Numerical Differentiation and Integration

Numerical Differentiation and IntegrationProblem Statement: If the values of a function f are given at a few points, say, x 0 , x 1 ,···, x n , we attempt to estimate a derivative f ′ (c) oranintegral ∫ ba f(x)dx.• Basics of Numerical Differentiation• Richardson Extrapolation• Basics of Numerical Integration• Quadrature Formulas• Trapezoidal Rule• Simpson’s 1 Rule 3♥ Simpson’s 3 Rule 8♥ Boole’s Rule

Basics of Numerical Differentiationf ′ (x) = limit h→0f(x + h) − f(x)h(1)f ′ (x) ≈ 1 [f(x + h) − f(x)] (2)hf(x + h) = f(x)+hf ′ (x)+ h22 f ′′ (ξ) (3)f ′ (x) = 1 h [f(x + h) − f(x)] − h 2 f ′′ (ξ) (4)f ′ (x) ≈ 1 [f(x + h) − f(x − h)] (5)2hf(x + h) = f(x)+hf ′ (x)+ h22 f(x)+h3 3! f (3) (ξ) (6)f(x − h) = f(x) − hf ′ (x)+ h2 h3f(x) −2 3! f (3) (τ) (7)f ′ (x) = 1h2[f(x + h) − f(x − h)] −2h 12 [f (3) (ξ)+f (3) (τ))] (8)Examples:1. f(x) =cos(x), evaluate f ′ (x) atx = π/4 withh =0.01, h =0.0052. g(x) =ln(1 + x), evaluate g ′ (x) atx =1withh =0.01, h =0.005.3. t(x) =tan −1 x,evaluatet ′ (x) atx = √ 2withh =0.01, h =0.005.• True Solutions1. f ′ (π/4) = − sin(π/4) = − 1 √2= −0.7071067812. g ′ (1) = 11+1 =0.5000000003. t ′ ( √ 2) =11+( √ 2) 2 = 1 3 =0.333333333

Approximation by Taylor ExpansionTheorem 1: Assume that f ∈ C 3 [a, b] andx − h, x, x + h ∈ [a, b] withh>0. Thenf ′ f(x + h) − f(x − h)(x) ≈2hFurthermore, ∃c ∈ [a, b] such that= D 0 (h) (9)f ′ (x) ≈f(x + h) − f(x − h)2h+ E 1 (f,h) (10)where E 1 (f,h) =− h2 f (3) (c)= O(h 2 ) is called the truncation error.6Theorem 2: Assume that f ∈ C 5 [a, b] andx ∓ 2h, x ∓ h, x ∈ [a, b] withh>0. Thenf ′ −f(x +2h)+8f(x + h) − 8f(x − h)+f(x − 2h)(x) ≈12hFurthermore, ∃c ∈ [a, b] such that= D 1 (h) (11)f ′ (x) ≈ −f 2 +8f 1 − 8f −1 + f −212h+ E 2 (f,h) (12)where E 2 (f,h) =− h4 f (5) (c)30= O(h 4 )h By Theorem 1 By Theorem 2 Richardson0.1 -0.716161095 -0.7173537030.01 -0.717344150 -0.7173561080.001 -0.717356000 -0.7173561670.0001 -0.717360000 -0.717360833f ′ (0.8) -0.717356091 -0.717356091 -0.717356091Table 1: Approximating the derivative of f(x) =cos(x) atx =0.8

Richardson’s ExtrapolationRecall thatThenf ′ (x 0 ) ≈ D 0 (h)+Ch 2 , f ′ (x 0 ) ≈ D 0 (2h)+4Ch 2 (13)f ′ (x 0 ) ≈ 4D 0(h) − D 0 (2h)3≈ −f 2 +8f 1 − 8f −1 + f −212h= D 1 (h) (14)Similarly,f ′ (x 0 )= −f 2 +8f 1 − 8f −1 + f −212h+ h4 f (5) (ξ)30≈ D 1 (h)+Ch 4 (15)Thenf ′ (x 0 )= −f 4 +8f 2 − 8f −2 + f −412h+ h4 f (5) (τ)30f ′ (x 0 ) ≈ 16D 1(h) − D 1 (2h)15≈ D 1 (2h)+16Ch 4 (16)(17)Richardson’s Extrapolation Theorem: Let D k−1 (h) andD k−1 (2h) be two approximationsof order O(h 2k )forf ′ (x 0 ), such thatf ′ (x 0 )=D k−1 (h)+c 1 h 2k + c 2 h 2k+2 + ··· (18)f ′ (x 0 )=D k−1 (2h)+2 2k c 1 h 2k +2 2k+2 c 2 h 2k+2 + ··· (19)Then an improved approximation has the formf ′ (x 0 )=D k (h)+O(h 2k+2 )= 4k D k−1 (h) − D k−1 (2h)4 k − 1+ O(h 2k+2 ) (20)

Differentiation Approximation Algorithms• Algorithm: Differentiation Using LimitsGenerate the numerical sequencef ′ (x) ≈ D j = f(x +2−j h) − f(x − 2 −j h)2 × (2 −j h)for j =0, 1, ···,nuntil |D n+1 − D n |≥|D n − D n−1 | or |D n − D n−1 |

Matlab Codes for Richardson Extrapolation%% Script file: richardson.m% Example: format long% richardson(@cos,0.8,0.00000001,0.00000001)% richardson(@sinh,1.0,0.00001,0.00001)%% Richardson Extrapolation for numerical differentiation% P.333 of John H. Mathews, Kurtis D. Fink% Input f is the function input as a string ’f’% delta is the tolerance error% toler is the tolerance for the relative error% Output D is the matrix of approximate derivatives% err is the error bound% relerr is the relative error bound% n is the coordinate of the best approximation%function [D, err, relerr, n]=richardson(f,x,delta,toler)err=1.0;relerr=1.0;h=1.0;j=1;D(1,1)=(feval(f,x+h)-feval(f,x-h))/(2*h);while (relerr>toler & err>delta & j

Basics of Numerical Integration• ∫ ∞01 √2πe −x2 /2 dx = 1 2• ∫ √20 1 − x2 dx = π 4 2• ∫ π0 sin(x)dx =2• ∫ 4 √1 xdx =143• erf(x) = 2 √ π∫ x0 e−t2 /2 dt, 0

Quadrature FormulasThe general approach to numerically compute the definite integral ∫ ba f(x)dx is by evaluatingf(x) at a finite number of sample and find an interpolating polynomial to approximatethe integrand f(x).Definition: Suppose that a = x 0

Approximating Integrals by Trapezoidal andSimpson’s RulesExample: The arc length of a curve f(x) = 2 3 x3/2 between x ∈ [0, 1] can be computed byα =• Trapezoidal Rule∫ 10√1+xdx =23 (2√ 2 − 1) = 1.21895142α ≈ h 2 [f(x 0)+2f(x 1 )+2f(x 2 )+···+2f(x n−1 )+f(x n )]where 0 = x 0

Matlab Codes for Simpson’s 1/3 Rule%% Simpson.m - Simpson’s 1/3 Rule for \sqrt(1+x)%format longn=20;h=1/n;x0=0; x1=x0+h; x2=x1+h;s=0;for i=0:2:n-2,f0=sqrt(1+x0);f1=sqrt(1+x1);f2=sqrt(1+x2);s=s+f0+4*f1+f2;x0=x2;x1=x2+h;x2=x2+h+h;ends=h*s/3.0;’Simpson Approximated Arc length is’, s