matlab习题 —— 符号运算相关练习-繁依Fanyi

matlab系列文章 目录

在这里插入图片描述

一、题目

1. 计算下列极限

  • (1)

    lim

    x

    0

    1

    +

    x

    1

    x

    1

    +

    x

    3

    1

    x

    3

    limlimits_{x to 0}frac{sqrt{1+x}-sqrt{1-x}}{sqrt[3]{1+x}-sqrt[3]{1-x}}

    x0lim31+x
    31x
    1+x
    1x

  • (2)

    lim

    x

    0

    (

    3

    x

    +

    2

    3

    x

    1

    )

    2

    x

    1

    limlimits_{x to 0}(frac{3x+2}{3x-1})^{2x-1}

    x0lim(3x13x+2)2x1

  • (3)

    lim

    x

    0

    (

    1

    x

    2

    1

    sin

    2

    x

    )

    limlimits_{x to 0}(frac{1}{x^2}-frac{1}{sin^2x})

    x0lim(x21sin2x1)

  • (4)

    lim

    x

    0

    (

    π

    2

    arctan

    x

    )

    1

    ln

    x

    limlimits_{x to 0}(frac{pi}{2}-arctan{x})^{frac{1}{ln{x}}}

    x0lim(2πarctanx)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}+tsin{t})\ y&=a(sin{t}-tcos{t}) end{aligned} right .

    {xy=a(cost+tsint)=a(sinttcost)

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

    02π1+sin2xcosxdx

  • (2)

    0

    a

    x

    2

    a

    x

    a

    +

    x

    d

    x

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

    0ax2a+xax
    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(x2x+1)23dx

4. 计算下列级数的和

  • (1)

    n

    =

    1

    1

    n

    2

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

    n=1n21

  • (2)

    n

    =

    1

    (

    1

    )

    n

    1

    n

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

    n=1n(1)n1

  • (3)

    n

    =

    1

    x

    2

    n

    1

    2

    n

    1

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

    n=12n1x2n1

二、解答

题一,计算下列极限

>> 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
>> 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
>> 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
>> 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

>> 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)
>> 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
>> 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))
>> 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

>> 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
>> 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
>> 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

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

在这里插入图片描述

© 版权声明
THE END
喜欢就支持一下吧
点赞11 分享
评论 抢沙发

请登录后发表评论

    暂无评论内容