matlab习题 —— 符号运算相关练习

一、题目

1. 计算下列极限

• (1) lim ⁡ x → 0 1 + x − 1 − x 1 + x 3 − 1 − x 3 \lim\limits_{x \to 0}\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt[3]{1+x}-\sqrt[3]{1-x}} x→0lim​31+x ​−31−x ​1+x ​−1−x ​​
• (2) lim ⁡ x → 0 ( 3 x + 2 3 x − 1 ) 2 x − 1 \lim\limits_{x \to 0}(\frac{3x+2}{3x-1})^{2x-1} x→0lim​(3x−13x+2​)2x−1
• (3) lim ⁡ x → 0 ( 1 x 2 − 1 sin ⁡ 2 x ) \lim\limits_{x \to 0}(\frac{1}{x^2}-\frac{1}{\sin^2x}) x→0lim​(x21​−sin2x1​)
• (4) lim ⁡ x → 0 ( π 2 − arctan ⁡ x ) 1 ln ⁡ x \lim\limits_{x \to 0}(\frac{\pi}{2}-\arctan{x})^{\frac{1}{\ln{x}}} x→0lim​(2π​−arctanx)lnx1​

2. 计算下列导数

• (1) 求 y ′ y^{'} y′, y = ( x + 1 ) arctan ⁡ x y=(\sqrt{x}+1)\arctan{x} y=(x ​+1)arctanx
• (2) 求 y ′ ′ y^{''} y′′， y = 1 + x 2 sin ⁡ x + cos ⁡ x y=\frac{1+x^2}{\sin{x}+\cos{x}} y=sinx+cosx1+x2​
• (3) 求 y ′ y^{'} y′, y = x + x + x y=\sqrt{x+\sqrt{x+\sqrt{x}}} y=x+x+x ​ ​ ​
• (4) { x = a ( cos ⁡ t + t sin ⁡ t ) y = a ( sin ⁡ t − t cos ⁡ t ) \left\{ \begin{aligned} x&=a(\cos{t}+t\sin{t})\\ y&=a(\sin{t}-t\cos{t}) \end{aligned} \right . {xy​=a(cost+tsint)=a(sint−tcost)​

3. 计算下列定积分

• (1) ∫ 0 π 2 cos ⁡ x 1 + sin ⁡ 2 x d x \int_{0}^{\frac{\pi}{2}}\frac{\cos{x}}{1+\sin^2x}dx ∫02π​​1+sin2xcosx​dx
• (2) ∫ 0 a x 2 a − x a + x d x \int_{0}^{a}x^2\sqrt{\frac{a-x}{a+x}}dx ∫0a​x2a+xa−x​ ​dx
• (3) ∫ 0 1 d x ( x 2 − x + 1 ) 3 2 \int_{0}^{1}\frac{dx}{(x^2-x+1)^{\frac{3}{2}}} ∫01​(x2−x+1)23​dx​

4. 计算下列级数的和

• (1) ∑ n = 1 ∞ 1 n 2 \sum_{n=1}^{\infty}{\frac{1}{n^2}} n=1∑∞​n21​
• (2) ∑ n = 1 ∞ ( − 1 ) n − 1 n \sum_{n=1}^{\infty}{\frac{(-1)^{n-1}}{n}} n=1∑∞​n(−1)n−1​
• (3) ∑ n = 1 ∞ x 2 n − 1 2 n − 1 \sum_{n=1}^{\infty}{\frac{x^{2n-1}}{2n-1}} n=1∑∞​2n−1x2n−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))