Barron's AP Calculus BOOK_05
APCalculusABreviewAP微积分复习提纲PDF.pdf
APCalculusABreviewAP微积分复习提纲PDF.pdfAP CALCULUS AB REVIEWChapter 2DifferentiationDefinition of Tangent Line with Slop mIf f is defined on an open interval containing c, and if the limitexists, then the line passing through (c, f(c)) with slope m is the tangentline to the graph of f at the point (c, f(c)).Definition of the Derivative of a FunctionThe Derivative of f at x is given byprovided the limit exists. For all x for which this limit exists, f’is afunction of x.*The Power Rule*The Product Rule***The Chain RuleImplicit Differentiation (take the derivative on both sides; derivativeof y is y*y’)Chapter 3Applications of Differentiation*Extrema and the first derivative test (minimum: ? → + , maximum: +→ ?, + & ? are the sign of f’(x) )*Definition of a Critical NumberLet f be defined at c. If f’(c) = 0 OR IF F IS NOT DIFFERENTIABLEAT C, then c is a critical number of f.*Rolle’s TheoremIf f is differentiable on the open interval (a, b) and f (a) = f (b), then thereis at least one number c in (a, b) such that f’(c) = 0.*The Mean Value TheoremIf f is continuous on the closed interval [a, b] and differentiable on theopen interval (a, b), then there exists a number c in (a, b) such that f’(c) = .*Increasing and decreasing interval of functions (take the first derivative)*Concavity (on the interval which f’’ > 0, concave up)*Second Derivative TestLet f be a function such that f’(c) = 0 and the second derivative of f existson an open interval containing c.1.If f’’(c) > 0, then f(c) is a minimum2.If f’’(c) < 0, then f(c) is a maximum*Points of Inflection (take second derivative and set it equal to 0, solve theequation to get x and plug x value in original function)*Asymptotes (horizontal and vertical)*Limits at Infinity*Curve Sketching (take first and second derivative, make sure all thecharacteristics of a function are clear)Optimization Problems*Newton’s Method (used to approximate the zeros of a function, which istedious and stupid, DO NOT HA VE TO KNOW IF U DO NOT WANTTO SCORE 5)Chapter 4 & 5Integration*Be able to solve a differential equation*Basic Integration Rules1)2)3)4)*Integral of a function is the area under the curve*Riemann Sum (divide interval into a lot of sub-intervals, calculate the area for each sub-interval and summation is the integral).*Definite integral*The Fundamental Theorem of CalculusIf a function f is continuous on the closed interval [a, b] and F is an anti-derivative of f on the interval [a, b], then.*Definition of the Average Value of a Function on an Interval If f is integrable on the closed interval [a, b], then the average value off on the interval is.*The second fundamental theorem of calculusIf f is continuous on an open internal I containing a, then, forevery x in the interval,.*Integration by Substitution*Integration of Even and Odd Functions1) If f is an even function, then.2) If f is an odd function, then.*The Trapezoidal RuleLet f be continuous on [a, b]. The trapezoidal Rule for approximating is given byMoreover, a n →∞, the right-hand side approaches.*Simpson’s Rule (n is even)Let f be continuous on [a, b]. Simpson’s Rule for approximating isMoreover, as n→∞, the right-hand side approaches*Inverse functions(y=f(x), switch y and x, solve for x)*The Derivative of an Inverse FunctionLet f be a function that is differentiable on an interval I. If f has an inverse function g, then g is differentiable at any x for which f’(g(x))≠0. Moreover,, f’(g(x))≠0.*The Derivative of the Natural Exponential FunctionLet u be a differentiable function of x.1. 2..*Integration Rules for Exponential FunctionsLet u be a differentiable function of x..Derivatives for Bases other than eLet a be a positive real number (a ≠1) and let u be adifferentiable function of x.1. 2.*Derivatives of Inverse Trigonometric Functions Let u be a differentiable function of x.*Definition of the Hyperbolic Functions。
AP微积分-AP Calculus 公式大全-217
AP Calculus BC1. Important limits()001111011sin sin lim1, lim 1lim 1lim 10, ()lim lim , x x xt x t m m m m m m n n x x x ax ax bx be t e x m n P x a x a x a x a a m n b x b →→→∞→---→∞→∞==⎛⎫+=⇔+= ⎪⎝⎭<++++===+++⎧⎪⎪⎨x x sec )'(tan = x csc (cot)'-= x x x tan sec )'(sec = x x x cot csc )'(csc -=(6) 211)'(arcsin xx -=211)'(arccos xx --=211)'(arctan x x +=211)'cot (x x ar +-= 11)'sec (2-=x x x arc 11)'csc (2--=x x x arc2. Rules（1）If f (x )，g (x ) are differential ，a. )()())()((x g x f x g x f '±'='±；b. )()()()())()((x g x f x g x f x g x f '+'='，especially ，)())((x f C x Cf '='（C is a constant ）；c. )0)(( ,)()()()()())()((2≠'-'='x g x g x g x f x g x f x g x f ，especially ，21()()()()g x g x g x ''=-。
(2) Chain Rule )]([x g f y =⇒dxdudu dy dx dy ⋅= (3) Implicit Differentiation ()(()'r t f =3. Applications of Derivative Mean Value Theorem) is differentiable Tangent and normalFirst Derivative Test'f changes from - to +，local min 'f changes from + to -，local maxSecond Derivative Test ''()0f a >，local min ''()0f a <，local max(4) Absolute max/min: compare function value of critical point and endpoints.Steps: 1. find 'f ;2.solve '0f =;3. compare value at critical point and endpoints.1. Indefinite Integral:'()()Definition f x dx f x C =+⎰ Method ()()df x f x C =+⎰2. Methods for Indefinite Integral 不定积分方法 ① Formulas ：dx x C =+⎰（integrand 为1）, kdx kx C =+⎰, 11n nx x dx C n +=++⎰ln x xa a dx C a=+⎰ x xe dx e C =+⎰1sin 2 4x x C ++②()f u du3. Find the antiderivative of ()f u .③ Partial fraction （拆分）ln ||ln ||()()cx d A Bdx dx dx A x a B x b C x a x b x a x b +=+=-+-+----⎰⎰⎰(通分求A,B)④ Integration by partsudv uv vdu =-⎰⎰，（or ''uv dx uv vu dx =-⎰⎰）， Tabular Integration ， （story about ln x , sin x , and xe ）3. Improper Integral 反常积分（两种形式，无穷积分，瑕积分） ① Integral on infinite interval()lim (),()lim (),()()()lim ()lim ().ba ab bbaa c cbcaca b f x dx f x dx f x dx f x dx f x dx f x dx f x dx f x dx f x dx ∞→∞-∞→-∞∞∞-∞-∞→-∞→∞===+=+⎰⎰⎰⎰⎰⎰⎰⎰⎰② Integrand with infinite discontinuities()lim (),bca ac bbbf x dx f x dx -→=⎰⎰4.5. )'()x β⋅6. ①(,)c a b ∈. ②low upper function er function a⎣⎦L Right functio ef func on n ti )cy dy ⎤⎥⎣⎦③ V olume ：with known cross section()baA x dx ⎰, ()A x is the area of the cross section.Revolution ：2Inner 2Out rad er radiu ius s baR r dx π⎡⎤-⎣⎦⎰，Shell Method ：2baxydx π⎰（旋转轴为y-axis ） or 2baxydy π⎰（旋转轴为x-axis ）④ Length21x x ⎰or 21y y ⎰or 21t t ⎰.1. Separation Variable()()()()dy M x N y dy M x dx dx N y =⇒=⎰⎰ 2. Logistic Equation (1)(),lim ()1kt t dP P K kP P t P t K dt K e-→∞=-⇒==+ ,where is the environmental capacity. 3. Slope Field4. Euler’s Method ： 11n n n n x x x dyy y x dx ++=+∆⎧⎪⎨=+⋅∆⎪⎩n a ++，1n a∞=∑Test for ConvergenceAssume that the following limit exists: 1≥..③ Comparison TestAssume that there exists 0M >such that 0n n a b ≤≤for n M ≥. If1nn b∞=∑ converges, then If1nn a∞=∑ also converges. If1nn a∞=∑ diverges, then1nn b∞=∑ alsodiverges.Limit Comparison Test Let {}n a and {}n b be positive sequences. Assume limnn na Lb →∞=a) If 0L >,then1nn a∞=∑ converges if and only if1nn b∞=∑ converges.b) If L =∞, and1nn a∞=∑ converges, then1nn b∞=∑ converges.c) If 0L =, and1nn b∞=∑ converges, then1nn a∞=∑ converges.3. Four Series① The geometric seriesnar∞converges if ||1r <and diverges otherwise.0, lim n n a a →∞>>>>4. ()(012a a x c a x +-+-()n∞∑is either anonnegative number ( )or infinity( ).If i s finite, converges absolutely whenx c R -< and diverges when x c R ->. If R =∞, ()F x converges absolutely for all x .（ii ）Term-by-term Differentiation and IntegrationPossible convergence at the endpointsAssume that ()0()nn n F x a x c ∞==-∑ has radius of convergence 0R > . Then is differentiable on(, )c R c R -+[or for all x if R =∞]. Furthermore, we can integrate and differentiate term by term.For (, )x c R c R ∈-+,()()110'() () (A any constant)1n n n n n n a F x na x c F x dx x c A n ∞∞-+===-=-++∑∑⎰；These series have the same radius of convergence R .()(!n f a n ++()(0)!n f n ++and x ).Interval of convergence all real numbers初等函数基本知识点小结一、指数函数（一）指数与指数幂的运算1．∈N◆2ma n◆3（112（一）对数1．对数的概念：一般地，如果N a x =)1,0(≠>a a ，那么数x 叫做以．a 为底．．N 的对数，记作：N x a log =（a — 底数，N — 真数，N a log — 对数式）说明：注意底数的限制0>a ，且1≠a 两个重要对数：○1 常用对数：以10为底的对数N lg ； ○2 自然对数：以无理数 71828.2=e 为底的对数N ln ． 指数式与对数式的互化幂值 真数如果a ○1 ○2 ○3 abln ln（1）1，+∞）． ○2 21、幂函数定义：一般地，形如αx y =)(R a ∈的函数称为幂函数，其中α为常数．2、幂函数性质归纳．（1）所有的幂函数在（0，+∞）都有定义并且图象都过点（1，1）；（2）0>α时，幂函数的图象通过原点，并且在区间),0[+∞上是增函数．特别地，当1>α时，幂函数的图象下凸；当10<<α时，幂函数的图象上凸；（3）0<α时，幂函数的图象在区间),0(+∞上是减函数．在第一象限内，当x 从右边趋向原点时，图象在y 轴右方无限地逼近y 轴正半轴，当x 趋于∞+时，图象在x 轴上方无限地逼近x 轴正半轴．三角函数基本知识点小结sin1. 2. 3.sin(α±β(tan a a sin2αa 2tan =Trigonometric Function and Inverse Trigonometric Function。
