matlab编程

E304731541

d=sym(3*x^3-2*x+1);
diff(d,2)
5 R2 k+ \) T" ?" W2 X  @& M, {7 Uezplot('cos(x)',[0,pi])//4.2
ezplot('(cos(t))^3','(sin(t))^3',[0,2*pi])//4.3
( l, @* U% K$ R7 Y' \ezplot('exp(x)+sin(x*y)',[-2,0.5,0,2])//4.4
fminbnd('x+3*(x^2+cos(x))',-1,1)  //直接求
f=sym('x+3*(x^2+cos(x))');//由导数为零来求解
  ff=solve(diff(f));
subs(f,x,ff),subs(f,x,-1),subs(f,x,-1)
% Q  l" f" A8 j  N5 o3 K% U  Jsyms x;
int(1/sin(x)^2*cos(x)^2)
syms x z;
# C: S( a1 b; X/ X3 D; |int(x/(1+z^2),x)
  |8 }! ]/ ?; r, G$ w3 k9 JA=[1,2,3;4,5,6;7,8,9;10,11,12]
A =
     1     2     3
     4     5     6
8 l7 z2 V0 ^+ }     7     8     9
/ f$ M, }6 K- G    10    11    12
B=[0,1,1;2,2,1;1,1,1];
A*B
A^3//错误，矩阵的幂必须矩阵的行列相同才行
rref(A)
9 l$ O3 a) ]8 n. T. W# N# Ox=linspace(0,2*pi,30);
1 {+ t1 z  ?. p& i; Wy=sin(x);
z=cos(x);
plot(x,y,'r',x,z,'pg')
+ a0 @9 l' |$ c" F* C% T4 b. }$ xx=linspace(0,2*pi,30);
6 h3 P0 t" X3 [2 z9 k6 dy=sin(x);
/ p4 e: v4 L' m* o& ?: t3 jz=cos(x);
9 X: ^( `  J4 ^9 {" ~xlabel('自变量x')
1 F3 v3 q+ T1 L8 ]ylabel('函数y')
' S; C- D4 H# I6 V  Mtitle('示意图')
plot(x,y,'r',x,z,'pg'),grid on;
x=linspace(0,2*pi,30);
y=sin(x);
z=cos(x);
plot(x,z,'pg');
8 R6 ], N# p3 c+ jhold on;
7 }# Y+ W$ l5 D1 K% x6 ?3 {) bplot(x,y,'r')
x=linspace(0,2*pi,100);
* e1 A$ c: C# e  }7 V3 jy=sin(x);
2 u% R3 c9 {: Z1 ez=cos(x);
7 q3 r0 A2 x( q4 mplot(x,y);
0 I  l" ~8 \( {$ o' P1 `3 ~9 g, Ctitle('sin(x)')
) Z+ K6 O* }  @/ r: o: w! S; f, m+ Fpause;
: ?0 j0 M$ U) S4 ifigure(2);
7 ?9 R! ?* e5 ]' r( c: G# L8 M6 V8 nplot(x,z,'pg');
title('cos(x)')
x=linspace(0,2*pi,30);
% p' E' x" P5 M& T# y, zy=sin(x);
0 L+ U5 q0 f4 Bplot(x,y,'r');
zoom on
p=[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。
7 R+ X7 N. X; L, V% W# Nr=solve('x^3-4*x^2+9*x-10=0')//简单方法
5 H- @3 f9 ~" I/ q& T* d3 H5 ]r=solve('x^2-a*x-4*b=0','x')
d=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

hahahahahahhaaaaaaaaaaaa

zhula

2# mrx
5 c$ p# X& o% I: w4 L! o3 p谢谢啊！！

zhula

2# mrx
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~~