非线性方程及非线性方程组解决

1、第八章非线性方程及非线性方程组解决1abx0 x1a1b2x*1区间对分法（二分法）1. 确定有根区间:2. 逐次对分区间：3. 取根的近似值:b1a22其误差为:根的近似值:abx0 x1a1b2x*b1a23用对分区间法求根步骤：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简单迭代法2.1 简单迭代法的一般形式及其几何意义6xyy = xxyy = xxyy = xxyy = xx*x*x*x*x0p0 x1p1x0p0 x1p1x0p0 x1p1x0p0 x1p17显然，此迭代序列发散。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的结果表明，对同一方程可构造不同的迭代格式，产生的迭代序列收敛性也

6、不同。迭代序列的收敛性取决于迭代函数在方程的根的邻近的性态。112.2 1.迭代法的收敛条件1213由定理8.2，迭代收敛142.3Steffensen (斯蒂芬森)方法简单迭代法的加速（一）收敛速度15（二）Steffensen(斯蒂芬森)方法将Aitken（埃特肯）加速技巧用于线性收敛的迭代序列，即得Steffensen方法，其计算过程为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)%达芬奇在1425年研究此方程%并得到1.368808107f=x.3+2*x.2.+10*x-2017 Newton法与弦截法3.1Newton法基本思想：将非线性方程线性化，以线性方程的解

9、逼近非线性方程的解。18由此导出迭代公式xyx*x019解：代入初值得：Newton法迭代公式为20由定理8.4，Newton法至少二阶收敛。21以上两种改进都是至少二阶收敛的。22在Newton公式中，用差商代替导数，即即得迭代公式3.2弦截法23x0 x1切线 /* tangent line */割线 /* secant line */切线斜率割线斜率24证明略，因弦截法非单步法，不能用定理8.4判别证明参考（关治，陆金甫 数值分析基础）。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类似割线法，过三点做f(x)的二次插值多项式4抛物线法（muller法）28y(x)xSecant linex1抛物线插值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