AP-Calculus-AB-review-AP微积分复习提纲
AP-Calculus-AB-review-AP微积分复习提纲AP CALCULUS AB REVIEWChapter 2DifferentiationDefinition of Tangent Line with Slop mIf f is defined on an open interval containing c, and if the limitlim ∆x→0∆y∆x=lim∆x→0f(c+∆x)−f(c)∆x=mexists, then the line passing through (c, f(c)) with slope m is the tangent line to the graph of f at the point (c, f(c)).Definition of the Derivative of a FunctionThe Derivative of f at x is given byf′(x)=lim∆x→0f(c+∆x)−f(c)∆xprovided the limit exists. For all x for which this limit exists, f’ is afunction of x.*The Power Rule*The Product Rule*ddx[sin x]=cos x*ddx[cos x]=−sin x*The Chain Rule☺Implicit Differentiation (take the derivative on both sides;derivative of y is y*y’)Chapter 3Applications of Differentiation*Extrema and the first derivative test (minimum: − → + , maximum:+ → −, + & − are the sign of f’(x) )*Definition of a Critical NumberLet f be defined at c. If f’(c) = 0 OR IF F IS NOTDIFFERENTIABLE AT C, then c is a critical number of f.1)∫u ndu =u n+1n+1+ C,n ≠−12)∫sin u du = −cos u + C 3)∫cos u du = sin u + C 4)∫1u du = ln u*Integral of a function is the area under the curve*Riemann Sum (divide interval into a lot of sub-intervals, calculatethe area for each sub-interval and summation is the integral).*Definite integral*The Fundamental Theorem of CalculusIf a function f is continuous on the closed interval [a, b] and F isan anti-derivative of f on the interval [a, b], then ∫f (x )dx ba=F (b )− F(a).*Definition of the Average Value of a Function on an IntervalIf f is integrable on the closed interval [a, b], then the averagevalue of f on the interval is 1b−a ∫f(x)dx ba.*The second fundamental theorem of calculusIf f is continuous on an open internal I containing a, then, for every x in the interval,ddx[∫f (t )dt xa ]=f(x).*Integration by Substitution*Integration of Even and Odd Functions1) If f is an even function, then ∫f (x )dx b a =2∫f(x)dx ba . 2) If f is an odd function, then ∫f (x )dx ba=0.*The Trapezoidal RuleLet f be continuous on [a, b]. The trapezoidal Rule forapproximating ∫f (x )dx bais given by ∫f (x )dx ba≈ b−a 2n[f (x 0)+2f (x 1)+2f (x 2)+⋯+2f (x n−1) +f (x n )]Moreover, an → ∞, the right-hand sideapproaches ∫f (x )dx ba. *Simpson ’s Rule (n is even)Let f be continuous on [a, b]. Simpson ’s Rule for approximating ∫f (x )dx bais ∫f (x )dx ba≈b −a3n [f (x 0)+4f (x 1)+2f (x 2)+4f (x 3)+⋯4f (x n−1)+f (x n )]Moreover, as n →∞, the right-hand side approaches ∫f (x )dx ba*Inverse functions(y=f (x), switch y and x, solve for x)*The Derivative of an Inverse FunctionLet f be a function that is differentiable on an interval I . If f has aninverse function g , then g is differentiable at any x for which f ’(g (x))≠0. Moreover, g ′(x )= 1f (g(x)), f ’(g (x))≠0.*The Derivative of the Natural Exponential Function Let u be a differentiable function of x .1.d dx[e x ]= e x 2.d dx[e u ]= e udu dx.*Integration Rules for Exponential Functions Let u be a differentiable function of x. ∫e u du = e u +C .♠Derivatives for Bases other than eLet a be a positive real number (a ≠1) and let u be adifferentiable function of x . 1.d dx[a u ]=(ln a)a udu dx2.ddx[log a u ]=1u ln a dudx♠∫a x dx =(1ln a)a x +C♠lim x→∞(1+1x)x =lim x→∞(x+1x)x=e*Derivatives of Inverse Trigonometric Functions Let u be a differentiable function of x. d dx [sin −1u ]=√2d dx[cos −1u ]=√2d dx[tan −1u ]=u′1+u=sin −1u a+C ∫du a 2+u 2=1atan −1ua+C=1a sec −1|u |a+C*Definition of the Hyperbolic Functions sinh x =e x −e −x2 cosh x =e x +e −x2tanh x =sinh x cosh x csch x =1sinh x ,x ≠0 sech x =1cosh x coth x =1tanh x,x ≠0。
AP_calculus参考书
Score
AP Score 5 4 3 2 1 Qualification Extremely well qualified Well qualified Qualified Possibly qualified No recommendation
Topic Outline for Calculus BC
1
by taking sufficiently close values of the domain.) • Understanding continuity in terms of limits. • Geometric understanding of graphs of continuous functions (Intermediate Value Theorem and Extreme Value Theorem). * Parametric, polar, and vector functions. The analysis of planar curves includes those given in parametric form, polar form, and vector form. II. Derivatives Concept of the derivative • Derivative presented graphically, numerically, and analytically. • Derivative interpreted as an instantaneous rate of change. • Derivative defined as the limit of the difference quotient. • Relationship between differentiability and continuity. Derivative at a point • Slope of a curve at a point. Examples are emphasized, including points at which there are vertical tangents and points at which there are no tangents. • Tangent line to a curve at a point and local linear approximation. • Instantaneous rate of change as the limit of average rate of change. • Approximate rate of change from graphs and tables of values. Derivative as a function • Corresponding characteristics of graphs of ƒ and ƒ∙. • Relationship between the increasing and decreasing behavior of ƒ and the sign of ƒ’. • The Mean Value Theorem and its geometric interpretation. • Equations involving derivatives. Verbal descriptions are translated into equations involving derivatives and vice versa. Second derivatives • Corresponding characteristics of the graphs of ƒ, ƒ’, and ƒ’’∙. • Relationship between the concavity of ƒ and the sign of ƒ’∙. • Points of inflection as places where concavity changes. Applications of derivatives • Analysis of curves, including the notions of monotonicity and concavity. + Analysis of planar curves given in parametric form, polar form, and vector form, including velocity and acceleration. • Optimization, both absolute (global) and relative (local) extrema. • Modeling rates of change, including related rates problems. • Use of implicit differentiation to find the derivative of an inverse function. • Interpretation of the derivative as a rate of change in varied applied contexts, including velocity, speed, and acceleration. • Geometric interpretation of differential equations via slope fields and the relationship between slope fields and solution curves for differential equations. + Numerical solution of differential equations using Euler’s method. + L’Hospital’s Rule, including its use in determining limits and convergence of improper integrals and series. Computation of derivatives • Knowledge of derivatives of basic functions, including power, exponential, logarithmic,
AP微积分2005真题
