解方程的二分法 matlab 程序

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解方程的二分法  matlab 程序  （ x^3-3*x^2+6*x-1=0）
* h2 t  `2 r  y; m3 C3 {
! [, U; G6 }; F% H) c; _% Bisection.m
function[k,x,f_value]=Bisection(f,a,b,eps1,eps2,N)
fprintf('k ,a ,b ,x, f\n')
5 v3 u. I. K0 d" {/ m5 |; c+ cfor k=1:N
0 b! }* }  p) c9 F. G    x=(a+b)/2;
2 R6 Q* I9 Q" f) ^% k% K    f_value=f(x);
    fprintf( '%3d, %10.9f, %10.9f, %10.9f,%10.9f,\n'...
: Z6 d& G* ^$ ~; ~. J# B      ,k ,a ,b ,x, f_value)
    if abs(f_value)< eps1||0.5*(b-a)<eps2
1 w% @& J( h9 w- |) w9 {) D    return
    else
        if f(x)*f(a)<0
8 S# e! O7 J* @% r          b=x;
        else
+ |0 c+ I6 u! z          a=x;
        end
        if k== N
' Q4 X8 R% m1 v0 j$ [2 M( i          warning ('算法超出最大迭代数!')   
        end
) K2 e( P9 n& s- C$ {, B    end

% examlpe7.m
a=0; b=1;
4 S# N' }2 p8 [' F1 P! m1 U0 Peps1=1e-4;eps2=1e-4;
N=300;
f=@ (x) (x^3-3*x^2+6*x-1);
Hfun=@Bisection;
3 Q% K; c: J' f. X% _% O$ _[k,x,f_value]= feval(Hfun, f,a,b,eps1,eps2,N);
( e3 E" P; h/ e+ _. \
1 Q8 Q/ B4 Y" c: N, A运行结果
# _  \' ?! L/ |7 d, L. j7 i1 n>> examlpe7
) ^8 x4 ~. @! S( mk ,a ,b ,x, f
0 B  R# h3 O" C) \  ?  1, 0.000000000, 1.000000000, 0.500000000,1.375000000,
  2, 0.000000000, 0.500000000, 0.250000000,0.328125000,
7 b) d7 @; n) [' @( E  3, 0.000000000, 0.250000000, 0.125000000,-0.294921875,
  4, 0.125000000, 0.250000000, 0.187500000,0.026123047,
  5, 0.125000000, 0.187500000, 0.156250000,-0.131927490,
  6, 0.156250000, 0.187500000, 0.171875000,-0.052295685,
  7, 0.171875000, 0.187500000, 0.179687500,-0.012936115,
  8, 0.179687500, 0.187500000, 0.183593750,0.006630838,
0 b( w3 K" L' K- T! r( j7 W  9, 0.179687500, 0.183593750, 0.181640625,-0.003143273,
& v& w. b/ v1 f. r- F 10, 0.181640625, 0.183593750, 0.182617188,0.001746121,
3 t! T- S& R  G: K* x, `* h3 c 11, 0.181640625, 0.182617188, 0.182128906,-0.000697991,
12, 0.182128906, 0.182617188, 0.182373047,0.000524211,
13, 0.182128906, 0.182373047, 0.182250977,-0.000086854,