บทที่ 2 รากของสมการตัวแปรเดียว

11/6/2012Overview5fx rfx rxrxrภOverview6f x a, b f a ∙ f b < 0f x = 0a, babababab3

11/6/2012Bisection Method7แนวคิดการสุ่มเลือกช่วงที่บรรจุค่าราก แล้วแบ่งช่วงนั้นออกเป็นสองส่วนเท่าๆกันตรวจสอบว่าค่ารากอยู่ทางซ้ายหรือขวาของจุดแบ่ง ท้าซ้าๆ เช่นนี้ไป จนจุดแบ่งเข้าใกล้ค่ารากมากขึ้นfyaf bax2x3x 1bxBisection Method: Example102e x + x − 4 = 0 0,1f af 0 = 2 + 0 − 4 = −2f 1 = 2e + 1 − 4 = 2.4x if bf 0 ∙ f 1 < 0 f x 0,1x 1 = 0+12 = 0.50,1x i = a+b24

11/6/2012Bisection Method: Example11f x 1 < 0 f a f x 1 > 0 x 1 , b = 0.5,1a = x 1 = 0.5x 1x 2 , x 3 , x 4 , …i a b f(a) f(b) xi f(xi) |x(i)-x(i-1)|1 0.000000000 1.000000000 -2.000000000 2.436563657 0.500000000 -0.2025574592 0.500000000 1.000000000 -0.202557459 2.436563657 0.750000000 0.984000033 0.2500000003 0.500000000 0.750000000 -0.202557459 0.984000033 0.625000000 0.361491915 0.1250000004 0.500000000 0.625000000 -0.202557459 0.361491915 0.562500000 0.072609314 0.0625000005 0.500000000 0.562500000 -0.202557459 0.072609314 0.531250000 -0.066635396 0.0312500006 0.531250000 0.562500000 -0.066635396 0.072609314 0.546875000 0.002565113 0.0156250007 0.531250000 0.546875000 -0.066635396 0.002565113 0.539062500 -0.032139781 0.0078125008 0.539062500 0.546875000 -0.032139781 0.002565113 0.542968750 -0.014813596 0.0039062509 0.542968750 0.546875000 -0.014813596 0.002565113 0.544921875 -0.006130820 0.00195312510 0.544921875 0.546875000 -0.006130820 0.002565113 0.545898438 -0.001784500 0.00097656311 0.545898438 0.546875000 -0.001784500 0.002565113 0.546386719 0.000389895 0.00048828112 0.545898438 0.546386719 -0.001784500 0.000389895 0.546142578 -0.000697405 0.00024414113 0.546142578 0.546386719 -0.000697405 0.000389895 0.546264648 -0.000153781 0.00012207014 0.546264648 0.546386719 -0.000153781 0.000389895 0.546325684 0.000118051 0.00006103515 0.546264648 0.546325684 -0.000153781 0.000118051 0.546295166 -0.000017867 0.000030518Bisection Method: Example12x 15 = 0.546295166f x 15 = 0.179 × 10 −415x 15 − x 14 = 0.305 × 10 −4x i − x i−1 x 15 − x 14 = 0.305 ×10 −4 ≤ 0.5 × 10 −3x 15 = 0.546295166x 15 = 0.5463 D.P.5

11/6/2012False Position Method13b, f bxa, f ayb,f b0ax 1x x2 3a,f a x , f 1 x 1bxFalse Position Method14y − y 1 = m(x − x 1 )y − f a =xf b −f ab−ax = a −x i =a, f ax − ay = 0f a b−af b −f aaf b −bf af b −f am = y 2−y 1x 2 −x 1b, f by =f b −f ab−a=af b −bf af b −f ax − a + f a6

11/6/2012False Position Method15x ix i x i f x i = 0f x i ∙ f af x i ∙ f a > 0 a x i ax if x i ∙ f a < 0 b x i bx ix i+1False Position Method: Example182e x + x − 4 = 00, 1x i =i = 1f a ∙ f b = f 0 ∙ f 1 = −2 ∙ 2.4 < 00,1af b −bf af b −f ax 1 =af b −bf af b −f a= 0.45079935, f x 1 = −0.41017

