英文版微积分

n 1
Example
A sequence {an } is defined recursively by the equation,
a0 a,1 1
n(n 1)an (n 1)(n 2)an 1 (n 3)an 2 , n 2,3,
n0
Determine whether the series an is convergent or divergent. And find the sum of this series.
The volume of the parallelepiped determined by the vector a , b , and c is the magnitude of their scalar triple product:
v a (b c )
Partial Derivatives
and z-axes.
a a1 , a 2 , a3
o
, , [0, ]
cos , cos , and cos are called the directio cosin of a
We have
a3 a1 a2 cos , cos , cos a a a
四、 Fill in the blanks:
( x a)n 1.If (1) n 1 diverges when x 0, n n 1

and converges at

x 0,
n 1
then
n
a
2. The sum of
(1)
n 1
( x 1) is n
Example
en (1) n n n2 x n 1

二、Find the sum of the following series.

n 1

1 n(n 1)( n n 1)
二、Find the sum of the following series.
x n 2n n 0

n 1
Choose the best answer for each of the following 1.If
lim u n a,
n
then
(a
n 1

n
an 1 )
A.
C.
0
u1 a
B.
a
D.Diverges
2.We have the following four statement:
.Is the converse true， prove it or give a counter-example.
z z . 1.If z (2 x 2 3 y ) y sin x , calculate and y x
2
x3 y3 2 2.If f ( x, y ) x y 2 0
j a2 b2 a3 b3
k a3 b3 i a1 b1 a3 b3 j a1 b1 a2 b2 k
Properties of the cross product The length of the cross product a b is equal to the area of the parallelogram determined by a and b . pærə'leləgræm
Review
Infinite Sequences and Series
一、Determine whether the series is convergent or divergent
2 n! (1) n n n 1
2
n n (2) ( ) n 1 2 n 1 1 (4) n 3 n ln n
Implicit Differentiation Example
3.If z ( x, y) is defined implicitly as a function of x and y by the equation
z z F ( x , y ) 0. y x
Directional Derivatives and the Gradient Vector
Example Given that
z f ( x, y)
2 2
is defined by
2
cos xyz ln( x y z )
, find
(1) z
(3) Du f (0,1)
(2) dz
Where, U=(1,1).
Example Given f(x, y) = x3 +y3 –3xy, find the extreme values of function f. Solution f has continuous second partial derivatives so the critical points of f are those at which f’x and f’y are 0. Since f’x(x, y) = 3x2 –3y and f’y(x, y) = 3y2 –3x, Solving for x, y from 3x2 –3y = 0, 3y2 –3x = 0, we obtain the critical points of f,
calculate
x y 0
2 2
x2 y2 0
f y (0,0).
f x (0,0)
and
Vectors and Geometry of Space
If is the angle between the nonzero vectors a and b then
a b cos ab


True or false
4.If an 0and
a
n
converges, then
lim n
5.If
an 1 1, an
an 1 1. an
then
a
n
is convergent.
2 an
6.If an 0and
an converges, then
is convergent.
The vector cos , cos , cos is a unit vector in the direction of a
Example Find scalar projection and the vector the projection of b 3i 2 j k onto a 2i j k.
2 2
Implicit Differentiation
z 1.Find dz and xy if z is defined implicitly as a function of x and y by the equation
Example
2
xyz x 2 y 2 z 2 2 .
Example
(19) 求幂级数
1 x 2 n 1 n 2n 1 n 1
n 1
的收敛域及和函数
(9) 若级数
s( x)
.
a
n 1

n
收敛,则级数 ( (A)
)
a
n 1


n
收敛 (B)
Байду номын сангаас
1
n 1
n
an
收敛 (C)
a a
n 1

n n 1
收敛 (D)
Direction Angles and Direction Cosins
The direction angles of a nonzero vector a are the angles , and , that a makes with the positive x, y ,
Solution The scalar projection is a b comp a b a The vector projection is a proja b comp a b a
a b a1 b1 a2 b2
i
z z . 1.If z (2 x 2 3 y ) y sin x , calculate and y x
2
x3 y3 2 2.If f ( x, y ) x y 2 0
calculate
x y 0
2 2
x2 y2 0
f y (0,0).
f x (0,0)
(1).If
n 1
(u2n 1 u2n )

is convergent,then
n 1
un

is convergent.
(2).If
n 1
un is convergent,then
u
n 1 n

100 is convergent.
un 1 lim 1 (3).If n un


(3)

n 1

2n 3n 1
2
3n 7 n 2 2

(1) n ln(1 n) (5) 1 n n 1
二、Find the radius of convergence and interval of Convergence of the following series.

,then
n 1
un

is divergent. (4).If
n 1
(un vn )
u
n 1 n
is convergent,then are both convergent.
u
n 1

n
and
Then the correct statement is A.(1) (2) B.(2)(3) C.(3)(4) D.(4)(1)
and
Chain rule Example Show that when Laplace’s equation
is written in spherical coordinates, it becomes
u u u 2 2 0 2 x y z
2 2 2
2
u 2 u cot u 1 u 1 u 2 2 2 0. 2 2 2 2 sin
二、Find the sum of the following series.
x (2n)! n0

2n
二、Find the sum of the following series and the Maclaurin Series for the sum.
(1) 2 n 1 (2n 1)!22n2 x n 1

收敛
an an 1 2 n 1

(19) (本题满分9分) 设级数
x4 x6 x8 ( x ) 2 4 2 46 2 468
的和函数为S(x). 求： (I) S(x)所满足的一阶微分方程； (II) S(x)的表达式.
求幂级数
1 (1) n
cn 1.Prove that lim 0(c 1) n n!
2. Suppose that an 0, (n 1,2,3). Show that the convergence of implies the convergence of

a
n 1

n
2 an n 1
Implicit Differentiation Example
dz dy 2.Find dx and dx if z z (x) and y y(x)
are both defined implicitly as function of by the equation system.
x
x y z z2 0 x y2 z z3 0
n 1

x 2n ( x 1) 2n
的和函数f(x)及其极值.
True or false
1. If lim an 0,then n
2.
a is convergent.
n
an is divergent ,then If divergent.

a
n
is
3.
an is If an bn for all n ,and divergent,then bn is divergent.