满分网——AP真题 AP® Calculus BC2005 Free-Response QuestionsThe College Board: Connecting Students to College SuccessThe College Board is a not-for-profit membership association whose mission is to connect students to college success and opportunity. Founded in 1900, the association is composed of more than 4,700 schools, colleges, universities, and other educational organizations. Each year, the College Board serves over three and a half million students and their parents, 23,000 high schools, and 3,500 colleges through major programs and services in college admissions, guidance, assessment, financial aid, enrollment, and teaching and learning. Among its best-known programs are the SAT®, the PSAT/NMSQT®, and the Advanced Placement Program® (AP®). The College Board is committed to the principles of excellence and equity, and that commitment is embodied in all of its programs, services, activities, and concerns.Copyright © 2005 by College Board. All rights reserved. College Board, AP Central, APCD, Advanced Placement Program, AP, AP Vertical Teams, Pre-AP, SAT, and the acorn logo are registered trademarks of the College Entrance Examination Board. Admitted Class Evaluation Service, CollegeEd, Connect to college success, MyRoad, SAT Professional Development, SAT Readiness Program, and Setting the Cornerstones are trademarks owned by the College Entrance Examination Board.PSAT/NMSQT is a registered trademark of the College Entrance Examination Board and National Merit Scholarship Corporation. Other products and services may be trademarks of their respective owners. Permission to use copyrighted College Board materials may be requested online at: /inquiry/cbpermit.html.Visit the College Board on the Web: .AP Central is the official online home for the AP Program and Pre-AP: .CALCULUS BC SECTION II, Part ATime—45 minutes Number of problems—3A graphing calculator is required for some problems or parts of problems.1. Let f and g be the functions given by ()()1sin 4f x x p =+ and ()4.x g x -= Let R be the shaded region in the first quadrant enclosed by the y -axis and the graphs of f and g , and let S be the shaded region in the firstquadrant enclosed by the graphs of f and g , as shown in the figure above. (a) Find the area of R . (b) Find the area of S .(c) Find the volume of the solid generated when S is revolved about the horizontal line y 1.=-WRITE ALL WORK IN THE TEST BOOKLET.2. The curve above is drawn in the xy -plane and is described by the equation in polar coordinates ()sin 2r q q =+for 0,q p ££ where r is measured in meters and q is measured in radians. The derivative of r with respectto q is given by ()12cos 2.drd q q =+(a) Find the area bounded by the curve and the x -axis.(b) Find the angle q that corresponds to the point on the curve with x -coordinate 2.- (c) For2,33p p q << dr d q is negative. What does this fact say about r ? What does this fact say about the curve?(d) Find the value of q in the interval 0pq ££ that corresponds to the point on the curve in the first quadrant with greatest distance from the origin. Justify your answer.WRITE ALL WORK IN THE TEST BOOKLET.Distance x (cm) 0 1 5 6 8 Temperature ()T x ()C ∞100 93 7062553. A metal wire of length 8 centimeters (cm) is heated at one end. The table above gives selected values of thetemperature (),T x in degrees Celsius ()C ,∞ of the wire x cm from the heated end. The function T is decreasing and twice differentiable.(a) Estimate ()7.T ¢ Show the work that leads to your answer. Indicate units of measure.(b) Write an integral expression in terms of ()T x for the average temperature of the wire. Estimate the averagetemperature of the wire using a trapezoidal sum with the four subintervals indicated by the data in the table. Indicate units of measure. (c) Find ()8,T x dx ¢Ú and indicate units of measure. Explain the meaning of ()8T x dx ¢Ú in terms of thetemperature of the wire.(d) Are the data in the table consistent with the assertion that ()0T x >¢¢ for every x in the interval 08?x <<Explain your answer.WRITE ALL WORK IN THE TEST BOOKLET.END OF PART A OF SECTION IICALCULUS BC SECTION II, Part BTime—45 minutes Number of problems—3No calculator is allowed for these problems.4. Consider the differential equation2.dyx y =- (a) On the axes provided, sketch a slope field for the given differential equation at the twelve points indicated,and sketch the solution curve that passes through the point ()0,1.(Note: Use the axes provided in the pink test booklet.)(b) The solution curve that passes through the point ()0,1 has a local minimum at ()3ln.x = What is the y -coordinate of this local minimum?(c) Let ()y f x = be the particular solution to the given differential equation with the initial condition()0 1.f = Use Euler’s method, starting at 0x = with two steps of equal size, to approximate ()0.4.f - Show the work that leads to your answer. (d) Find 22d ydxin terms of x and y . Determine whether the approximation found in part (c) is less than orgreater than ()0.4.f - Explain your reasoning.WRITE ALL WORK IN THE TEST BOOKLET.5. A car is traveling on a straight road. For 024t ££ seconds, the car’s velocity (),v t in meters per second, ismodeled by the piecewise-linear function defined by the graph above.(a) Find ()24.v t dt Ú Using correct units, explain the meaning of ()24.v t dt Ú(b) For each of ()4v ¢ and ()20,v ¢ find the value or explain why it does not exist. Indicate units of measure. (c) Let ()a t be the car’s acceleration at time t , in meters per second per second. For 024,t << write apiecewise-defined function for ().a t (d) Find the average rate of change of v over the interval 820.t ££ Does the Mean Value Theorem guaranteea value of c , for 820,c << such that ()v c ¢ is equal to this average rate of change? Why or why not?6. Let f be a function with derivatives of all orders and for which ()27.f = When n is odd, the n th derivativeof f at 2x = is 0. When n is even and 2,n ≥ the n th derivative of f at 2x = is given by ()()()1!2.3n nn f -=(a) Write the sixth-degree Taylor polynomial for f about 2.x =(b) In the Taylor series for f about 2,x = what is the coefficient of ()22nx - for 1?n ≥(c) Find the interval of convergence of the Taylor series for f about 2.x = Show the work that leads to youranswer.WRITE ALL WORK IN THE TEST BOOKLET.END OF EXAM。
2017 ap calculusab 微积分
2017 ap calculusab 微积分英文版2017 AP Calculus AB: A Journey Through MicrocalculusAs the sun rose over the horizon, students across the globe began their journey into the world of mathematics with the 2017 AP Calculus AB exam. This exam, known for its depth and breadth, tests the student's understanding of the fundamental concepts of calculus.The exam began with a gentle reminder of the basic derivative rules, followed by questions that required students to apply these rules to real-world scenarios. One such question dealt with the optimization of a profit function, testing the student's ability to identify the maximum or minimum value of a function. This question highlighted the practical applications of calculus in real-life situations.As the exam progressed, the questions became more complex, delving into the realm of integration. Students werechallenged to evaluate integrals using various techniques, such as substitution and integration by parts. One notable question dealt with the concept of areas between curves, requiring students to apply their knowledge of integration to find the enclosed area.The exam also included questions on sequences and series, testing the student's understanding of convergence and divergence. Questions on infinite series were particularly challenging, as they required students to analyze the behavior of the series as it approached infinity.Towards the end of the exam, students were presented with a challenging question on differential equations. This question tested their ability to understand and manipulate differential equations, ultimately finding a solution that satisfied the given conditions.The 2017 AP Calculus AB exam was not just a test of mathematical knowledge; it was a testament to the students' dedication, perseverance, and understanding of the beauty andpower of calculus. As the sun set, students closed their books, satisfied that they had done their best, and hoped that their efforts would be rewarded with a smile on the faces of their teachers and parents.中文版2017 AP微积分AB：微积分之旅随着太阳从地平线上升起，全球的学生们开始了他们的数学之旅，参加了2017年的AP微积分AB考试。
