微积分英文课件PPT (7)
高等数学-微积分第1章（英文讲稿）
高等数学-微积分第1章（英文讲稿）C alc u lus (Fifth Edition)高等数学- Calculus微积分（双语讲稿）Chapter 1 Functions and Models1.1 Four ways to represent a function1.1.1 ☆Definition(1-1) function: A function f is a rule that assigns to each element x in a set A exactly one element, called f(x), in a set B. see Fig.2 and Fig.3Conceptions: domain; range (See fig. 6 p13); independent variable; dependent variable. Four possible ways to represent a function: 1)Verbally语言描述(by a description in words); 2) Numerically数据表述(by a table of values); 3) Visually 视觉图形描述(by a graph);4)Algebraically 代数描述(by an explicit formula).1.1.2 A question about a Curve represent a function and can’t represent a functionThe way ( The vertical line test ) : A curve in the xy-plane is the graph of a function of x if and only if no vertical line intersects the curve more than once. See Fig.17 p 171.1.3 ☆Piecewise defined functions (分段定义的函数)Example7 (P18)1－x if x ≤1f(x)＝﹛x2if x＞1Evaluate f(0)，f(1)，f(2) and sketch the graph.Solution：1.1.4 About absolute value (分段定义的函数)⑴∣x∣≥0；⑵∣x∣≤0Example8 (P19)Sketch the graph of the absolute value function f(x)＝∣x∣.Solution：1.1.5☆☆Symmetry ，(对称) Even functions and Odd functions (偶函数和奇函数)⑴Symmetry See Fig.23 and Fig.24⑵①Even functions: If a function f satisfies f(－x)＝f(x) for every number x in its domain，then f is call an even function. Example f(x)＝x2 is even function because: f(－x)＝ (－x)2＝x2＝f(x)②Odd functions: If a function f satisfie s f(－x)＝－f(x) for every number x in its domain，thenf is call an odd function. Example f(x)＝x3 is even function because: f(－x)＝(－x)3＝－x3＝－f(x)③Neither even nor odd functions:1.1.6☆☆Increasing and decreasing function (增函数和减函数)⑴Definition(1-2) increasing and decreasing function:① A function f is called increasing on an interval I if f(x1)＜f(x2) whenever x1＜x2 in I. ①A function f is called decreasing on an interval I if f(x1)＞f(x2) whenever x1＜x2 in I.See Fig.26. and Fig.27. p211.2 Mathematical models: a catalog of essential functions p251.2.1 A mathematical model p25A mathematical model is a mathematical description of a real-world phenomenon such as the size of a population, the demand for a product, the speed of a falling object, the concentration of a product in a chemical reaction, the life expectancy of a person at birth, or the cost of emission reduction.1.2.2 Linear models and Linear function P261.2.3 Polynomial P27A function f is called a polynomial ifP(x) ＝a n x n＋a n－1x n－1＋…＋a2x2＋a1x＋a0Where n is a nonnegative integer and the numbers a0，a1，a2，…，a n－1，a n are constants called the coefficients of the polynomial. The domain of any polynomial is R＝(－∞,＋∞).if the leading coefficient a n≠0, then the degree of the polynomial is n. For example, the function P(x) ＝5x6＋2x5－x4＋3x－9⑴Quadratic function example: P(x) ＝5x2＋2x－3 二次函数（方程）⑵Cubic function example: P(x) ＝6x3＋3x2－1 三次函数（方程）1.2.4Power functions幂函数P30A function of the form f(x) ＝x a，Where a is a constant, is called a power function. We consider several cases：⑴a＝n where n is a positive integer ，(n＝1，2，3，…，)⑵a＝1/n where n is a positive integer，(n＝1，2，3，…，) The function f(x) ＝x1/n⑶a＝n－1 the graph of the reciprocal function f(x) ＝x－1 反比函数1.2.5Rational function有理函数P 32A rational function f is a ratio of two polynomials:f(x)＝P(x) /Q(x)1.2.6Algebraic function代数函数P32A function f is called algebraic function if it can be constructed using algebraic operations ( such as addition，subtraction，multiplication，division，and taking roots) starting with polynomials. Any rational function is automatically an algebraic function. Examples: P 321.2.7Trigonometric functions 三角函数P33⑴f(x)＝sin x⑵f(x)＝cos x⑶f(x)＝tan x＝sin x / cos x1.2.8Exponential function 指数函数P34The exponential functions are the functions the form f(x) ＝a x Where the base a is a positive constant.1.2.9Transcendental functions 超越函数P35These are functions that are not a algebraic. The set of transcendental functions includes the trigonometric，inverse trigonometric，exponential，and logarithmic functions，but it also includes a vast number of other functions that have never been named. In Chapter 11 we will study transcendental functions that are defined as sums of infinite series.1.2 Exercises P 35-381.3 New functions from old functions1.3.1 Transformations of functions P38⑴Vertical and Horizontal shifts (See Fig.1 p39)①y＝f(x)＋c，（c＞0）shift the graph of y＝f(x) a distance c units upward.②y＝f(x)－c，（c＞0）shift the graph of y＝f(x) a distance c units downward.③y＝f(x＋c)，（c＞0）shift the graph of y＝f(x) a distance c units to the left.④y＝f(x－c)，（c＞0）shift the graph of y＝f(x) a distance c units to the right.⑵ V ertical and Horizontal Stretching and Reflecting (See Fig.2 p39)①y＝c f(x)，（c＞1）stretch the graph of y＝f(x) vertically bya factor of c②y＝(1/c) f(x)，（c＞1）compress the graph of y＝f(x) vertically by a factor of c③y＝f(x/c)，（c＞1）stretch the graph of y＝f(x) horizontally by a factor of c.④y＝f(c x)，（c＞1）compress the graph of y＝f(x) horizontally by a factor of c.⑤y＝－f(x)，reflect the graph of y＝f(x) about the x-axis⑥y＝f(－x)，reflect the graph of y＝f(x) about the y-axisExamples1: (See Fig.3 p39)y＝f( x) ＝cos x，y＝f( x) ＝2cos x，y＝f( x) ＝（1/2）cos x，y＝f( x) ＝cos（x/2），y＝f( x) ＝cos2xExamples2: (See Fig.4 p40)Given the graph y＝f( x) ＝( x)1/2，use transformations to graph y＝f( x) ＝( x)1/2－2，y＝f( x) ＝(x－2)1/2，y＝f( x) ＝－( x)1/2，y＝f( x) ＝2 ( x)1/2，y＝f( x) ＝(－x)1/21.3.2 Combinations of functions (代数组合函数)P42Algebra of functions: Two functions (or more) f and g through the way such as add, subtract, multiply and divide to combined a new function called Combination of function.☆Definition(1-2) Combination function: Let f and g be functions with domains A and B. The functions f±g，f g and f /g are defined as follows: （特别注意符号(f±g)( x) 定义的含义）①(f±g)( x)＝f(x)±g( x)，domain ＝A∩B②(f g)( x)＝f(x) g( x)，domain ＝A∩ B③(f /g)( x)＝f(x) /g( x)，domain ＝A∩ B and g( x)≠0Example 6 If f( x) ＝( x)1/2，and g( x)＝(4－x2)1/2，find functions y＝f(x)＋g( x)，y＝f(x)－g( x)，y＝f(x)g( x)，and y＝f(x) /g( x)Solution: The domain of f( x) ＝( x)1/2 is [0，＋∞)，The domain of g( x) ＝(4－x2)1/2 is interval [－2，2]，The intersection of the domains of f(x) and g( x) is[0，＋∞)∩[－2，2]＝[0，2]Thus，according to the definitions, we have(f＋g)( x)＝( x)1/2＋(4－x2)1/2，domain [0，2](f－g)( x)＝( x)1/2－(4－x2)1/2，domain [0，2](f g)( x)＝f(x) g( x) ＝( x)1/2(4－x2)1/2＝(4 x－x3)1/2domain [0，2](f /g)( x)＝f(x)/g( x)＝( x)1/2/(4－x2)1/2＝[ x/(4－x2)]1/2 domain [0，2)1.3.3☆☆Composition of functions (复合函数)P45☆Definition(1-3) Composition function: Given two functions f and g the composite func tion f⊙g (also called the composition of f and g ) is defined by(f⊙g)( x)＝f( g( x)) （特别注意符号(f⊙g)( x) 定义的含义）The domain of f⊙g is the set of all x in the domain of g such that g(x) is in the domain of f . In other words, (f⊙g)(x) is defined whenever both g(x) and f (g (x)) are defined. See Fig.13 p 