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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。

微分方程差分方程(英文版)(托马斯微积分)

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嘉兴学院
6 March 2020
第十章常微分方程与差分方程
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(3) y f ( y, y) 型
特点不显含自变量 x. 解法令 y P( x), y P dp ,
dy 代入原方程, 得 P dp f ( y, P).
dy
４.线性微分方程解的结构
（1）二阶齐次方程解的结构:
形如 y P( x) y Q( x) y 0
(1)
嘉兴学院
6 March 2020
第十章常微分方程与差分方程
第12页
定理 1 如果函数 y1( x)与 y2 ( x)是方程(1)的两个
解,那末 y C1 y1 C2 y2 也是(1)的解.（C1, C2 是常数）
定理 2：如果 y1( x)与 y2 ( x)是方程(1)的两个线性
嘉兴学院
6 March 2020
第十章常微分方程与差分方程
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定义2
含有未知函数两个或两个以上时期的符号 yx , yx1 , 的方程，称为差分方程.
形式：F ( x, yx , yx1, , yxn ) 0 或G( x, yx , yx1, , yxn ) 0 (n 1)
定理 3 设 yx* 是 n 阶常系数非齐次线性差分方程
yxn a1 yxn1 an1 yx1 an yx f x 2
的一个特解, Yx 是与(2)对应的齐次方程(1)的通
解,
那么
yx
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y
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是
n
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方程(2)的通解.
嘉兴学院

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

初等函数在定义区间上有原函数
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定理 . 原函数都在函数族
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( C 为任意常数 ) 内 .
即
又知
[(x) F(x)] (x) F(x) f (x) f (x) 0
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(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
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x
1
C
( 1)
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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
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d
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2
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例. 求解:
dx a 1 (ax)2
d
(
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)
1
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d u arcsinu C 1u2
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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。

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 satisfies 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 by a 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 function 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 is called 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 functionIf 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。

微积分知识点英语版

简介微积分数学就像一种奇妙的幻想，但这种奇妙幻想最终还是会真实的体现在现实中。

做数学运算有一种在做一个想象的发明的感觉，但它确定是强化我们的洞察力的过程，所以我们在周围任何地方都可以发现那样的情景模式，我们数学教育的目标是为了飞跃到现实的脚步之前并分享数学运算带来的理智愉悦的体验。

微积分的发展史是数学的重要做成部分。

极限、函数、导数、积分和无穷级数等内容在微积分中所体现。

微积分这门学科依然在现代数学教育史中占据一席之地。

微分学和积分学统称为微积分学。

这两大部分间起桥梁作用的是著名的微积分学基本定理。

微积分研究的主要内容是丰富、变化的。

微积分课程是先进思想的传播课程，后来人们也将微积分称为数学分析。

微积分作为工具广泛用于在科学，经济学，工程等领域，用于解决许多问题，这是代数这门学科独自解决是不能满足的。

一、微积分希腊数学家阿基米德是第一个找到切线方向的曲线，除了一个圆圈，在一个方法类似于微积分。

当研究螺旋时，他一个点的运动分开成两个部分，一个径向运动部件和一个圆周运动部件，然后继续增加双组分在一起从而找到切线运动的曲线。

印度数学家及天文学家阿雅巴塔在499年为解决无穷小天文问题，采用了新的观念和表达方式，创造性的利用了一种基本微分方程形式。

Manjula，在世纪十周年开个玩会，详细阐述了该微分方程一个评论。

该方程skara最终导致Bha 二世时12世纪发展一种衍生为代表无穷小观念的改变，而他描述早期版本的“罗尔定理”。

在15世纪，一个早期版本中值定理parameshvara是在天文学的喀拉拉学校里他的评定中和数学登顶，巴卡拉II被首次描述（1370-1460）。

在17世纪，欧洲数学家撒向后拉线，皮埃尔、德、费，布莱斯帕斯科，约翰沃利斯和其他学者讨论了概念的衍生体系，特别是，在Methodus广告disquirendam maximan最小风险，在tangentibus等linearum curvarum，开发出一种方法测定费最大值、最小值，并对各种曲线的切线相当于分化。

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
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Direct Substitution Property
If f is a polynomial or a rational function
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7. lim c c xa
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In fact, f(x) need not even be defined when x=a. the only thing that matters is how f is defined near a.