11/6/2012False Position Method: Example19f x 1 ∙ f a = −0.4101 ∙ −2 > 0 x 1α ∈ x 1 , b a x 1 a = x 1a b f(a) f(b) xi f(xi) |x(i)-x(i-1)|0.00000000 1.00000000 -2.00000000 2.43656366 0.45079935 -0.410068030.45079935 1.00000000 -0.41006803 2.43656366 0.52991377 -0.07251459 0.079114420.52991377 1.00000000 -0.07251459 2.43656366 0.54349968 -0.01245460 0.013585910.54349968 1.00000000 -0.01245460 2.43656366 0.54582124 -0.00212822 0.002321550.54582124 1.00000000 -0.00212822 2.43656366 0.54621759 -0.00036335 0.000396360.54621759 1.00000000 -0.00036335 2.43656366 0.54628525 -0.00006202 0.000067660.54628525 1.00000000 -0.00006202 2.43656366 0.54629680 -0.00001059 0.000011557 x 7 − x 6 = 0. 115 × 10 −4 < 0.5 × 10 −4 4 D.P.x 15Modified method of false position20แนวคิดขจัดข้อจ้ากัดของระเบียบวิธีการวางตัวผิดที่ค่าขอบด้านใดด้านหนึ่งถูกตรึงไว้ โดยใช้เทคนิคการลดความชันของเส้นตรงโดยลดความสูงของขอบลงครึ่งหนึ่ง ทุกครั้งที่ค่ารากบรรจุอยู่ในช่วงที่ซ้ากันเป็นครั้งที่สองyb,f b0ax x21a, f a x , f 1 x 1x 3 1 b , f b 2 1 b , f b 4 bx8

11/6/2012Modified method of false position233. α ∈ x 2 , b ( )x 2 (a = x 2 )x 3 =x 31af b −bf a41f b −f a4α ∈ x 1 , x 2 ( x 2 )x 2 (b = x 2 )( 1)Modified method of false position244. α ∈ a, x 1 x 2 x 2α ∈ a, x 2f ax 2 (b = x 2 )x 35.x 3 = af b −1 bf a2f b − 1 f a210

11/6/2012Modified method of false position25( ) f x r ∙ f x r−1 > 0f x r ∙ f x r−1 > 0a x 028Modified method of false position:Example2e x + x − 4 = 0 0,1f a ∙ f b = −2 × 2.4 < 0 0,1x 1x 1 =af b −bf af b −f a= 0.450799347 f x 1 = −0.4101f a ∙ f x 1 > 0 a x 1 a = x 1x 1 =1af b −bf a21 = 0.589104778 f x2 f b −f a 2 = 0.19385311411

11/6/201229Modified method of false position:Examplef x 1 ∙ f x 2 < 0 ( ) b = x 2x 3x 3 =0.450799347f 0.589104778 − 0.589104778f 0.450799347f 0.589104778 − f 0.450799347= 0.544710011f x 3 = −0.007073315f x 2 ∙ f x 3 < 0 ( ) a = x 3x 4x 4 =0.544710011f 0.589104778 − 0.589104778f 0.544710011f 0.589104778 − f 0.544710011= 0.546272862f x 4 = −0.00011720030Modified method of false position:Examplef x 3 ∙ f x 4 > 0a = x 4x 5 (f b = 1 f b )21∙ 0.546272862f 0.589104778 − 0.589104778f 0.546272862x 5 =21∙ f 0.589104778 − f 0.5462728622= 0.546324590f x 5 = 0.00011318112