44 Example7 If f (g)＝( g)1/2 and g(x)＝(4－x3)1/2find composite functions f⊙g and g⊙f Solution We have(f⊙g)(x)＝f (g (x) ) ＝( g)1/2＝((4－x3)1/2)1/2(g⊙f)(x)＝g (f (x) )＝(4－x3)1/2＝[4－((x)1/2)3]1/2＝[4－(x)3/2]1/2Example8 If f (x)＝( x)1/2 and g(x)＝(2－x)1/2find composite function s①f⊙g ②g⊙f ③f⊙f④g⊙gSolution We have①f⊙g＝(f⊙g)(x)＝f (g (x) )＝f((2－x)1/2)＝((2－x)1/2)1/2＝(2－x)1/4The domain of (f⊙g)(x) is 2－x≥0 that is x ≤2 {x ︳x ≤2 }＝(－∞，2]②g⊙f＝(g⊙f)(x)＝g (f (x) )＝g (( x)1/2 )＝(2－( x)1/2)1/2The domain of (g⊙f)(x) is x≥0 and 2－( x)1/2x ≥0 ，that is( x)1/2≤2 ，or x ≤ 4 ，so the domain of g⊙f is the closed interval[0，4]③f⊙f＝(f⊙f)(x)＝f (f(x) )＝f((x)1/2)＝((x)1/2)1/2＝(x)1/4The domain of (f⊙f)(x) is [0，∞)④g⊙g＝(g⊙g)(x)＝g (g(x) )＝g ((2－x)1/2 )＝(2－(2－x)1/2)1/2The domain of (g⊙g)(x) is x－2≥0 and 2－(2－x)1/2≥0 ，that is x ≤2 and x ≥－2，so the domain of g⊙g is the closed interval[－2，2]Notice: g⊙f⊙h＝f (g(h(x)))Example9Example10 Given F (x)＝cos2( x＋9)，find functions f，g，and h such that F (x)＝f⊙g⊙h Solution Since F (x)＝[cos ( x＋9)] 2，that is h (x)＝x＋9 g(x)＝cos x f (x)＝x2Exercise P 45-481.4 Graphing calculators and computers P481.5 Exponential functions⑴An exponential function is a function of the formf (x)＝a x See Fig.3 P56 and Fig.4Exponential functions increasing and decreasing (单调性讨论)⑵Lows of exponents If a and b are positive numbers and x and y are any real numbers. Then1) a x+y＝a x a y2) a x－y＝a x / a y3) (a x)y＝a xy4) (ab)x+y ＝a x b x⑶about the number e f (x)＝e x See Fig. 14，15 P61Some of the formulas of calculus will be greatly simplified if we choose the base a .Exercises P 62-631.6 Inverse functions and logarithms1.6.1 Definition(1-4) one-to-one function: A function f iscalled a one-to-one function if it never takes on the same value twice；that is，f (x1)≠f (x2)，whenever x1≠x2( 注解：不同的自变量一定有不同的函数值，不同的自变量有相同的函数值则不是一一对应函数) Example: f (x)＝x3is one-to-one function.f (x)＝x2 is not one-to-one function, See Fig.2,3,4 ☆☆Definition(1-5) Inverse function:Let f be a one-to-one function with domain A and range B. Then its inverse function f －1(y)has domain B and range A and is defined byf－1(y)＝x f (x)＝y for any y in Bdomain of f－1＝range of frange of f－1＝domain of f( 注解：it says : if f maps x into y, then f－1maps y back into x . Caution: If f were not one-to-one function，then f－1 would not be uniquely defined. )Caution: Do not mistake the－1 in f－1for an exponent. Thus f－1(x)＝1/ f(x) Because the letter x is traditionally used as the independent variable, so when we concentrate on f－1(y) rather than on f－1(y), we usually reverse the roles of x and y in Definition (1-5) and write as f－1(x)＝y f (x)＝yWe get the following cancellation equations:f－1( f(x))＝x for every x in Af (f－1(x))＝x for every x in B See Fig.7 P66Example 4 Find the inverse function of f(x)＝x3＋6Solution We first writef(x)＝y＝x3＋6Then we solve this equation for x:x3＝y－6x＝(y－6)1/3Finally, we interchange x and y：y＝(x－6)1/3That is, the inverse function is f－1(x)＝(x－6)1/3( 注解：The graph of f－1 is obtained by reflecting the graph of f about the line y＝x. ) See Fig.9、8 1.6.2 Logarithmic function If a＞0 and a≠1，the exponential function f (x)＝a x is either increasing or decreasing and so it is one-to-one function by the Horizontal Line Test. It therefore has an inverse function f－1，which is called the logarithmic function with base a and is denoted log a，If we use the formulation of an inverse function given by (See Fig.3 P56)f－1(x)＝y f (x)＝yThen we havelogx＝y a y＝xThe logarithmic function log a x＝y has domain (0,∞) and range R.Usefully equations:①log a(a x)＝x for every x∈R②a log ax＝x for every x＞01.6.3 ☆Lows of logarithms :If x and y are positive numbers, then①log a(xy)＝log a x＋log a y②log a(x/y)＝log a x－log a y③log a(x)r＝r log a x where r is any real number1.6.4 Natural logarithmsNatural logarithm isl og e x＝ln x ＝ythat is①ln x ＝y e y＝x② ln(e x)＝x x∈R③e ln x＝x x＞0 ln e＝1Example 8 Solve the equation e5－3x＝10Solution We take natural logarithms of both sides of the equation and use ②、③ln (e5－3x)＝ln10∴5－3x＝ln10x＝(5－ln10)/3Example 9 Express ln a＋(ln b)/2 as a single logarithm.Solution Using laws of logarithms we have:ln a＋(ln b)/2＝ln a＋ln b1/2＝ln(ab1/2)1.6.5 ☆Change of Base formula For any positive number a (a≠1), we havel og a x＝ln x/ ln a1.6.6 Inverse trigonometric functions⑴Inverse sine function or Arcsine functionsin－1x＝y sin y＝x and －π/2≤y≤π / 2，－1≤x≤1 See Fig.18、20 P72Example13 ① sin－1 (1/2) or arcsin(1/2) ② tan(arcsin1/3)Solution①∵sin (π/6)＝1/2，π/6 lies between －π/2 and π / 2，∴sin－1 (1/2)＝π/6② Let θ＝arcsin1/3，so sinθ＝1/3tan(arcsin1/3)＝tanθ＝s inθ/cosθ＝(1/3)/(1－s in2θ)1/2＝1/(8)1/2Usefully equations:①sin－1(sin x)＝x for －π/2≤x≤π / 2②sin (sin－1x)＝x for －1≤x≤1⑵Inverse cosine function or Arccosine functioncos－1x＝y cos y＝x and 0 ≤y≤π，－1≤x≤1 See Fig.21、22 P73Usefully equations:①cos－1(cos x)＝x for 0 ≤x≤π②cos (cos－1x)＝x for －1≤x≤1⑶Inverse Tangent function or Arctangent functiontan－1x＝y tan y＝x and －π/2＜y＜π / 2 ，x∈R See Fig.23 P73、Fig.25 P74Example 14 Simplify the expression cos(ta n－1x).Solution 1 Let y＝tan－1 x，Then tan y＝x and －π/2＜y＜π / 2 ，We want find cos y but since tan y is known, it is easier to find sec y first:sec2y＝1 ＋tan2y sec y＝(1 ＋x2 )1/2∴cos(ta n－1x)＝cos y ＝1/ sec y＝(1 ＋x2)－1/2Solution 2∵cos(ta n－1x)＝cos y∴cos(ta n－1x)＝(1 ＋x2)－1/2⑷Other Inverse trigonometric functionscsc－1x＝y∣x∣≥1csc y＝x and y∈(0，π / 2]∪(π，3π / 2]sec－1x＝y∣x∣≥1sec y＝x and y∈[0，π / 2)∪[π，3π / 2]cot－1x＝y x∈R cot y＝x and y∈(0，π)Exercises P 74-85Key words and PhrasesCalculus 微积分学Set 集合Variable 变量Domain 定义域Range 值域Arbitrary number 独立变量Independent variable 自变量Dependent variable 因变量Square root 平方根Curve 曲线Interval 区间Interval notation 区间符号Closed interval 闭区间Opened interval 开区间Absolute 绝对值Absolute value 绝对值Symmetry 对称性Represent of a function 函数的表述（描述）Even function 偶函数Odd function 奇函数Increasing Function 增函数Increasing Function 减函数Empirical model 经验模型Essential Function 基本函数Linear function 线性函数Polynomial function 多项式函数Coefficient 系数Degree 阶Quadratic function 二次函数（方程）Cubic function 三次函数（方程）Power functions 幂函数Reciprocal function 反比函数Rational function 有理函数Algebra 代数Algebraic function 代数函数Integer 整数Root function 根式函数（方程）Trigonometric function 三角函数Exponential function 指数函数Inverse function 反函数Logarithm function 对数函数Inverse trigonometric function 反三角函数Natural logarithm function 自然对数函数Chang of base of formula 换底公式Transcendental function 超越函数Transformations of functions 函数的变换Vertical shifts 垂直平移Horizontal shifts 水平平移Stretch 伸张Reflect 反演Combinations of functions 函数的组合Composition of functions 函数的复合Composition function 复合函数Intersection 交集Quotient 商Arithmetic 算数。
微积分CALCULUS.ppt
Solution:
a. Since g 32 ft/sec2 (9.8m/s2 ), V0 96 ft/sec and H0 112 ft, the height of the ball above the ground at time t is H (t) 16t 2 96t 112 feet. The velocity at time t is
c. Set v(t)=0, solve t=3. Thus, the ball is at its highest point when t=3 seconds.
