用matlab 解超越方程~

用matlab 解超越方程~

使用二分法 求解上面超越方程
下面是二分法的函数文件,你直接设置输入参数就可以了
function [c,err,yc]=bisect(f,a,b,delta)
%Input - f is the function
% - a and b are the left and right endpoints
% - delta is the tolerance
%Output - c is the zero
% - yc= f(c)
% - err is the error estimate for c
%If f is defined as an M-file function use the @ notation
% call [c,err,yc]=bisect(@f,a,b,delta).
%If f is defined as an anonymous function use the
% call [c,err,yc]=bisect(f,a,b,delta).
% NUMERICAL METHODS:Matlab Programs
% (c) 2004 by John H.Mathews and Kurtis D.Fink
% Complementary Software to accompany the textbook:
% NUMERICAL METHODS:Using Matlab,Fourth Edition
% ISBN:0-13-065248-2
% Prentice-Hall Pub.Inc.
% One Lake Street
% Upper Saddle River,NJ 07458
ya=f(a);
yb=f(b);
if ya*yb > 0,return,end
max1=1+round((log(b-a)-log(delta))/log(2));
for k=1:max1
c=(a+b)/2;
yc=f(c);
if yc==0
a=c;
b=c;
elseif yb*yc>0
b=c;
yb=yc;
else
a=c;
ya=yc;
end
if b-a < delta,break,end
end
c=(a+b)/2;
err=abs(b-a);
yc=f(c);
结果
[answer,error,value]=bisect(@(x)log(x)-cos(x),1,5,1e-8)
answer =
1.303
error =
7.4506e-009
value =
-5.2774e-009
[answer,error,value]=bisect(@(x)log(x)-cos(x+pi/8),1,5,1e-8)
answer =
1.0909
error =
7.4506e-009
value =
-2.4967e-010
[answer,error,value]=bisect(@(x)log(x)-cos(x+pi/4),0,10,1e-8)
answer =
0.89573
error =
9.3132e-009
value =
-5.8866e-009
answer =
0.72301
error =
9.3132e-009
value =
2.5559e-009