Derivatives, Finding Roots, Interpolating and Integration

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1 Derivtives, Finding Roots, Interpolting nd Integrtion Doing Numericl Derivtives Altoug tking derivtives is very esy opertion for ll of you, tere re situtions were it must e done numericlly. For exmple, suppose you re given experimentl dt points, rter tn known mtemticl function. Or, suppose you ve complicted mtemticl function ( Bessel function, or te like were you don t know te derivtive formul or it is too difficult to work out (lter wen we study Mtemtic we will se tt it cn symoliclly work out mny derivtives An lgoritm is te specific procedure you follow to clculte desired result, i.e., wen you find te roots x 1 nd x 2 of x 2 + x + c = 0 y clculting x 1, x 2 = ± 2 4c 2 you re using n lgoritm. Using guesswork nd intuition to put te eqution in te form (x x 1 (x x 2 = 0 would e different lgoritm. Te word comes from ook y gret Persin mtemticin of te 9t century, Au l-kwrizmi; we lso get te word "lger" from im. Imgine tt you ve n experiment tt produces some discrete vlues of te position of prticle. So tere is some unknown function x(t for wic you ve only numericl vlues t discrete set of points. You would, owever, like to clculte te speed of te prticle v(t - tis mens tt you need to do numericl derivtive. Suppose tt we re interested in te derivtive t x = 0, x'(0, of some function x(t nd lso suppose tt, s stted ove. we only know te function x(t on n eqully spced lttice of t-vlues. x n = x(t n, t n = n, n = 0,1,2,3... Pge 1

2 We egin y using Tylor series to expnd te function x(t in te neigorood of t = 0. We ten ve x(t = x(0 + x'(0t + x''(0 2! x +1 = x(t 1 = x( = x(0 + x'( ! t 2 + x'''(0 t ! x''( ! x'''(0 + O(4 x +2 = x(t 2 = x(2 = x(0 + 2x'( x''( x'''(0 + O(4 x 1 = x(t 1 = x( = x(0 x'(0 + 2 x 2 = x(t 2 = x( 2 = x(0 2x'( x''( x'''(0 + O(4 were O( 4 mens terms of order 4 or iger ( smll ve 2! x''(0 3 3! x'''(0 + O(4 een neglected. We ten ve x'(0 = x x 1 0 x( x(0 + O( = + O( wic is clled 2-point formul for te derivtive t t = 0. It is lso clled te forwrd difference lgoritm. Alterntively we ve x'(0 = x x 0 1 x(0 x( + O( = + O( Using tese two results we cn derive more ccurte 3-point formul or centrl difference lgoritm. or 1 2 x 1 x 0 x 1 x x 0 x 1 = x'(0 = (x(0 + x'(0 + x''(0... (x(0 x'(0 + x''( ! 2! = x'(0 + O( 2 x'(0 = x x O( 2 2 In similr mnner, it is possile to improve on te 3-point formul y relting x'(0 to grid points futer removed from x = 0. Te 5-point formul is x'(0 = 1 ( 12 x 8x + 8x x O( 4 Pge 2

3 Formuls for iger derivtives cn e constructed y tking pproprite comintions, for exmple, x''(0 = 1 2 ( x 1 2x 0 + x 1 + O( 2 Finlly, we give tle of iger-order derivtives (5-point formuls using centrl-difference formuls: x'(0 = 1 ( 12 x 8x + 8x x x''(0 = 1 ( x x 1 30x x 1 x 2 x'''(0 = 1 ( x x x 1 13x 1 + 8x 2 x 3 x''''(0 = 1 ( x x 2 39x x 0 39x x 2 x 3 Derivtives t oter vlues of t ( oter tn t = 0 re otined y trnsltion: x( x( x(t + x(t x'(0 = x'(t = 2 2 x''(0 = 1 2 ( x( 2x(0 + x( x''(t = 1 ( x(t + 2x(t + x(t 2 Smple Code for Numericl derivtives function y = fp(x y= sin(x*exp(x; % fivepoint derivtives =input('enter stepsize - (0 < < 0.1: ' ; x = 2; d1 = (fp(x-2*-8*fp(x-+8*fp(x+-fp(x+2*/(12*; d2 = (-fp(x-2*+16*fp(x--30*fp(x+16.0*fp(x+... -fp(x+2*/(12*^2; d3 = (fp(x-3*-8*fp(x-2*+13*fp(x--13*fp(x *fp(x+2*-fp(x+3*/(8*^3; d4 = (-fp(x-3*+12*fp(x-2*-39*fp(x-+56*fp(x *fp(x++12*fp(x+2*-fp(x+3*/(6*^4; [d1,d2,d3,d4] Pge 3