d. The ball starts at H(0)=112 feet and rises to a maximum height of H(3)=256. So: The total distance travelled=2(256-112)+112 =400 feet.
acceleration acting on the object is the constant
downward acceleration g due to gravityistance is negligible). Thus, the height of the object
at time t is given by the formula
H
(t )
1 2
gt 2
V0t
H0
where H0 and V0 are the initial height and velocity of the object, respectively.
Example 10 Suppose a person standing at the top of a building 112 feet high throws a ball vertically upward with an initial velocity of 96 ft/sec.
微积分英文版课件
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定理 . 原函数都在函数族
证: 1)
( C 为任意常数 ) 内 .
即
又知
[(x) F(x)] (x) F(x) f (x) f (x) 0
故
(x) F(x) C0 (C0 为某个常数)
即 (x) F(x) C0 属于函数族 F(x) C .
( k 为常数)
(2)
x dx
1
1
x
1
C
( 1)
(3)
dx x
ln
x
C
x 0时 ( ln x ) [ ln(x) ] 1
x
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(4)
1
dx x
2
arctan
x
C
或 arccot x C
(5)
dx arcsin x C 1 x2
或 arccos x C
想到公式
1
d
u u
2
arctan u C
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例. 求 解:
dx a 1 (ax)2
d
(
x a
)
1
(
x a
)2
想到
d u arcsinu C 1u2
f [(x)](x)dx f ((x))d(x)
(直接配元)
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例4. 求 解:
例1. 求
解: 令 u ax b ,则 d u adx , 故
原式 = um 1 d u 1 1 um1 C a a m1
注: 当
时
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微积分一些相关PPT
微分学
微分学主要研究的是在函数自变量变化时如
何确定函数值的瞬时变化率（或微分）。换 言之，计算导数的方法就叫微分学。微分学 的另一个计算方法是牛顿法，该算法又叫应 用几何法，主要通过函数曲线的切线来寻找 点斜率。费马常被称作“微分学的鼻祖”。
积分学
积分学是微分学的逆运算，即从导数推算出
原函数,又分为定积分与不定积分。一个一元 函数的定积分可以定义为无穷多小矩形的面 积和，约等于函数曲线下包含的实际面积。 因此，我们可以用积分来计算平面上一条曲 线所包含的面积、球体或圆锥体的表面积或 体积等。而不定积分的用途较少，主要用于 微分方程的解。
牛顿
牛顿在1671年写了《流数法和 无穷级数》，这本书直到1736 年才出版，它在这本书里指出： 变量是由点、线、面的连续运动产生的，否定了以 前自己认为的变量是无穷小元素的静止集合。他把 连续变量叫做流动量，把这些流动量的导数叫做流 数。牛顿在流数术中所提出的中心问题是：已知连 续运动的路径，求给定时刻的速度（微分法）；已 知运动的速度求给定时间内经过的路程（积分法）。
2：微积分的创立
微积分学的建立 从微积分成为一门学科来说，是在十七世纪，但是，微分和 积分的思想在古代就已经产生了。 极限的产生 公元前三世纪，古希腊的阿基米德在研究解决抛物弓形的面 积、球和球冠面积、螺线下面积和旋转双曲体的体积的问题 中，就隐含着近代积分学的思想。作为微分学基础的极限理 论来说，早在古代以有比较清楚的论述。比如中国的庄周所 著的《庄子》一书的“天下篇”中，记有“一尺之棰，日取 其半，万世不竭”。三国时期的刘徽在他的割圆术中提到 “割之弥细，所失弥小，割之又割，以至于不可割，则与圆 周和体而无所失矣。”这些都是朴素的、也是很典型的极限 概念。
微积分教学资料-calculusi
Therefore , by the definition of a limit,
lim [f(x)g(x)] lim f(x)lim g(x)
x a
x a
x a
LM
Direct Substitution Property
If f is a polynomial or a rational function
x a
n a
posi n titveeger
If lif m (x ) Lan lid g m (x ) M
x a
x a
both exist, then
li [ f m ( x ) g ( x ) ] li f ( m x ) li g ( m x ) L M
x a
lim sin x 1 x0 x
Caution: Notice the phrase “but x≠a” in the definition of limit.
This means that in finding the limit of f(x) as x approaches a, we never consider x=a.
7. lim c c xa
8.lim x a xa
9. lim x n a n , where n is a positive integer xa
10 . lim n x n a , where n is a positive integer xa
1.1li.m n f(x)nlim f(x) whenriesa
In fact, f(x) need not even be defined when x=a. the only thing that matters is how f is defined near a.
微积分的名称ppt课件市公开课金奖市赛课一等奖课件
窮竭法
當矩形數目愈來愈多， 它們面積之和會愈來 愈迫近曲邊形面積。
第7页
不可分元法
對一個平面片而言，其“不可分元”是 指它一條弦(chord)
對一個立體而言，其“不可分元”是指 它一個平面截面
第8页
不可分元法
第9页
Democritus(德謨克利特 460~370BC)
根據不可分元想法，推出稜錐(或 圓錐)體積是含有同樣底和高稜 柱(或圓柱)體積三分之一。
26 歲出任駐巴黎大使，結識 Huygens，並開
始鑽研數學，研讀 Descartes 及 Pascal 的著作
27 歲出使倫敦
30 歲任 Elector of Hanover 的顧問及圖書館
長
1700 年服務於 Elector of Brandenburg 宮廷
1716 年去世
萊布尼茲十分博學，對法律、歷史、神學、
但書寫得非常艱澀。 1703 年任英國皇家學會主席 62 歲放棄教授職位，轉任倫敦大英造幣廠，同年被封為爵士 死時年 85 歲
第25页
第26页
萊布尼茲(Leibnitz 1646~1716)
第27页
萊布尼茲(Leibnitz 1646~1716)貢獻
1684年發表《一種求極大極小和切 線新 办法，適用於分式和無窮量，以及這種 新办法奇妙類型計算》
《莊子天下篇》曰：「一尺之棰，日取 其半，萬世不竭。」
劉徽創割圓術，謂「割之彌細，所失彌 少。割之又割，以至於不可割，則與圓 合體而無所失矣。」
祖氏父子：『夫疊棋成立積，緣冪勢既 同，則積不容異。』(注：“冪”指截面 面積，“勢”指高度)
第15页
問題引路
第一類問題：已知距離表為時間函數，求 速度和加速度。反過來，已知加速度表為 時間函數，求距離和速度。 (例：Galileo曾探討此類問題)
《微积分》PPT课件
公式.
微积分Ⅰ
第九章
重积分
10
说明: ① 使用公式 (1) 必须是 X－ 型域, 使用公式 (2) 必 须是 Y － 型域. ② 若积分区域既是 X － 型区域又是 Y－ 型区域,
则有
f ( x, y ) d x d y
dx
a
d
y
y 2 ( x)
D b
x 1 ( y)
微积分Ⅰ
第九章
重积分
6
在 [a, b] 上任意取定一点 x0, 作平行于 yOz 面的平
面 x = x0, 则该平面截曲顶柱体所得的截面是一个以区 间 [ 1 (x0), 2 (x0) ] 为底、曲线 z = f (x0 , y) 为曲边的 曲边梯形.
z
z f ( x, y)
y
A( x0 )
2
R
它的底为 D {( x, y ) | 0 y R2 x 2 , 0 x R},
微积分Ⅰ
第九章
重积分
23
∴所求体积为
8
R
0
R 2 x 2 dx
R2 x 2
0
dy
8 ( R 2 x 2 )dx
0
R
16 3 R . 3
微积分Ⅰ
第九章
重积分
24
1 x
y x
1
微积分Ⅰ
第九章
重积分
21
说明: ① 计算二重积分时, 选择积分次序是比较重要的 一步, 积分次序选择不当, 可能会使计算繁琐, 甚至无
法计算. 一般地, 既要考虑积分区域 D 的形状, 又要考
虑被积函数 f (x, y) 的特性. ② 应遵循 “能积分, 少分快, 计算简” 的原则.
高等数学(微积分)ppt课件
目录•绪论•函数与极限•导数与微分•微分中值定理与导数的应用•不定积分与定积分•微分方程与级数绪论01020304古代数学算术、几何与代数的起源与发展中世纪数学数学与哲学的交织文艺复兴时期数学解析几何与微积分的萌芽现代数学抽象化、公理化与结构化的趋势数学的发展历程微积分的创立与意义01微积分的创立牛顿与莱布尼兹的贡献02微积分的意义解决现实问题的有力工具，推动科学技术的发展03微积分的应用领域物理学、工程学、经济学等高等数学的研究对象与内容研究对象01函数、极限、连续、微分、积分等基本概念与性质研究内容02一元函数微积分学、多元函数微积分学、常微分方程等高等数学与其他学科的联系03为其他数学分支提供基础，为其他学科提供数学工具函数与极限函数定义设$x$和$y$是两个变量，$D$是一个数集。
如果存在一种对应法则$f$，使得对于$D$中的每一个数$x$，在数集$M$中都有唯一确定的数$y$与之对应，则称$f$为定义在$D$上的函数，记作$y=f(x),x in D$。
函数的性质包括有界性、单调性、奇偶性、周期性等。
常见函数类型一次函数、二次函数、幂函数、指数函数、对数函数、三角函数等。
010203函数的概念与性质设函数$f(x)$在点$x_0$的某个去心邻域内有定义。
如果存在常数$A$，对于任意给定的正数$epsilon$(无论它多么小)，总存在正数$delta$，使得当$x$满足不等式$0<|x-x_0|<delta$时，对应的函数值$f(x)$都满足不等式$|f(x)-A|<epsilon$，那么常数$A$就叫做函数$f(x)$当$x to x_0$时的极限，记作$lim_{x tox_0}f(x)=A$或$f(x) to A(x to x_0)$。
极限的性质唯一性、局部有界性、保号性、保不等式性、迫敛性等。
极限定义极限的定义与性质VS极限的运算法则极限的四则运算法则若两个函数的极限存在，则它们的和、差、积、商（分母不为零）的极限也存在，且等于这两个函数极限的和、差、积、商。
《微积分英文》课件 (2)
Types of Limits
One-sided limits
Limits approached
from one direction
Limits at infinity
Behavior of functions at
infinity
● 02
第2章 Limits and Continuity
01 Definition of a limit
Explanation of what a limit is
02 Properties of limits
Key characteristics of limits
03 Calculating limits algebraically
Graphing functions by analyzing their derivatives and key points
Higher Order Derivatives
Second derivative
Rate of change of the rate of
change
nth derivative
● 03
第3章 Differentiation
Derivatives and Rates of
Change
A derivative is defined as the rate of change of a function at a given point. Notation for derivatives includes symbols such as f'(x) or dy/dx. Derivatives can be interpreted as rates of change in various realworld applications.
《微积分英文版》课件
Limits and continuity
Definition: A limit is the value that a function approaches as the input approaches a certain point Continuity means that the function doesn't have any breaks or jumps at any point
Course structure
03
The course is divided into several modules, each focusing on a specific topic in calculus Learners can complete the course at their own pace and in any order of the modules
Properties: One side limits, absolute continuity, uniform continuity, etc
Differentiation
Definition: The derivative of a function at a point is the slope of the tangent line to the graph of the function at that point It can be used to find the rate of change of a function
Integral definition: The integral of a function is a measure of the area under its curve It is calculated by finding the limit of the sum of areas of rectangles under the curve as the width of the rectangles approaches zero
微积分(下)英文教材
Chapter 1 Infinite SeriesGenerally, for the given sequence,.......,......,3,21n a a a a the expressionformed by the sequence ,.......,......,3,21n a a a a.......,.....321+++++n a a a ais called the infinite series of the constants term, denoted by ∑∞=1n n a , that is∑∞=1n n a =......., (32)1+++++n a a a aWhere the nth term is said to be the general term of the series, moreover, the nth partial sum of the series is given by=n S......321n a a a a ++++1.1 Determine whether the infinite series converges or diverges.Whil e it’s possible to add two numbers, three numbers, a hundred numbers, or even a million numbers, it’s impossible to add an infinite number of numbers.To form an infinite series we begin with an infinite sequence of real numbers:.....,,,3210a a a a , we can not form the sum of all the ka (there isan infinite number of the term), but we can form the partial sums∑===000k ka a S∑==+=1101k ka a a S∑==++=22102k ka a a a S∑==+++=332103k ka a a a a S……………….∑==+++++=nk kn n a a a a a a S 03210.......Definition 1.1.1If the sequence {n S } of partial sums has a finite limit L, We write∑∞==k ka Land say that the series ∑∞=0k k a converges to L. we call L the sum ofthe series.If the limit of the sequence {n S } of partial sums don’t exists, we say that the series ∑∞=0k k a diverges.Remark it is important to note that the sum of a series is not a sum in the ordering sense. It is a limit.EX 1.1.1 prove the following proposition: Proposition1.1.1: (1) If 1<xthen the ∑∞=0k k a converges, and ;110xx k k -=∑∞=(2)If,1≥xthen the ∑∞=0k kx diverges.Proof: the nth partial sum of the geometric series ∑∞=0k k a takes theform1321.......1-+++++=n n xx x x S ① Multiplication by x gives).......1(1321-+++++=n n xx x x x xS =nn xxx x x +++++-1321.......Subtracting the second equation from the first, we find thatnn xS x -=-1)1(. For ,1≠x this givesxxS nn --=11 ③If,1<x then→nx ,and this by equation ③.xxxS nn n n -=--=→→1111limlimThis proves (1).Now let us prove (2). For x=1, we use equation ① and device that,n S n =Obviously, ∞=∞→n n S lim, ∑∞=0k k a diverges.For x=-1 we use equation ① and we deduce If n is odd, then 0=n S ,If n is even, then.1-=n SThe sequence of partial sum nS like this 0,-1,0,-1,0,-1………..Because the limit of sequence }{n S of partial sum does not exist. By definition 1.1.1, we have the series ∑∞=0k K x diverges. (x=-1).For 1≠x with,1>x we use equation ③. Since in this instance, wehave -∞=--=∞→∞→xxS nn n n 11limlim . The limit of sequence of partial sum not exist,the series ∑∞=0k kxdiverges.Remark the above series is called the geometric series. It arises in so many different contexts that it merits special attention.A geometric series is one of the few series where we can actually give an explicit formula for n S ; a collapsing series is another.Ex.1.1.2 Determine whether or not the series converges ∑∞=++0)2)(1(1k k kSolution in order to determine whether or not this series converges we must examine the partial sum. Since2111)2)(1(1+-+=++k k k kWe use partial fraction decomposition to write2111111 (4)1313121211)2111()111(..............)4131()3121()2111()2)(1(1)1(1 (3).212.11+-+++-++-+-+-=+-+++-++-+-+-=++++++⨯+⨯=n n n n n n n n n n n n S nSince all but the first and last occur in pairs with opposite signs, the sum collapses to give211+-=n S nObviously, as.1,→∞→n S n this means that the series converges to 1.1)211(lim lim =+-=∞→∞→n S n n ntherefore 1)2)(1(10=++∑∞=n k kEX.1.1.3 proves the following theorem:Theorem 1.1.1 the kth term of a convergent series tends to 0; namely if∞=0k ka Converges, by definition we have the limit of the sequence}{n S ofpartial sums exists. NamelylaS nk kn n n ==∑=∞→∞→0limlimObviously.limlim 01l aS nk kn n n ==∑=∞→-∞→since1--=n n s s a n , we