2 SolutionofNonlinearEquationsf(x)=007-2.4 Newton-RaphsonandScantMethodsNumericalMethodsResearchBuilding 710#, UESTC
2 Solution of Nonlinear Equations f(x)=0 07-2.4 Newton-Raphson and Scant Methods Research Building 710#, UESTC
Slope Methods for Finding RootsIf f(x),f'(x),andf(x)are continuous near a rootp.The extra informationregardingthenatureoff(x)canbeusedym=f'(po)y=f(x)Theslopeequation0-f(po)mPr-PoPoP1f(po)P2Pi=Po(P1,f(pi))f'(po)(Po. f(Po)f(p)Pz = Pf'(p)Figure2.13Thegeometric constructionof p1andP2fortheNewton-Raphsonmethodf(Pk-1)Pk = g(Pk-1) = Pk-If'(pk-l)9/22/20265:58AM
9/22/2026 5:58 AM 2 ⚫ If 𝒇 𝒙 ,𝒇 ′ 𝒙 , 𝒂𝒏𝒅 𝒇 ′′ 𝒙 are continuous near a root 𝒑. The extra information regarding the nature of 𝒇(𝒙) can be used. ⚫ The slope equation 0 '( ) m f p = 0 1 0 0 f p( ) m p p − = − 0 1 0 0 ( ) '( ) f p p p f p = − 1 2 1 1 ( ) '( ) f p p p f p = − 1 1 1 1 ( ) ( ) '( ) k k k k k f p p g p p f p − − − − = = −
SlopeMethods forFindingRootsTheorem 2.5 Newton-RaphsonTheorem.AssumefEC2[a,bl,wheref(p)=0. If f'(p)+0 then there exists a s>0 such that the sequence(pk)k=o defined by the iterationf(pk-1)Pk= g(Pk-)= p)f'(pk-1)willconvergencetopforanyinitialapproximationpoE[p-S,p+]Remarkf(x)g(x)=xf'(x)Newton-Raphsoniterationfunctionf(p)= 0= g(x)=xProof.Taylorpolynomial9/22/20265:58AM
9/22/2026 5:58 AM 3 ⚫ Theorem 2.5 Newton-Raphson Theorem. Assume 𝒇 ∈ 𝑪𝟐[𝒂, 𝒃] , where 𝒇 𝒑 = 𝟎. If 𝒇′(𝒑) ≠ 𝟎 then there exists a 𝜹>0 such that the sequence {𝒑𝒌}𝒌=𝟎 ∞ defined by the iteration will convergence to 𝒑 for any initial approximation 𝒑𝟎 ∈ [𝒑 − 𝜹, 𝒑 + 𝜹] Remark Newton-Raphson iteration function Proof. Taylor polynomial 1 1 1 1 ( ) ( ) '( ) k k k k k f p p g p p f p − − − − = = − ( ) ( ) - '( ) f x g x x f x = f p( ) = 0 = g x x ( )
Slope Methods for Finding RootsCorollary2.2Newton'sIterationforFindingSguare Roots,AssumeA>0howtocomputeADProof.f(x)= x? -Af(x)g(x)=xf'(x)2xg(x)=±+A / x29/22/20265:58AM
9/22/2026 5:58 AM 4 ⚫ Corollary 2.2 Newton’s Iteration for Finding Square Roots, Assume 𝑨 > 𝟎, how to compute 𝑨 ⚫ Proof. 1 1 2 k k k A p p p − − + = 2 f x x A ( ) = − 2 2 ( ) ( ) - '( ) f x x A g x x x f x x− = = − 2 / ( ) x A x g x + =
Slope Methods for Finding RootsEx.2.11 Use Newton's square-root algorithm to find 5Example2.11.UseNewton'ssquare-rootalgorithmtofind/5Startingwithpo=2andusingformula(11),wecompute2+5/2=2.25Pi=2Ex.2.12Projectileisfired2.25+5/2.25=2.236111111P2=22.236111111+5/2.2361111112.236067978y=v,t-16t2, and x=vrP322.36067978+5/2.2360679782.236067978p4=2ConsidertheairresistanceFurther iterations produce pk2.236067978 fork≥4,so we see that convergenceaccuratetoninedecimalplaceshasbeenachievedy= f(t)=(Cu, +32C2)(1-e-t/c)-32Ctx=r(t)=Cu. (1-e-"C)b。 = 45°,U, = U, = 160 ft / sec,C = 10f(8) = 83.22, f(9) =-31.5Po= 8, f'(p)=-104.3283.22f(x)=8.7977=pi=8g(x)= x-104.32f'(x)9/22/20265:58AM
9/22/2026 5:58 AM 5 ⚫ Ex. 2.11 Use Newton’s square-root algorithm to find 𝟓 ⚫ Ex. 2.12 Projectile is fired ⚫ Consider the air resistance 2 , 16 y x y t t and x = − = ( ) ( ) 2 32 1 32 1 / / ( ) ( ) ( ) t C y t C x y f t C C e Ct x r t C e − − = = + − − = = − 0 0 45 160 10 , / sec, y x b ft C = = = = f f ( ) . , ( ) . 8 83 22 9 31 5 = = − 0 0 p f p = = − 8 104 32 , '( ) . 1 83 22 8 8 7977 104 32 ( ) . ( ) - . '( ) . f x g x x p f x = = − = −