非線性方程及非線性方程組解決

1、第八章非線性方程及非線性方程組解決1abx0 x1a1b2x*1區(qū)間對(duì)分法（二分法）1. 確定有根區(qū)間:2. 逐次對(duì)分區(qū)間：3. 取根的近似值:b1a22其誤差為:根的近似值:abx0 x1a1b2x*b1a23用對(duì)分區(qū)間法求根步驟：4 f=inline(x3+10*x-20,x); x,err=bisection(f,1,2,0.00005,15) n x err 1.000 1.500 0.500 2.000 1.750 0.250 3.000 1.625 0.125 4.000 1.56250000000000 0.000 5.000 1.59375000000000 0.000 6.0

2、00 1.600 0.00 7.000 1.60 0.000 8.000 1.59765625000000 0.000 9.000 1.59570312500000 0.000 10.000 1.59472656250000 0.000 11.000 1.59423828125000 0.000 12.000 1.59448242187500 0.500 13.000 1.59460449218750 0.250 14.000 1.59454345703125 0.625 15.000 1.59457397460938 0.813x = 1.59457397460938err =3.0000e

3、-00552簡(jiǎn)單迭代法2.1 簡(jiǎn)單迭代法的一般形式及其幾何意義6xyy = xxyy = xxyy = xxyy = xx*x*x*x*x0p0 x1p1x0p0 x1p1x0p0 x1p1x0p0 x1p17顯然，此迭代序列發(fā)散。8 f=inline(x3+11*x-20,x); iteration(f)enter initial guess x1=1.5allowable tolerance tol=0.01maximum number of iterations max=3zero not found to desired tolerance step x 1.0000 1.5000 2

4、.0000 -0.1250 3.0000 -21.3770ans = 1.5000 -0.1250 -21.37709 f=inline(20/(x2+10),x); iteration(f)enter initial guess x1=1.5allowable tolerance tol=0.0000005maximum number of iterations max=20iteration method has convergedstep x 1.000 1.500 2.000 1.63265306122449 3.000 1.579 4.000 1.685 5.000 1.592 6.

5、000 1.59559279984346 7.000 1.59414421311147 8.000 1.59473154634776 9.000 1.59449342271545 10.000 1.59458996763346 11.000 1.59455082476108 12.000 1.59456669477999 13.000 1.59456026047567 14.000 1.59456286918682 15.000 1.59456181151656 16.000 1.5945622403361610代入初值得：例2的結(jié)果表明，對(duì)同一方程可構(gòu)造不同的迭代格式，產(chǎn)生的迭代序列收斂性也

6、不同。迭代序列的收斂性取決于迭代函數(shù)在方程的根的鄰近的性態(tài)。112.2 1.迭代法的收斂條件1213由定理8.2，迭代收斂142.3Steffensen (斯蒂芬森)方法簡(jiǎn)單迭代法的加速（一）收斂速度15（二）Steffensen(斯蒂芬森)方法將Aitken（埃特肯）加速技巧用于線性收斂的迭代序列，即得Steffensen方法，其計(jì)算過(guò)程為16 iter_steffen(leonardo_iter)enter initial guess x1=2allowable tolerance tol=0.000001maximum number of iterations max=10iterati

7、on method has convergedstep x 1 2.000 2 1.192 3 1.35956267950143 4 1.36878079644430 5 1.36880810758180 6 1.36880810782137 iteration(leonardo_iter)enter initial guess x1=2allowable tolerance tol=0.0001maximum number of iterations max=10zero not found to desired tolerancestep x 1 2.000 2 0.400 3 1.961

8、60000000000 4 0.47562601431040 5 1.94399636218919 6 0.586 7 1.93485140020862 8 0.52692937882011 9 1.92983865092369 10 0.53641914395953 function f=leonardo_iter(x)f=(20-x.3-2*x.2)/10function f=leonardo(x)%達(dá)芬奇在1425年研究此方程%并得到1.368808107f=x.3+2*x.2.+10*x-2017 Newton法與弦截法3.1Newton法基本思想：將非線性方程線性化，以線性方程的解

9、逼近非線性方程的解。18由此導(dǎo)出迭代公式xyx*x019解：代入初值得：Newton法迭代公式為20由定理8.4，Newton法至少二階收斂。21以上兩種改進(jìn)都是至少二階收斂的。22在Newton公式中，用差商代替導(dǎo)數(shù)，即即得迭代公式3.2弦截法23x0 x1切線 /* tangent line */割線 /* secant line */切線斜率割線斜率24證明略，因弦截法非單步法，不能用定理8.4判別證明參考（關(guān)治，陸金甫 數(shù)值分析基礎(chǔ)）。25 newton(leonardo,leonardo_pr)enter initial guess x1=2allowable tolerance t

10、ol=0.000001maximum number of iterations max=10Newton method has convergedstep x y 1 2.000 16.0000 2 1.46666666666667 2.123851851851853 3 1.372 0.4323 4 1.36881022263390 0.6963 5 1.36880810782267 0.7313 6 1.36880810782137 0.0000ans = 2.0000 1.4667 1.3715 1.3688 1.3688 1.3688function f=leonardo_pr(x)f

11、=3*x.2.+4*x+10function f=leonardo(x)f=x.3+2*x.2.+10*x-2026 secant(leonardo)enter initial guess x0=1enter initial guess x1=2allowable tolerance tol=0.000001maximum number of iterations max=10secant method has converged step x y 1 1.000 -7.0000 2 2.000 16.0000 3 1.396 -1.334757951836934 4 1.3579123046

12、5787 -0.2290 5 1.369 0.2105 6 1.36880745972192 -0.000 7 1.36880810778288 -0.2143ans = 1.0000 2.0000 1.3043 1.3579 1.3690 1.3688 1.368827類似割線法，過(guò)三點(diǎn)做f(x)的二次插值多項(xiàng)式4拋物線法（muller法）28y(x)xSecant linex1拋物線插值x2x3Parabola29 mullerEnter function f(x)=x.3+10*x-20enter the initial guess xr=1.75enter the interval l

13、ength h=0.25enter the toleration tol=0.000001maximum number of iteration max=10 step x y 1 1.5949003375 0.0059626661 2 1.5945609031 -0.0000213915 3 1.5945621166 0.0000000001 4 1.5945621166 0.0000000000muller method has converged30 mullerEnter function f(x)=x.3+2*x.2+10*x-20enter the initial guess xr=1.5enter the interval length h=0.5enter the toleration tol=0.000001maximum number of iteration max=10 step x y 1 1.3702416528 0.0302548158 2 1.3688024583 -0.0001191819 3