matlab函数求极值.ppt

函数的极值,1、一元函数的极值,函数命令：fminbnd,调用格式：x,feval,exitflag,output=fminbnd(fun,x1,x2,options),%求fun在区间(x1,x2)上的极值. 返回值： x:函数fun在(x1,x2)内的极值点 feval:求得函数的极值 exitflag: exitflag0,函数收敛于解x处 exitflag=0,已达最大迭代次数 exiflag0,函数在计算区间内不收敛.,例1：求函数,在,上的极小值.,fun=inline(x+pi)*exp(abs(sin(x+pi),x,feval,exitflag,output=fminbnd(fun,-pi/2,pi/2),fun = Inline function: fun(x) = (x+pi)*exp(abs(sin(x+pi) x = -1.2999e-005 feval = 3.1416 exitflag = 1 output = iterations: 21 funcCount: 22 algorithm: golden section search, parabolic interpolation message: 1x112 char,xx=-pi/2:pi/200:pi/2; yxx=(xx+pi).*exp(abs(sin(xx+pi); plot(xx,yxx) xlabel(x),grid on % 可以用命令xx,yy=ginput(1) 从局部图上取出极值点及相应函数值,例2：求解函数humps的极小值.,type humps %humps 是一个Matlab提供的M函数文件,function out1,out2 = humps(x) %HUMPS A function used by QUADDEMO, ZERODEMO and FPLOTDEMO. % Y = HUMPS(X) is a function with strong maxima near x = .3 % and x = .9. % % X,Y = HUMPS(X) also returns X. With no input arguments, % HUMPS uses X = 0:.05:1. % % Example: % plot(humps) % % See QUADDEMO, ZERODEMO and FPLOTDEMO. % Copyright 1984-2002 The MathWorks, Inc. % $Revision: 5.8 $ $Date: 2002/04/15 03:34:07 $ if nargin=0, x = 0:.05:1; end,y = 1 ./ (x-.3).2 + .01) + 1 ./ (x-.9).2 + .04) - 6; if nargout=2, out1 = x; out2 = y; else out1 = y; end,x,y=fminbnd(humps,0.5,0.8) x = 0.6370 y = 11.2528 xx=0:0.001:2; yy=humps(xx); plot(xx,yy),例3：求,在(0,1)内的极小值.,type myfunmin1 %显示M文件内容 function f=myfunmin1(x) f=x.x; x,y=fminbnd(myfunmin1,0,1) x = 0.3679 y = 0.6922 xx=0:0.001:1; yy=myfunmin1(xx); plot(xx,yy),x,y=fminbnd(x.x,0,1) x = 0.3679 y = 0.6922,2、 多元函数的极值,函数命令：fminsearch 调用格式：x,feval,exitflag,output=fminsearch(fun,x0,optipons) % 求在x0附近的极值,例4：求,的极小值.,type myfunmin2,function f=myfunmin2(v) x=v(1); y=v(2); f=100*(y-x.2).2+(1-x).2;,sx,sfeval=fminsearch(myfunmin2,1 1),sx = 1 1 sfeval = 0 x0=-5,-2,2,5;-5,-2,2,5 sx,sfeval=fminsearch(myfunmin2,x0),x0 = -5 -2 2 5 -5 -2 2 5 sx = 1.0000 -0.6897 0.4151 8.0886 1.0000 -1.9168 4.9643 7.8004 sfeval = 2.4112e-010 disp(myfunmin2(sx(:,1),myfunmin2(sx(:,2),myfunmin2(sx(:,3), myfunmin2(sx(:,4) 1.0e+005 * 0.0000 0.0058 0.0230 3.3211,3、函数零点,函数命令：fzero 求一元函数的零点 调用格式：x,feval,exitflag,output=fzero(fun,x0,options),例5：求,的零点。,（1）采用符号计算 S=solve(sin(t)2*exp(-0.1*t)-0.5*abs(t),t) S = 0.,y=inline(sin(t).2.*exp(-0.1*t)-0.5*abs(t),t); t=-10:0.01:10; Y=y(t); plot(t,Y,r); % r 用红色作图 hold on %在原来基础上作图 plot(t,zeros(size(t),k); %y横为0，黑色 xlabel(t);ylabel(y(t) hold off,（2）数值法求解,zoom on %获局部放大图 tt,yy=ginput(5);zoom off %用鼠标获5个零点燃猜测值 tt yy,tt = -2.0046 -0.5300 -0.1152 0.6221 1.6820 yy = -0.0088 -0.0088 -0.0088 -0.0088 -0.0088 t1,y1=fzero(y,-0.1) -0.1附近找 t1 = -0.5198 y1 = 0 t2,y2=fzero(y,2) t2 = 1.6738 y2 = 2.2204e-016,非线性方程组的求解：fsolve solve 代数方程解 语法：x,fval=fsolve(fun,x0),例6：求解非线性方程组：,function f=myfunsolve1(v) %v是一个数组 x=v(1); y=v(2); z=v(3); f(1)=sin(x)+y+z2.*exp(x)-4; f(2)=x+y*z; f(3)=x*y*z; f=fsolve(myfunsolve1,1,1,1),Optimization terminated: relative function value changing by less than max(options.TolFun2,eps) and sum-of-squares of function values is less than sqrt(options.TolFun). f = 0.0005 -0.0002 1.9995 yy=myfunsolve1(f) yy = 1.0e-006 * -0.2636 0.1200 -0.2131,例7：求一个3阶方阵A，使A3=1 2 3;4 5 6;7 8 9,function F=myfsolve2(A) F=A3-1 2 3;4 5 6;7 8 9; A=fsolve(myfsolve2,ones(3) % ones（3）初始矩阵，直全为1 myfsolve2(A),4、数值积分,函数命令：quad quadl odbquad 调用格式: q1=quad(fun,a,b,tol) %采用自适应Simpson算法计算积分 q1=quadl(fun,a,b,tol) %采用自适应Lobatto算法计算积分 q2=dblquad(fun,xmin,xmax,ymin,ymax,tol) %二重积分 q3=triplequad(fun,xmin,xmax,ymin,ymax,zmin,zmax,tol) %三重积分,例1：求I=,.,(1)利用quad,quadl计算 format long I=quad(exp(-x.2),0,1,1e-8) Il=quadl(exp(-x.2),0,1,1e-8) I = 0.74682413285445 Il = 0.74682413398845,(2) 采用编程计算:,function I=myquad1(a,b,n) x=linspace(a,b,n); %把ab区间平均分成n等份 y=exp(-x.2)*(b-a)/n; %高底=每个取边梯形的面积 I=sum(y); I1=myquad1(0,1,10000) I2=myquad1(0,1,100000) I1 = 0.74681784375801 I2 = 0.74682350396218,例2：计算,利用dblquad s1=dblquad(x.y,0,1,1,2) , %被积函数写成数组的形式 db_(double) 二重积分，即两次积分 s1 = 0.40546626724351 s2=dblquad(x,y)x.y,0,1,1,2) %采用匿名函数表示被积函数 x y的区间 s2 = 0.40546626724351 s3=dblquad(inline(x.y),0,1,1,2) s3 = 0.40546626724351,求s1=,s2=,s3=,例3：,syms x x符号变量 f1=1/(1+x2); f2=1/(x2+2*x+3); f3=1/(x2+2*x-3); s1=int(f1,1,inf) 1到正无穷 s2=int(f2,-inf,inf) int符号积分 s3=int(f3,-inf,inf) s1 = 1/4*pi s2 = 1/2*pi*2(1/2) s3 = NaN 不确定的结果,s4=qu