Civil Engineering Computation

Transcription

1 Civil Egieerig Computatio Fidig Roots of No-Liear Equatios March 14, 1945 World War II The R.A.F. first operatioal use of the Grad Slam bomb, Bielefeld, Germay. Cotets 2 Root basics Excel solver Newto-Raphso Numerical derivatives Secat rule Two simultaeous equatios Geeral sets of o-liear equatios Multiple equatios with secat rule 1

2 Root Basics Root of f(x) = 0: value of x that satisfies equatio Example: f(x) = 2*x - 6 = 0 x = 3 is root Example: f(x) = x2-10*x + 21 = 0 two roots, x = 3 ad x = 7 Roots occur whe curves touch or cross x axis as: 3 a a Root Basics f(x) first f(x) secod f(x) X 4 2

3 Liear, quadratic equatios easy to solve What if equatio is o-liear like f(x) = ta(x)*si(x) , x i radias Alterative to closed-form solutio is iteratio ( guess ad correct ) 5 The Excel Solver Ope Excel solver from Tools meu Place startig value for ukow i cell, equatio i terms of that cell i aother cell Example: f(x) = ta(x)*si(x) -2.15, x i radias Put x i say C2 I C3 type = ta(c2)*si(c2)

4 Oe Equatio Select Solver from Tools meu (If ot istalled go to Add-is i Tools ad activate the solver) Solver will automatically maipulate x by chagig value i C2 Value of f(x) i C3 is target value. Whe equatio solved should be early to 0 Cotrol settigs with Solver dialog box: 7 Solver Dialog Box 8 4

5 Newto-Raphso Method Amog best-kow methods is Newto-Raphso: Begi with startig value (x0); Fid subsequet values with (7.2.1) x+1 = x! 9 f ( x) f " ( x) Requires calculatio of fuctio derivative -- may be tedious Stop iteratio whe x chage less tha certai amout Newto-Raphso Method 10 3 requiremets: Startig value must be reasoably close to desired root; otherwise could diverge or fid differet root First derivative must ot be 0 whe evaluated; Otherwise divisio by 0 Secod derivative must ot be too large; 5

6 Newto-Raphso Method Apply Newto-Raphso to: f(x) = ta(x)*si(x) -2.15, x i radias Let x0 = 1.47 f (x) = sec2(x)*si(x) + ta(x)*cos(x) f(1.47) = 7.687, f'(1.47) = x1 = /99.25 = 1.39 Ea = (x1-x0)/x1 *100 = ( )/1.39 *100 = 5.74% 11 Example Results 12 First 5 iteratios: iter x f(x) f '(x) εa, %

7 Secat Rule Alterate to Fidig Derivatives Approximate first derivative as: f! ( x) " f (x ) # f (x#1 ) x # x#1 Replace f'(x) i the Newto-Raphso method ad simplify f ( x ) f ( x ) f (x )(x x 1 ) x +1 = x = x = x f ( x ) f ( x 1 ) f (x ) f ( x ) f ( x 1 ) x x 1 13 Secat Rule Alterate to Fidig Derivatives 14 This is secat rule Have replaced derivative of f(x) by approximatio based o chage of f(x) with x betwee 2 cosecutive iteratios Requires always kowig 2 values Hece 2 startig values eeded 7

8 Modified Secat Rule Alterate to Fidig Derivatives Not a difficulty if 1 is kow Make a small chage i the first value to get a 2d value Will geerally ot coverge as quickly as NewtoRaphso but easier to use 15 Secat Rule Example Previous problem with secat rule iter x f(x) εa, % Seve iteratios beyod x1 were required to get same aswer obtaied with Newto-Raphso i five But o differetiatio required 8

9 Geeral Numerical Derivative lim "f "x # 0 "x Choose some small chage! ad evaluate the fuctio at x+! ad x-! f! ( x) = "f = f ( x +! ) $ f ( x $! ) f! ( x) % f (x +! ) $ f (x $! ) 2! 17 Example of Numerical Derivative For f (x) = sec2(x)*si(x) + ta(x)*cos(x) At x* = 1.47 f (x) = Take small δ, say x*+δ = , x*- δ = Usig these values i derivative formula f (x) = ( )/ = Close to the actual value 18 9

10 Estimatig Startig Values Newto-Raphso ad secat rule easy to use ad work well oce startig value(s) selected Startig value suggested by problem, e.g. voltage may be ear 120, room temperature ear 72 of, etc Use such physical uderstadig whe possible If ot obvious, approximate equatio ad solve simpler form to get startig value 19 Equatios with Power Terms Pipe flow problem had pressure drop equatio: ΔP - 5.4*V1.8 = 0, ΔP = 1440 Need startig value for V Approximate V1.8 as V2 The ΔP 5.4*V2 = 0 yields V = 16.3 Good startig value Root at V =

11 Polyomial Roots Polyomials: ax + bx-1 + cx zx + cost = 0 To estimate largest root cosider oly highest power term ad costat To estimate smallest root cosider oly liear term ad costat 21 Multiple Equatios with Secat Rule Secat rule may be applied to 2 equatios f(x,y) = 0 ad g(x,y) = 0 Solvig f(x.y) for x ad g(x,y) for y alterately util covergece or divergece) results (7.7.1a) x =x +1 ( f (x, y ) * (x x ) ( f ( x, y ) f ( x, y )) (7.7.1b) y =y (g(x, y ) * ( y y ) ( g ( x, y ) g ( x, y ))

12 Two Simultaeous Equatios (7.4.1) (7.4.2) f(x,y) = x*si(y) ex = 0 g(x,y) = y*cos(x) + ey = 0 x, y i radias 23 Polyomial Example Smallest root: 0.900xs = xs = Actual root at x = This a good example; ot all work so well If oe approximatio attempt fails, try somethig else; No guaratees Each polyomial eq. root foud reduces equatio order by 1 24 x3 2.55x2 +0.9x = 0 Largest root: xl3 = xl = Actual root at x =

13 Two-equatio Newto-Raphso ivolves 4 partial derivatives # & "g "f! g ( x, y ) % f ( x, y ) "y "y x,y (' $ x,y x+1 = x! "f "g "f "g! "x x,y "y x,y "y x,y "x x,y # "f "g &! f ( x, y ) % g ( x, y ) "y "y x,y (' $ x,y y+1 = y! "f "g "f "g! "x x,y "y x,y "y x,y "x x,y 25 Two-Equatio NewtoRaphso Example Applyig this equatio: iter x y f g * *10-8 f(x,y) = x*si(y) ex = 0 g(x,y) = y*cos(x) + ey =

14 Solver for Multiple Equatios Oly oe target cell possible, eve with multiple equatios Wat both f(x,y) ad g(x,y) = 0 Not eough to make sum 0 sice plus ad mius fuctio values could offset oe aother Solutio is to square f ad g ad require f2 + g2 = 0. Make this target value 27 For f(x,y) = x*si(y) ex = 0 g(x,y) = y*cos(x) + ey = 0, 28 Put x i C2 ad y i C3 f(x,y) i C4, g(x,y) i C5 ad their sum i C6 Put appropriate labels i colum B Ivoke solver Target cell value (C5) is 0 by maipulatig C2:C3 rage 14

15 Solver Results Fid x= y= f(x,y) = g(x,y) = f^2+g^2= e