4 >> fivepoint Enter stepsize - (0 < < 0.1:.5 ns = >> fivepoint Enter stepsize - (0 < < 0.1:.1 ns = >> fivepoint Enter stepsize - (0 < < 0.1:.01 ns = Clerly we re seeing convergence. Te exct nswers re: Finding Zeroes or Roots Te solution of n ritrry eqution f(x = 0 is commonly clled te root or "zero" of te function f(x. More specificlly, if f(x is continuous on n intervl [, ] nd te signs of f( nd f( differ, ten for some x in [, ], sy x = z, f(z = 0 nd z is te root of te function. Tere re vriety of itertive metods for finding roots of functions nd we sll illustrte few of tem elow: Crude Home-In Metod (CHIM In tis metod we need to know n pproximte vlue of te root. Te procedure ten is: (1 pick x < xroot (guess it nd clculte f(x (2 let x x + D, clculte f(x (3 continue tis process until f(x cnges sign (4 ck up 1 step, decrese te step size, D D/10 (5 return to step (2 (6 decide wen to stop If we cn compute(nlyticlly te derivtive f (x, ten noter tecnique is used. Pge 4

5 Newton - Rpson Tngent Metod (NRTM Tis metod genertes sequence of vlues x i converging to x root. Using te digrm elow NRTM uses Tngent Metod tngent x root Te NRTM uses te eqution x x i+1 i or y tngent line = f (x i + f '(x i (x x i y root = 0 = f (x i + f '(x i (x i+1 x i x i+1 = x i f (x i f '(x i Any vlue x 0 cn e used to strt te process. A second version does not require n nlytic formul for te derivtive. Newton - Rpson Secnt Metod (NRSM Tis metod genertes sequence of vlues x i converging to x root. Using te digrm elow, NRSM uses f '(x i = f (x i f (x i 1 x i x i 1 (s n pproximtion for te slope in NRTM metod x i+1 = x i f (x i x i x i 1 f (x i f (x i 1 Pge 5

6 Secnt Metod x root secnt x i+1 x i x i-1 Any two vlues x 1 nd x 2 cn e used to strt te process Wen te function is dly eved ner its root, i.e., tere is n inflection point or severl nery roots, te NR metods cn fil to converge t ll or converge to te wrong nswer deping on your strting point. A sfe procedure is to use te CHIM metod to get close nd ten us one of te NR metods to get te exct result. Smple Code for Finding Roots function z=nucler(x z=tn(0.1*x-9.2*exp(-x; % CHIM metod % 0,10,1,.001 xmin=input('enter xmin : ' ; xmx=input('enter xmx : ' ; step=input('enter step : ' ; tol=input('enter tol : ' ; Pge 6

7 wile ((xmx-xmin >= tol x=xmin; c0=sign(nucler(x; c=c0; wile (c == c0 x=x+step; c0=c; c=sign(nucler(x; xmx=x; xmin=x-step; step=step/10; [xmin,xmx,nucler(xmin] function z=nucler(x z=tn(0.1*x-9.2*exp(-x; function z=dnucler(x z=0.1*(1.0/cos(0.1*x^2+9.2*exp(-x; % NRTM Metod % 0, x1=input('enter x1 : ' ; tol=input('enter tol : ' ; xnew = x1; xold = xnew-10*tol; wile (s(xold-xnew >= tol xold=xnew; xnew=xold-nucler(xold/dnucler(xold; [xold,nucler(xold] function z=nucler(x z=tn(0.1*x-9.2*exp(-x; % NRSM Metod % 0,10,0.001 x0=input('enter x0 : ' ; x1=input('enter x1 : ' ; tol=input('enter tol : ' ; xold = x1; xoldold = x0; wile (s(xoldold-xold >= tol xnew=xoldold-nucler(xoldold*(xold-xoldold/... (nucler(xold-nucler(xoldold; Pge 7

