and relative error and relative error Thus we have no and 5 significant digits. But rationalizing numerators, c x b + b 4ac and x c b b 4ac, so c.3334, b + b 4ac 6.5, and b x b.0054 and x.3334, undefined. 0 4ac 0, giving Relative error for x is.3 0 4, giving 4 significant digits. Male. See nested.mw and/or nested.df

11 4. ERRORS IN SCIENTIFIC COMPUTATION algorithm a rocedure that unambiguously describes a finite, ordered sequence of stes to solve a roblem or aroximate a solution. stable algorithm small changes in initial data roduce small changes in the final results unstable algorithm not stable conditionally stable algorithm stable only for certain choices of initial data Growth of round-o error: Suose an error E0 > 0 is introduced at some stage of the calculations and the error after n subsequent oerations is En. linear growth of (relative) error En CnE0, where C is a constant indeendent of n. This reflects a stable algorithm. exonential growth of (relative) error En C ne0 for some C >. This reflects an unstable algorithm. Male. See convergence.mw and/or convergence.df.

12 . MATHEMATICAL PRELIMINARIES AND ERROR ANALYSIS Rates of convergence for sequences: Our algorithms often generate sequences of values. Suose { n }! and n! 0. Recall { n },, 3,..., n,... and { n }! means lim n! n. If there exists K > 0 such that n ale K n for large n, we say { n } converges to with a rate of convergence O( n ) ( big oh of or { n }! with rate of convergence O( n ). Usually, n for some n > 0. We want the largest such value so that n + O. n Examle. Consider n + 3 n + 7 so n + 3 n O n n n + 3 o n + 7! and n n o n 3 + 7!. n + 3 n n n + 7 n n n n ) n + 3 n 4 n + 7 n + 7 ale n + 7 n ( n n ) and the rate of convergence is O since lim n n! n 0.

13 4. ERRORS IN SCIENTIFIC COMPUTATION 3 n n n3 + 3 n 3 4 n n n ale n 3 ( n n 3) so n3 + 3 n O n 3 and rate of convergence is O n 3 Problem (.9 #0b). Find the rate of convergence of We know lim n! 0. From calculus, lim n h!0 sin h h since lim n! n 0. 3 n sin n o. ) lim sin k ) k!0 k letting k sin n lim n n! <. Then, for large n, sin n < n. n Since lim n! n 0, sin n 0 + O, so rate of convergence is O n. n We can also use the big oh notation to show how some divergent sequences grow as n becomes large. If ositive constants and K exist with n Kn for all large values of n, we say that { n }! with rate of convergence O(n ). Finally, we can use big oh for functions. Suose lim h!0 G(h) 0 and lim h!0 F (h) L. If there is K > 0 such that F (h) L ale K G(h) for h small enough, then F (h) L + O(G(h)). Usually, we like to use G(h) h, and want the largest such that F (h) L + O(h ).

14 4. MATHEMATICAL PRELIMINARIES AND ERROR ANALYSIS Recall. sin h h cos h e h + h + h + + hn n! + e (n + )! hn+ h 3 3! + h5 5! h! + h4 4! + ( )n h n+ (n + )! + ( )n h n (n)! + ( )n+ cos h n+3 (n + 3)! + ( )n+ cos h n+ (n + )! + h h + h + ( ) n h n + ( )n+ ( + ) n+hn+ e h Problem (.9 #b). Find the rate of convergence for lim h!0 h e h + h + e h ) e h h e h h + e h < Examle. Find lim h!0 cos h + h h 4 cos h e h ) eh h h for h small enough. Thus eh h h + cos 4 h4 70 h6 for 0 < < h ) cos h + h 4 h4 cos 70 h6 ) cos h + h h 4 4 {z} Looks like L cos 70 h {z } +O(h ) ) and its rate of convergence. e h ). +O(h). cos h + h h 4 4 cos 70 h ale 70 h.

15 4. ERRORS IN SCIENTIFIC COMPUTATION 5 Thus lim h!0 cos h + h h 4 4 with rate of convergence O(h ) or cos h h 4 + h 4 + O(h ).