MATLAB结果: (1)lim(3?9)
x??xx1x>> syms x;
>> f=(3^x+9^x)^(1/x); >> L=limit(f,x,inf) L = 9
(2)limx?0y?0xyxy?1?1
>> syms x y ;
>> f=(x*y)/(((x*y)+1)^(1/2)-1); >> L=limit(limit(f,x,0),y,0) L = 2
(3)limx?0y?01?cos(x2?y2)(x?y)e22x2?y2
>> syms x y ;
f=(1-cos(x^2+y^2))/((x^2+y^2)*exp(x^2+y^2)); >> L=limit(limit(f,x,0),y,0) L = 0
?x?lncostd2ydy8、已知参数方程?,试求出和2dxdxy?cost?tsint?MATLAB结果:
t??/3
>> syms t;x=log(cos(t));y=cos(t)-t*sin(t);f1=diff(y,t)/diff(x,t)
f1 =
-(-2*sin(t)-t*cos(t))/sin(t)*cos(t)
>> f2=diff(y,t,2)/diff(x,t,2);subs(f2,t,sym(pi/3)) ans =
3/8-1/24*pi*3^(1/2) 9、假设f(x,y)??MATLAB结果:
>> syms x y t;f=int(exp(-t^2),t,0,x*y);
F=(x/y)*(diff(f,x,2))-2*(diff(diff(f,x,1),y,1))+diff(f,y,2) F =
2*x^2*y^2*exp(-x^2*y^2)-2*exp(-x^2*y^2)-2*x^3*y*exp(-x^2*y^2)
xy0x?2f?2f?2fedt,试求?2?2。 2y?x?x?y?y?t210、试求出下面的极限。
?1?111 (1)lim?2?2?2????; 2n??2?14?16?1(2n)?1??(2)limn(n??1111?????)。 2222n??n?2?n?3?n?n?MATLAB结果:
?1?111(1)lim?2?2?2???? 2n??2?14?16?1(2n)?1??>> syms n m;limit(symsum(1/((2*m)^2-1),m,1,n),n,inf) ans = 1/2
(2)limn(n??1111?????) 2222n??n?2?n?3?n?n?>> syms m n;limit(symsum(n*(1/(n^2+m*pi)),m,1,n),n,inf) ans = 1
11、试求出以下的曲线积分。
(1)?(x2?y2)ds,l为曲线x?a(cost?tsint),y?a(sint?tcost),
l (0?t?2?)。
(2)?(yx3?ey)dx?(xy3?xey?2y)dy,其中l为a2x2?b2y2?c2正向上半
l椭圆。
MATLAB结果:
(1)
>> syms t;syms a positive;x=a*(cos(t)+t*sin(t));y=a*(sin(t)-t*cos(t)); >> I=int((x^2+y^2)*sqrt(diff(x,t)^2+diff(y,t)^2),t,0,2*pi) I =
2*a^3*pi^2+4*a^3*pi^4 (2)
>> syms t x y;syms a c b positive;x=c*cos(t)/a;y=c*sin(t)/b; F=[y*x^3+exp(y),x*y^3+x*exp(y)-2*y];ds=[diff(x,t);diff(y,t)]; >> I=int(F*ds,t,pi,0) I =
2/15*c*(-2*c^4+15*b^4)/a/b^4
?a4?4?b12、试求出Vandermonde矩阵A??c4?4?d?e4?形式显示结果。 MATLAB结果: >> syms a b c d e;
a3b3c3d3e3a2b2c2d2e2a1??b1?c1?的行列式,并以最简的
?d1?e1??A=[a^4,a^3,a^2,a,1;b^4,b^3,b^2,b,1;c^4,c^3,c^2,c,1;d^4,d^3,d^2,d,1;e^4,e^3,e^2,e,1];
simple(det(A)) ans =
(a-c)*(b-c)*(b-a)*(d-c)*(d-a)*(d-b)*(-c+e)*(e-a)*(e-b)*(e-d)
??20.5?0.50.5??0?1.50.5?0.5??进行Jordan变换,13、试对矩阵A??并得出变换矩阵。
?20.5?4.50.5???21?2?2??MATLAB结果:
>> A=[-2 0.5 -0.5 0.5;0 -1.5 0.5 -0.5;2 0.5 -4.5 0.5;2 1 -2 -2]; [V,J]=jordan(A) V =
0 0.50000000000000 0.50000000000000 -0.25000000000000 0 0 0.50000000000000 1.00000000000000 0.25000000000000 0.50000000000000 0.50000000000000 -0.25000000000000 0.25000000000000 0.50000000000000 1.00000000000000 -0.25000000000000 J =
-4 0 0 0 0 -2 1 0 0 0 -2 1 0 0 0 -2
14、试用数值方法和解析方法求取下面的Sylvester方程,并验证得出的结果。
?6?40?3?142?2???63?67???13100?11?403?05?4??3?X??0?4????21?1??12??3?21??4???X??2?92???5?61?
?????2?196?4?4???????66?3??MATLAB结果:
数值解:
>> A=[3 -6 -4 0 5;1 4 2 -2 4;-6 3 -6 7 3;-13 10 0 -11 0;0 4 0 3 4]; B=[3 -2 1;-2 -9 2;-2 -1 9];C=-[-2 1 -1;4 1 2;5 -6 1;6 -4 -4;-6 6 -3]; X=lyap(A,B,C),norm(A*X+X*B+C) X =
4.05689641914739 14.51278207738230 -1.56531072870960 -0.03555750760035 -25.07427777300226 2.74080695262868 -9.48864527098864 -25.93232765032886 4.41773341615773 -2.69692229648079 -21.64500921402601 2.88508413748940 -7.72287238941912 -31.90996053309000 3.76340871020125 ans =
2.791709573159607e-013
15、假设已知矩阵A如下,试求出eAt,sinAt,eAtsin(A2eAtt)。
0.5?1.5???4.50??0.5?40.5?0.5?? A???1.51?2.51.5????1?1?3??0
百度搜索“77cn”或“免费范文网”即可找到本站免费阅读全部范文。收藏本站方便下次阅读,免费范文网,提供经典小说综合文库东北大学 matlab实验(2)在线全文阅读。
相关推荐: