matlab系列文章: 目录
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}}
- (2)
lim
x
→
0
(
3
x
+
2
3
x
−
1
)
2
x
−
1
limlimits_{x to 0}(frac{3x+2}{3x-1})^{2x-1}
- (3)
lim
x
→
0
(
1
x
2
−
1
sin
2
x
)
limlimits_{x to 0}(frac{1}{x^2}-frac{1}{sin^2x})
- (4)
lim
x
→
0
(
π
2
−
arctan
x
)
1
ln
x
limlimits_{x to 0}(frac{pi}{2}-arctan{x})^{frac{1}{ln{x}}}
2. 计算下列导数
- (1) 求
y
′
y^{‘}
y
=
(
x
+
1
)
arctan
x
y=(sqrt{x}+1)arctan{x}
- (2) 求
y
′
′
y^{”}
y
=
1
+
x
2
sin
x
+
cos
x
y=frac{1+x^2}{sin{x}+cos{x}}
- (3) 求
y
′
y^{‘}
y
=
x
+
x
+
x
y=sqrt{x+sqrt{x+sqrt{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 .
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
- (2)
∫
0
a
x
2
a
−
x
a
+
x
d
x
int_{0}^{a}x^2sqrt{frac{a-x}{a+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}}}
4. 计算下列级数的和
- (1)
∑
n
=
1
∞
1
n
2
sum_{n=1}^{infty}{frac{1}{n^2}}
- (2)
∑
n
=
1
∞
(
−
1
)
n
−
1
n
sum_{n=1}^{infty}{frac{(-1)^{n-1}}{n}}
- (3)
∑
n
=
1
∞
x
2
n
−
1
2
n
−
1
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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