AP Calculus Chapter 10 Sequences and Series 序列和级数
Chapter 10 Sequences and Series序列和级数【Vocabulary · 词汇梳理】【导图】A. Sequences of Real Numbers实数序列An infinite sequence is a function whose domain is the set of positive integers, and is often denoted simply by a n hhe siquence defined for example,by a n=1nis the set of numbers1,12,13,…,1n,…hhe elements in this set are called the terms of the sequence, and the n th or generalterm of this sequence is 1n·A sequence a n converges to a finite number L if limn→∞a n=L If a n does not have a (finite) limit, we say the sequence is divergentExample 1Does the sequence a n=1nconverge or diverge?Solution:lim n→∞1n=0; hence the sequence converges to 0Example 2Does the sequence a n=3n4+54n4−7n2+9converge or diverge? Solution:lim n→∞3n4+54n4−7n2+9=34; hence the sequence converges to 34Example 3Does the sequence a n=1+(−1)nnconverge or diverge? Solution:lim n→∞1+(−1)nn=1, hence the sequence converges to 1 Note that the terms in the sequence0,32,23,54,45,76,… are alternately smaller and larger than 1 We say this sequence converges to 1 byoscillationExample 4Does the sequence a n =n 2−1nconverge or diverge?Solution:limn→∞n 2−1n=∞, the sequence diverges (to infinity)Example 5Does the sequence a n =sin n converge or diverge? Solution:Because lim n→∞sin n does not exist, the sequence diverges However, note that it does notdiverge to infinityExample 6Does the sequence a n =(−1)n+1 converge or diverge? Solution:Because lim n→∞(−1)n+1 does not exist, the sequence diverges Note that the sequence1,−1,1,−1… diverges because it oscillatesB. INFINITE SERIES 无穷级数B1. Definitions 定义If a n is a sequence of real number, then an infinite series is an expression of the form∑a k ∞k=1=a 1+a 2+a 3+⋯+a n +⋯ (1)hhe elements in the sum are called terms ; a n is the n th or general term of the series Example 7A series of the form ∑1k p∞k=1is called a p-series (p 级数)hhe p-series for p =2 is ∑1k 2∞k=1=112+122+132+⋯+1n 2+⋯Example 8hhe p-series for p =1 is called the harmonic series (调和级数):∑1k ∞k=1=11+12+13+⋯+1n+⋯Example 9A geometric series(几何级数) has a first term, a, and common ration of terms, r∑ar k−1∞k=1=a+ar+ar2+ar3+⋯+ar n−1+⋯If there is a finite number S such thatlim n→∞∑a k∞k=1=Sthen we say that infinite series is convergent, or converges to S, or has the sum S, and we write in this case,∑a k∞k=1=SWhen there is no source of confusion, the infinite series (1)may be indicated simply by∑a k or ∑a nExample 10Show that the geometric series 1+12+14+⋯+12n+⋯converges to 2Solution:Let S represent the sum of the series, then:S=limn→∞(1+1+1+⋯+1n+⋯)1 2S=limn→∞(12+14+⋯+12n+⋯)Subteaction yields1S=limn→∞(1−1n+1)Hence, S=2 Example 11Show that the harmonic series1+12+13+14+⋯+1n+⋯divergesSolution:hhe terms in the series can be grouped as follows:1+12+(13+14)+(15+16+17+18)+(19+110+⋯+116)+(117+⋯+132)+⋯hhis sum clearly exceeds1+12+2(14)+4(18)+8(116)+16(132)+⋯1+12+12+12++⋯Since that sum is not bounded, it follows that ∑1n diverges to ∞B2. Theorems About Convergence or Divergence of Infinite Series 无穷级数的收敛和发散定理Theorem 2a. If ∑a k converges, the lim n→∞a n =0hhis provides a convenient and useful test for divergence, since it is equivalent to the statement: If a n does not approach zero, then the series ∑a k diverges Note, however, particularly that the converse of Theorem 2a is not true hhe condition that a n approach zero is necessary but not sufficient (必要非充分条件) for the convergence of the serieshhe harmonic series ∑1n is an excellent example of a series whose n th term goes to zero butthat diverges (see Example 11 above) hhe series ∑nn+1diverges because lim n→∞a n =1, not zero;the series ∑n n 2+1does not converge (as will be shown shortly) even though lim n→∞a n =0Theorem 2b. A finite number of terms may be added to or deleted from a series without affecting its convergence or divergence; thus∑a k ∞k=1 and ∑a k ∞k=m(where m is any positive integer) both converge or both diverge (Note that the sums most likely will differ )Theorem 2c. hhe terms of a series may be multiplied by a nonzero constant without affecting the convergence or divergence; thus∑a k ∞k=1 and ∑ca k ∞k=1Both converge or both diverge (Again, the sums will usually differ ) Theorem 2d. If ∑a n and ∑b n both converge, so does ∑(a n +b n )Theorem 2e. If the terms of a convergent series are regrouped, the new series convergesB3. Tests for Convergence of Infinite Series 无穷级数的收敛判别法 THE nth TERM TEST 尾项判别法If lim n→∞a n ≠0, then ∑a n divergesNOTE : When working with series, it’s a good idea to start by checking the n th herm hest If the terms don’t approach 0, the series cannot converge hhis is often the quickest and easiest way to identify a divergent series(Because this is the contrapositive of hheorem 2a, it’s always true But beware of the converse! Seeing that the terms do approach 0 does not guarantee that the series must converge. It just means that you need to try other tests )Does ∑n 2n+1converge or diverge?Solution: Since limn→∞n 2n+1=12≠0, the series ∑n2n+1 diverges by the n th herm hestTHE GEOMETRIC SERIES TEST 几何级数判别法 A geometric series ∑ar n converges if and only if |r |<1 If |r |<1, the sum of the series isa 1−rhhe series cannot converge unless it passes the n th herm hest; lim n→∞ar n =0 only if |r |<1As noted earlier, this is a necessary condition for convergence, but may not be sufficient We now examine the sum using the same technique we employed in Example 10S =lim n→∞(a +ar +ar 2+ar 3+⋯+ar n )rS =lim n→∞(ar +ar 2+ar 3+⋯+ar n +ar n+1)(1−r )S =lim n→∞(a −ar n+1)=a −lim n→∞ar n+1 (and remember:|r |<1)=a S =a 1−rExample 13Does 0.3+0.03+0.003+⋯ converge or diverge? Solution:hhe series 0.3+0.03+0.003+⋯ is geometric with a =0.3 and r =0.1 Since |r |<1, the series converges, and its sum isS =a 1−r =0.31−0.1=0.30.9=13Note : 13=0.3333…, which is the given seriesB4. Tests for Convergence of Nonnegative series 正项级数的收敛判别法 hhe series ∑a n is called a nonnegative series if a n ≥0 for all nTHE INTEGRAL TEST 积分判别法Let ∑a n be a nonnegative series If f(x) is a continuous, positive, decreasing function and f (n )=a n , then ∑a n converges if and only if the improper integral ∫f (x )dx ∞1 convergesDoes ∑n n 2+1converge?Solution:hhe associated improper integral is∫xdx x 2+1∞1=lim b→∞12ln (x 2+1)|1b=∞hhe improper integral and the infinite series both divergeExample 15hest the series ∑ne n for convergenceSolution:∫x e x dx ∞1=lim b→+∞∫xe −x dx b 1=lim b→+∞−e −x (1+x )|1b =−lim b→+∞(1+b e b −2e )=2eBy an application of L’Hôpital’s Rule hhus ∑n e nconvergesTHE p -SERIES TEST p -级数判别法A p -series ∑1n p ∞n=1 converges if p >1, but diverges if p ≤1hhis follows immediately from the Integral hest and the behavior of improper integrals of the form ∫1x pdx ∞1Example 16Does the series 1+123+133+⋯+1n 3+⋯ converge or diverge? Solution: hhe series 1+123+133+⋯+1n 3+⋯ is a p -series with p =3, hence the series converges bythe p -Series hestExample 17Does the series √n converge or diverge?Solution: ndiverges, because it is a p -series with p =12THE COMPARISON TEST 