北邮高等数学英文版课件Lecture 10-2
北京邮电大学国际学院线性代数讲义Lecture 02

1 0 0 0 0
and we end up with
1 0 0 0 0
1 1 0 0 0 1 0 0 0 0
1 1 0 0 0 1 0 0 0 0
1 1 2 0 1 3 pivotal row 1 1 1 0 1 1 0 0 0 1 1 0 0 0 1 1 2 0 1 3. 0 4 0 3
in row k 1 is greater that the number of leading zero entries in row k . nonzero entries.
Row Echelon Form
Definition: The process of using row operation I, II and III to transform a linear system into one whose augmented matrix is in row echelon form is called Gaussian elimination. Remark: Row operation II is necessary in order to scale the rows so that lead coefficients are all 1. Remark: If the row echelon form of the augmented matrix contains a row of the form 0 0 0 1 The system has no solution set. Otherwise, the system will have solution set. Definition: If a system has solution set, we will be referred as consistent and if a system has no solution, we will be referred as inconsistent.
高等数学 北邮第2版

高等数学北邮第2版介绍《高等数学北邮第2版》是由北京邮电大学出版社出版的教材,主要面向高等院校的理工科专业学生。
本教材是高等数学领域的经典教材之一,经过多年的使用和改进,逐渐发展成为一本综合性的高等数学教材。
本教材分为多个部分,涵盖了高等数学的各个方面,包括数列与极限、函数与连续、导数与微分、定积分与定积分应用、常微分方程等内容。
每个章节都以理论知识和例题相结合的形式,旨在帮助学生深入理解高等数学的概念和原理,并培养学生的数学思维和解题能力。
目录1.数列与极限2.函数与连续3.导数与微分4.定积分与定积分应用5.常微分方程6.二重积分与曲线积分7.空间解析几何数列与极限在数学中,数列是将自然数映射到数域的一个函数。
数列的极限是数列中接近某一特定值的值。
数列与极限是高等数学中的重要概念,它们贯穿了高等数学的各个领域。
本章介绍了数列的基本概念和性质,包括数列的定义、极限的定义和性质、极限存在准则等等。
通过学习本章,学生将掌握数列的极限计算方法,并能够应用数列的极限解决实际问题。
函数与连续函数是数学中的一个基本概念,它描述了两个数集之间的对应关系。
连续是函数在某一点处没有跳跃或间断的特性,连续性是函数中的一个重要概念。
本章介绍了函数的基本概念和性质,包括函数的定义、函数的运算、反函数与复合函数等。
同时,本章还介绍了连续函数的定义和性质,并通过例题讲解了如何判断函数的连续性以及如何应用连续函数解决实际问题。
导数与微分导数是函数在某一点处的变化率,微分是函数在某一点处的近似变化量。
导数与微分是高等数学中的核心概念,它们在数学和物理等领域具有广泛的应用。
本章介绍了导数的定义和性质,包括导数的计算公式、导数的几何意义、高阶导数等。
同时,本章还介绍了微分的定义和性质,包括微分的近似计算、微分中值定理等。
通过学习本章,学生将掌握导数和微分的计算方法,并能够应用导数和微分解决实际问题。
定积分与定积分应用定积分是计算曲线下面积的数学工具,定积分应用是将定积分应用于实际问题的过程。
《信号与系统》第十章课件(英文版)

9The z-transform reduces to the Fourier transform for values of z Unit circle on the unit circle.
Im z=ejω
z-plane
ω
1
Re
5
Sichuan University <Signals and Systems> Ch 10 The z-Transform
Example 10.3 Consider a signal that is the sum of two real
exponentials: x[n] = 7⎜⎛ 1 ⎟⎞ n u[n] − 6⎜⎛ 1 ⎟⎞ n u[n].
⎝3⎠
⎝2⎠
The z-transform is then
∑ ∑ X ( z) = 7 +∞ ⎜⎛ 1 ⎟⎞n u[n]z −n − 6 +∞ ⎜⎛ 1 ⎟⎞ n u[n]z −n
z z-transform expand the application in which Fourier analysis can be used.
2
Sichuan University <Signals and Systems> Ch 10 The z-Transform
10.1 The z-Transform
Transform 2. The Inverse z-Transform 3. Geometric Evaluation of the Fourier Transform from the Pole-
Zero Plot 4. Properties of the z-Transform and some Common z-Transform
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)
高数Lecture_2

Geometric Series
The sum of the first n terms of this geometric progression is expressed as
qn −1 a( )(q ≠ 1) q −1
Where a is the first term of a geometric progression, and q is the geometric ratio
Mathematics for Economics and Business
Lecture 2: Mathematics of Finance Ⅱ
Terms
Simple interest Deposit Compound interest Principal Future value Geometric progression Geometric ratio Sinking fund Mortgage payments Outstanding debt Instalment 单利 存款 复利 现值(本金) ( 终值 等比数列 公比 偿债基金(零存整取) 分期付款 有待偿还的债务 分期付款
r n S = P(1+ ) 100
where the future value, or the final sum, is denoted by S, the principal, or the original sum of money, is denoted by P, and the interest rate is denoted by r
Lecture Goals
After completing this lecture, you should be able to: Calculate compound interest Apply geometric series to calculate the total investment and instalments
高数英文课程

1 variable t x ( x ) ; hence we obtain
is also a composite function of x with intermediate
d y
d dy d dy dt 2 dx dx dx dt dx dx & d y( t ) 1 &t ) x ( t ) & dt x ( & y( & & & & x( t ) & t ) x( t ) y( t ) 1 & 2 & [ x ( t )] x( t ) & & & & & & xy xy 3 & x
x x(t ) y y(t )
(1) If the functions x x(t ) and y y(t ) are both derivable with respect to t in ( , ) and x( t ) 0, then &
dy dx & y & x ;
y y( t )
is called the function defined by the parametric equations.
For example, if the frictional damping of air is omitted, the locus of motion of a projectile may be represented by a parametric equation:
12
2
Finish.
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
(高等数学英文课件)Some exercises of CHAPTER 2

