实验三使用Matlab解决微积分问题

【实验学时】2学时

【实验重点】MATLAB符号变量与符号表达式创建方法和命令使用 【实验难点】用符号微积分方法和命令解决数学问题

【目的要求】掌握MATLAB中求函数极限命令；掌握MATLAB导数和微分命令；掌握MATLAB

不定积分和定积分命令；掌握MATLAB级数求和与泰勒级数展开命令。

【实验内容】

cosxex0x4x21. 求下列函数的极限： (1). limsyms x

f=(cos(x)-exp(x/2))/(x^4) limit(f,x,0) ans =NaN

2t (2)lim1

xxsyms x t

limit(((1+(2*t)/x)^(3*x)),x,Inf) ans =exp(6*t) (3) limx03x1 xsyms x

limit(1/x,x,0,'right') ans =Inf

5xlnsinxesinx (4) limx3syms x

limit(5*x+log(sin(x)+exp(sin(x))),x,3,'left')

ans =15+log(sin(3)+exp(sin(3)))

2. 按要求实现下面的求导运算： (1).已知ye2xln(x21)tan(x)，求y，y syms x y

y=exp(2*x)*log(x^2+1)*tan(-x) diff(y)

ans =

-2*exp(2*x)*log(x^2+1)*tan(x)-2*exp(2*x)*x/(x^2+1)*tan(x)-exp(2*x)*log(x^2+1)*(1+tan(x)^2) diff(y,3) ans =

-8*exp(2*x)*log(x^2+1)*tan(x)-12*exp(2*x)*log(x^2+1)*(1+tan(x)^2)-24*exp(2*x)*x/(x^2+1)*tan(x)-12*exp(2*x)/(x^2+1)*tan(x)-24*exp(2*x)*x/(x^2+1)*(1+tan(x)^2)+24*exp(2*x)*x^2/(x^2+1)^2*tan(x)-12*exp(2*x)*log(x^2+1)*tan(x)*(1+tan(x)^2)+12*exp(2*x)/(x^2+1)^2*tan(x)*x+12*exp(2*x)*x^2/(x^2+1)^2*(1+tan(x)^2)-16*exp(2*x)*x^3/(x^2+1)^3*tan(x)-2*exp(2*x)*log(x^2+1)*(1+tan(x)^2)^2-4*exp(2*x)*log(x^2+1)*tan(x)^2*(1+tan(x)^2)-6*exp(2*x)/(x^2+1)*(1+tan(x)^2)-12*exp(2*x)*x/(x^2+1)*tan(x)*(1+tan(x)^2)

x2y2xy(2)已知z(xy)e22z2z2z，求, 2,

xxxysyms x y

z=(x^2+y^2)*exp((x^2+y^2)/(x*y)) diff(z) ans =

2*x*exp((x^2+y^2)/x/y)+(x^2+y^2)*(2/y-(x^2+y^2)/x^

2/y)*exp((x^2+y^2)/x/y)

diff(z,x,2) ans =

2*exp((x^2+y^2)/x/y)+4*x*(2/y-(x^2+y^2)/x^2/y)*exp(

(x^2+y^2)/x/y)+(x^2+y^2)*(-2/x/y+2*(x^2+y^2)/x^3/y)*exp((x^2+y^2)/x/y)+(x^2+y^2)*(2/y-(x^2+y^2)/x^2/y)^2*exp((x^2+y^2)/x/y)

t=diff(z,x,1) t =

2*x*exp((x^2+y^2)/x/y)+(x^2+y^2)*(2/y-(x^2+y^2)/x^

2/y)*exp((x^2+y^2)/x/y)

diff(t,y,1) ans =

2*x*(2/x-(x^2+y^2)/x/y^2)*exp((x^2+y^2)/x/y)+2*y*(2

/y-(x^2+y^2)/x^2/y)*exp((x^2+y^2)/x/y)+(x^2+y^2)*(-2/y^2-2/x^2+(x^2+y^2)/x^2/y^2)*exp((x^2+y^2)/x/y)+(x^2+y^2)*(2/y-(x^2+y^2)/x^2/y)*(2/x-(x^2+y^2)/x/y^2)*exp((x^2+y^2)/x/y)

