MATLAB 例题解答

1、1.说出以下4条指令产生的结果各属于哪种数据类型，是“双精度”对象，还是“符号”对象?3/7+0.1; sym(3/7+0.1) ; vpa(sym(3/7+0.1),4) ; vpa(sym(3/7+0.1)syms aa=3/7+0.1class(a) a =0.5286ans =double syms bb=sym(3/7+0.1)class(b) b =37/70ans =sym syms cc=vpa(sym(3/7+0.1),4)class(c) c =0.5286ans =sym syms dd=vpa(sym(3/7+0.1)class(d) d =0.528571428571

2、42857142857142857143ans =sym 2在不加专门指定的情况下，以下符号表达式中的哪一个变量被认为是独立自由变量。sym('sin(w*t)');sym('a*exp(-X)');sym('z*sxp(j*th)')symvar(sym('sin(w*t)') ans = t, w symvar(sym('a*exp(-X)') ans = X, a symvar(sym('z*sxp(j*th)') ans = j, th, z 3现有方程f1(x)=x3+(4-a)*x2+(

3、3-4*a)*x-3*a=0和f2(x)=x2-a*x+a2=0，且在这两个方程中，都有a>0的前提。请写出实现以下要求的程序：（1）只求f1(x)=0的正实根，而不出现其他形式的根；syms x a positivef1=x3+(4-a)*x2+(3-4*a)*x-3*arf=solve(f1,x) f1 =x3 + (4 - a)*x2 + (3 - 4*a)*x - 3*arf =a （2）求f2(x)=0的全部根。syms x clearsyms a positivesyms xf2=x2-a*x+a2rf=solve(f2,x) f2 =a2 - a*x + x2rf = a/

4、2 + (3(1/2)*a*i)/2 a/2 - (3(1/2)*a*i)/2 4观察一个数（在此用记述）在以下四条不同指令作用下的异同：a=，b=sym()，c=sym(,'d')，d=sym('')在此分别代表7/3，pi/3，pi*3(1/3)；而异同通过vpa(abs(a-d)，vpa(abs(b-d)，vpa(abs(c-d)等来观察。cleara=7/3b=sym(7/3)c=sym(7/3,'d')d=sym('7/3')vpa(abs(a-d)vpa(abs(b-d)vpa(abs(c-d) a =2.3333b

5、=7/3c =2.3333333333333334813630699500209d =7/3ans =0.0ans =0.0ans =0.00000000000000014802973661668756666666667788716 a=pi/3b=sym(pi/3)c=sym(pi/3,'d')d=sym('pi/3')vpa(abs(a-d)vpa(abs(b-d)vpa(abs(c-d) a =1.0472b =pi/3c =1.047197551196597631317786181171d =pi/3ans =0.0ans =0.0ans =0.0000

6、0000000000011483642827992216762806615818554 a=pi*3(1/3)b=sym(pi*3(1/3)c=sym(pi*3(1/3),'d')d=sym('pi*3(1/3)')vpa(abs(a-d)vpa(abs(b-d)vpa(abs(c-d) a =4.5310b =1275352044764433/281474976710656c =4.5309606547207899041040946030989d =pi*3(1/3)ans =0.0000000000000002660111416629094437484239

7、3221638ans =0.00000000000000026601114166290944374842393221638ans =0.0000000000000002660111416629094726767991785515 5.求符号矩阵A=的行列式值和逆，所得结果应采用“子表达式置换”简洁化。syms a11 a12 a13 a21 a22 a23 a31 a32 a33A=a11,a12,a13;a21,a22,a23;a31,a32,a33DA=det(A) IA=inv(A) IAw=subexpr(IA,'w') A = a11, a12, a13 a21, a

8、22, a23 a31, a32, a33DA =a11*a22*a33 - a11*a23*a32 - a12*a21*a33 + a12*a23*a31 + a13*a21*a32 - a13*a22*a31IA = (a22*a33 - a23*a32)/(a11*a22*a33 - a11*a23*a32 - a12*a21*a33 + a12*a23*a31 + a13*a21*a32 - a13*a22*a31), -(a12*a33 - a13*a32)/(a11*a22*a33 - a11*a23*a32 - a12*a21*a33 + a12*a23*a31 + a13*a2

9、1*a32 - a13*a22*a31), (a12*a23 - a13*a22)/(a11*a22*a33 - a11*a23*a32 - a12*a21*a33 + a12*a23*a31 + a13*a21*a32 - a13*a22*a31) -(a21*a33 - a23*a31)/(a11*a22*a33 - a11*a23*a32 - a12*a21*a33 + a12*a23*a31 + a13*a21*a32 - a13*a22*a31), (a11*a33 - a13*a31)/(a11*a22*a33 - a11*a23*a32 - a12*a21*a33 + a12*a