have0lim lim )(lim lim 11=-=-=-=-∞→∞→-∞→∞→l l S S S S a n n n n n n n n nA change in notation gives 0lim =∞→n k a .The next result is an obviously, but important, consequence of theorem1.1.1. Theorem 1.1.2 (A diverges test) iflim ≠∞→k k a , or ifn k a ∞→lim does not exist, then the series ∑∞=0k ka diverges.Caution, theorem 1.1.1 does not say that iflim =∞→k k a , and then∑∞=0k ka converge. In fact, there are divergent series for which 0lim=∞→k k a . Forexample, the series .....1 (2)11111++++=∑∞=nkk . Since it issequence }{n S of partial sum nn n n S n =>+++=1 (2)111}{ is unbounded. So∞===∞→∞→n S n n n limlim , therefore the series diverges.But01limlim ===∞→∞→k a k k kEX.1.1.3 determine whether or not the series:..........5443322101+++++=+∑∞=k k kConverges.Solution since 01111lim1limlim≠=+=+==∞→∞→∞→kk k a k k k k , this series diverges.EX.1.1.4 Determine whether or not the series ∑∞=021k kSolution1 thegivenseries is a geometric series.121,)21(0<==∑∑∞=∞=x and xk k k k,by proposition 1.1.1 we know that seriesconverges. Solution 2 ,21 (412)111-++++=n n S ① ,2121 (2)1212121132nn n S +++++=-②①-② (1-21))211(2,211nn nnS S-=-=.2)211(2lim lim =-=∞→∞→nn n n SBy definition of converges of series, this series converges.EX.1.1.5 proofs the following theorem:Theorem 1.1.2 If the series ∑∑∞=∞=0k k k k b and a converges, then (1))(0∑∞=+k k kb aalso converges, and is equal the sum of the two series.(2) If C is a real number, then ∑∞=0k k Ca also converges. Moreover iflak k=∑∞=0then ClCa k k=∑∞=0.Proof let∑∑====nk knnk knbS aS20)1(,∑∑===+=nk knnk k knCa Sb aS40)3(,)(Note that )1()4()2()1()3(nnnnnCSS andS S S =+=Since (),lim ,lim )2(1m S l S nn n n ==∞→∞→Thenm l S S S S S nn n n n n n nn +=+=+=∞→∞→∞→∞→)2()1()2()1()3(lim lim )(lim lim.lim lim lim )1()1()4(Cl S C CSS nn nn nn ===∞→∞→∞→Theorem 1.1.4 (squeeze theorem) Suppose that}{}{n n c and a both converge to l and thatnn n c b a ≤≤ for,k n ≥(k is a fixed integer), then }{n b also converges to l .Ex.1.1.6 show that 0sinlim3=∞→n nn .Solution For ,1≥n ,1)sin(13nnnn≤≤-since,0)1(lim ,0)1(lim ==-∞→∞→n and nn nthe result follows by the squeeze theorem.For sequence of variable sign, it is helpful to have the following result.EX1.1.7 prove that the following theorem holds.Theorem 1.1.5 If 0lim ,0lim==∞→∞→n n n n a then a ,Proof since,n n n a a a ≤≤-from the theorem 1.1.4Namely the squeeze theorem, we know the result is true.Exercise 1.1(1) An expression of the form 123a a a +++…is called(2) A series123a a a +++…is said to converge if the sequence {}S n converges, whereS n=1. The geometric series 2a ar ar +++…converges if; in this casethe sum of the series is 2. Iflim 0n n a →∞≠, we can be sure that the series1nn a∞==∑3. Evaluate 0(1),02k k r r r ∞=-<<∑.4. Evaluate 0(1),11k k k x x ∞=--<<∑.5. Show that 1ln 1k k k ∞=+∑diverges.Find the sums of the series 6-11 6. 31(1)(2)k k k ∞=++∑7.112(1)k k k ∞=+∑8.11(3)k k k ∞=+∑9.0310kk ∞=∑10.0345kkkk ∞=+∑11.3023k kk +∞=∑12. Derivethefollowingresultsfromthegeometricseries 221(1),||11k kk x x x∞=-=<+∑.Test the following series for convergence: 13. 11n nn∞=+∑ 14.312k k ∞+=∑1.2 Series With Positive T erms1.2.1 The comparison TestThroughout this section, we shall assume that our numbers n a are x≥,then the partial sum12n nS a a a =+++… are increasing, i.e.1231n n S S S S S +≤≤≤≤≤≤……If they are to approach a limit at all, they cannot become arbitrarily large. Thus in that case there is a number Bsuch thatn S B≤ for all n.Such a number Bis called an upper bound. By a least upper bound we mean a number Swhich is an upper bound, and such that every upperboundBis S ≥. We take for granted that a least upper bound exists. Thecollection of numbers {}n S has therefore a least upper bound, i.e., there is a smallest numbers such that n S S≤ for all n. In that case, the partialsumsn SapproachSas a limit. In other words, given any positivenumber 0ε>, we have n S S Sε-≤≤ for all n sufficiently large.This simply expresses the factSis the least of all upper bounds forour collection of numbers n S . We express this as a theorem.Theorem 1.2.1 Let {}(1,2,n a n =…) be a sequence of numbers≥and let 12n n S a a a =+++…. If the sequence of numbers {}n S is bounded,then it approaches a limit S , which is its least upper bound.Theorem 1.2.2 A series with nonnegative terms converges if and only if the sequence of partial sums is bounded above.Theorem 1.2.1 and 1.2.2 give us a very useful criterion to determine when a series with positive terms converges.The convergence or divergence of a series with nonnegative terms isusually deduced by comparison with a series of known behavior.S 1 S 2 S n STheorem 1.2.3(The Ordinary Comparison Test) Let 1n n a ∞=∑and1nn b ∞=∑be two series, with0n a ≥for all n and0n b ≥for all n. Assume thatthere is a numbers 0c >, such thatn na cb ≤ for all n, and that 1n n b ∞=∑converges, then 1n n a ∞=∑converges, and 11nnn n a c b ∞∞==≤∑∑.Proof: We have1212121()n n n nn a a a cb cb cb c b b b c b ∞=+++≤+++=+++≤∑……….This means that 1n n c b ∞=∑ is a bound for the partial sums 12n a a a +++….The least upper bound of these sums is therefore 1n n c b ∞=≤∑, thus proving ourtheorem.Theorem 1.2.3 has an analogue to show that a series does not converge.Theorem 1.2.4(Ordinary Comparison Test) Let 1n n a ∞=∑ and 1n n b ∞=∑ betwo series, withna and 0nb ≥ for all n. Assume that there is a number0c >such that n n a cb ≥for all n sufficiently large, and 1n n b ∞=∑ does notconverge, then 1n n a ∞=∑ diverges.Proof. Assume n na cb ≥forn n ≥, since 1n n b ∞=∑diverges, we canmake the partial sum0001Nn n n Nn n b b b b +==+++∑…arbitrarily large as N becomes arbitrarily large. But 0NNNn n nn n n n n n a cb c b ===≥=∑∑∑.Hence the partial sum 121NnNn a a a a ==+++∑… are arbitrarily large as Nbecomes arbitrarily large, are hence 1n n a ∞=∑ diverges, as was to be shown.Remark on notation you have easily seen that for each 0j ≥, 0k k a ∞=∑converges iff 1kk j a ∞=+∑converges. This tells us that, in determining whetheror not a series converges, it does not matter where we begin the summation, where detailed indexing would contribute nothing, we will omit it and write ∑without specifying where the summation begins. For instance, it makes sense to you that 21k∑converges and 1k∑diverges without specifying where we begin the summation. But in the convergent case it does, however, affect the sum. Thus for example122kk ∞==∑,1112kk ∞==∑,21122kk ∞==∑, and so forth.Ex 1.2.1 Prove that the series 211n n∞=∑converges.Solution Let us look at the series:22222222211111111112345781516+++++++++++………We look at the groups of terms as indicated. In each group of terms, if we decrease the denominator in each term, then we increase the fraction. We replace 3 by 2 , then 4,5,6,7 by 4, then we replace the numbers from 8 to 15 by 8, and so forth. Our partial sums therefore less than or equal to222222221111111112244488++++++++++………and we note that 2 occurstwice, 4 