3. 完成下列积分运算： (1) 求不定积分x3exdx, syms x

f = x^3*exp(-x^2) int(f,x) ans =

-(exp(-x^2)*(x^2 + 1))/2

syms x

f=1/(x*sqrt(x^2+1)) int(f,x) ans =

log(x) - log((x^2 + 1)^(1/2) + 1)

2xdxx12

（2）求定积分：syms x;

f=x/(sin(x))^2;

34x2dx，sin4xcos2xdx； 20sinxint(f,x,pi/4,pi/3) ans =

pi/4 + log(6^(1/2)/2) - (pi*3^(1/2))/9 syms x;

f=(sin(x))^4*(cos(x))^2; int(f,x,0,pi/2) ans = pi/32

（3）求多重积分syms x y; f=x*sin(x);

int(int(f,x,y,sqrt(y)),y,0,1) ans =

5*sin(1) - 4*cos(1) – 2

syms x y z f=x*y*z

int(int(int(f,z,0,x*y),y,0,x),x,0,1) ans = 1/64

10yyxsinxdxdy；

001xxy0xyzdzdydx

4. 试求解无穷级数的和S1111 1447710(3n2)(3n1)syms n;

f=1/((3*n-2)*(3*n+1)); S=symsum(f,n,1,inf) S = 1/3

5. 试求出函数f(x)sinx的麦克劳林幂级数展开式的前9项，并求出关于x2的2x4x3Taylor幂级数展开式的前5项。

syms x;

f=sin(x)/(x^2+4*x+3); r1=taylor(f,0,1) r1 =

0

>> r2=taylor(f,0,2)-r1 r2 = x/3

>> r3=taylor(f,0,3)-r2-r1 r3 =

(4*x^2)/9

>> r4=taylor(f,0,4)-r3-r2-r1 r4 =

(23*x^3)/54

>> r5=taylor(f,0,5)-r4-r3-r2-r1 r5 =

- (34*x^4)/81

>> r6=taylor(f,0,6)-r5-r4-r3-r2-r1 r6 =

(4087*x^5)/9720

>> r7=taylor(f,0,7)-r6-r5-r4-r3-r2-r1 r7 =

- (3067*x^6)/7290

>> r8=taylor(f,0,8)-r7-r6-r5-r4-r3-r2-r1 r8 =

(515273*x^7)/1224720

>> r9=taylor(f,0,9)-r8-r7-r6-r5-r4-r3-r2-r1

r9 =

- (386459*x^8)/918540 >> r1=taylor(f,2,1) r1 =

sin(1)/8 + (x - 1)*(cos(1)/8 - (3*sin(1))/32)

>> r2=taylor(f,2,2)-r1 r2 =

sin(2)/15 - sin(1)/8 - (x - 1)*(cos(1)/8 - (3*sin(1))/32) + (x - 2)*(cos(2)/15 - (8*sin(2))/225)

>> r3=taylor(f,2,3)-r2-r1 r3 =

sin(3)/24 - sin(2)/15 - (x - 2)*(cos(2)/15 - (8*sin(2))/225) + (x - 3)*(cos(3)/24 - (5*sin(3))/288)

>> r4=taylor(f,2,4)-r3-r2-r1 r4 =

sin(4)/35 - sin(3)/24 - (x - 3)*(cos(3)/24 - (5*sin(3))/288) + (x - 4)*(cos(4)/35 - (12*sin(4))/1225)

>> r5=taylor(f,2,5)-r4-r3-r2-r1 r5 =

sin(5)/48 - sin(4)/35 + (x - 5)*(cos(5)/48 - (7*sin(5))/1152) - (x - 4)*(cos(4)/35 - (12*sin(4))/1225)