matlab习题 —— 符号运算相关练习

繁依Fanyi

发布于 2023-05-07 19:13:31

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一、题目

1. 计算下列极限

\lim\limits_{x \to 0}\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt[3]{1+x}-\sqrt[3]{1-x}}

\lim\limits_{x \to 0}(\frac{3x+2}{3x-1})^{2x-1}

\lim\limits_{x \to 0}(\frac{1}{x^2}-\frac{1}{\sin^2x})

\lim\limits_{x \to 0}(\frac{\pi}{2}-\arctan{x})^{\frac{1}{\ln{x}}}

2. 计算下列导数

(1) 求

y^{'}

y=(\sqrt{x}+1)\arctan{x}

(2) 求

y^{''}

，

y=\frac{1+x^2}{\sin{x}+\cos{x}}

(3) 求

y^{'}

y=\sqrt{x+\sqrt{x+\sqrt{x}}}

\left\{ \begin{aligned} x&=a(\cos{t}+t\sin{t})\\ y&=a(\sin{t}-t\cos{t}) \end{aligned} \right .

3. 计算下列定积分

\int_{0}^{\frac{\pi}{2}}\frac{\cos{x}}{1+\sin^2x}dx

\int_{0}^{a}x^2\sqrt{\frac{a-x}{a+x}}dx

\int_{0}^{1}\frac{dx}{(x^2-x+1)^{\frac{3}{2}}}

4. 计算下列级数的和

\sum_{n=1}^{\infty}{\frac{1}{n^2}}

\sum_{n=1}^{\infty}{\frac{(-1)^{n-1}}{n}}

\sum_{n=1}^{\infty}{\frac{x^{2n-1}}{2n-1}}

二、解答

题一，计算下列极限

①

>> syms x
>> y = ((1+x)^(1/2)-(1-x)^(1/2))/((1+x)^(1/3)-(1-x)^(1/3))
 
y =
 
((x + 1)^(1/2) - (1 - x)^(1/2))/((x + 1)^(1/3) - (1 - x)^(1/3))
 
>> limit(y,x,0)
 
ans =
 
3/2

②

>> syms x
>> y = ((3*x+2)/(3*x-1))^(2*x-1)
 
y =
 
((3*x + 2)/(3*x - 1))^(2*x - 1)
 
>> limit(y,x,0)
 
ans =
 
-1/2

③

>> syms x
>> y = ((1/x^2)-1/(sin(x)^2))
 
y =
 
1/x^2 - 1/sin(x)^2
 
>> limit(y,x,0)
 
ans =
 
-1/3

④

>> syms x
>> y = (pi/2 - atan(x))^(1/(log(x)))
 
y =
 
(pi/2 - atan(x))^(1/log(x))
 
>> limit(y,x,0)
 
ans =
 
1

题二，计算下列导数

①

>> y = (x^(1/2)+1)*atan(x)
 
y =
 
atan(x)*(x^(1/2) + 1)
 
>> diff(y,x)
 
ans =
 
atan(x)/(2*x^(1/2)) + (x^(1/2) + 1)/(x^2 + 1)

②

>> y = (1+x^2)/(sin(x)+cos(x))
 
y =
 
(x^2 + 1)/(cos(x) + sin(x))
 
>> y1 = diff(y,x)
 
y1 =
 
(2*x)/(cos(x) + sin(x)) - ((x^2 + 1)*(cos(x) - sin(x)))/(cos(x) + sin(x))^2
 
>> y2 = diff(y1,x)
 
y2 =
 
(x^2 + 1)/(cos(x) + sin(x)) + 2/(cos(x) + sin(x)) + (2*(x^2 + 1)*(cos(x) - sin(x))^2)/(cos(x) + sin(x))^3 - (4*x*(cos(x) - sin(x)))/(cos(x) + sin(x))^2

③

>> y = (x+(x+(x)^(1/2))^(1/2))^(1/2)
 
y =
 
(x + (x + x^(1/2))^(1/2))^(1/2)
 
>> diff(y,x)
 
ans =
 
((1/(2*x^(1/2)) + 1)/(2*(x + x^(1/2))^(1/2)) + 1)/(2*(x + (x + x^(1/2))^(1/2))^(1/2))

④

>> syms x y t a
>> x = a*(cos(t)+t*sin(t))
 
x =
 
a*(cos(t) + t*sin(t))
 
>> y = a*(sin(t)-t*cos(t))
 
y =
 
a*(sin(t) - t*cos(t))

>> dx = diff(x,t)
 
dx =
 
a*t*cos(t)
 
>> dy = diff(y,t)
 
dy =
 
a*t*sin(t)
 
>> dy/dx
 
ans =
 
sin(t)/cos(t)

题三，计算下列定积分

①

>> syms x y
>>  y = cos(x)/(1+(sin(x))^2)
 
y =
 
cos(x)/(sin(x)^2 + 1)
 
>> int(y,x,0,pi/2)
 
ans =
 
pi/4

②

>> syms x y a
>> y = (x^2)*((a-x)/(a+x))^(1/2)
 
y =
 
x^2*((a - x)/(a + x))^(1/2)
 
>> int(0,a)
 
ans =
 
0
 
>> int(y,x,0,a)
 
ans =
 
(a^3*(3*pi - 8))/12

③

>> syms x y a
>> y = 1/((x^2-x+1)^(3/2))
 
y =
 
1/(x^2 - x + 1)^(3/2)
 
>> int(y,x,0,1)
 
ans =
 
4/3

题四，计算下列级数的和

①

>> syms x y n
>> y = 1/(n^2)
 
y =
 
1/n^2
 
>> symsum(y,n,1,Inf)
 
ans =
 
pi^2/6

②

>> y = (-1)^(n-1)/n
 
y =
 
(-1)^(n - 1)/n
 
>> symsum(y,n,1,Inf)
 
ans =
 
log(2)

③

>> y = (x^(2*n-1))/(2*n-1)
 
y =
 
x^(2*n - 1)/(2*n - 1)
 
>> symsum(y,n,1,Inf)
 
ans =
 
piecewise(abs(x) < 1, atanh(x))

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