《微积分英文》课件 (2)

Methods for finding limits using algebra
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

Properties: Continuity, differentiation, integrity, etc
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

《高等数学课件-全英文版(英语思维篇)》

Fundamental Theorem of Calculus
Discover the Fundamental Theorem of Calculus and its significance in integration.
Riemann Sums
Explore Riemann sums as a method for approximating definite integrals.
Functions and Graphs
Types of Functions
Discover the different types of functions and their graphical representations.
Graph Plotting
Learn how to plot and analyze functions using mathematical tools and software.
Differentiation
1
Derivative Definition
Learn the definition and basic rules
Chain Rule
2
of differentiation.
Discover how to differentiate
composite functions using the
Work and Energy
Explore how integration is used to calculate work and energy in various scenarios.
Differential Equations
1
Introduction to Differential

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
添加标题
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(下)英文教材

Chapter 1 Infinite SeriesGenerally, for the given sequence ,.......,......,3,21n a a a a the expression formed 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 na, thatis∑∞=1n na=.......,.....321+++++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 ++++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 k a (there is an infinite number of the term), but we can form the partial sums∑===0000k k a a S∑==+=1101k k a a a S∑==++=22102k k a a a a S∑==+++=332103k k a a a a a S……………….∑==+++++=nk k n 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 ∑∞==0k k a Land say that the series ∑∞=0k kaconverges 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 kadiverges.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<x then the ∑∞=0k k a converges, and;110xx k k -=∑∞= (2)If ,1≥x then the∑∞=0k kxdiverges.Proof: the nth partial sum of the geometric series∑∞=0k katakes theform 1321.......1-+++++=n n x x x x S ① Multiplication by x gives).......1(1321-+++++=n n x x x x x xS =n n x x x x x +++++-1321.......Subtracting the second equation from the first, we find that n n x S x -=-1)1(. For ,1≠x this givesxx S nn --=11 ③If ,1<x then 0→n x ,and this by equation ③.x x x S n n n n -=--=→→1111lim lim 00 This proves (1).Now let us prove (2). For x=1, we use equation ① and device that,n S n =Obviously,∞=∞→nn Slim ,∑∞=0k kadiverges.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 n S 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 Kxdiverges. (x=-1).For 1≠x with ,1>x we use equation ③. Since in this instance, wehave -∞=--=∞→∞→xx S nn n n 11lim lim . 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........................41313121211)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 n Since all but the first and last occur in pairs with opposite signs, the sum collapses to give 211+-=n S n Obviously, as .1,→∞→n S n this means that the series converges to 1.1)211(lim lim =+-=∞→∞→n S n n n therefore 1)2)(1(1=++∑∞=n k k1.1.3 p the following theorem:Theorem 1.1.1 the kth term of a convergent series tends to 0; namely if∑∞=0k kaConverges, by definition we have the limit of the sequence }{n S ofpartial sums exists. Namelyl a S nk k n n n ==∑=∞→∞→0lim limObviously.lim lim 01l a S nk k n 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 (A diverges test) if 0lim ≠∞→k k a , or if n k a ∞→lim doesnot exist, then the series ∑∞=0k k a diverges.Caution, theorem 1.1.1 does not say that if 0lim =∞→k k a , and then∑∞=0k kaconverge. 