1、3101301218 邱振涛练习三：符号运算用 MATLAB符号运算做下列各题1、 求极限 ， ，0ln()l(imhxLlim1nnaM1/0()lixxePsyms x hL=limit(log(x+h)-log(x)/h,h,0)L =1/xsyms a nM=limit(1-a/n)n,n,inf)M =exp(-a)syms xP=limit(exp(1)-(1+x)(1/x)/x,x,0)P =NaN2、 求和 ， ，21nkS21k32(1)nxSsyms k nS1=symsum(k2,k,1,n)S1 =1/3*(n+1)3-1/2*(n+1)2+1/6*n+1/6syms
2、k nS2=symsum(1/k2,k,1,inf)S2 =1/6*pi2syms x nS3=symsum(xn/(n*(n-1),n,2,inf)S3 =-(x-1)*log(1-x)+x3、 求导数 并化简、求偏导数2l()yxa(5)y32(1,3)2sinxyzsyms x adiff(log(x+sqrt(a2+x2),x,5)=(105/(a2+x2)(9/2)*x5-150/(a2+x2)(7/2)*x3+45/(a2+x2)(5/2)*x)/(x+(a2+x2)(1/2)-5*(-(5)y15/(a2+x2)(7/2)*x4+18/(a2+x2)(5/2)*x2-3/(a2+
3、x2)(3/2)/(x+(a2+x2)(1/2)2*(1+1/(a2+x2)(1/2)*x)+20*(3/(a2+x2)(5/2)*x3-3/(a2+x2)(3/2)*x)/(x+(a2+x2)(1/2)3*(1+1/(a2+x2)(1/2)*x)2-10*(3/(a2+x2)(5/2)*x3-3/(a2+x2)(3/2)*x)/(x+(a2+x2)(1/2)2*(-1/(a2+x2)(3/2)*x2+1/(a2+x2)(1/2)-60*(-1/(a2+x2)(3/2)*x2+1/(a2+x2)(1/2)/(x+(a2+x2)(1/2)4*(1+1/(a2+x2)(1/2)*x)3+30*(-
4、1/(a2+x2)(3/2)*x2+1/(a2+x2)(1/2)2/(x+(a2+x2)(1/2)3*(1+1/(a2+x2)(1/2)*x)+24*(1+1/(a2+x2)(1/2)*x)5/(x+(a2+x2)(1/2)5化简得 =3*(3*a4-24*a2*x2+8*x4)/(a2+x2)(9/2)()ysyms x y z f f1 f2 f3 f4f=diff(sin(x2*y*z),x,2)f1=diff(f,y,1)f2=subs(f1,x,1)f3=subs(f2,y,1)f4=subs(f3,z,3)= 88.278432(1,3)2sinxyz4、求不定积分 ，2d()Ix
5、 d()(lnl)xJababsyms x ii=int(x2+1)/(x2-2*x+2)2,x)i =3/2*atan(x-1)+1/4*(2*x-6)/(x2-2*x+2)syms a b x jj=int(1/(x*(sqrt(a+log(x)+sqrt(b+log(x),x)j=-2/3/(-b+a)*(b+log(x)(3/2)+2/3/(-b+a)*(a+log(x)(3/2)5、求定积分 ，含参积分20dxIe 3()sin(2)dxIyxysyms x II=int(exp(-x2),0,inf)I =1/2*pi(1/2)syms x y ff=int(x-y)3*sin(x
6、+2*x),y)subs(f,x)-subs(f,-x)I(x)= -1/4*(x-y)4*sin(3*x)-1/4*(-x-y)4*sin(3*x)6、试求正弦函数 的 次麦克劳林展开式 ，在同一幅图上画出正弦函数以及它的 次麦克劳林多项式的图像，观察逼近情况。sinyx(3,579,1)nn-6 -4 -2 0 2 4 6-2-1012Taylor Series AproximationTN(x) =x-1/6 x3-6 -4 -2 0 2 4 6-2-1012Taylor Series AproximationTN(x) =x-1/6 x3+1/120 x5-6 -4 -2 0 2 4
7、6-2-1012Taylor Series AproximationTN(x) =x-1/6 x3+1/120 x5-1/5040 x7-6 -4 -2 0 2 4 6-2-1012Taylor Series AproximationTN(x) =x-1/6 x3+1/120 x5-1/5040 x7+1/36280 x9-6 -4 -2 0 2 4 6-2-1012Taylor Series AproximationTN(x) =x-1/6 x3+1/120 x5-1/5040 x7+1/36280 x9-1/391680 x17、试求方程或方程组的解：(1) ; (2) 32410x2uv
8、pqsyms x x=solve(4*x4-2*x2-3*x+1)x = 1/6*(37+9*17(1/2)(1/3)-1/3/(37+9*17(1/2)(1/3)-1/3-1/12*(37+9*17(1/2)(1/3)+1/6/(37+9*17(1/2)(1/3)-1/3+1/2*i*3(1/2)*(1/6*(37+9*17(1/2)(1/3)+1/3/(37+9*17(1/2)(1/3)-1/12*(37+9*17(1/2)(1/3)+1/6/(37+9*17(1/2)(1/3)-1/3-1/2*i*3(1/2)*(1/6*(37+9*17(1/2)(1/3)+1/3/(37+9*17(1
9、/2)(1/3)syms u v p q u,v=solve(u2+v2=2*p*q,u-v=p+q,p,q)u = 1/2*u-1/2*v-1/2*i*(u+v)1/2*u-1/2*v+1/2*i*(u+v)v = 1/2*u-1/2*v+1/2*i*(u+v)1/2*u-1/2*v-1/2*i*(u+v)8、试解常微分方程：(1) (2) sin,/2)0.xt2,()1(/a)0.y(3) , 是 的函数。,()()yxxt(1) syms x t x=dsolve(Dx=x+sin(t),x(pi/2)=0)x =-1/2*cos(t)-1/2*sin(t)+1/2*exp(t)/(cosh(pi)+sinh(pi)(1/2)(2) syms y ay=dsolve(D2y=-a2*y,y(0)=1,Dy(pi/a)=0)y =cos(a*t)(3) syms y x t ss=dsolve(Dx=y,Dy=x,x(0)=0,y(0)=1)x=s.x,y=s.ys = x: 1x1 symy: 1x1 symx =1/2*exp(t)-1/2*exp(-t)y =1/2*exp(t)+1/2*exp(-t)
