The total error in numerical differentiation
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1 AMS 147 Computational Metods and Applications Lecture 08 Copyrigt by Hongyun Wang, UCSC Recap: Loss of accuracy due to numerical cancellation A B 3, 3 ~10 16 In calculating te difference between A and B, te relative error is magnified by a factor of A + B A + B. Wen A and B are very close to eac oter, can be very large. A B A B fl( A fl( B= ( A B 1 + A + B Ad oc ways of avoiding numerical cancellation 4 sources of error Statistical uncertainty in an estimated parameter umerical differentiation metods Te total error in numerical differentiation Consider te first order metod f( x+ f( x = f( x + e, umerical e = O Discretization error Question: How sould we select? Answer: In exact aritmetic: te smaller is, te better. In IEEE double precision: we need to study te total error. First, we point out tat f ( x + f ( x is te numerical result in exact aritmetic. f ( x + f ( x is OT te numerical result we get from a computer! Te numerical result we get from a computer is - 1 -
2 AMS 147 Computational Metods and Applications fl( f( x+ fl( f( x umerical result from a computer Let us examine te numerical result we get from a computer. fl( f ( x + = f ( x + ( 1 + 1, 1 ~10 16 fl( f ( x = f x, ~ fl( f ( x + fl( f ( x = f ( x + ( f ( x ( 1 + = f ( x + f ( x + f ( x + 1 f ( x = f x + e + f x+, 3 ~10 16 Te discretization error: Discretization error e C 0 as 0 Te effect of round-off error: ( + f( x 3 Effect of round-off error ( f( x+ + f( x 3 ~ f( x+ + f( x Te total error is defined as fl( f ( x + fl( f ( x = f ( x umerical result from a computer 1016 as 0 = e + f x+ Discretization error ( + f( x 3 Effect of round-off error were e ~ C, 3 ~10 16 Let us consider te simplified situation were te total error is given by =
3 AMS 147 Computational Metods and Applications Let us find te minimum of (. = = 1 ( Solving = 0, we get c =10 8 For te first order metod, we ave argmin ( = 10 8 min = 10 8 Terefore, in te first order numerical differentiation metod, we sould use ~10 8 ote: min is te minimum total error tat can be acieved using te given numerical metod and te given finite precision aritmetic. In oter words, given te numerical metod and te finite precision aritmetic, we cannot make te total error smaller tan min no matter wat we use for! Total error in te second order metod f( x+ f( x = f( x + e Te total error is were = umerical Discretization error fl( f( x+ fl( f( x f( x umerical result from a computer = e + f x+ 3 Discretization error e C, 3 ~10 16 In a simplified situation, te total error is + f( x Effect of round-off error - 3 -
4 AMS 147 Computational Metods and Applications = (Skip te derivation of arg min Let us find te minimum of (. in lecture. = 1016 = Solving = 0, we get c = For te second order metod, we ave arg min min = = Terefore, in te second order numerical differentiation metod, we sould use ~10 5 ote: Te minimum total error of te second order metod is smaller tan tat of te first order metod. Tat is, by using te second order metod wit a suitable of, we can acieve a lower total error tan wat can be acieved by using te first order metod. Tis is an advantage of iger order metods. Total error in te fourt order metod In a simplified situation, te total error is given by = (Skip te derivation of arg min Let us find te minimum of (. in lecture. = = Solving = 0, we get c =
5 AMS 147 Computational Metods and Applications For te fourt order metod, we ave arg min min = = Terefore, in te fourt order numerical differentiation metod, we sould use ~10 3 Comparison of te tree metods: First order metod Second order metod Fourt order metod argmin ( = 10 8 arg min = arg min = min = 10 8 min = min = Advantage of a iger order metod: It is clear tat wen we use a iger order metod, we can acieve a smaller total error. umerical integration Goal: To approximate b a f x dx Strategy: Divide [ a, b] into subintervals of te size = b a. Let = a + i, i = 0,1,,,. (Draw te real axis to sow a, b b a f ( x dx = f ( x dx 1 [ ] and = a + i, i = 0,1,,, To approximate b f ( x dx, we only need to approximate f ( x dx. a 1-5 -
6 AMS 147 Computational Metods and Applications We describe two metods for approximating f ( x dx : 1 * trapezoidal rule and * Simpson s rule. Te derivation and te error analysis of te two metods is included in te Appendix. Trapezoidal rule: ( f + f i 1 i = f ( x dx + e i (, x i1 Error otation: umerical f i = f( Te discretization error is = O 3 e i We will derive te trapezoidal rule and analyze te discretization error in te Appendix. (Draw a trapezoid to sow te geometric meaning of trapezoidal rule Composite trapezoidal rule: Sum from i = 1 to i =, we obtain ==> ( f i1 + f i = f ( x dx + e i i =1 1 1 b f + f + f 0 i = f ( x dx + E, a Error umerical Te discretization error is E= e i = O ( 3 = O( 3 = O ( b a Here we ave used = = ( b a= O(. 1 Te composite trapezoidal rule is a second order metod. ote:
