Nonlinear Equations How can we solve these equations? ! = 40 %/' - PowerPoint PPT Presentation

Nonlinear Equations

How can we solve these equations? ! = 40 %/' • Spring force: ! = # $ What is the displacement when - I ! = 2N? F- = Kx o ( XI 0.05M = I = -2N -0.05M x 40 Nlm k

How can we solve these equations? • Drag force: * ! = 0.5 %-/' ! = 0.5 + ( , - . ) = / ( . ) = What is the velocity when ! = 20N ? ⇐ ÷H ← , to ⇐ 6.5 mls TRA - IFA F - give → it - Ff ⇒ E- 6.3 m/s → v -

E- µ if ⇒ FIZ - O * ! = 0.5 %-/' f- ( v ) = 0 0 . = / ( . ) −! = 0 IT Find the root (zero) of the nonlinear equation 0 . tf@loQ6-3m1s0O0r7ooo.s → Nonlinear Equations in 1D Goal: Solve 0 $ = 0 for 0: ℛ → ℛ iterative numerical - Often called Root Finding =

interval that * Define Bisection method ghastlier interval :[ a. b ] • a=o;b l O ) > to - lb - a 1=10 fan l is 1/1 KH q * Define midpoint : f softa m-ts.az - )l0 l * check the signs to Bo beto ! B A .lt//sinmkqItesIffa).f( iffca.fm ) > o : A -0 | tan :c :[ miss igneous else b) so = = Ia , my new b. = M

to = lb - al a- to Bisection method ti-lb-al-e.to , 2 2 i - tf te tf l ma - f l - tf = tf l l ts ! I , i i j ! i ! : its A=0 b- to fritz / to = 10 - * every iteration , the interval 2 ! divided by is

Convergence An iterative method converges with rate 5 if: ||2 !"# || lim ||2 ! || $ = +, 0 < + < ∞ 5 = 1: linear convergence .→0 Linear convergence gains a constant number of accurate digits each step (and + < 1 matters!) =kjm4fItY¥7 For example: Power Iteration constant -8 → linear convergence → show convergence → constant f I & Xz - Xi = at → faster convergence as X , = A Xz C → increases a

Convergence An iterative method converges with rate 5 if: E' " " " " ||> .34 || lim ||> . || 5 = +, 0 < + < ∞ .→0 5 = 1: linear convergence 2 5 > 1: superlinear convergence L re l 5 = 2: quadratic convergence Linear convergence gains a constant number of accurate digits each step (and + < 1 matters!) Quadratic convergence doubles the number of accurate digits in each step (however it only starts making sense once ||> . || is small (and + does not matter much)

X * is the root Convergence → ta errox* * • The bisection method does not estimate $ . , the approximation of the desired root $. It instead finds an interval smaller than a given tolerance that contains the root. ¥ → stop to I tr L t :i÷÷÷÷:¥÷÷ . ' ⑤ ← at the lr-Itc-Ige-adienea.nu 1¥ .

in general : ties lol Example: 1bjgIcto12k2lb-alTJeogzfbIoaT@fkslogzf8z.Ig Consider the nonlinear equation 0 $ = 0.5$ ) − 2 and solving f x = 0 using the Bisection Method. For each of the initial intervals below, how many iterations are required to ensure the root is accurate within 2 67 ? ) -7.3 flb ) flat ) .f( b) so → ok ! ( 8 iterations ) ffa A) [−10, −1.8] > logzf5.fi/- Muffat .fCbI2o → hot0k!k B) [−3, −2.1] 6.56 f- (a) fasho → on ! G- iterations ) C) [−4, 1.9]

Bisection method ' ' Ii iii. . . i . Algorithm: 1.Take two points, ! and " , on each side of the root such that #(!) and #(") have opposite signs. 2.Calculate the midpoint & = !"# $ 3. Evaluate #(&) and use & to replace either ! or " , keeping the signs of the endpoints opposite.

Bisection Method - summary q The function must be continuous with a root in the interval L, M = q Requires only one function evaluations for each iteration! o The first iteration requires two function evaluations. e q Given the initial internal [L, M] , the length of the interval after # iterations is 869 - ) ! q Has linear convergence

Newton’s method of f Cx ) • Recall we want to solve ! " = 0 for !: ℛ → ℛ linear approximation EE - • The Taylor expansion: = Ich ) ! " ! + ℎ ≈ ! " ! + !′ " ! ℎ - - gives a linear approximation for the nonlinear function ! near " ! . f- ( xn th ) = O ' Au ) h = O Ich ) ⇒ → fan ) t f × . = random tininess ) # → Newton update fh=-s¥;I → Newtons Neawetgoritnm : f- Go ) , f' Go ) → h → ftp.ti-Xkth

Find x # s .t . f # too Newton’s method Stx ) ( ft - line atxn tangent - - • - slope .IS#-O---ftxn)J g. fed - - - - - - - - t - Xktl Xin ' ° f' Gr ) . . / x Yeti × * *FfHH= × k-f¥ITy O : " g

= ? Example X , O Xo = Consider solving the nonlinear equation ffxf-2ettx-5.gl/kti=Xrthh=-fCXn)DftXI=2e+zxf'Cxn 5 = 2.0 / " + " # ↳ → What is the result of applying one iteration of Newton’s method for solving nonlinear equations with initial starting guess " $ = 0, i.e. what is " % ? A) −2 ) B) 0.75 C) −1.5 D) 1.5 h=¥÷fI Xo → ffto ) - I -5=-3 E) 3.0 six , - a h -4.5 x , = Xo th - Ot 1.5 -71 × 1=1.57

Newton’s Method - summary q Must be started with initial guess close enough to root (convergence is - only local). Otherwise it may not converge at all. - q Requires function and first derivative evaluation at each iteration (think - about two function evaluations) q Typically has quadratic convergence ④ re ||> .34 || lim ||> . || ) = +, 0 < + < ∞ .→0 q What can we do when the derivative evaluation is too costly (or difficult to evaluate)?

If ⇒ approximation for f ' ( x ) Secant method Also derived from Taylor expansion, but instead of using 0′ $ . , it approximates the tangent with the secant line: → Xkt , = Xk - f(X# $ .34 = $ . − 0 $ . /0′ $ . df ( § xn ) : : : ÷÷÷¥ 1 × 0,57 * ÷ ⇒ ;/ Tzpointss ! , XkH=Xk-f§¥÷#¥ - t far ) -9 - - - - - - t ! • y* Xk - l Xk re

Secant Method - summary q Still local convergence q Requires only one function evaluation per iteration (only the first iteration requires two function evaluations) q Needs two starting guesses q Has slower convergence than Newton’s Method – superlinear convergence ||> .34 || lim ||> . || 5 = +, 1 < 5 < 2 .→0 ' → df f

1D methods for root finding: Method Update Convergence Cost Check signs of & ' and Linear (/ = 1 and c = 0.5) Bisection One function evaluation per & ( iteration, no need to compute derivatives ) ! = |( − '| 2 ! Superlinear / = 1.618 , Secant 6 !"# = 6 ! + ℎ One function evaluation per local convergence properties, iteration (two evaluations for ℎ = −& 6 ! /&′ 6 ! convergence depends on the the initial guesses only), no initial guess need to compute derivatives Quadratic / = 2 , local Newton 6 !"# = 6 ! + ℎ Two function evaluations per convergence properties, iteration, requires first order ℎ = −& 6 ! />&' convergence depends on the derivatives initial guess >&' = & 6 ! − & 6 !$# 6 ! − 6 !$#