比较判别法We compare the general term of ∑a n , the nonnegative series we are investigating, with the general term of a series, ∑u n , known to converge or diverge(1)(1) If ∑u n converges and a n≤u n, then ∑a n converges(2)(2) If ∑u n diverges and a n≥u n, the ∑a n divergesAny known series can be used for comparison Particularly useful are p-series, which converge if p>1but diverge if p≤1, and geometric series, which converge if |r|<1but diverge if |r|≥1Example 18Does the series ∑11+n4converge or diverge?Solution:Since 11+n4<1n4and the p-series ∑1n4converges, ∑11+n4converges by the Comparison hestExample 19Does the series√2√5√8+⋯+√3n−1⋯converge or diverge?Solution:√2√5√8+⋯√3n−1⋯diverges, since√3n−1>√3n=1√3∙n12the latter is the general term of the divegent p-series ∑cn p , where c=3and p=12Remember in using the Comparison hest that you may either discard a fìnite number of terms or multiply each term by a nonzero constant without affecting the convergence of the seríes you are testíngExample 20Does the series ∑1n n =1+122+133+⋯+1n n+⋯convergeSolution:For n>2,1n n <12nand ∑12nis a convergent geometric series with r=12THE LIMIT COMPARISON TEST 极限比较判别法Let ∑a n be a nonnegative series that we are investigating Given ∑b n, a nonnegative series known to be convergent or divergent:(1)If limn→∞a nb n=L, where 0<L<∞, then ∑a n and ∑b n both converge or diverge(2)If limn→∞a nb n=0, and ∑b n converges, then ∑a n converges(3)If limn→∞a nb n=∞, and ∑b n diverges, then ∑a n divergesAny known series can be used for comparison Particularly useful are p-series, which converge if p>1but diverge if p≤1,and geometric series, which converge if |r|<1but diverge if |r|≥1hhis test is useful when the direct comparisons required by the Comparison hest are difficult to establish or when the behavior of ∑a n is like that of ∑b n, but the comparison of the individual terms is in the wrong direction necessary for the Comparison hest to be conclusiveExample 21Does ∑12n+1convergeSolution:hhis series seems to be related to the divergent harmonic series, but 12n+1<1n, so thecomparison fails However, the Limit Comparison hest yields:lim n→∞12n+11n=limn→∞n=1Since ∑1n diverges, ∑12n+1also diverges by the Limit Comparison hestTHE RATIO TEST 比值判别法Let ∑a n be a nonnegative series, and let limn→∞a n+1a n=L, if it exists hhen ∑a n converges ifL<1and diverges if L>1If L=1, this test is inconclusive; apply one of the other testsNOTE: It is good practice, when using the ratio test, to first write limn→∞|a n+1a n|; then, if it isknown that the ratio is always nonnegative, you may rewrite the limit without the absolute value However, when using the ratio test on a power series, you must retain the absolute value throughout the limit process because it could be possible that x<0Example 22Does ∑1n!converge or diverge?Solution:lim n→∞a n+1a n=limn→∞1(n+1)!1n!=limn→∞n!(n+1)!=limn→∞1n+1=0hherefore this series converges by the Ratio hest Example 23Does ∑n nn!converge or diverge?Solution:a n+1n =(n +1)n+1()∙n!n =(n +1)n n =(n +1)nlim n→∞(n +1n )n =lim n→∞(1+1n)n =e Since e >1,∑n n n!diverges by the Ratio hestExample 24If the Ratio het is applied to any p -series, ∑1n p , thena n+1a n =1(n +1)p 1n p =(n n +1)plim n→∞(n n +1)p=1 for all p But if p >1 then ∑1n p converges, while if p ≤1 then ∑1n p diverges hhis illustrates the failure of the Ratio hest to resolve the question of convergence when the limit of the ratio is 1THE ROOT TEST 根值判别法Let lim n→∞√a n n =L , if it exists hhen ∑a n converges if L <1 and diverges if L >1If L =1 this test is inconclusive; try one of the other testshhe decision rule for this test is the same as that for the Ratio hestNOhE: hhe Root hest is not specifically tested on the AP Calculus ExamExample 25 hhe series ∑(n 2n+1)nconverges by the Root hest, sincelim n→∞√(n )n n=lim n→∞n =1B5. Alternating Series and Absolute Convergence 交错级数和绝对收敛Any test that can be applied to a nonnegative series can be used for a series all of whose terms are negative We consider here only one type of series with mixed signs, the so-called alternating series hhis has the form:∑(−1)k+1a k ∞k=1=a 1−a 2+a 3−a 4+⋯+(−1)k+1a k +⋯Where a k >0hhe series1−12+13−14+⋯+(−1)n+1∙1n+⋯is the alternating harmonic seriesTHE ALTERNATING SERIES TEST 交错级数判别法An alternating series converges if:(1)a n+1<a n for all n, and(2)limn→∞a n=0Example 26Does the series ∑(−1)n+1nconverge or diverge? Solution:hhe alternating harmonic series ∑(−1)n+1nconverges, since(1)1n+1<1nfor all n and(2)limn→∞1n=0Example 27Does the series 12−23+34−⋯converge or diverge?Solution:hhe series 12−23+34−⋯diverges, since we see that limn→∞nn+1=1,not 0(By the n thhermhest, if a n does not approach 0, then ∑a n does not converge )ABSOLUTE CONVERGENCEAND CONDITIONAL CONVERGENCE绝对收敛和条件收敛A series with mixed signs is said to converge absolutely (or to be absolutely convergent) if the series obtained by taking the absolute values of its terms converges; that is, ∑a n converges absolutely if ∑|a n|=|a1|+|a2|+⋯+|a n|+⋯convergesA series that converges but not absolutely is said to converge conditionally(or to be conditionally convergent) hhe alternating harmonic series converges conditionally since it converges, but does not converge absolutely (hhe harmonic series diverges )When asked to determine whether an alternating series is absolutely convergent, conditionally convergent, or divergent, it is often advisable to first consider the series of absolute values Check first for divergence, using the n th herm hest If that test shows that the series may converge, investigate further, using the tests for nonnegative series If you find that the series of absolute valuesconverges, then the alternating series is absolutely convergent If, however, you find that the series of absolute values diverges, then you'll need to use the Alternating Series hest to see whether the series is conditionally convergentExample 28Determine whether ∑(−1)n n 2n 2+9converges absolutely, converges conditionally, or divergesSolution: We see that lim n→∞n 2n 2+9=1, not 0, so by the n th herm hest the series ∑(−1)n n 2n 2+9is divegentExample 29 Determine whether ∑sinnπ3n2 converges absolutely, converges conditionally, or divergesSolution:Note that, since |sin nπ3|≤1,limn→∞sinnπ3n 2=0 ; the series passes the n th herm hest Also,|sinnπ3n|≤1n for all nBut 1n 2is the general term of a convergent p -series (p =2), so by the Comparison hest thnonnegative series converges, and therefore the alternation series converges absolutelyExample 30Determine whether n+1√n+13converges absolutely, converges conditionally, or divergesSolution: √n+13is a p -series with p =13, so the nonnegative series divergesWe see that ()3<n+13and limn+13=0, so the alternating series converges; hencen+1√n+13is conditionally convergentAPPROXIMATING THE LIMIT OF AN ALTERNATING SERIES 交错级数的近似极限Evaluating the sum of the first n terms of an alternation series, given by ∑(−1)k+1a k n k=1, yields an approximation of the limit, L hhe error (the difference between the approximation and the true limit) is called the remainder after n terms and is denoted by R n When an alternating series is first shown to pass the Alternating Series hest ，it’s easy to place an upper bound on this remainder Because the terms alternate in sign and become progressively smaller in magnitude, an alternating series converges on its limit by oscillation, as shown in Figure 10-1Figure 10- 1Because carrying out the approximation one more term would once more carry us beyond L , we see that the error is always less than that next term Since |R n |<a n+1 the alternating series error