11/6/201231Modified method of false position:Examplei a b f(a) f(b) x f(x) |xi-x(i-1)|1 0.000000000 1.000000000 -2.000000000 2.436563657 0.450799347 -0.4100680282 0.450799347 1.000000000 -0.410068028 2.436563657 0.589104778 0.193853114 0.1383054313 0.450799347 0.589104778 -0.410068028 0.193853114 0.544710011 -0.007073315 0.0443947674 0.544710011 0.589104778 -0.007073315 0.193853114 0.546272862 -0.000117200 0.0015628515 0.546272862 0.589104778 -0.000117200 0.193853114 0.546324590 0.000113181 0.0000517286 0.546272862 0.546324590 -0.000117200 0.000113181 0.546299177 -0.000000001 0.0000254137 0.546299177 0.546324590 -0.000000001 0.000113181 0.546299178 0.000000000 0.000000000x 6f x 6 = 0.1 × 10 −8 < 0.5 × 10 −88 D.P. x 7f x 7 = 0.1 × 10 −4 < 0.5 × 10 −44 D.P.Newton-Raphson Methodx i f x ixx i+1x , f x 1 1x 0x2x113

11/6/2012Newton-Raphson Methodf ′ x r f x rx 0 y = f x .x 0 , f x 0y − f x 0 = f ′ x 0 x − x 0y = 0 x ( x 1 )x 1 = x 0 − f x 0f ′ x 0f ′ x 0 ≠ 0xNewton-Raphson Methodi + 1x i+1 = x i − f x if ′ x iy = f x f ′ x i ≠ 0- x 0x 1 , x 2 , … a, b (OPEN Method)14

11/6/2012Newton-Raphson Method: Example2e x + x − 4 = 0 0,1 -f x = 2e x + x − 4 f ′ x = 2e x + 1 x- x i+1 = x i − f x if ′ x ix i+1 = x i − 2ex +x−42e x +1x 0 = 0 f x 0 = −2 f ′ x 0 = 3x 1 = x 0 − f x 0= 0 − −2f ′ x 0 3= 0.666666667Newton-Raphson Method: Examplef x 1 f ′ x 1f x 1 = 0.562134748776 f ′ x 1 = 4.895468082109x 2 = x 1 − f x 1f ′ x 1= 0.666666667 − 0.5621347487764.895468082109 = 0.551839087496x i − x i−1f x i15

11/6/2012Newton-Raphson Method: Examplei x i f x i f ′ x i x i − x i−10 0.000000000000 -2.000000000000 3.0000000000001 0.666666666667 0.562134748776 4.895468082109 0.6666666666672 0.551839087496 0.024726197014 4.472887109518 0.1148275791713 0.546311070265 0.000052966290 4.453741896025 0.0055280172314 0.546299177728 0.000000000244 4.453700822516 0.0000118925375 0.546299177673 0.000000000000 4.453700822327 0.000000000055Convergence Analysisx if α = 0 f ′′ α0 = f α = f x i + f ′ x i α − x i + f′′ ξξ x i α f ′ x i ≠ 0α − x i + f x if ′ x i = − f′′ ξ2f ′ x i α − x i 2x i+1 = x i − f x if ′ x iα − x i+1 = − f′′ ξ2f ′ x i α − x i 22α − x i 216

11/6/2012Convergence Analysisf ′′ M α x iα − x i+1 = f′′ ξ2f ′ x i α − x i 2 ≤ M2 f ′ x iα − x i 2x r+1x rNewton-Raphson Method: Examplex = 3 −x8 D.P.f x = x − 3 −xf f 0 = −1 f 1 = 2 3f 0 ∙ f 1 < 0 f x 0,1 x 0x 0 = 0.5f ′ x = 1 + 3 −x ln 3-x i+1 = x i −x i − 3 −x i1 + 3 −x iln 317

11/6/2012Newton-Raphson Method: Examplei x i f x i f ′ x i x i − x i−10 0.500000000000 -0.077350269190 1.6342841005981 0.547329756902 -0.000767135404 1.602145981268 0.0473297569022 0.547808574321 -0.000000075819 1.601829314883 0.0004788174193 0.547808621654 0.000000000000 1.601829283588 0.000000047333Convergence Analysis (more)- f ′ α ≠ 0α f ′ -f ′ α ≠ 0 α f xg xf x = x − α g x lim x−α g x ≠ 0-18