8 if (s(xnew-xold > s(xnew-xoldold xoldold =xoldold; xold=xnew; else xoldold=xold; xold=xnew; [xoldold,xold,nucler(xold] Interpolting Dt Lgrnge Interpoltion Te metod of Lgrnge Interpoltion cn e used to pproximte function everywere even if vlues of te function re only known t finite set of points. We derive Lgrnge s tree-point interpolting polynomil p(x from Tylor series y expressing te function t x 1 nd x 2 in terms of te function nd its derivtives: f (x 1 = f (x + (x 1 x f '(x +... f (x 2 = f (x + (x 2 x f '(x +... We would like to truncte te series nd retin only te first two terms. But in doing so, te equlity would e destroyed. We cn, owever, introduce pproximtions to te function nd its derivtives suc tt te equlity is retined. Consider te reltions f (x 1 = p(x + (x 1 xp'(x f (x 2 = p(x + (x 2 xp'(x were we must ten ve p(x 1 = f (x 1 nd p(x 2 = f (x 2 Some lger ten gives p(x = x x 2 f (x 1 + x x 1 f (x 2 x 1 x 2 x 1 x 2 wic is liner function of x. Tis function is te eqution of te line pssing troug te points (x 1, f (x 1,(x 2, f (x 2. Note ow te weigting fctors empsize ec term t different plces. Higer pproximtions (non-liner re otined y keeping nd pproximting more terms in te Tylor series. Pge 8

9 Tus, if we know te function t tree points, f (x 1 = f (x + (x 1 - x f '(x + (x 1 x2 2 f (x 2 = f (x + (x 2 - x f '(x + (x 2 x2 2 f ''(x +... f ''(x +... f (x 3 = f (x + (x 3 - x f '(x + (x 3 x2 f ''(x Truncting tese equtions(s ove nd solving we get p(x = (x x 2 (x x 3 (x 1 x 2 (x 1 x 3 f (x 1 + (x x 1 (x x 3 (x 2 x 1 (x 2 x 3 f (x 2 + (x x 1 (x x 2 (x 3 x 1 (x 3 x 2 f (x 3 Tese results cn e generlized to give generl n-point interpolting polynomil of order (n-1 s were n p(x = C j,n (x f (x j j =1 (x x 1 (x x 2...(x x j 1 (x x j +1...(x x n C j,n (x = (x j x 1 (x j x 2...(x j x j 1 (x j x j +1...(x j x n were C j,n (x i = 1 i=j 0 i j = δ ij Smple Code for Lgrnge Interpoltion % interpolting t = [94.0, 205.0, 371.0]; ro = [929.0, 902.0, 860.0]; x = 251.0; p1 = ((x-t(2*(x-t(3*ro(1/((t(1-t(2*(t(1-t(3; p2 = ((x-t(1*(x-t(3*ro(2/((t(2-t(1*(t(2-t(3; p3 = ((x-t(1*(x-t(2*ro(3/((t(3-t(1*(t(3-t(2; dens=p1 + p2 + p3 Numericlly Doing Integrls Now we re interested in numericlly clculting te definite integrl of function f(x etween two limits <. We define N = Pge 9