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 nnn nS n =>+++=1 (2)111}{ is unbounded. So∞===∞→∞→n S n n n lim lim , therefore the series diverges.But 01lim lim ===∞→∞→ka k k kEX.1.1.3 determine whether or not the series: (54)433221010+++++=+∑∞=k k k Converges.Solution since 01111lim 1limlim ≠=+=+==∞→∞→∞→kk k a k k k k , this series diverges. EX.1.1.4 Determine whether or not the series ∑∞=021k kSolution1the given series is a geometric series.121,)21(00<==∑∑∞=∞=x and xk k k k, by proposition 1.1.1 we know that seriesconverges.Solution 2 ,21 (4)12111-++++=n n S ①,2121.........21212121132n n n S +++++=-② ①-② (1-21))211(2,211n n n n S S -=-=.2)211(2lim lim =-=∞→∞→n n n n SBy definition of converges of series, this series converges.1.1.5 p the following theorem: Theorem 1.1.2 If the series∑∑∞=∞=0k kk k band 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 kCaalso converges. Moreover ifl ak k=∑∞=0thenCl Cak k=∑∞=0.Proof let ∑∑====nk k nnk k nb S a S20)1(,∑∑===+=nk k nnk k k nCa S b a S40)3(,)(Note that )1()4()2()1()3(n n n n n CS S and S S S =+= Since (),lim ,lim )2(1m S l S n n n n ==∞→∞→Then m l S S S S S n n n n n n n n n +=+=+=∞→∞→∞→∞→)2()1()2()1()3(lim lim )(lim lim.lim lim lim )1()1()4(Cl S C CS S n n n n n n ===∞→∞→∞→Theorem 1.1.4 (squeeze theorem)Suppose that }{}{n n c and a both converge to l and that n n 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 0sin lim3=∞→nnn . Solution For ,1≥n ,1)sin (13n n n n ≤≤- since ,0)1(lim ,0)1(lim ==-∞→∞→n and nn n the 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 1.1.4 Namely the squeeze theorem, we know the result is true.Exercise(1) An expression of the form 123a a a +++…is called (2) A series 123a a a +++…is said to converge if the sequence{}S n converges, where S n =1. The geometric series 2a ar ar +++…converges if ; in this case the sum of the series is2. If lim 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 1ln1k kk ∞=+∑diverges. Find the sums of the series 6-116. 31(1)(2)k k k ∞=++∑ 7.112(1)k k k ∞=+∑ 8.11(3)k k k ∞=+∑ 9.0310k k ∞=∑10.0345k k k k ∞=+∑ 11.3023k k k +∞=∑12. Derivethe following results from the geometricseries 221(1),||11k k k x x x∞=-=<+∑. Test the following series for convergence:13. 11n n n ∞=+∑ 14.3012k k ∞+=∑1.2 Series With Positive Terms1.2.1 The comparison TestThroughout this section, we shall assume that our numbers n a are x 0≥, then the partial sum12n n S a a a =+++… areincreasing, .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 B such that n S B ≤ for all n. Such a number B is called an upper bound. By a least upper bound we mean a number S which is an upper bound, and such that every upper bound B is S ≥. We take for granted that a least upper bound exists. The collection of numbers {}n S has therefore a least upper bound, ., there is a smallest numbers such that n S S ≤ for all n. In that case, the partial sums n S approach S as a limit. In other words, given any positive number 0ε>, we have n S S S ε-≤≤ for all n sufficiently large.This simply expresses the fact S is the least of all upper bounds for our 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 numbers0≥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 give us a very useful criterion to determine when a series with positive terms converges.The convergence or divergence of a series with nonnegative termsS 1 S 2 S n Sis usually deduced by comparison with a series of known behavior.Theorem 1.2.3(The Ordinary Comparison Test) Let1nn a∞=∑and1nn b∞=∑betwo series, with 0n a ≥ for all n and 0n b ≥ for all n. Assume that there is a numbers 0c >, such that n n a cb ≤ for all n, and that 1nn b∞=∑ converges,then1nn a∞=∑converges, and11nn n n ac b ∞∞==≤∑∑.Proof: We have1212121()n n n n n 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 provingour theorem.Theorem 1.2.3 has an analogue to show that a series does not converge.Theorem 1.2.4(Ordinary Comparison Test) Let1nn a∞=∑ and1nn b∞=∑ betwo series, with n a and 0n b ≥ for all n. Assume that there is a number0c > such that n n a cb ≥ for all n sufficiently large, and1nn b∞=∑ does notconverge, then1nn a∞=∑ diverges.Proof. Assume n n a cb ≥for 0n n ≥, since1nn b∞=∑diverges, we canmake the partial sum0001Nnn n N n n bb b b +==+++∑…arbitrarily large as N becomes arbitrarily large. But 0NNNn n n n n n n n n a cb c b ===≥=∑∑∑.Hence the partial sum121NnN n aa a a ==+++∑… are arbitrarily large as Nbecomes arbitrarily large, are hence 1nn a∞=∑ diverges, as was to beshown.Remark