北京邮电大学高等数学ppt

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北京邮电大学国际学院高等数学(下)幻灯片讲义(无穷级数)Le.

1(x 2
−
2) +
1(x 4
−
2)2
=
3−
3x 2
+
x2 4
#
8
Convergence of Power Series
Abel’s Theorem (1)
(2)
∞
∑ Consider the series an xn. n=0
If it converges at x0 , x0 ≠ 0 , then it must converge
but converges for| x − x0 |< R. The series may or may not converge at either of the endpointsx = x0 − Randx = x0 + R. 2. The series converges for every x(R = ∞) . 3. The series converges at x = x0 and diverges elsewhere (R = 0).
n=1
13
Finding the Interval of Convergence Using the Ratio Test
∑ Solution Apply the Ratio Test to the series | un | , where un
is the nth term of the series in question.
divergence points is called the divergence domain of the series.
1
2
Series of Functions
Definition (Sum function and Convergence)

北京邮电大学国际学院高等数学(下)幻灯片讲义(无穷级数)-Lecture 1

is called the partial sum of the series. The partial sum of the series form a sequence
s1 = a1 , s2 = a1 + a2 ,
, sn = ∑ ak ,
k =1 n
n approaching infinite, we say that the series converges to the sum S, and we write
1+ 1 1 1 1 + + + + 2 4 8 16
Infinite Series
∑a
n =1
∞
n
=1 +
1 1 1 1 + + + + 2 4 8 16
Partial Sum First: Second: Third:
…
Value
s1 = 1
1 s2 = 1 + 2
21
2 2 1 2 1 4
The way to do so is not to try to add all the terms at once (we cannot) but rather to add the terms one at a time from the beginning and look for a pattern in how these "partial sum" grow.
k =1 n
Convergence and Divergence
Definition (Convergence and Divergence of a series) If the sequence of partial sums of a series

北京邮电大学高等数学教学课件-7二阶常系非齐

Q(x) (2 p )Q (x )(2pq)Q (x)Pm(x)
(1) 若不是特征方程的根, 即 2pq0,则取
Q (xe)为x[mQ次(x待) 定(系2 数多p 项)式Q (Qxm) (x(),2 从而p得到q 特)解Q(x)]
形式e为xPy m*( x)exQ m(x).
Q(x) (2 p )Q (x )(2pq)Q (x)Pm(x)
例. 求y 方 y x 程 c2 o x的s 一个特解 .
解: 0,2, Pl(x)x, P ~n(x)0,
特征方程
r210
i 2i不是特征方程的根, 故设特解为
y * ( a x b ) c 2 x o ( c x s d ) s2 x in
代入方程得
( 3 a x 3 b 4 c ) c 2 x ( 3 o c x 3 d s 4 a ) s 2 x i x c 2 n x
比较系数, 得
2b 30 b03b 311
b0 1,
b1
1 3
于是所求特解为 y*x1. 3
例1 求方 y 3 y程 2 yx2x e 的.通解
解特征方程 r23r20,
特征根 r11， r22，
对应齐次方程通解 Yc1exc2e2x,
2是单根设，y x (A B x ) e 2 x ,
1 a b 0
比较系数得 2ac 1 a b 0
a0 b1 c2
故原方程为 yy2ex
yexxex
对应齐次方程通解: YC1exC2ex
原方程通解为 yC1exC2ex x e x
三、小结
(待定系数法)
(1)f(x)exPm(x),(可以是复数）
yxkexQ m (x);

北邮高等数学英文课件Lecture 12-2

then, one of the normal vector of this surface is ( z x , z y ,1). Notice that γ is the angle between this vector and the positive direction of z – axis, 1 then cos . So, 2 2 zx z y 1 1 2 S d z x z2 y 1d |cos | ( xy ) ( xy )
(S)

2 f ( x ( y , z ), y , z )dS f ( x , y , z ) x 2 y x z 1dydz . ( )
Integrating Over a Surface
Example Integrate f ( x , y , z ) xyz over the surface of the cube cut from the first octant by the planes x 1, y 1 and z 1. Solution We integrate xyz over each of the
Similarly, if the surface can be expressed as
y f ( x , z ), ( x , z ) ( xz ) R 2 , or x f ( y , z ), ( y , z ) ( yz ) R 2 ,
we have
S
( )
y y( x , z ), ( x , z ) ( xz ) R 2 ,
then the surface integral of first type of f on (S) can be reduce to double integral