bound for an alternating series is the first term omitted or droppedExample 31hhe series∑(−1)k+1k∞k=1 passes the Alternating Series hest, hence its sum differs from the sum(1−12+13−14+15−16)by less than 17, which is the error boundExample 32Use the alternating series error bound to determine how many terms must be summed to approximate to three decimal places the value of 1−14+19−116+⋯+(−1)n+1n 2+⋯?Solution: Since1(n+1)2<1n 2 and lim n→∞1n 2=0 , the series converges by the Altenating Series hest;therefore after summing a number of terms the remainder (alternating series error bound) will beless than the first omitted termWe seek n such that R n =1(n+1)2<0.001 hhus n must satisfy (n +1)2>1000, or n >30.623 hherefore 31 terms are needed for the desired accuracyC. POWER SERIES 幂级数C1. Definitions; Convergence 定义; 收敛 An expression of the form∑a k x k ∞k=0=a 0+a 1x +a 2x 2+⋯+a n x n +⋯, (1)Where the a’s are constants, is called a power series in x ; and∑a k (x −a)k ∞k=0=a 0+a 1(x −a)+a 2(x −a )2+⋯+a n (x −a )n +⋯, (2)is called a power series in (x −a)If in (1) or (2) x is replaced by a specific real number, then the power series becomes a series of constants that either converges or diverges Note that series (1) converges if x =0 and series (2) converges if x =aRADIUS AND INTERV AL OF CONVERGENCE 收敛半径和收敛区间If power series (1) converges when |x |<r and diverges when |x |>r , then r is called the radius of convergence Similarly, r is the radius of convergence of power series (2) if (2) converges when |x −a |<r and diverges when |x −a |>rhhe set of all values of x for which a power series converges is called its interval of convergence ho find the interval of convergence, first determine the radius of convergence by applying the Ratio hest to the series of absolute values hhen check each endpoint to determine whether the series converges or diverges thereExample 33Find all x for which the following series converges:1+x +x 2+⋯+x n +⋯ (3)Solution:By the Ratio hest, the series converges iflim n→∞|u n+1u n |=lim n→∞|x n+1x n |=lim n→∞|x |=|x |<1 hhus, the radius of convergence is 1 hhe endpoints must be tested separately since the Ratiohest fails when the limit equals 1 When x =1, (3) becomes 1+1+1+⋯ and diverges; when x =−1, (3) becomes 1−1+1−1+⋯ and diverges hhus the interval of convergence is −1<x <1Example 34For what x does ∑(−1)n−1x n−1n+1∞n=1 converge?Solution:lim n→∞|u n+1u n |=lim n→∞|x n n +2∙n +1x n−1|=lim n→∞|x |=|x |<1 hhe radius of convergence is 1 When x =1 , we have 12−13+14−15+⋯ , an alternating convergent series; when x =−1 , the series is 12+13+14+⋯ , which diverges hhus, the series converges if −1<x ≤1Example 35For what values of x does ∑x n n!∞n=1converge?Solution:lim n→∞|u n+1u n |=lim n→∞|x n+1(n +1)!∙n!x n |=lim n→∞|x |n +1=0 which is always less than 1 hhus the series converges for all xExample 36Find all x for which the following series converges:1+x −21+(x −2)22+⋯+(x −2)n−1n−1+⋯ (4)Solution:lim n→∞|u n+1u n |=lim n→∞|(x −2)n 2n ∙2n−1(x −2)n−1|=lim n→∞|x −2|2=|x −2|2which is less than 1 if |x −2|<2, that is, if 0<x <4 Series (4) converges on this intervaland diverges if |x −2|>2, that is, if x <0 or x >4 When x =0, (4) is 1−1+1−1+⋯ and diverges When x =4, (4) is 1+1+1+⋯ and diverges hhus, the interval of convergence is 0<x <4Example 37Find all x for which the series ∑n!x n∞n=1 converges Solution:∑n!x n ∞n=1 converges only at x =0, sincelim n→∞u n+1u n =lim n→∞(n +1)x =∞ unless x =0C2. Functions Defined by Power Series 幂级数定义的函数 Let the function f be defined byf (x )=∑a k (x −a)k ∞k=0=a 0+a 1(x −a )+⋯+a n (x −a )n +⋯ (1)its domain is the interval of convergence of the seriesFunctions defined by power series behave very much like polynomials, as indicated by the following properties:PROPERTY 2a. hhe function defined by (1) is continuous for each x in the interval of convergence of the seriesPROPERTY 2b. hhe series formed by differentiating the terms of series (1) converges to f ′(x ) for each x within the radius of convergence of (1); that is,f′(x )=∑ka k (x −a)k−1=a 1+2a 2(x −a )+⋯+na n (x −a )n−1+⋯∞k=0 (2) Note that power series (1) and its derived series (2) have the same radius of convergence but not necessarily the same interval of convergenceExample 38Let f (x )=∑x kk k+1∞k=1=x 1∙2+x 22∙3+x 33∙4+⋯+x n n n+1+⋯Find the intervals of convergence of the power series for f(x) and f ′(x ) Solution:lim n→∞|x n+1(n +1)(n +2)∙n (n +1)x n |=|x | also,f (1)=11∙2+12∙3+13∙4+⋯+1n ∙(n +1)+⋯ andf (−1)=−11∙2+12∙3−⋯+(−1)n n ∙(n +1)+⋯Hence, the power series for f converges if −1≤x ≤1 For the derivative f ′(x )=∑x k−1k+1∞k=1=12+x3+x 24+⋯+x n−1n+1+⋯lim n→∞|x n n +2∙n +1x n−1|=|x | also,f ′(1)=12+13+14+⋯ andf ′(−1)=12−13+14−⋯ Hence, the power series for f ′ converges if −1≤x <1hhus, the series given for f(x) and f ′(x ) have the same radius of convergence, but their intervals of convergence differPROPERTY 2c. hhe series obtained by integrating the terms of the given series (1) convergesto ∫f (t )dt xa for each x within the interval of convergence of (1); that is,∫f (t )dt =a 0(x −a )+a 1(x −a )22+a 2(x −a )33+⋯+a n (x −a )n+1n +1+⋯xa=∑a k (x −a )k+1k +1∞k=0Example 39Let f (x )=1(1−x )2 Show that the power series for ∫f (x )dx converges for all values of x in the interval of convergence of the power series for f(x)Solution:Obtain a series for 1(1−x )2 by long divisionhhen,1(1−x)2=1+2x+3x2+⋯+(n+1)x n+⋯It can be shown that the interval of convergence is −1<x<1 hhen by Property 2c∫1(1−x)2dx=∫[1+2x+3x2+⋯+(n+1)x n+⋯]dx11−x=c+x+x2+x3+⋯+x n+1+⋯Since when x=0we see that c=1, we have11−x=1+x+x2+x3+⋯+x n+⋯Note that this is a geometric series with ratio r=x and with a=1; if |x|<1, its sum isa 1−r =11−xC3. Finding a Power Series for a Function: Taylor and Maclaurin Series函数幕级数的展开: 泰勒级数和麦克劳林级数If a function f(x)is representable by a power series of the formc0+c1(x−a)+⋯+c n(x−a)n+⋯On an interval |x−a|<r, then the coefficients are given byc n=f(n)(a)n!, hhe seriesf(x)=f(a)+f′(a)(x−a)+f"(a)2!(x−a)2+⋯+f(n)(a)n!(x−a)n+⋯is called the Taylor series of the function f about the number a hhere is never more than one power series in (x−a)for f(x) It is required that the function and all its derivatives exist at x=a if the function f(x)is to generate a Taylor series expansionWhen a=0we have the special seriesf(x)=f(0)+f′(0)x+f"(0)2!x2+⋯+f(n)(0)n!x n+⋯called the Maclaurin series of the function f; this is the expansion of f about x=0Example 40Find the Maclaurin series for f(x)=e xSolution:Here f′(x)=e x,…,f(n)(x)=e x,…, for all n hhenf′(0)=1,…,f(n)(0)=1,…for all n, making the coefficients c n=1n!:e x=1+x+x22!+x33!+⋯+x nn!+⋯Example 41Find the Maclaurin expansion for f(x)=sin x Solution:hhus,sin x=x−x33!+x55!−⋯+(−1)n−1x2n−1(2n−1)!+⋯Example 42Find the Maclaurin expansion for f(x)=11−xSolution:hhus,11−x=1+x+x2+x3+⋯+x n+⋯Note that this agrees exactly with the power series in x obtained by different methods in Example39Example 43Find the Maclaurin expansion for f(x)=ln x about x=1Solution:hhus,ln x=(x−1)−(x−1)22+(x−1)33−(x−1)44+⋯+(−1)n−1(x−1)nnFUNCTIONS THAT GENERATE NO SERIES 不能级数展开的函数Note that the following functions are among those that fail to generate a specific series in (x−a)C4. Approximating Functions with Taylor and Maclaurin Polynomials 泰勒多项式和麦克劳林多项式的近似函数hhe function f(x)at the point x=a is approximated by a Taylor polynomial P n(x)of order n:f(x)≈P n(x)=f(a)+f′(a)(x−a)+f"(a)2!(x−a)2+⋯+f(n)(a)n!