matlab编程

E304731541

d=sym(3*x^3-2*x+1);
diff(d,2)
ezplot('cos(x)',[0,pi])//4.2
ezplot('(cos(t))^3','(sin(t))^3',[0,2*pi])//4.3
2 l  T6 ?6 _0 s) t3 Jezplot('exp(x)+sin(x*y)',[-2,0.5,0,2])//4.4
fminbnd('x+3*(x^2+cos(x))',-1,1)  //直接求
( D( \  h6 v4 N- Z5 E/ o3 z f=sym('x+3*(x^2+cos(x))');//由导数为零来求解
3 e$ {1 g2 v9 W/ G" J6 v' a  ff=solve(diff(f));
/ C7 S+ D8 d" d  B' Z, s1 zsubs(f,x,ff),subs(f,x,-1),subs(f,x,-1)
( [. j" x& T# Q0 c) ?/ A! vsyms x;
int(1/sin(x)^2*cos(x)^2)
6 m5 l7 L; H3 L9 k+ g% n1 lsyms x z;
% ?! v4 A" O: }  E/ K2 kint(x/(1+z^2),x)
A=[1,2,3;4,5,6;7,8,9;10,11,12]
A =
     1     2     3
     4     5     6
     7     8     9
    10    11    12
. q1 ?8 v/ E7 T: SB=[0,1,1;2,2,1;1,1,1];
" @! a- s8 a1 u. @( J& m$ _/ c4 }A*B
1 c- g0 T* ]  t  e* H3 ?A^3//错误，矩阵的幂必须矩阵的行列相同才行
rref(A)
x=linspace(0,2*pi,30);
& e# Y5 z4 X; S' y# m! ly=sin(x);
z=cos(x);
plot(x,y,'r',x,z,'pg')
x=linspace(0,2*pi,30);
y=sin(x);
/ j- D# p$ s, K. o2 Y  _; z8 \z=cos(x);
xlabel('自变量x')
4 e; j; R& w7 K3 C2 d" ?ylabel('函数y')
title('示意图')
: W" i" R% `# U5 F/ [# Tplot(x,y,'r',x,z,'pg'),grid on;
x=linspace(0,2*pi,30);
y=sin(x);
" D+ m# s. J2 z+ {$ M; lz=cos(x);
plot(x,z,'pg');
8 w  z# v6 ?! E9 @hold on;
plot(x,y,'r')
/ B, A0 `$ G/ t0 f; lx=linspace(0,2*pi,100);
  l" ?8 F- c: i, `2 @y=sin(x);
: B; {0 n  H& Y7 b, V; S8 x/ C' Qz=cos(x);
$ H) f2 P. I. lplot(x,y);
title('sin(x)')
0 X( Q3 k- D: w0 s1 F( F- I6 @pause;
figure(2);
plot(x,z,'pg');
, N' m( c+ ]0 `9 E! utitle('cos(x)')
$ l5 E$ y" q: W# Mx=linspace(0,2*pi,30);
, L  r) V+ o, v/ a6 v8 c& ~y=sin(x);
2 G9 X6 H# c6 q7 b8 ]7 W  e1 Gplot(x,y,'r');
zoom on
/ G- k6 B* b" \" Q9 D' I1 ^, a+ Qp=[1 -4 9 -10];
r=roots(p)//次数小于等于4，若大于4则用该命令不一定能求出所有根的准确形式。
4 A, L( N+ p# S, T! ^r=solve('x^3-4*x^2+9*x-10=0')//简单方法
r=solve('x^2-a*x-4*b=0','x')
: d+ _! f- m* Fd=solve('p*sin(x)-r','x')
r=fzero('x-(cos(x))^2',1)//其中必须输入的是方程对应的函数名
[x,y]=solve('x+y=1','x-11*y=5','x','y')

mrx

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qingqing20072

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qingqing20072

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qingqing20072

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xxj198712

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zhula

2# mrx
谢谢啊！！

zhula

2# mrx
/ N5 @: k) j# i, u( \) ]. r太感谢了

zhula

2# mrx
最近在做有关促销的文章，可以借鉴，３Ｑ

byit

Thanks you~~