北京邮电大学《高等数学教学课件》5-习题课

4 (cos x sin x)dx
0
2(sin x cos x)dx
4
2 2 2.
例5
求
1
2 1
2
[
sin x x8 1
ln2(1 x)]dx.
1
解
原式 0
2 1
ln(1
x)dx
2
0
1
1 ln(1 x)dx
2 ln(1 x)dx
0
2
3 ln 3 ln 1 . 22 2
x
x
x 0 f (u)du 0 uf (u)du
所以
x
0
u 0
f
( x)dxdu
x
( x u) f (u)du ．
0
1/ 2
例设 f ( x ) 在[0，1]可微，且满足 f (1) 2 xf ( x)dx 0
证明：（0，1）使
f ( ) f ( )
0
分析：变形为： f ( ) f ( ) 0, [ xf ( x)] 0
0
1 x
5、 1 dx ；
1
1
1 2x
7、 2
dx
；
1 x 3x2 2x 1
2、 a
dx
；
0 x a2 x2
4、 5 x 2 2x 3 dx ； 2
6、
x 2
dx 4x
； 9
8、
1
dx .
1 x x1
五、设 f ( x)在 0 , 1 上有连续导数， f (0) 0 ,
且0 f ( x) 1,试证：
f (u a2 ) du u 2u
1 a2 21
f (u
a2 ) du uu

北京邮电大学高等数学6-xiti市公开课获奖课件省名师示范课获奖课件

2）设想把区间[a, b]分成n 个小区间，取其中任
一小区间并记为[ x, x dx]，求出相应于这小区
间的部分量U 的近似值．如果U 能近似地表
示为[a, b]上的一个连续函数在x 处的值 f ( x) 与
dx的乘积，就把 f ( x)dx 称为量U 的元素且记作
dU ，即dU f ( x)dx ；
三、设曲边梯形是由连续曲线y f ( x) ( f ( x) 0) ， x 轴与两直线x a , x b 所围成的，求证：存在
直线x ( ( a , b ))将曲边梯形的面积平分 .
四、求摆线
x y
a a
( t sin t ) ，( ( 1 cos t )
0
t
2
)
1、绕 x 轴旋转一周所成曲面的面积；
3）以所求量U 的元素 f ( x)dx 为被积表达式，在
区间[a, b]上作定积分，得U
b
a
f
( x)dx
，
即为所求量U ．
5、定积分应用旳常用公式
(1) 平面图形旳面积
直角坐标情形
y
y f (x)
y
y f2(x)
A
oa
b
b
A a f ( x)dx
A
y f1( x)
x oa
bx
A
b
a [
f2(x)
f1( x)]dx
参数方程所表达旳函数
假如曲边梯形旳曲边为参数方程
x (t)
y
(t
)
曲边梯形旳面积
A
t2 t1
(t) (t)dt
（其中t1和t2 对应曲线起点与终点的参数值）
在[t1,t2 ]（或[t2 ,t1 ]）上x (t ) 具有连续导数， y (t)连续.

北京邮电大学高等数学1-1-37页PPT资料

5.绝对值: a aa
a0 a0
运算性质:
abab;
(a 0)
a

a ;
bb
a b a b a b .
绝对值不等式:
xa(a0)
axa ;
xa(a0)
xa或 x a;
二、函数概念
例圆内接正多边形的周长
S3
S4

Sn
2nrsin n
n3,4,5,
因变量
自变量
当 x 0 D 时 ,称 f(x 0)为函 x 0 处数的在 . 函点
函数值全体组成的数集 W{yyf(x),xD}称为函数的 . 值域
函数的两要素: 定义域与对应法则.
( x D x0)
对应法则f
(
W
y f (x0)
自变量
)
因变量
约定: 定义域是自变量所能取的使算式有意义的一切实数值.
o
x
I
设函 f(x)的数定D ,义区I域间 D , 为如果 I上对任于 x 1 及 x 2 意 ,区当 x 1 两 x 2 间时 , 点恒 ( 2 )f有 ( x 1 ) f(x 2 ), 则称函 f(x)在数区 I上间是单调 ; 减少的
y
y f (x)
f ( x1)
数集分类: N----自然数集 Z----整数集 Q----有理数集 R----实数集
数集间的关系: N Z ,Z Q ,Q R . 若 A B ,且 B A ,就称 A 与 B 相集 .(A等合 B) 例如 A{1,2},
C{xx23x20},则AC. 不含任何元素的集合称为空集. (记作 ) 例如, {xxR ,x210}

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2arcsin( ln x )
e
. 6
1 Example: Find dx . (a 0) 2 2 0 x a x Solution: Let x a sin t , dx a cos tdt , x a t , x 0 t 0, 2
x
7
Example: Find
3 e4
e
3 e4
dx . x ln x(1 ln x )
Solution

e
d (ln x ) ln x(1 ln x ) d (ln x ) 2 ln x (1 ln x )
3 e4
3 e4
3 e4
e
e
d ln x 1 ( ln x )2