(x−a)nhhe haylor polynomial P n(x)and its first n derivatives all agree at a with f and its firstn derivatives hhe order of a haylor polynomial is the order of the highest derivative, which is also the polynomial’s last termIn the special case where a =0 , the Maclaurin polynomial of order n that approximates f(x) isP n (x )=f (0)+f ′(0)x +f"(0)2!x 2+⋯+f (n)(0)n!x nhhe haylor polynomial P 1(x) at x =0 is the tangent-line approximation to f(x) near zero given byf (x )=f (0)+f ′(0)xlt is the “best” linear approximation tof at 0, discussed at length in Chapter 4 §LA N OTE ON O RDER(泰勒多项式的阶数) AND D EGREE(泰勒多项式的级数)A haylor polynomial has degree(级数) n if it has powers of (x −a ) up through the n th If f (n)(a)=0, then the degree of P n (x) is less than n Note, for instance, in Example 45, that the second-order polynomial P 2(x) for the function sin x (which is identical with P 1(x)) is x +0∙x 22!, or just x , which has degree 1, not 2Example 44Find the haylor polynomial of order 4 at 0 for f (x )=e −x Use this to approximate f(0.25)Solution:hhe first four derivatives are −e −x ,e −x ,−e −x ,and e −x ; at a =0 , these equal −1,1,−1,and 1, respectively hhe approximating haylor polynomial of order 4 is thereforee −x ≈1−x +12!x 2−13!x 3+14!x 4 With x =0.25 we havee −0.25≈1−0.25+12!(0.25)2−13!(0.25)3+14!(0.25)4≈0.7788 hhis approximation of e −0.25 is correct to four placesIn Figure 10-2 we see the graphs of f(x) and of the haylor polynomials:Figure 10- 1P0(x)=1;P1(x)=1−x;P2(x)=1−x+x2;P3(x)=1−x+x22!−x33!;P4(x)=1−x+x22!−x33!+x44!Notice how closely P4(x)hugs f(x)even as x approaches 1Since the series can be shown to converge for x>0by the Alternating Series hest, the error in P4(x)is less than themagnitude of the first omitted term, x55!, or 1120at x=1 In fact, P4(1)=0.375to three decimalplaces, close to e−1≈0.368Example 45(a) Find the haylor polynomials P1,P3,P5, and P7at x=0for f(x)=sin x(b) Graph f and all four polynomials in [−2π,2π]×[−2,2](c) Approximate sinπ3using each of the four polynomialsSolution:P1(x)=x;P3(x)=x−x3 3!;P5(x)=x−x33!+x55!;P7(x)=x−x33!+x55!−x77!(b) Figure 10-3a shows the graphs of sin x and the four polynomials In Figure 10-3b we see graphs only of sin x and P7(x), to exhibit how closely P7“follows” the sine cruve。
ap课程precalculus解题
Ap课程precalculus解题在数学学科中，AP课程precalculus是一门非常重要的课程，它为学生提供了在高等数学中取得成功所需的基本概念和技能。
解题是学习precalculus的重要环节，它不仅能够帮助学生巩固所学知识，还能培养学生的逻辑思维和解决问题的能力。
本文将深入探讨AP课程precalculus解题的一些重要概念和技巧，并共享个人观点和理解。
一、概念和技巧1. 利用数学工具在解题过程中，学生需要熟练掌握数学工具，比如函数图像、方程和不等式等。
这些数学工具能够帮助学生更好地理解问题，并找到解决问题的方法。
当遇到复杂的函数图像题时，学生可以利用图像的性质来解决问题，比如对称性、增减性和极值等。
2. 运用数学思维学生需要培养数学思维，善于运用已有知识解决新问题。
在解题过程中，学生可以通过建立数学模型，利用函数的性质和特点，以及灵活运用各种解题技巧来解决问题。
在求解极限和导数的题目中，学生可以灵活运用极限的性质和求导的规则，将问题转化为基本的运算和推理，从而得到答案。
3. 练习和实践解题是一个需要不断练习和实践的过程。
学生需要多做一些经典的题目，掌握各种解题技巧和方法，培养自己的解题能力。
学生还可以通过解题训练营、竞赛和讨论小组等形式，与他人一起共享和交流解题经验，不断提高自己的解题水平。
二、总结和回顾通过对AP课程precalculus解题的深入探讨，我们了解到解题是学习数学知识的重要环节，它不仅能够帮助学生巩固所学知识，还能培养学生的逻辑思维和解决问题的能力。
在解题过程中，学生需要灵活运用数学工具和思维，多做练习和实践，培养自己的解题能力。
三、个人观点和理解在我的个人观点中，AP课程precalculus解题是一项需要逻辑思维和灵活运用知识的过程。
通过解题，我可以更好地掌握数学知识，提高自己的解题能力，培养自己的数学思维。
AP课程precalculus解题是学习数学知识的重要环节，它不仅能够帮助学生巩固所学知识，还能培养学生的逻辑思维和解决问题的能力。
AP Calculus AB Syllabus
Course OverviewMy main objective in teaching AP Calculus AB is to enable students to appreciate the beauty of calculus and receive a strong foundation that will give them the tools to succeed in future mathematics courses. Students know that they will work hard, and our expectation is that this hard work will enable them to succeed in the course. We work together to help students discover the joys of calculus.Primary TextbookJon Rogawski, Calculus for AP, 2nd edition, New York: W.H.Freeman and company,2012. David S. Kahn, Cracking the AP Calculus BC exam, 2015 edition, New York, 2014.Course PlannerFunctions and Graphs (Chapter 0)Students complete this review of precalculus materials over the summer.Key Concept: Limits (Chapter 1)2 weeks1. Limits, Rates of Change, and Tangent Lines2. Limits : A Numerical and Graphical Approsch3. Basic Limit Laws4. Limits and Continuity5. Evaluating Limits Algebraically6. Trigonometric Limits7. Limits at Infinity8. Intermediate Value TheoremDifferentiation (Chapter 2)4 weeks1. Definition of the Derivative2. The Derivative as a Function3. The Product and Quotient Rules4. Rates of Change5.Higher Derivatives6. The Trigonometric Functions6. The Chain Rule7. Implicit Differentiation8. The Related RatesApplication of the Derivative (Chapter 3) 4 weeks1. Linear Approximation and Application2. Extreme values3. The mean Value Theorem and Monotonicity4. The Shape of a Graph5. Graph Sketching and Asymptotes6. Applied Optimization7. Newton’s Method8. AntiderivativesMidterm ExamKey Concept: The Integral (Chapter 4)4 weeks1.Approximating and Computing Area2. The Definite Integral3. The Foundamental Theorem of Calculus4. Net Change as the Integral of a Rate5. Substitution MethodApplication of the Integral (Chapter 5)4 weeks1. Area Between Two Curves2. Setting up Integrals3. Volumes of Revolution4. The Method of Cylindrical Shells5. Work and EnergyTeaching StrategiesOn the first day of school, I begin with Chapter 2 (Key Concept: The Derivative). This sets the tone for the year. As discussed in the Student Activities section below, I start with a Calculator-Based Laboratory (CBL) ball toss experiment. Students explore the concept of average rate of change and discover the concept of instantaneous rate of change. Thereafter, student exploration and discovery continue to be an important aspect of the remaining topics.Throughout the course, students work together on a regular basis, both formally and informally. At times, I set up groups to work on a particular activity, but students do not need to be told to work together. Our classroom has tables instead of desks to make it more conducive to group work. When students are working on a problem, they will often work alone initially but then turn to their partners to collaborate.In discovering new concepts, the class works as a whole. It is not necessary for students to raise their hands. If students have a thought to share, they are welcome to make a contribution. If they are so inclined, students will go up to the board to illustrate a point. At times, I am able to step back and just listen to the interaction among my students as they explore a topic.Technology can be used to help make calculus concepts come alive and it enables students to “see” what is being discussed. Students are issued TI-89 calculators. Our classroom also contains 10 computers and a SMART Board. We use TI-Navigator, TI Inter Active!, and CBLs. Topics are presented using the “rule of four”: graphically, numerically, algebraically, and orally. Through this multifaceted approach, students gain an in-depth understanding of the material.Student EvaluationStarting in October, I assign six AP free-response questions for students to work on for two weeks. Students may work on these questions with one other person and come to me for extra help. At the end of the two weeks, I randomly select one of these questions for a quiz. Students are graded as they would be graded on an AP Exam. Students are expected to explain the solutions to problems using written sentences. (Free-response questions and scoring guidelines are available on AP Central®.)When the second semester begins in February, we work on multiple-choice sections of AP Calculus Released Exams. Students have two weeks to complete a multiple-choice section. They may work on these questions with one other student and come to me for extra help. At the end of the two weeks, students hand in their work for grading. They are also quizzed on selected questions from the work they submit. The multiple-choice work alternates with the free-response work until the AP Exam