matlab arccos uint8,《高等应用数学问题的MATLAB求解》——第3章习题代码

(1)求极限

[ lim_{xrightarrow infty} (3^x+9^x)^{1/x},

lim_{xrightarrowinfty}frac{(x+2)^{x+2}(x+3)^{x+3}}{(x+5)^{2x+5}},

lim_{x rightarrow a}{left(frac{tan x}{tan a}right)^{cot(x-a)}},\lim_{xrightarrow 0}left[frac{1}{ln(x+sqrt{1+x^2})}-frac{1}{ln(1+x)}right],\lim_{xrightarrow infty}left[sqrt[3]{x^3+x^2+x+1}-sqrt{x^2+x+1}frac{ln(e^x+x}{x}right]]>> syms x,a;

>> f1=(3^x+9^x)^(1/x);

>> f2=(x+2)^(x+2)*(x+3)^(x+3)/(x+5)^(2*x+5);

>> f3=(tan(x)/tan(a))^cot(x-a);

>> f4=1/log(x+sqrt(1+x^2))-1/log(1+x);

>> f5=(x^3+x^2+x+1)^(1/3)-sqrt(x^2+x+1)*log(exp(x)+x)/x;

>> limit(f1,x,inf)

>> limit(f2,x,inf)

>> limit(f3,x,a)

>> limit(f4,x,0)

>> limit(f5,x,inf)

(2)

(5)(y(t)=sqrt{frac{(x-1)(x-2)}{(x-3)(x-4)}})的4阶导数

>> syms x;y=sqrt((x-1)*(x-2)/((x-3)*(x-4)));

>> tic,diff(y,x,4),toc

(6)

(8)直接求极限与洛必达对比：

[lim_{xrightarrow 0}frac{ln(1+x)ln(1-x)-ln(1-x^2)}{x^4}

]>> syms x;f=(log(1+x)*log(1-x)-log(1-x^2))/x^4;

>> f1=log(1+x)*log(1-x)-log(1-x^2);f2=x^4;

>> y1=limit(f,x,0)

>> y2=diff(f1,x,4)/diff(f2,x,4);subs(y2,x,0)

(9)参数方程(begin{cases}x=ln(cos t)\y=cos t-tsin tend{cases})，计算(frac{text{d}y}{text{d}x},frac{text{d}^2y}{text{d}x^2})(例题的paradiff)

>> syms t;x=log(cos(t));y=cos(t)-t*sin(t);

>> paradiff(y,x,t,1)

>> paradiff(y,x,t,2)

(11)(u=arccossqrt{frac{x}{y}})验证(frac{partial^2u}{partial xpartial y}=frac{partial^2u}{partial ypartial x})

>> syms x y;

>> u=acos(sqrt(x/y));

>> diff(diff(u,y,1),x,1)-diff(diff(u,x,1),y,1)

(14)计算(frac{x}{y}frac{partial^2 f}{partial x^2}-2frac{partial^2 f}{partial xpartial y}+frac{partial^2 f}{partial^2 y}) 其中$$f(x,y)=int_0^{xy} e^{-t^2}text{d}t$$

>> syms x y t;

>> f(x,y)=int(exp(-t^2),0,x*y);

>> %f(x,y)=int(exp(-t^2),t,0,x*y);

>> g=diff(f,x,2)*x/y-2*diff(diff(f,x,1),y,1)+diff(f,y,2)

(15)计算(frac{text{d}y}{text{d}x},frac{text{d}^2y}{text{d}x^2},frac{text{d}^3y}{text{d}x^3})

[begin{cases}x=e^{2t}cos^2t\y=e^{2t}sin^2tend{cases};begin{cases}x=frac{arcsin t}{sqrt{1+t^2}}\ y=frac{arccos t}{sqrt{1+t^2}}end{cases}

]>> syms t x1 y1 x2 y2;

>> x1=exp(2*t)*cos(t)^2;

>> y1=exp(2*t)*sin(t)^2;

>> x2=asin(t)/sqrt(1+t^2);

>> y2=acos(t)/sqrt(1+t^2);

>> paradiff(y1,x1,t,1)

>> paradiff(y1,x1,t,2)

>> paradiff(y1,x1,t,3)

>> paradiff(y2,x2,t,1)

>> paradiff(y2,x2,t,2)

>> paradiff(y2,x2,t,3)

(16)写题目浪费时间，专心代码

>> syms x y;f=x^2-x*y+2*y^2+x-y-1;

>> subs(impldiff(f,x,y,1),{x,y},{0,1})

>> subs(impldiff(f,x,y,2),{x,y},{0,1})

>> subs(impldiff(f,x,y,3),{x,y},{0,1})

(17)

>> syms x y z;

>> f=[3*x+exp(y)*z, x^3+y^2*sin(z)];

>> jacobian(f,[x y z])

(18)

>> syms x y;

>> u=x-y+x^2+2*x*y+y^2+x^3-3*x^2*y-y^3+x^4-4*x^2*y^2+y^4;

>> ux4=diff(u,x,4); ux3y1=diff(diff(u,x,3),y,1); ux2y2=diff(diff(u,x,2),y,2);

(19)

>> syms x y;

>> u=x-y+x^2+2*x*y+y^2+x^3-3*x^2*y-y^3+x^4-4*x^2*y^2+y^4;

>> laplacian(u,[x,y])

(20)

(21)

>> syms x y z t psi(z);

>> z=x^2+y^2;t=psi(z);

>> y*diff(t,x)-x*diff(t,y)

(22)

>> syms x y z u psi(z) phi(z);

>> z=x+y;u=x*phi(z)+y*psi(z);

>> diff(u,x,2)-2*diff(diff(u,x),y)+diff(u,y,2)

(23)

%1

>> syms x y v(x,y);

>> v(x,y)=[5*x^2*y 3*x^2-2*y];

>> divergence(v(x,y),[x,y]),curl(v(x,y),[x,y])

%2

>> syms x y z v;

>> v=[x^2*y^2 1 z];

>> divergence(v,[x,y,z]),curl(v,[x,y,z])

%3

>> syms x y z v;

>> v=[2*x*y*z^2 x^2*z^2+z*cos(y*z) 2*x^2*y*z+y*cos(y*z)]

>> divergence(v,[x,y,z]),curl(v,[x,y,z])

(25)

>> syms a b c x t;

>> I1=-(3*x^2+a)/(x^2*(x^2+a)^2);

>> I2=sqrt(x*(x+1))/(sqrt(x)+sqrt(1+x));

>> I3=x*exp(a*x)*cos(b*x);

>> I4=exp(a*x)*sin(b*x)*sin(c*x);

>> I5=(7*t^2-2)*3^(5*t+1);

>> int(I1,x),int(I2,x),int(I3,x),int(I4,x),int(I5,t)

>> %I2积不出

(26)

>> syms x, n;

>> f1=cos(x)/sqrt(x);

>> f2=(1+x^2)/(1+x^4);

>> f3=abs(cos(log(1/x)));

>> int(f1,x,0,inf),int(f2,x,0,1),int(f3,x,exp(-2*pi*n),1)

>> subs(int(3,x),x,1)-subs(int(3,x),x,exp(-2*pi*n))

>> %f3定积分积不出，先不定积分处理再利用牛顿莱布尼茨公式

>> %这里使用的是解析函数，使用subs赋值，如果直接函数格式，代入即可

(27)

原文：https://www.cnblogs.com/Math-Nav/p/13367308.html