occurs four times, 8 occurs eight times, and so forth. Our partialsum are therefore less than or equal to222222221111111112244488++++++++++………and we note that 2 occurs twice, 4 occurs four times, 8 occurs eight times, and so forth. Hence the partial sums are less than or equal to2222124811124848+++++++1 (1)2Thus our partial sums are less than or equal to those of the geometric series and are bounded. Hence our series converges.Generally we have the following result: The series 1111111234pppppn nn∞==++++++∑……, wherepis a constant,is called a p-series.Proposition1.2.1. If 1p >, the p-series converges; and if 1p ≤, thenthe p-series diverges.Ex 1.2.2 Determine whether the series 2311n nn ∞=+∑converges.Solution We write2323111(1)1111nn nn nn==++++. Then we see that23111122nn nn ≥=+. Since 11n n∞=∑ does not converge, it follows that the series 2311n nn∞=+∑ does not converge either. Namely this series diverges.Ex 1.2.3 Prove the series 241723n n n n ∞=+-+∑converges.Proof :Indeed we can write2222424334477(1)171331123(2())2()n n nnn n n n n n n n+++==-+-+-+For n sufficiently large, the factor23471312()nn n+-+ is certainly bounded,and in fact is near 1/2. Hence we can compare our series with 21n∑ tosee converges, because ∑21nconverges and the factor is bounded.Ex.1.2.5 Show that 1ln()k b +∑diverges.Solution 1 We know that ask →∞,ln 0k k→. It follows that ln()0k b k b+→+, and thus thatln()ln()0k b k b k b kk bk+++=→+. Thus forksufficiently large, ln()k b k+< and11ln()kk b <+. Since 1k∑diverges,we can conclude that 1ln()k b +∑diverges.Solution 2 Another way to show that ln()k b k+< for sufficiently largekis to examine the function()ln()f x x x b =-+. At3x = thefunction is positive:(3)3ln 93 2.1970f =-=->Since '1()10f x x b=->+ for all 0x >, ()0f x > for all3x >. It followsthatln()x b x+< for all 3x ≥.We come now to a somewhat more comparison theorem. Our proof relies on the basic comparison theorem.Theorem 1.2.5(The Limit Comparison Test) Let k a ∑ and k b ∑ beseries with positive terms. If lim()k k ka lb →∞=, where l is some positivenumber, then k a ∑ and k b ∑converge or diverge together.Proof Choose ε between 0 and l , sincekka lb →, we know for allksufficiently large (for all k greater than some 0k ) ||k ka lb ε-<.For such k we have k ka l lb εε-<<+, and thus()()k k kl b a l b εε-<<+this last inequality is what we needed.(1) If k a ∑converges, then ()k l b ε-∑converges, and thuskb ∑converges.(2) If k b ∑converges, then ()k l b ε+∑converges, and thuska ∑converges.To apply the limit comparison theorem to a series k a ∑, we must first find a series k b ∑of known behavior for which k ka b converges to apositive number.Ex 1.2.6 Determine whether the series sin kπ∑converges ordiverges.Solution Recall that as sin 0,1x x x→→.As,k kπ→∞→ and thussin 1k kππ→. Sincekπ∑diverges, so sin()kπ∑diverges.Ex 1.2.7Determine whether the series 100∑converges ordiverges.Solution For large value of k,dominates the numeratorand22kthe denominator, thus, for such k,differs252k=. Since2210051020012k kk÷==→And 2255122kk=∑∑converges, this series converges.Theorem 1.2.6 Let k a ∑ and k b ∑ be series with positive terms and suppose thus0k ka b →, then(1) If k b ∑converges, then k a ∑converges. (2) If k a ∑diverges, then k b ∑diverges.(3) If k a ∑converges, then k b ∑may converge or diverge. (4) If k b ∑diverges, then k a ∑may converge or diverge. [Parts (3) and (4) explain why we stipulated 0l >in theorem 1.2.5]1.2.2 The root test and the ratio testTheorem 1.2.7 (the root test, Cauchy test) let ∑k a be a series with nonnegative terms and suppose thatρ==∞→∞→k k k kk k a a 1lim lim, ifρ<1, ∑k a converges, ifρ>1,∑ka diverges, ifρ=1, the test is inconclusive.Proof we suppose firstρ<1 and chooseμso that 1<<u ρ. Sinceρ→k k a 1)(, we have μ<kk a 1, for all k sufficiently large thus kk a μ< for allk sufficiently large since ∑k μ converges (a geometric series with 0<1<μ), we know by theorem 1.2.5 that∑ka converges.We suppose now that1>ρand chooseμso that 1>>u ρ. sinceρ→k k a 1)(, we haveμ>k k a 1)( for all k sufficiently large. Thuskk a μ>for all k sufficiently large.Since ∑k μ diverges (a geometric series with 1>μ ) the theorem1.2.6 tell us that ∑k a diverges.To see the inconclusiveness of the root test when 1=ρ, note that1)(1→k k a for both:112∑∑kandk,11)1()1()(221121=→==kk k k kka 11)1()(11→==k k k k kk aThe first series converges, but the second diverges. EX.1.2.7 Determine whether the series ∑kk )(ln 1converges ordiverges.Solution For the series ∑kk )(ln 1, applying the root test we have0ln 1lim)(lim 1==∞→∞→ka k kk k , the series converges.EX.1.2.8 Determine whether series ∑3)(2k k converges or diverges.Solution For the series ∑kk )3(2, applying the root test, we have1212]1[2)1(.2)(3331>=⨯→==kk k k kka . So the series diverges.EX1.2.9 Determines whether the series kk∑-)11(converges ordiverges.Solution in the case of kk∑-)11(, we have 111)(1→-=ka kk . Ifapplying the root test, it is inconclusive. But since kkka )11(-=convergestoe1 and not to 0, the series diverges.We continue to consider only series with terms≥. To comparesuch a series with a geometric series, the simplest test is given by the ratio test theoremTheorem 1.2.8 (The ratio test, DAlembert test) let ∑k a be a series with positive terms and suppose thatλ=+∞→kk k a a 1lim,If ,1<λ∑k a converges, if,1>λ∑ka diverges.If the,1=λthe test is inconclusive.Proof we suppose first that,1<λsince1lim1<=+∞→λkk k a aSo there exists some integer N such that if n ≥NCa a nn ≤+1 ThenNN N N N a C Caa Caa 212,1≤≤≤+++ and in general byinduction,N kk N a C a ≤+Thusca c c c c a a c a c ca a aNkN N kN N N kN Nn n-≤++++≤++++≤∑+=11)........1( (3)22Thus in effect, we have compared our series with a geometric series, and we know that the partial sums are bounded. This implies that our series converges.The ratio test is usually used in the case of a series with positive terms nasuch that.1lim1<=+∞→λnn n a aEX.1.2.10 show that the series ∑∞=13n n nconverges.Solution we let ,3nn n a =then31.13.3111n n n n a a nn nn +=+=++,this ratioapproaches∞→n as 31, and hence the ratio test is applicable: the seriesconverges.EX1.2.11 show that the series ∑!k kkdiverges.Solution we have kkkk nn kkk kk k k a a )11()1(!)!1()1(11+=+=++=++Soe ka a kk nn n =+=∞→+∞→)11(lim lim1Since 1>e , the series diverges. EX.1.2.12 proves the series ∑+121k diverges.Solution sincekk k k k k a a k k 32123212112.1)1(211++=++=+++=+ 13212limlim1=++=∞→+∞→kk a a k kk k .Therefore the ratio test is inconclusive. We have to look further. Comparison with the harmonic series shows that the series diverges:∑++=+>+)1(21,11.21)1(21121k k k k dverges.Exercise 1.21. The ordinary comparison test says that if ____ and if ∑i b converges. Then ∑k a also converges.2. Assume that 00>≥k kb and a . Thelimit comparison Test says that if0<____<+∞ then ∑k a and ∑k b converges or diverge together. 