7 AMS 147 Computational Metods and Applications Te trapezoidal rule is for approximating f ( x dx. 1 Te composite trapezoidal rule is for approximating b f( xdx. a Simpson s rule: ( 6 f + 4 f + f i 1 i 1/ i = f ( x dx + e i, x i1 Error otation: umerical f i = f (, f i1/ = f ( x, x = a + i 1 i 1/ i 1/ Te discretization error is = O 5 e i We will derive te Simpson s rule and analyze te discretization error in te Appendix. Draw a parabola to sow te geometric meaning of Simpson s rule Composite Simpson s rule: Sum from i = 1 to i =, we obtain 6 ( f i1 + 4 f i1/ + f i = f ( x dx + e i 1 i =1 ==> 1 b 6 f + f + f 0 i + 4 f i1/ = i =1 f ( x dx + E, a Error umerical = e i E = O 5 = O 5 = O ( 4 Te composite Simpson s rule is a fourt order metod. Remark: In bot te numerical differentiation and numerical integration, tere is penalty for using very small step size
8 AMS 147 Computational Metods and Applications Using very small in numerical differentiation ----> Loss of accuracy due to numerical cancellation Using very small in numerical integration ----> Hig computational cost (umber of function evaluations in numerical integration is proportional to = ( b a In bot cases, iger order metods ave advantages over lower order metods. Using a iger order metod in numerical differentiation ----> Acieving a smaller total error wit a carefully selected Using a iger order metod in numerical integration ----> Acieving a smaller error wit lower computational cost. Go troug sample codes in assignment # Run Matlab codes to illustrate te beavior of te total error Appendix: Derivation of trapezoidal rule (skip te derivation in lecture Consider a special interval of size :,. We can always sift [ 1, ] to,. Let us introduce a linear operator on function f. I [ f ]= a f 1 + a 1 f We use I f / [ ] to approximate f x [ ] I f umerical /. dx. / = f ( x dx + e / Error - 8 -
9 AMS 147 Computational Metods and Applications Question #1: How to determine a 1 and a 1? Question #: Wat is te order of e(? To determine a 1 and a 1, we require tat te numerical be exact for two special functions: f(x = 1 and f(x = x. []= 1 I 1 / / / dx I[ x]= x dx / It is exact for f ( x =1 = ==> a 1 + a 1 It is exact for f ( x = x ==> a 1 + a 1 = 0 ==> ==> ==> I f a 1 + a 1 =1 a 1 a 1 = 0 a 1 = 1 a 1 = 1 [ ] = f + f To answer te second question, we use te Taylor expansion of f ( x. f ( x = f ( 0 + otice tat bot I f f ( 0x + f 0 / [ ] and f ( x / / / e = I[ f ] f ( x dx = I f ( 0 + f ( 0x + / = f ( 0 I[ 1] 1dx / x + dx are linear operators on f. / f ( 0 x + f ( 0 + f ( 0 x + f ( 0 x + dx + f 0 Ix / / [ ] x dx + f 0 / / Ix [ ] x dx + / - 9 -
10 AMS 147 Computational Metods and Applications Te first two terms on te rigt and side are zero. Te tird term is / Ix [ ] x dx = 1 / ==> e = f = O 3 Tus, we obtain te trapezoidal rule: + 1 x 3 / 3 / ( f + f i 1 i = f ( x dx + e i (, x i1 Error e i umerical = O 3 = 3 6 Derivation of Simpson s rule (skip te derivation in lecture Let I [ f ]= a f 1 + a 0 f ( 0+ a 1 f We use I f / [ ] to approximate f x [ ] I f umerical / dx. /. = f ( x dx + e / Error Question #1: How to determine a 1, a 0 and a 1? Question #: Wat is te order of e(? To determine a 1, a 0 and a 1, we require tat te numerical be exact for tree special functions: f(x = 1, f(x = x and f(x = x. []= 1 I 1 / / / dx I[ x]= x dx / / I x = x dx /
11 AMS 147 Computational Metods and Applications It is exact for f ( x =1 = ==> a 1 + a 0 + a 1 It is exact for f ( x = x ==> a 1 + a 1 = 0 It is exact for f ( x = x ==> a 1 ==> ==> ==> I f a 1 + a 0 + a 1 = 1 a 1 a 1 = 0 a 1 + a 1 = 1 3 a 1 = 1 6 a 0 = 4 6 a 1 = 1 6 [ ] = 6 f + a 1 = f 0 + f To answer te second question, we use te Taylor expansion of f ( x. f ( x = f ( 0 + f ( 0x + otice tat bot I f f ( 0 / [ ] and f ( x / / / e = I[ f ] f ( x dx / = f ( 0 I[ 1] 1dx / + f ( f 0 / x + f ( 3 ( 0 6 x 3 + f ( 4 ( 0 x dx are linear operators. Ix Ix [ 3 ] x 3 dx / / [ ] x dx + f 0 / + f ( / / Ix [ ] x dx / Ix [ 4 ] x 4 dx + / Te first tree terms on te rigt and side are zero. Te fourt term is also zero because x 3 is an odd function. Te fift term is
12 AMS 147 Computational Metods and Applications / Ix [ 4 ] x 4 dx = 1 6 / ==> e = f ( 4 ( = O x / 5 / = 5 10 Tus, we obtain te Simpson s rule: ( 6 f i1 + f i1/ + f i = f( x dx + e i, x i1 Error e i umerical = O 5-1 -
AMS 147 Computational Methods and Applications Lecture 09 Copyright by Hongyun Wang, UCSC. Exact value. Effect of round-off error.
Lecture 09 Copyrigt by Hongyun Wang, UCSC Recap: Te total error in numerical differentiation fl( f ( x + fl( f ( x E T ( = f ( x Numerical result from a computer Exact value = e + f x+ Discretization error
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