10 suc tt N = n even integer nd = step size. Alterntely, we cn coose N(even nd compute te corresponding step size. Now we cn write (y reking up te rnge of integrtion into mny smller rnges f (xdx = f (xdx + f (xdx f (xdx + f (xdx Terefore we need only clculte te integrl g(0 = f (xdx nd te generl formul cn e clculted from tis result using te reltion 2 g(q = f (xdx q wic follows from trnsltion nd gives q+ f (xdx = g( + + g( g( g( 3 + g( Te sic ide ere(tis is te so-clled Newton-Cotes metod is to pproximte f(x etween - nd y function tt cn e integrted exctly so we cn evlute g(0 exctly nd ten use tt exct result to generte n pproximtion for te entire integrl. For exmple, te simplest pproximtion is given y considering te intervls [-,0] nd [0,] seprtely nd ssuming tt te function f(x is constnt in ec region of tese intervls. In tis cse we ve g(0 = f (xdx = f (xdx + f (xdx = f ( dx + f (0dx 0 ( = f ( + f (0 wic is te well-known "rectngle" rule ecuse we re summing rectngulr res. Tis gives for te entire integrl or f (xdx = g( + + g( g( g( 3 + g( = f ( + f ( + 0 ( + ( f ( f ( ( f ( f ( ( f ( 4 + f ( 3 + ( f ( 2 + f ( 0 0 Pge 10

11 f (xdx = ( f ( + f ( + + f ( f ( f ( f ( f ( 4 + f ( 3 + f ( 2 + f ( Te next simplest pproximtion cn e d y considering te intervls [-,0] nd [0,] seprtely nd ssuming tt te function f(x is liner in ec region of tese intervls. Te pproximte integrl is ten 0 f (0 f ( f ( f (0 g(0 = f (xdx = f ( + (x + dx + f (0 + x dx ( = f ( + 2 f (0 + f ( 2 wic is te well-known "trpezoidl" rule ecuse we re summing trpezoidl res. Tis gives or f (xdx = g( + + g( g( g( 3 + g( = ( f ( f ( + + f ( 2 + ( f ( f ( f ( f (xdx = + ( f ( f ( 3 + f ( ( f ( f ( + f ( 2 ( f ( + 2 f ( f ( f ( f ( f ( f ( f ( f ( + f ( An even etter pproximtion cn e d y relizing tt Tylor series cn provide n improved interpoltion of te function f(x We ve wic gives f (x = f (0 + f ( f ( x + 0 f ( 2 f (0 + f ( 2 2 x 2 + O(x 3 f (xdx = ( f ( + 4 f (0 + f ( 3 Tis is "Simpson s" Rule. It is ccurte to two orders iger tn te trpezoid rule. It corresponds to pproximting f(x y prol. Using tis result we get Pge 11

12 or f (xdx = ( f ( f ( + + f ( 3 + ( f ( f ( f ( f (xdx = + ( f ( f ( 3 + f ( ( f ( f ( + f ( 3 ( f ( + 4 f ( f ( f ( f ( f ( f ( f ( f ( + f ( Rememer, te numer of intervls must e even(or te numer points is odd. Smple Codes to Implement Rectngle Rule function y=tfofx(x % integrtion function y= x.*exp(x; % rectngle rule using for loop n=100;=2;=0;=(-/n;intrect=0; for x=0::(n-1* intrect=intrect+tfofx(x*; intrect % rectngle rule - no loops (vectorized n=100;=2;=0;=(-/n;x=*(0:n-1; f=tfofx(x; intrect=*sum(f; intrect Vectorized Trpezoid nd Simpson Rule(two wsys Codes % Numericl Integrtion n=100; =2; =0; =(-/n; x=*(0:n; f=tfofx(x; % trpezoid rule wtrp=[1,2*ones(1,n-1,1]; inttrp=(/2*sum(wtrp.*f; % simpson rule w2=2*ones(1,n+1; Pge 12

13 w2(1:2:n+1=zeros(1,n/2+1; wsimp=[1,2*ones(1,n-1,1]+w2; intsimp=(/3*sum(wsimp.*f; [inttrp,intsimp] % Numericl Integrtion n=1000; =2; =0; =(-/n; x=*(0:n; f=tfofx(x; %trpezoid rule wtrp=[1,2*ones(1,n-1,1]; inttrp=(/2*sum(wtrp.*f; % simpson rule wsimp=[1,respe((ones(1,(n-2/2'*[4,2]',1,n-2,4,1]; intsimp=(/3*sum(wsimp.*f; [inttrp,intsimp] Pge 13