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P159 40.
Analysis
f0limf0hf0...
h 0
h
f 0 02 f 0 0
f0 lim f0 h f0 lim fh
h 0
h
h 0 h
h2 f h h 2
h
h
h
目录 上页 下页 返回 结束
P159 40.
Analysis
f0limf0hf0
3. Horizontal Asymptotes
alxi m fxx,blxi m fxax
yaxb
目录 上页 下页 返回 结束
P123
60. Asymptotes
f
x
x3 x2 1 x2 1
1. Horizontal Asymptotes
limf x ...
x
2. Vertical Asymptotes
rx2000011x
1. 销量为100台的边际收益.
rx20000x2 r100 2
2. 用收益函数的导数来估计当销量从每周100台增加 到每周101台时所产生的额外的收益.
r r100
3. 计算极限并解释经济意义.
lim rxlim 2 0 0 0 0x20
x
x
目录 上页 下页 返回 结束
x
2. Vertical Asymptotes
limf x x0
x?
3. Horizontal Asymptotes
a
lim x
f
x
x
lxim
2sinx x
1 x2
0
目录 上页 下页 返回 结束
P132 6.
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Finding Local Extreme Values
Example Find the local extreme values of
f ( x , y ) xy x 2 y 2 2 x 2 y 4.
Solution The function is defined and differentiable for all x and y and
3. f ( P0 ) is not an extreme lue of the function f if AC B 2 0; 4. Can not be determined if AC B 2 0.
8
Note
2 The expression AC B 2 or f xx f yy f xy is called the discriminant or
order partial derivatives in a neighbourhood of the point P0 ( x0 , y0 ), and P0 is a stationary point of the function f.
A f xx ( P0 ),
Let
B f xy ( P0 ), C f yy ( P0 ).
its domain has no boundary points. The function therefore has extreme
values only at the points where fx and fy are simultaneously zero. These leads to
Example Find the local extreme values of
f ( x , y ) xy x 2 y 2 2 x 2 y 4.
Solution (continued) To see if it does so, we calculate f xx 2, The discriminant of f at (-2,-2) is
5
Examples for Saddle Point
z z
y
x x
y
Function point (0,0).
xy( x 2 y 2 ) z at the 2 2 x y
Function
point (0,0).
z y 2 y 4 x 2 at the
6
Finding Local Extreme Values
Theorem Suppose that both partial derivatives of the function f ( x , y )
exist at the point ( x0 , y0 ) , and ( x0 , y0 ) is an extreme point of the function f. Then
Section 10.2
Extreme Value Problems of Multivariable Functions
1
Unrestricted Extreme Values
Definition Suppose the that the function f ( x , y ) is defined in U ( x0 , y0 ).
f xx 0, giving rise to a local maximum,and upwards if f xx 0, giving
rise to a local minimum. On the other hand, if the discriminant is negative at P0 ( x0 , y0 ), then the surface curves up in some directions and down in others, so we have a saddle point.
1. Interior points where f x f y 0.
2. Interior points where one or both of f x and f y do not exist. Definition An interior point of the domain of a function f ( x , y ) where both fx and fy are zero or where one or both fx and fy do not
exist everywhere. where Therefore, local extreme values can occur only
fx 2x 0
and
f y 2 y 0.
z
The only possibility is the origin, where the value of f is zero. Since f is never
f x y 2 x 2 0, f y x 2 y 2 0,
or
x y 2.
Therefore, the point (-2,-2) is the only point f may take on an extreme value.
10
Finding Local Extreme Values
For example, the function
z x 2 y 2 attains a minimum at the point
(0,0) and z 1 x 2 y 2 attains a maximum at the point (0,0).
The Necessary Condition for An Extreme Value
Example Find the local extreme values of f ( x , y ) x 2 y 2 .
Solution The domain of f is the entire plane ( so there are no
boundary points ) and the partial derivatives f x 2 x and f y 2 y
Hessian of f. It is sometimes easier to remember it in determinant form,
f xx f yy f
2 xy
f xx f xy
f xy f yy
.
The last theorem says that if the discriminant is positive at the point P0 , then the surface curves the same way in all directions: downwards if
exist is a critical point of f.
4
Saddle Point
Thus, the only points where a function f ( x , y ) can assume extreme values are critical points and boundary points. As with differentiable functions of a single variable, not every critical point gives rise to a local extremum. A differentiable function of a single variable might have a point of inflection. A differentiable function of two variables might have a saddle point.
Then
1. f ( P0 ) is a minimum value of the function f if A 0 and AC B 2 0; 2. f ( P0 ) is a maximum value of the function f if A 0 and AC B 2 0;
2 f xx f yy f xy ( 2)( 2) (1)2 4 1 3.
f yy 2,
f xy 1.
The combination
f xy 0 and
negative, we see that the origin gives a
local minimum.
x
y
7
Sufficient Condition for Extreme Values
Theorem Suppose that function z f ( x , y ) has continuous second
Definition
A differentiable function f ( x , y ) has a saddle point at a
f ( x , y ) f (a , b ) and domain points (x,y)
stationary point (a,b) if in every open disk centered at (a,b) there are domain points (x,y) where where f ( x , y ) f (a , b ). The corresponding point (a,b,f(a,b)) on the surface z f ( x , y ) is called a saddle point of the surface.
3
f ( x , y ) x 2 y 2 , the point (0,0) is a stationary