on notation you have easily seen that for each 0j ≥, 0kk a∞=∑converges iff1kk j a∞=+∑ converges. This tells us that, in determiningwhether or 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 summationbegins. For instance, it makes sense to you that21k ∑ converges and1k ∑ diverges without specifying where we begin the summation. But inthe convergent case it does, however, affect the sum. Thus for example0122kk ∞==∑, 1112kk ∞==∑, 21122kk ∞==∑, and so forth. Ex 1.2.1 Prove that the series211n 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 occurs twice, 4 occurs four times, 8 occurs eight times, andso forth. Our partial sum 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+?2 Thus 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 1111111234p p p p p n n n ∞==++++++∑……, where p is a constant, is called a p-series.Proposition1.2.1. If 1p >, the p-series converges; and if 1p ≤, then the p-series diverges.Ex 1.2.2 Determine whether the series 2311n n n ∞=+∑ converges.Solution We write 2323111(1)1111n n n n n n ==++++. Then we see that 23111122n n n n≥=+. Since 11n n∞=∑ does not converge, it follows that the series 2311n n n ∞=+∑ 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 n n n n n n n n n n+++==-+-+-+ For n sufficiently large, the factor23471312()n n n+-+ is certainly bounded, and in fact is near 1/2. Hence we can compare our series with 21n ∑ tosee converges, because∑21n converges and the factor is bounded.Ex.1.2.5 Show that1ln()k b +∑ diverges.Solution 1 We know that as k →∞,ln 0kk→. It follows that ln()0k b k b +→+, and thus that ln()ln()0k b k b k bk k b k+++=→+. Thus for k sufficiently large, ln()k b k +< and 11ln()k k b <+. Since 1k ∑ diverges,we can conclude that1ln()k b +∑ diverges.Solution 2 Another way to show that ln()k b k +< for sufficiently large k is to examine the function ()ln()f x x x b =-+. At 3x = thefunction is positive:(3)3ln93 2.1970f =-=->Since '1()10f x x b=->+ for all 0x >, ()0f x > for all 3x >. It follows thatln()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) Letka∑ andkb∑ beseries with positive terms. If lim()k k ka lb →∞=, where l is some positivenumber, thenka∑ andkb∑converge or diverge together.Proof Choose ε between 0 and l , since k ka lb →, we know forall k sufficiently large (for all k greater than some 0k ) ||kka lb ε-<. For such k we have kka l lb εε-<<+, and thus ()()k k k l b a l b εε-<<+ this last inequality is what we needed. (1) Ifka∑converges, then()kl b ε-∑converges, and thuskb∑converges.(2) Ifkb∑converges, then()kl bε+∑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 whichkka b converges to a positive number.Ex 1.2.6 Determine whether the series sinkπ∑converges ordiverges.Solution Recall that as sin 0,1x x x →→. As ,0k kπ→∞→ and thus sin 1k kππ→. Sincek π∑diverges, sosin()k π∑diverges.Ex 1.2.7 Determine whether theseries convergesor diverges.Solution For large value of k, dominates the numeratorand2k the denominator, thus, for such k,differs252k=. Since2512k ÷==→And2255122k k =∑∑converges, this series converges.Theorem 1.2.6 Let ka∑ andkb∑ be series with positive termsand suppose thus0kka b →, then (1) If kb∑converges, then ka∑converges.(2) If ka ∑diverges, thenkb ∑diverges.(3) If ka ∑converges, then kb∑may converge or diverge.(4) Ifkb∑diverges, thenka∑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 test Theorem 1.2.7 (the root test, Cauchy test) let ∑kabe a series withnonnegative terms and suppose thatρ==∞→∞→k kk k k k a a 1lim lim ,if ρ<1,∑kaconverges, if ρ>1,∑kadiverges, if ρ=1, the test is inconclusive.Proof we suppose first ρ<1 and choose μ so that 1<<u ρ. Sinceρ→kk a 1)(, we have μ<k ka 1, for all k sufficiently large thus k k a μ< for allk sufficiently large since∑kμconverges (a geometric series with0<1<μ), we know by theorem 1.2.5 that∑kaconverges.We suppose now that 1>ρand choose μso that 1>>u ρ. sinceρ→kk a 1)(, we have μ>kk a 1)( for all k sufficiently large. Thus k k a μ>for all k sufficiently large.Since∑kμdiverges (a geometric series with 1>μ ) the theorem1.2.6 tell us that∑kadiverges.To see the inconclusiveness of the root test when 1=ρ, note that1)(1→kk a for both:112∑∑k and k ,11)1()1()(221121=→==kk kk k ka 11)1()(11→==k k kk 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 have 0ln 1lim)(lim 1==∞→∞→ka k kk k , the series converges. EX.1.2.8 Determine whether series ∑3)(2k kconverges or diverges.Solution For the series ∑k k)3(2, applying the root test, we have1212]1[2)1(.2)(3331>=⨯→==k k kk k k a . 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 k k . Ifapplying the root test, it is inconclusive. But since k k ka )11(-=converges toe1and not to 0, the series diverges. We continue to consider only series with terms 0≥. To compare such 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 ∑kabe a serieswith positive terms and suppose thatλ=+∞→kk k a a 1lim, If ,1<λ∑kaconverges, if ,1>λ∑kadiverges.If the ,1=λthe test is inconclusive.Proof we suppose first that ,1<λ since 1lim1<=+∞→λkk k a a So there exists some integer N such that if n ≥NC a a nn ≤+1Then N N N N N a C Ca a Ca a 212,1≤≤≤+++ and in general byinduction ,N k k N a C a ≤+Thusca c c c c a a c a c ca a aNk N N k N N N k N Nn n-≤++++≤++++≤∑+=11)........1( (322)Thus 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 termsn a such that .1lim1<=+∞→λnn n a a EX.1.2.10 show that the series∑∞=13n n nconverges. Solution we let ,3n n na = then 31.13.3111n n n n a a n n n n +=+=++, this ratioapproaches ∞→n as 31, and hence the ratio test is applicable: the series converges.EX1.2.11 show that the series ∑!k k kdiverges.Solution we have kk k k n n kk k k k k k a a )11()1(!)!1()1(11+=+=++=++So e ka a k k n n n =+=∞→+∞→)11(lim lim1Since 1>e , the series diverges. 1.2.12 p the series∑+121k diverges.Solution since kkk k k k a a k k 32123212112.1)1(211++=++=+++=+ 13212limlim 1=++=∞→+∞→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. Exercise1. The ordinary comparison test says that if ____ and if ∑ib converges.Then∑kaalso converges.2. Assume that 00>≥k k b and a . The limit comparison Test says that if 0<____<+∞ then ∑kaand∑kbconverges or diverge together.3. Let nn n a a 1lim+∞→=ρ. The ratio Test says that a series ∑kaof positive termsconverges if ___, diverges if ____and may do either if ___. Determine whether the series converges or diverges 4.∑+13k k5.∑+2)12(1k 6.∑+11k 7.∑-kk 2218. ∑+-1tan 21k k 9.∑321k10. ∑-k )43( 11.∑k kln 12.∑!10k k13. ∑k k 114.∑k k 100!15.∑++k k k 6232 16.kk ∑)32( 17.∑+k 11.18.∑k k 410!19. Let }{n a be a sequence of positive number and assume thatna a n n 111-≥+ for all n. show that the series ∑nadiverges.Alternating series, Absolute convergence and conditional convergence In this section we consider series that have both positive and negative terms.1.3.1 Alternating series and the tests for convergenceThe series of the form .......4321+-+-u u u u is called the alternating series, where 0>n u for all n, here two example:∑∞=--=+-+-+-11)1(....61514131211n n n ,11)1( (65544332211)+-=+-+-+-∑∞=n n nWe see from these examples that the nth term of an alternating series is the form n n n n n n u a or u a )1()1(1-=-=-, where n u is a positive number (in fact n n a 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 seriesnn nu∑∞=-1)1(satisfy:(1) 1+≥n n u u (n=1,2………); (2) 0lim =∞→n n u ,then the series converges. Moreover, it is sum 1u s ≤, and the errorn r make by using n s of the first n terms to approximate the sum s of theseries is not 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 find 2s we subtract 2u , so 2s is the left of 1s . Then to find 3s we add 3u , so 3s is to the right of 2s . But, since 3u <2u , 3s is to the left of 1s . Continuing in this manner, we see that the partial sums oscillate back and forth. Since 0→n u , the successive steps are becoming smaller and smaller. The even partial sums ,........,,642s s s are increasing and the odd partial sums ,........,,531s s s are decreasing. Thus it seems plausible that both are converging 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 sSince 