3. Let nn n a a 1lim+∞→=ρ. The ratio Test says that a series ∑k a of positive termsconverges if ___, diverges if ____and may do either if ___. Determine whether the series converges or diverges 4. ∑+13kk 5. ∑+2)12(1k 6. ∑+11k 7. ∑-kk2218. ∑+-1tan 21kk9. ∑321k10. ∑-k)43( 11. ∑k kln 12. ∑!10k k13. ∑kk1 14. ∑kk 100! 15. ∑++kk k623216. kk ∑)32( 17.∑+k11.18. ∑kk 410!19. Let}{n a be a sequence of positive number and assume thatna a nn 111-≥+ for all n. show that the series ∑n a diverges.1.3 Alternating series, Absolute convergence and conditional convergenceIn this section we consider series that have both positive and negative terms.1.3.1 Alternating series and the tests for convergence The series of the form .......4321+-+-u u u u iscalled the alternatingseries, where 0>n ufor all n, here two example:∑∞=--=+-+-+-11)1( (6)1514131211n n n,11)1( (6)5544332211+-=+-+-+-∑∞=n nnWe see from these examples that the nth term of an alternating series is the form n n n n n nu a or u a )1()1(1-=-=-,wherenu is a positive number (infact nna u =.)The following test says that if the terms of an alternating series decrease toward 0 in absolute value, then the series converges. Theorem 1.3.1 (Leibniz Theorem) If the alternating series n n nu ∑∞=-1)1(satisfy:(1)1+≥n n u u(n=1,2………); (2)lim =∞→n n u ,then the series converges. Moreover, it is sum 1u s ≤, and the error nr makeby usingn s of the first n terms to approximate the sum s of the series isnot more than 1+n u , that is, 1+≤n n u r namely 1+≤-=n n n u s s r .Before giving the proof let us look at figure 1.3.1 which gives a picture of the idea behind the proof. We first plot 11u s =on a number line.To find2swe subtract 2u , so 2s is the left of 1s . Then to find3s weadd 3u , so 3s is to the right of 2s . But, since3u <2u ,3s is to the left of 1s .Continuing in this manner, we see that the partial sums oscillate back and forth. Since 0→nu , the successive steps are becoming smaller and smaller.The even partial sums ,........,,642s s s areincreasing and the odd partialsums,........,,531s s s aredecreasing. Thus it seems plausible that both areconverging to some number s, which is the sum of the series. Therefore, in the following proof we consider the even and odd partial sums separatelyWe give the following proof of the alternating series test. We first consider the even partial sums:,0212≥-=u u s Since 12u u ≤,)(24324s u u s s ≥-+= since uu ≤4 In general, 22212222)(---≥-+=n n n n n s u u s s since122-≤n n u uThus.........................02642≤≤≤≤≤≤n s s s sBut we can also writenn n n u u u u u u u u s 21222543212)(....)()(--------=--Every term in brackets is positive, so 12u s n ≤for all n. therefore, thesequence }{2n s of even partial sums is increasing and bounded above. It is therefore convergent by the monotonic sequence theor em. Let’s call it is limit s, that is, s s n n =∞→2limNow we compute the limit of the odd partialsums:sconditionby s u s u s s n n n n n n n n n =+=+=+=+∞→∞→+∞→+∞→))2((0lim lim )(lim lim 12212212Since both the even and odd partial sums converge to s, wehave s s n n =∞→lim , and so the series is convergent.EX.1.3.1 shows that the following alternating harmonic series is convergent:.)1(..........413121111∑∞=--=+-+-n n nSolution the alternating harmonic series satisfies (1)nu n u n n 1111=<+=+; (2)1limlim ==∞→∞→nu n n nSo the series is convergent by alternating series Test. Ex. 1.3.2 Test the series ∑∞=--1143)1(n nn n for convergence and divergence.Solution the given series is alternating but043143lim143limlim ≠=-=-=∞→∞→∞→nn n u n n n nSo condition (2) is not satisfied. Instead, we look at the limit of the nthterm of the series:143)1(limlim --=∞→∞→n n a n n n This limit does not exist, so the series diverges bythe test for divergence. EX.1.3.3 Test the series ∑∞=+-121)1(n nn for convergence or divergence.Solution the given series is alternating so we try to verify conditions (1) and (2) of the alternating series test.Unlike the situation in example 1.3.1, it is not obvious the sequence given by12+=n n u n is decreasing. If we consider the related function1)(2+=x x x f ，we easily find that10)1(1)1(21)(22222222'><+-=+-+=xwhenver x xx x x x f .Thus f is decreasing on [1,∞) and so )1()(+>n f n f .Therefore, }{n u isdecreasingWe may also show directly that nn u u <+1, that is11)1(122+<+++n n n nThis inequality it equivalent to the one we get by cross multiplication:nn n n n n n n n n n n n n n n +<⇔++<+++⇔++<++⇔+<+++2232322221221]1)1[()1)(1(11)1(1Since 1≥n , we know that the inequality12>+n n istrue. Therefore,n n u u <+1and }{n u is decreasing.Condition (2) is readily verified:。
微积分英文版课件
极限和连续性的关系：极限是连续 的必要条件，但不是充分条件
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添加标题
添加标题
添加标题
连续性：函数在某点或某区间上的 连续性
极限和连续性的应用：在微积分中， 极限和连续性是解决许多问题的基 础
导数：函数在 某一点的斜率， 表示函数在该
点的变化率
微分：函数在 某一点的增量， 表示函数在该
点的变化量
定义：含有两个未知函数 及其导数的方程
形式：ax^2+bx+c=0
解：通过求解特征方程得 到
应用：广泛应用于物理、 工程等领域
高阶微分方程：含有未知函数及其高阶导数的方程 线性微分方程组：含有未知函数及其导数的线性方程组 求解方法：包括积分法、幂级数法、拉普拉斯变换法等 应用领域：广泛应用于物理、化学、工程等领域
级数的形式
应用：在微积 分、数学分析、 物理等领域有
广泛应用
例子：泰勒级 数在求解微分 方程、积分方 程、傅里叶变 换等方面有重
要应用
感谢您的观看
汇报人：PPT
物理概念：力、速度、加速度、质量、能量等
几何概念：直线、平面、曲线、曲面、体积、面积等
物理和几何的结合：力与运动的关系、力与能量的关系、力与几何形状的关系等
微积分在物理和几何中的应用：微积分在力学、光学、电磁学等领域的应用，以及在几何学、 拓扑学等领域的应用。
微积分基本概念
极限：函数在某点或某区间上的极 限值
微积分在物理中 的应用：微积分 在物理中的应用 广泛，如力学、 电磁学、热力学 等
微积分在工程中 的应用：微积分 在工程中的应用 广泛，如建筑、 机械、电子等
微分方程
定义：含有一个未 知函数和一个未知 函数的导数的方程
微积分英文版课件
Applications of Derivatives
Local Extrema
Discover how derivatives help identify local maximums and minimums of functions.
Mean Value Theorem
Explore the mean value theorem and its applications in calculus.
Gradients and Directional Derivatives
2
derivatives and their applications in multivariable calculus.
Learn about gradients and
directional derivatives for
Derivatives
1
Definition of a Derivative
Uncover the definition and
Differentiability and Continuity
2
fundamental properties of derivatives.
Understand the relationship
Discover the conditions for a function to be continuous and its implications.
Explore the different types of discontinuities and their characteristics.
Conclusion
Review of Key Concepts
微积分英文课件PPT (7)
Definition:
A critical number of a function f is a number c in the domain of f such that either f (c) 0 or f (c) does not exist.
Example Find the absolute maximum and minimum values of the function
f (x) x3 3x2 1
1 x4
2
Solution: Since f is continuous on the given closed
interval, we can use the Closed Interval Method:
f (x) 0
For example:
f (x) x3 at x 0
2)There may be an extreme value even when f (c) does not exist.
For example: f (x) x at x 0
Fermat’s Theorem does suggest that we should at least start looking for extreme values of f at the
1) f is continuous on the closed interval [a,b]. 2) f is differentiable on the open interval (a,b). 3) f (a) = f (b) Then there is a number c in (a,b) such that
数学分析 高等数学 微积分 英语课件 上海交通大学Chapter7b
function into partial fractions is the key step to integrate
the rational function.