11/6/2012Newton-Raphson Method: Examplef x = e x − x − 1 x = 0 f 0 = 0f ′ 0 = 0-x i+1 = x i − ex i−x i −1e x i−1Newton-Raphson Method: Examplei x i f x i f ′ x i x i − x i−10 1.000000000000 0.718281828459 1.7182818284591 0.581976706869 0.207595689973 0.789572396842 0.4180232931312 0.319055040911 0.056772008685 0.375827049596 0.26292166595913 0.000173588892 0.000000015067 0.000173603959 0.00017356880514 0.000086796957 0.000000003767 0.000086800724 0.00008679193515 0.000043399107 0.000000000942 0.000043400049 0.00004339785019

11/6/2012Secant Methodf f ff x = x cos x 2 e x ln 3x-fx ix = x i−1 ,f ′ x i = limx→xif x − f x ix − x if ′ x i ≈ f x i−1 − f x ix i−1 − x i= f x i − f x i−1x i − x i−1Secant Methodx i+1 = x i − f x if ′ x i = x i −-f x i= xf x i − f x i − f x i x i − x i−1i−1 f x i − f x i−1x i − x i−1x i+1 = x i − f x i x i − x i−1f x i − f x i−12 x 0 x 1x 2 x 0 , f x 0 x 1 , f x 120

11/6/2012Secant Methodyx 1 , f x 1x 0x 2 x 3 x 4x 1xx 0 , f x 0Secant Method: Example2e x + x − 4 = 0 x 0 = 0x 1 = 1x i+1 = x i − f x ix i −x i−1f x i −f x i−1x 2 = x 1 − f x 1 x 1 − x 0f x 1 − f x = 1 − f 1 1 − 00 f 1 − f 0x 2 = 1 − f 1 1 − 0f 1 − f 0 = 0.4507993521

11/6/2012Secant Method: Examplei a f(x) |xi-x(i-1)|0 0.500000000 0.3775825621 0.785398163 -0.078291382 0.2853981632 0.736384139 0.004517719 0.0490140253 0.739058139 0.000045177 0.0026740004 0.739085149 -0.000000027 0.0000270105 0.739085133 0.000000000 0.000000016Fixed Point Methodf x = 0x = F xx r+1 = F x rF x rx r+123

11/6/2012Fixed Point Method x , x 1 1x0, Fx 0x0, x 1 x 2 x1x0f x = 0x = F xFixed Point Method: Exampleα 1 = −1, α 2 = 4)x 2 − 3x − 4 = 0 (1 F x = 3x + 4x r+1 = F x r = 3x r + 4x 0 = 5x 1 = 3 5 + 4 = 4.3588984x 2 = 3 4.3588984 + 4 = 4.1323960224

11/6/2012Fixed Point Method: Examplei xi f(xi) |xi-x(i-1)|0 5.00000000 6.000000001 4.35889894 1.92330317 0.641101062 4.13239602 0.67950878 0.226502933 4.04934415 0.24915560 0.083051874 4.01846145 0.09264809 0.030882705 4.00691706 0.03463317 0.011544396 4.00259306 0.01297202 0.004324017 4.00097228 0.00486234 0.001620788 4.00036459 0.00182307 0.000607699 4.00013672 0.00068361 0.0002278710 4.00005127 0.00025635 0.0000854511 4.00001923 0.00009613 0.0000320412 4.00000721 0.00003605 0.0000120213 4.00000270 0.00001352 0.0000045114 4.00000101 0.00000507 0.0000016915 4.00000038 0.00000190 0.00000063Fixed Point Method: Example2 x 2 − 3x − 4 = 0x 2 − 3x = 4 x(x − 3) = 4x 0 = 5x = 4x−3x r+1 = F x r = 4x r −3x 1 = 45 − 3 = 225

11/6/2012Fixed Point Method: Examplei xi f(xi) |xi-x(i-1)|0 5.00000000 6.000000001 2.00000000 -6.00000000 3.000000002 -4.00000000 24.00000000 6.000000003 -0.57142857 -1.95918367 3.428571434 -1.12000000 0.61440000 0.548571435 -0.97087379 -0.14478273 0.149126216 -1.00733496 0.03672862 0.036461187 -0.99816962 -0.00914857 0.009165358 -1.00045781 0.00228924 0.002288199 -0.99988556 -0.00057218 0.0005722410 -1.00002861 0.00014305 0.0001430511 -0.99999285 -0.00003576 0.0000357612 -1.00000179 0.00000894 0.0000089413 -0.99999955 -0.00000224 0.0000022414 -1.00000011 0.00000056 0.0000005615 -0.99999997 -0.00000014 0.00000014Fixed Point Method: Example3 x 2 − 3x − 4 = 0x 0 = 5x = 1 3 x2 − 4x r +1 = F x r = 1 3 x r 2 − 4i xi f(xi) |xi-x(i-1)|0 5.00 6.001 7.00 24.00 2.002 15.00 176.00 8.003 73.67 5201.78 58.674 1807.59 3261964.20 1733.935 1089128.99 1186198697305.98 1087321.4026