12u u ≤,)(24324s u u s s ≥-+= since u u ≤4In general, 22212222)(---≥-+=n n n n n s u u s s since 122-≤n n u u Thus .........................02642≤≤≤≤≤≤n s s s s But we can also writen n 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, the sequence }{2n s of even partial sums is increasing and bounded above. It is therefore convergent by the monotonic sequence theorem. Let’s call it is limit s, that is, s s n n =∞→2lim Now we compute the limit of the oddpartial sums:scondition by 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, we have 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( (41312111)1∑∞=--=+-+-n n n Solution the alternating harmonic series satisfies (1) nu n u n n 1111=<+=+; (2) 01lim lim ==∞→∞→n u n n n So the series is convergent by alternating series Test.Ex. 1.3.2 Test the series ∑∞=--1143)1(n n n nfor convergence and divergence.Solution the given series is alternating but043143lim 143lim lim≠=-=-=∞→∞→∞→nn n u n n n n So condition (2) is not satisfied. Instead, we look at the limit of the nth term of the series:143)1(lim lim --=∞→∞→n na n n n This limit does not exist, so the series diverges by the 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 by 12+=n nu n is decreasing. If we consider the related function 1)(2+=x xx f ，we easily find that 10)1(1)1(21)(22222222'><+-=+-+=x whenver x x x x x x f . Thus f is decreasing on [1,∞) and so )1()(+>n f n f . Therefore, }{n u is decreasingWe may also show directly that n n u u <+1, that is11)1(122+<+++n n n n This inequality it equivalent to the one we get by cross multiplication:nn n n n n n n n n n n n nn n +<⇔++<+++⇔++<++⇔+<+++2232322221221]1)1[()1)(1(11)1(1Since 1≥n , we know that the inequality 12>+n n is true. Therefore,n n u u <+1and }{n u is decreasing.Condition (2) is readily verified:011lim 1lim lim 2=+=+=∞→∞→∞→nn n n nu n n n n , thus the given series is convergent bythe Alternating series Test.1.3.2 Absolute and conditional convergenceIn this section we consider series that have both positive and negative terms. Absolute and conditional convergence. Definition 1.3.1 suppose that the series ∑∞=1k kais not series with positiveterms, if the series∑∞=1k kaformed with the absolute value of the termsn a converges, the series∑∞=1k kais called absolutely convergent. The series∑∞=1k kais called conditionally convergent, if the series∑∞=1k kaconvergesbut∑∞=1k kadiverges.Theorem 1.3.2 if∑kaconverges, the ∑k a converges.Proof for each k, k k k a a a ≤≤-, and therefore k k k a a a 20≤+≤.if∑kaconverges, then∑∑=k ka a22converges, and therefore, bytheorem 1.2.3(the ordinary comparison theorem),∑+)(k ka aconverges. Since k k k k a a a a -+=)(by the theorem (1), wecan conclude that∑kais convergence.The above theorem we just proved says that Absolutely convergent series are convergent.As well show presently, the converse is false. There are convergent series that are not absolutely convergent; such series are called conditionally convergent.1.3.4 P the following series is absolutely convergent (5)141312112222++-+-Proof If we replace term by it’s absolute value, we obtain the series (41312112)22++++This is a P series with P=2. It is therefore convergent. This means that the initial series is absolutely convergent.1.3.5 p that the following series is absolutely convergent: (2)12121212121212118765432+--+--+--Proof if we replace each term by its absolute value, we obtain the series: (2)12121212121212118765432+++++++=+This is a convergent geometric series. The initial series is therefore absolutely convergent.1.3.6 p that the following series is only conditionally convergent:∑∞=-=++-+-+-1)1(.............61514131211n nnProof the given series is convergent. Since (1) ,1111nu n u n n =<+=+(2) 01lim lim ==∞→∞→n u n n n , So this series is convergent by the alternating series test, but it is not absolutely.Convergent: if we replace each term by it is absolute value, we obtain the divergent harmonic series:......................4131211++++ So this series is only conditionally convergent.。