When Q(x) contains factor (x a)k , the partial fractions
contain A1 A2 Ak .
Example: reciprocal substitution
dx
Ex. Evaluate
.
x 3x2 2x 1
Sol. Let x 1, then t
dx
1 t2
dt.
x
dx 3x2
2x
1
dt 3 2t t2
dt 22 (t 1)2
1 2
dt arcsin t 1 C arcsin x 1 C.
f (x) P(x) S(x) R(x) ,
Q(x)
Q(x)
where S and R are also polynomials and degree of R less than degree of Q.
Technique for partial fraction
For example, by long division, we can derive
changes from a to 2a, t changes from 0 to / 3.
2a a
x2 a2 x4
dx
3 0
a a4
tan t sec4
t
a
sec t
tan tdt
1
3 sin2 t cos tdt
1
sin3 t 3
3.
a2 0
微积分英文版9
y
1 y x
其含义可理解为
A lim
0
1
dx 1 lim 2 x x 0
A
0
lim 2(1 ) 2
0
x
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定义2. 设 f ( x) C (a , b] , 而在点 a 的右邻域内无界,
若极限 存在 , 则称此极限为函
b
1
b
1 dx lim 2 b x 1 x
1 y 2 x A
1
b
1 lim 1 1 b b
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定义1. 设 f ( x) C [a , ) , 取 b a , 若
存在 , 则称此极限为 f (x) 的无穷限反常积分, 记作
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例. 计算反常积分
解:
[ arctan x ]
y
y
( ) 2 2
1 1 x 2
o
x
思考: 分析: 原积分发散 !
注意: 对反常积分, 只有在收敛的条件下才能使用 “偶倍奇零” 的性质 否则会出现错误 , .
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例2. 证明第一类 p 积分 时发散 .
即
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柯西定理的几何意义:
弦的斜率
切线斜率
x F (t ) y f (t )
注意:
y
f (b) f (a)
d y f (t ) d x F (t )
《微积分》课件
微积分的历史背景
01
微积分的发展可以追溯到古代数 学，如希腊数学家阿基米德在求 面积和体积时已经有了积分学的 萌芽。
02
微积分的真正奠基人是牛顿和莱 布尼茨，他们分别独立地发展出 了微积分的基本理论，为后来的 数学发展奠定了基础。
《微积分》PPT课件
contents
目录
• 微积分的定义与历史 • 微积分的基本概念 • 微积分的应用 • 微积分的解题技巧 • 微积分的重点与难点解析 • 微积分的习题与答案解析
01
微积分的定义与历史
微积分的定义
微积分是研究函数、极限和连续性的 数学分支，通过微分和积分的方法来 研究函数的性质和变化规律。
极限的运算性质与法则
1 2
极限的运算性质
极限的四则运算法则、复合函数的极限运算法则 等。
极限的法则
极限的保号性、极限的局部有界性等。
3
注意事项
理解极限的运算法则和性质是解决极限问题的关 键，需要注意运算过程中的等价变换和放缩技巧 。
导数的几何意义与运算性质
导数的几何意义
切线的斜率、函数图像的变化率等。
习题一：极限的运算
$lim_{x to infty} frac{1}{x}$
判断下列叙述是否正 确，并说明理由
$lim_{x to 0} (1 + x)^{1/x}$
习题一：极限的运算
$lim_{x to 0} frac{sin x}{x} = 1$
$lim_{x to infty} frac{1}{x} = 0$
$lim_{x to 0} (1 + x)^{1/x} = e$
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s f (t)
f (t) is called the position function of the object
o f (a)
f (a h) s
Average velocity displacement time
f (a h) f (a) h
The instantaneous velocity at t=a
m pQ
f (x) f (a) . xa
Q(x,f(x) f(x)
P(a,f(a)) 0
Let Q get closer to P and Q≠P.
The slope m of the tangent line l is the limit of the slopes of the secant lines, i.e
lim f (x) L
xa
is f (x) L as x a which is usually read “f(x) approaches L as x approaches a”
Example Guess the value of lim sin x x0 x
X
Sinx/x
±1.0 0.84147098 ±0.5 0.95885108 ±0.4 0.97354586 ±0.3 0.98506736 ±0.2 0.99334665 ±0.1 0.99833417 ±0.05 0.99998339 ±0.01 0.99998333 ±0.005 0.99999583 ±0.001 0.99999983
g(x) M
2
whenever
0 x a 2
Let min 1,2. Notice that
if 0 x a then
( f (x) g(x)) (L M ) f (x) L g(x) M
22
Therefore , by the definition of a limit,
(x2 9) x 3 x 3
If x 3 1, then 2 x 4, so x 3 7,
x 3 x 3 7 x 3 whenever 0 x 3
Therefor,there are two restrictions on x 3
namely x 3 1, and
Let
min(1, ).
7
x3 ,
7
2.Showing that this works.
given 0,
Let
min(1, )
7
If
0 x3 ,
then x2 9
Therefore , by the definition of a limit, lim x2 9 x3
Solution
lim
x2 4
lim（x2 4）
x1
x1 x3 5x 2 lim（x3 5x 2）
x1
5 2
Example: Eliminating zero denominators algebraically
x2 9 lim
0
x3 x 3
0
Solution
lim x2 9 lim (x 3)(x 3)
Chapter1 Limits and Continuity
1.1 Rates of Change and Limits
The Tangent Problem
Let f be a function and let P(a, f(a)) be a point on the graph of f. To find the slope m of the tangent line l at P(a, f(a)) on the graph of f, we first choose another nearby point Q(x, f(x)) on the graph (see Figure 1) and then compute the slope mPQ of the secant line PQ.
Example lim (x3 3cosx sin x 4) x0
Solution
lim (x3 3cosx sin x 4)
x0
lim x3 3lim cosx lim sin x lim 4
x0
x0
x0
x0
0 31 0 4 1
Example
lim x 2 4 x1 x3 5x 2
7.lim c c xa
8.lim x a xa
9.lim xn a n , where n is a positive integer xa
10.lim n x n a , where n is a positive integer xa
11..lim n f (x) n lim f (x) where n is a
then (4x 5) 7
Therefore
,
by
the
definition
of
a
limit,
lim (4x
x3
5)
7
Example Prove that lim x2 9 x3 Proof 1.Guessing a value for . Let be a given positive number.
l Q(x,f(x) f(x)
P(a,f(a)) 0
m lim f (x) f (a) . xa x a
The velocity problem
Suppose an object moves along a straight line according to an equation of motion
1.2 Finding Limits and One-Sided Limits
Calculating Limits Using the Limit Laws
Limit Laws Suppose that c is a constant and limits
lim f (x) and lim g(x)
lim
2
x1 x 1
３
２ yA
１
x
The limit value does not depend on
how the function is defined at X0 !!!
lim (x 1) 2
x1
２A
１
x
Limits may fail to exist
0, x 0
1) U (x) 1,
lithe phrase “but x≠a” in the definition of limit.
This means that in finding the limit of f(x) as x approaches a, we never consider x=a.
xa
xa
exist. then
1.lim[ f (x) g (x)] lim f (x) lim g (x)
xa
xa
xa
2.lim[ f (x) g (x)] lim f (x) lim g(x)
xa
xa
xa
3.lim[ f (x)g (x)] lim f (x) lim g (x)
The Unit step function
x0
2)
g
(
x)
1/ 0,
x,
x0 x0
3) lim sin .
x0
x
The Precise Definition of a Limit
Definition Let f be a function defined on some open interval that contains that number a ,except possibly at a itself. Then we say that the limit of f(x) as x approaches a is L, and we write
Let be a given positive number.
(4x 5) 7 4 x 3 whenever 0 x 3
This
suggests
that
we
should
choose
4
2.Showing that this works.
given 0,
Let
.
4
If
0 x3 ,
equals L” meaning: If we can make the
values of f(x) arbitrarily close to L(as close to L as we like) by taking x to be sufficiently close to a( on either side of a) but not equal to a. An alternative notation for
Since 0 and lim f (x) L ,
2
xa
there exists a number1 0 such that f (x) L whenever 0 x a 1
2
Similarly, since lim g(x) M xa
there exists a number2 0 such that
xa
such that
if 0 x a then f (x) L
Geometric interpretation of limit
f (x)
L
L
L a a a
Example Prove that lim(4x 5) 7 x3
Proof 1.Guessing a value for .