2
when x 1 ; thus by the formula, we have

1
0
1 x 2 dx 2 cos 2 tdt
0

1 1 2 t sin 2t 2 2 0

.
4
5
Integration by Substitutions for definite integrals
17
Integration by Parts
Integration by parts for definite integrals
When u and v are differentiable functions of x on the interval [a , b]. Then

Example: Evaluate
2 2
f (cos t )dt f (cos x )dx;
0 0
例若 f ( x ) 在[0,1]上连续，证明
0

xf (sin x )dx f (sin x )dx . 2 0
x t dx dt ,
Proof: Let

x 0 t ,
School of Science, BUPT
Integration by substitution and by parts in definite integrals
2
Integration by Substitutions for definite integrals
When we want to evaluate the value of a definite integral can find the corresponding indefinite integral
d dv du ( uv ) u v dx dx dx
Integrating both sides with respect to x and rearranging leads to the integral equation
dv d du u dx ( uv ) dx v dx dx dx dx du uv v dx . dx
2
0
2 cos n tdt
0

2 cos n xdx .
0

6
Integration by Substitutions for definite integrals
1 Example: Find lim 2 x 0 x sin xt 0 t dt du u . t 0 u 0; t x u x 2 . Solution Let xt u, then t , dt x x Then x 2 sin u du x sin xt x 2 sin u 0 t dt 0 u x 0 u du x So 0 2 x sin u 0 sin x 2 0 u du x 2 2 x 1 x sin xt lim 2 dt lim 1. lim 2 0 x 0 x x 0 x 0 t x 2x
偶倍奇零
(1) If (2) If Proof
a 0
then

a
a
f ( x )dx 2 f ( x )d x
0
a
then

a
a
a
f ( x )d x 0
a
a f ( x) dx a f ( x) dx 0 f ( x) dx
f (t ) d t f ( x) dx [ f ( x ) f ( x ) ] dx
b
a
udv uv a vdu
b a
b

4
0
e x dx .
Solution Let
2 x t , then x t , dx 2tdt . Thus
4

0
e dx e 2tdt 2 tde 2 te
x t t 0 0
2
2

t 2 0
e t dt
0
2 0

2

0
f (sin x )dx
Integration by Parts
Integration by parts for indefinite integrals
When u and v are differentiable functions of x, the Product Rule for differentiation tell us that

b
a
f ( x )dx , if we
f ( x )dx , then we can use
the Newton-Leibniz formula to obtain the value immediately.
Theorem (Integration by substitution for definite integrals) Suppose that the function x ( t ) is continuously differentiable on the interval [ , ] , ( ) a , ( ) b , and that f is continuous in the range of
0 0 a 0 a
Let x t

f ( x ) f ( x ) f ( x ) f ( x )
机动目录上页下页返回结束
Example: Find Solution:
1
1
2 x 2 x cos x dx . 2 1 1 x
1
1
1
1 x cos x 2x2 dx dx 1 2 2 1 1 x 1 1 x
, v x,

1 2
0
arcsin xdx x arcsin x 0
1 2
1 2
xdx 1 x2
0
1 1 1 1 2 d (1 x 2 ) 2 6 2 0 1 x2
, I . Then

b
a
f ( x )dx f ( t ) ( t )dt .

3
Integration by Substitutions for definite integrals
Proof Let F be an antiderivative of f on the interval I, then
例若 f ( x ) 在[0,1]上连续，证明
0
2
f (sin x )dx f (cos x )dx .
0
2
Proof （1）设 x t 2 x 0 t , 2
dx dt,
0
2
x t 0, 2 0 f (sin x )dx f sin t dt 2 2
b
Since

a
f ( x )dx F (b) F (a )
d F ( t ) f ( t ) ( t ), dt
we have

f ( t ) ( t )dt F ( t )

F ( ) F ( )
F (b) F (a )
1 d (cos x ) arctan(cos x )0 2 2 0 1 cos x 2 2 ( ) . 4 2 4 4
xf (sin x )dx f (sin x )dx . 2 0

0
(1) If f ( x )=f ( x )， then
0
T

a
0
f ( x )dx f (a x )dx
0
a
(let t a x )

2 0
f (sin x )dx 2 f (cos x )dx
0

0
f (sin x )dx 2 2 f (sin x )dx
0

(7)

0
xf (sin x )dx f (sin x )dx

0
0
xf (sin x )dx f (sin x )dx. 2 0
0
Example: Find
Solution
0

x sin x dx 2 1 cos x
0

x sin x sin x dx dx 2 2 2 0 1 cos x 1 cos x