数学分析高等数学微积分英语课件上海交通大学chapter11b

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《微积分》PPT课件

x x0
f (x)
f
(x0 )
何时函数f(x)在点处间断？
(1)f(x)在点 x0 处无定义；
(2)f(x)在点
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时，函数f（x)以常数A为极限，记作
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在一个正数M，使得当x＞M（x＜-M）时，
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y=arcsinx x [1,1], y [ , ]
22
y=arccos x [-1,1], y [0, ]
y=arctanx X R, y ( , ) 22
y=arccotx X R，y （0，）
1.4 初等函数（三角函数）
正弦函数和余弦函数
正切函数和余切函数
正割函数与余割函数
三角函数的基本关系式：
xx0
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2.4
被迫性定理若在某个变化过程中，
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两个重要极限（必考）
单调有界定理
单调有界的数列
必有极限
} 单调增 + 有上界
单调减+有下界
数列收敛
定理 2.12
定义 2.9
定理 2.13
若数列 {an}满足 an an1(或an an1）（n N）则称数列 {an}为单调增加(或单调减少)数列。
当x 0时，等价无穷小量：
sinx~x tanx~x
arcsinx~x 1-cosx~x2

微积分英文教材

Chapter 1 Infin ite SeriesGen erally, for the give n seque nee aLa2,a3……a n, .......................... ,the expressi on formed by the seque nee a「a2,a3……a n, ............. ,a i a 2 a 3 a n ,is called the infinite series of the eonstants term, denotedby a n, that isn 1a n =a i a2 a3 ••… a n ........................ ,n 1Where the nth term is said to be the general term of the series, moreover, the nth partial sum of the series is give n byS n a i a2 a3 ……a n •1.1 Determi ne whether the infin ite series conv erges 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 sequenee of real numbers: a0,a1,a2,a3••…,we can not form the sum of all the a k(there is an infinite number of the term), but we can form the partial sumsS o a0 a k(1) If x 1 then thea k converges, andX kS na 0a ia 2a 3na n a kk 0and say that the series S n } of partial sums don ' tDefi niti on 1.1.1If the sequenee { s n } of partial sums has a finite limit L, We writea kk 0a k converges to L. we call L thek 0sum of the series.If the limit of the seque nee { exists, we say that the seriesa k diverges.k 0Remark it is important to note that the sum of a series is not a sum in the orderi ng sen se. It is a limit.EX 1.1.1 prove the followi ng propositi on: Propositio n1.1.1: S 2a 。