10、23*a31 + a13*a21*a32 - a13*a22*a31), -(a11*a23 - a13*a21)/(a11*a22*a33 - a11*a23*a32 - a12*a21*a33 + a12*a23*a31 + a13*a21*a32 - a13*a22*a31) (a21*a32 - a22*a31)/(a11*a22*a33 - a11*a23*a32 - a12*a21*a33 + a12*a23*a31 + a13*a21*a32 - a13*a22*a31), -(a11*a32 - a12*a31)/(a11*a22*a33 - a11*a23*a32 - a12

11、*a21*a33 + a12*a23*a31 + a13*a21*a32 - a13*a22*a31), (a11*a22 - a12*a21)/(a11*a22*a33 - a11*a23*a32 - a12*a21*a33 + a12*a23*a31 + a13*a21*a32 - a13*a22*a31)w = 1/(a11*a22*a33 - a11*a23*a32 - a12*a21*a33 + a12*a23*a31 + a13*a21*a32 - a13*a22*a31)IAw = w*(a22*a33 - a23*a32), -w*(a12*a33 - a13*a32), w*

12、(a12*a23 - a13*a22) -w*(a21*a33 - a23*a31), w*(a11*a33 - a13*a31), -w*(a11*a23 - a13*a21) w*(a21*a32 - a22*a31), -w*(a11*a32 - a12*a31), w*(a11*a22 - a12*a21) 6.求的符号解，并进而用该符号解求的准确值（提示：注意subs的使用）。syms x kf=xks=symsum(f,k,0,inf)s1=subs(s,'x','-1/3')s2=subs(s,'x','1/pi')

13、s3=subs(s,'x','3') f =xks =piecewise(1 <= x, Inf, abs(x) < 1, -1/(x - 1)s1 =3/4s2 =-1/(1/pi - 1)s3 =Inf 7.对于x>0，求。（提示：理论结果为ln x；注意限定性假设。）syms x positivesyms kf=(2/(2*k+1)*(x-1)/(x+1)(2*k+1)s=symsum(f,k,0,inf) f =(2*(x - 1)/(x + 1)(2*k + 1)/(2*k + 1)s =2*atanh(x - 1)/(x + 1)

14、8.（1）通过符号计算求y(t)=|sin t|的导数。（2）然后根据此结果，求和。syms t dy_psyms d positivey_p=sin(t) dy_p=limit(subs(y_p,t,t+d)-y_p)/d,d,0) y_n=-sin(t) dy_n=limit(y_n-subs(y_n,t,t-d)/d,d,0) y_p =sin(t)dy_p =cos(t)y_n =-sin(t)dy_n =-cos(t) dy_n0=limit(subs(y_n,t,0)-subs(y_n,t,-d)/d,d,0) dy_p_=limit(subs(y_p,t,pi/2+d)-subs

15、(y_p,t,pi/2)/d,d,0) dy_n0 =-1dy_p_ =0 9.求出的具有64位有效数字的积分数值。clearsyms xf=(exp(-abs(x)*abs(sin(x);l=ezplot(f,-5*pi,1.7*pi);grid onhold onset(l,'color','r','LineWidth',5)title('f=exp(-abs(x)*abs(sin(x)')hold off s=int(f,x,-5*pi,1.7*pi)Vs=vpa(s) s =2*exp(-pi) + exp(-2*pi)

16、+ exp(-3*pi) + exp(-4*pi) + exp(-5*pi)/2 - exp(-(17*pi)/10)/8 - (5(1/2)*exp(-(17*pi)/10)/8 + (2(1/2)*exp(-(17*pi)/10)*(5 - 5(1/2)(1/2)/8 + 1Vs =1.0878494172555037011026337644989 10.计算二重积分。syms x yF=int(int(x2+y2,y,1,x2),x,1,2)VF=vpa(F) F =1006/105VF =9.5809523809523809523809523809524 11.在0,2区间，画出y(x

17、)=曲线，并计算y(4.5)。（提示：int，subs。）syms x t y(t)=int(sin(t)/t,t,0,x)l=ezplot(y(t),0,2*pi)grid onhold onset(l,'color','r','LineWidth',5)title('y(t)=int(sin(t)/t,t,0,x)')hold offf=int(sin(t)/t,t,0,4.5) y(t) =sinint(x)l = 174.0144f =sinint(9/2) 12.在n>0的限制下，求y(n)=的一般积分表达式，并计

18、算y(1/3)的32位有效数字表达。syms n positivesyms xy=int(sin(x)n,x,0,pi/2)vpa(subs(y,n,1/3),32) y =beta(1/2, n/2 + 1/2)/2ans =1.2935547796148951782413405453553 13.有序列,(在此)，求这两个序列的卷积。syms k n a bxk=akhk=bkxh_n=subs(xk,k,k-n)*subs(hk,k,n)yk=simple(simple(symsum(xh_n,n,0,k) xk =akhk =bkxh_n =a(k - n)*bnyk =piecewise(a = b, bk*(k + 1), a = b, (a*ak - b*bk)/(a - b) 20.求方程x2+y2=1，x*y=2的解。（提示：正确使用solve。）clearsyms x yS