11/6/2012Fixed Point Method: Examplex 0 = 0i xi f(xi) |xi-x(i-1)|0 0.00000000 -4.000000001 -1.33333333 1.77777778 1.333333332 -0.74074074 -1.22908093 0.592592593 -1.15043439 0.77480243 0.409693644 -0.89216691 -0.52753748 0.258267485 -1.06801274 0.34468941 0.175845836 -0.95311627 -0.23222059 0.114896477 -1.03052313 0.15354730 0.077406868 -0.97934069 -0.10286972 0.051182439 -1.01363060 0.06833880 0.0342899110 -0.99085100 -0.04566129 0.0227796011 -1.00607143 0.03039402 0.0152204312 -0.99594009 -0.02028306 0.01013134Condition of Convergencex r+1 = F x rF x r x r+1 x 0 , x 1 , …x1 x2x 0x1 x 0 x227

11/6/2012Condition of Convergencef xx = αf α = α − F αα = F αx r +1 = F x rx r +1 − α = F x r − αx r+1 −αx r −αcε x r , α= F x r −αx r −αF α x rF ′ c = F x r −αx r −αCondition of ConvergenceF ′ c = x r+1−αx r −αF ′ c x r − α = x r+1 − αM F ′ c ≤ M x r+1 − α ≤ M x r − αM < 1 x r+1 α x rF ′ c < 128

11/6/2012Condition of Convergence: Example323x = 2x 3 = 2 f x = x 3 − 21 x = x 3 + x − 2F x = x 3 + x − 2 x 0 = 1.2F ′ 1.2 1F ′ 1.2 = 3 1.2 2 + 1 = 5.32 > 12 x = 2−x 3 +5xF ′ 1.2 =55−3 1.2 25= 0.136 < 1F ′ x = 5−3x 25Condition of Convergence: Examplei xi f(xi) |xi-x(i-1)|0 1.2000000000 -0.27200000001 1.2544000000 -6.1896806400 0.05440000002 1.2596354630 -6.1922248893 0.00523546303 1.2599074059 -6.1923555463 0.00027194294 1.2599204009 -6.1923617861 0.00001299495 1.2599210190 -6.1923620829 0.00000061826 1.2599210484 -6.1923620970 0.00000002947 1.2599210498 -6.1923620977 0.000000001429

11/6/2012Condition of ConvergenceF x = x −- x r+1 = x r − f x rf ′ x rf xf ′ xF ′ x = 1 − f′ x 2 −f x f ′′ xF ′ x = f x f′′ xf ′ x 2 f ′ x 2f x f ′′ xf ′ x 2 < 1 f x f′′ x < f ′ x 2-Condition of Convergence: Examplex 4 − x = 10 -x 0 = 2f x = x 4 − x − 10f ′ x = 4x 3 − 1 f ′′ x = 12x 2-x 4 − x − 10 12x 2 < 4x 3 − 1 2x 0 = 22 4 − 2 − 10 12 2 2 = 192 < 31 2 = 4 2 3 − 1 230

11/6/2012Condition of Convergence: Example-x r+1 = x r − f x rf ′ x r = x r − x r 4 − x r − 104x r 3 − 1i xi f(xi) |xi-x(i-1)|0 2.000000000 4.0000000001 1.870967742 0.382674568 0.1290322582 1.855780702 0.004818128 0.0151870403 1.855584561 0.000000795 0.0001961414 1.855584529 0.000000000 0.0000000325 1.855584529 0.000000000 0.00000000031