MATLAB学习（四）线性方程求解，多项式运算，函数局部最优解

>> A=[2 1 3;1 3 2];B=[5;10]

B =

     5
    10

>> X=lsqnonneg(A,B)

X =

         0
    2.8571
    0.7143

>> AB

ans =

         0
    2.8571
    0.7143

　　

>> p=[1 -7 10] %表示x^2-7x+10

p =

     1    -7    10

>> r=roots(p)

r =

     5
     2

>> p=poly(r)%由根创建多项式

p =

     1    -7    10

>> A=ones(3),poly(A)

A =

     1     1     1
     1     1     1
     1     1     1


ans =

    1.0000   -3.0000   -0.0000   -0.0000

>> d=eig(A),[v,d]=eig(A)  %求方阵A的特征值d与特征向量

d =

   -0.0000
   -0.0000
    3.0000


v =

    0.4082    0.7071    0.5774
    0.4082   -0.7071    0.5774
   -0.8165         0    0.5774


d =

   -0.0000         0         0
         0   -0.0000         0
         0         0    3.0000

>> a=[1 -2],b=[1 -5],c=polyder(a,b) %多项式a和b乘积的导数

a =

     1    -2


b =

     1    -5


c =

     2    -7

　　

>> x=[1 2 3 4 5],y=[5.5 43.1 128 290.7 498.4]

x =

     1     2     3     4     5


y =

    5.5000   43.1000  128.0000  290.7000  498.4000

>> plot(x,y,'o')
>> p = polyfit(x,y,3)

p =

   -0.1917   31.5821  -60.3262   35.3400

>> xi=0:0.02:6;
>> yi=polyval(p,xi);
>> plot(x,y,'ro',xi,yi,'b-')
>> 

>> fun=@(x) 2*sin(x)-1

fun = 

    @(x)2*sin(x)-1

>> [x,f]=fminbnd(fun,3,6) %y=f(x)在指定区间[a,b]上的局部极小值指令为： [x,f]=fminbnd(fun,a,b)；返回取极小值时自变量值x与函数值f_

x =

    4.7124


f =

   -3.0000

>> funf=@(x) x(1)^2+2.5*sin(x(2))-x(1)*x(2)^2*x(3)^2

funf = 

    @(x)x(1)^2+2.5*sin(x(2))-x(1)*x(2)^2*x(3)^2

>> [x,f]=fminsearch(funf,[1,-1,0])  %猜一个初始值，从这个点开始找局部最小值

x =

   -0.0000   -1.5708    0.0008


f =

   -2.5000

　　

>> z=@(x,a,b) a*sin(x(1))+b*cos(x(2))

z = 

    @(x,a,b)a*sin(x(1))+b*cos(x(2))

>> [x,f]=fminsearch(@(x) z(x,2,1),[0,0]) %将一个新匿名函数做参数传入

x =

   -1.5708    3.1416


f =

   -3.0000

>>