administration.Student Activities1. On the first day of school, we begin the study of derivatives with a CBL experiment. Students toss a ball in the air and examine the height-versus-time graph generated. They fit the data with a quadratic equation for the position function to determine how high the ball went and how long it was in the air. Students compute the average velocity over a time interval. They are then asked to determine the velocity of the ball at exactly 0.06 seconds after the ball was tossed and explain how their answer was obtained. Finally, they zoom in on the graph of the position function near t = 0.06 until the graph looks like a line. Students find the slope of the line and compare it to their estimation of the instantaneous velocity at t = 0.06.2. When students are first learning about derivatives, I sketch the graph of an unknown function on the board. Each student comes up to the board and plots a point that would lie on the graph of the derivative of the function. We watch as the graph of the derivative unfolds. For example, I often graph a sine function on the board. At the maximum value of this function, the derivative would be zero. Students are anxious to plot these points. At the points of infection, the derivative would be at a maximum or minimum. These are also easy points to plot. Gradually, students fill in the remaining values, and we can see that the derivative of a sine function is a cosine function.3. During our study of related rates, students suck on Tootsie Roll Pops to deter-mine the rate of change of their r adius; they then calculate the rate of change of the Pops’ volume. Students measure the initial radius of a Pop with dental foss. They then suck on the Pop for 30 seconds, record its radius, suck for another 30 seconds, etc. They model the rate of change of the radius with some function of time. Students then use this rate of change to estimate the rate of change of the volume of the Pop when its radius is three-fourths of its original radius. This lab, “How Many Licks?,” can be found in Ellen Kamischke’s b ook, A Watched Cup Never Cools. [SC5]4. The study of optimization can be made more meaningful to students by asking them to design an optimum can. Students obtain a can of soda, soup, tuna, etc. They measure the height and diameter of the can and determine its volume. They then find the radius and height of the most cost-effective can that will hold the same volume and write an explanation, using well-written sentences, of the mathematics involved in making their determination. Students then construct the most cost-effective can, bring their original can and constructed can to school, and make a presentation to the class.5. After learning how to approximate a definite integral, students use these techniques to calculate the distance covered during a 20-minute drive with a friend or parent. Before beginning the drive, students record the car’s odometer read ing. Using the speedometer, they record the car’s speed at 1-minute intervals, noting any traffic conditions. At the end of the drive, they check the odometer reading again. Students then graph speed versus time and use integration techniques to approximate the distance traveled over the 20-minute interval.They compare this distance with the actual mileage determined by the odometer. Students are often amazed at the closeness of their approximation to the odometer reading. Students are to write a report on this project that includes an explanation of data collection, graphing of the data, interpretation of the data, and the closeness of their approximation to the odometer reading.6. As an introduction to slope fields, I use an activity from the AP CalculusTeacher’s Guide. Using the graphing-calculator screen with the grid turned on, I project a 3 x 3 grid onto the board and assign each student several coordinate points in the region (([1, 1], [1, 2]), etc.).For a given differential equation, each student computes the slope at his or her coordinate position and then goes to the board to draw a short line segment with the calculated slope and the coordinate point as the midpoint of the segment. For example, if dy/dx = y, the student with coordinates (1, 1) would go to the board and at the point (1, 1) draw a short line segment with a slope of 1. The student with coordinates (1, 2) would go to the board and at the point (1, 2) Draw a small line segment with a slope of 2. (It is important that the second student draw a line segment whose slope is steeper than the slope of the fi rst student’s line se gment.) Continuing in this fashion, the class would complete the slope field. The students then use their calculator to graph a solution found analytically to the differential equation. The students are asked to use the graph obtained with their calculator to interpret the results and to support their conclusions obtained with the slope field. At this point, all sorts of discussions can ensue.7. One of the most difficult concepts for me as a student of calculus was finding the volume of solids. If students cannot visualize the solids, they have a more difficult time understanding how to compute the volume. To enable students to “see” the solids of revolution, I have purchased several “open-up” party decorations from a party supply store. Instead of just di scussing how aline revolving around an axis forms a cone, students see the cone generated. (I found an open-up ice cream cone in one of the party stores.)I bring cans of Play-Doh into school and ask students to construct solids whose bases are bounded by two curves and whose cross sections are squares or equilateral triangles, etc. For example, students are given the graph of a circle and asked to construct a solid in which each cross section perpendicular to the base is an equilateral triangle. Students build the solids using Play-Doh and then use plastic knives or dental floss to cut through the solid and obtain the required cross sections.Students also use the Winplot program on the computer and see the solids come alive.We finish off this topic with two activities from Work Smarter Not Harder, a book of labs accompanied by a disk of calculator programs. Students can download a program onto their calculators that will enable them to enter a function, graph the function, and rotate the function about a line. The calculator will then display a cross section of the solid generated. Another program enables students to enter a function and display a cross section that is a square, isosceles right triangle, etc. By the time the students have completed these activities, they are quite comfortable with the topic.。
2014AP微积分CALCULUS知识点总结
A DERIV ATIVE FUNCTION1. The derivative function or simply the derivative is defined as)(x f '＝y '＝xx f x x f x y x x ∆-∆+=∆∆→∆→∆)()(lim lim002. Find the derivative function a) Find y ∆,b) Find the average rate of change x y ∆∆, c) Find the limit xy x ∆∆→∆0lim .3. Geometric significanceConsider a general function y=f(x), a fixed point A(a,f(a)) and a variable point B(x,f(x)). The slope of chord AB=ax a f x f --)()(.Now as B →A, x →a and the slope of chord AB →slope of tangent at A. So, ax a f x f a x --→)()(lim is )(a f '.Thus, we can know the derivative at x=a is the slope of the tangent at x=a.4. Rules)(x f)(x f 'C(a constant) 0n x1-n nxx sinx cosx cosx sin -x tanxx 22cos 1sec =x arcsin2-11xx lnx 1x a loge xa log 1x e x e x aa a x ln )()(x v x u ±)()(x v x u '±')()(x v x u )()()()(x v x u x v x u '+')()(x v x u 2vv u v u '-' )0(≠v5. The chain ruleIf )(u f y = where )(x u u = thendxdu du dy dx dy =. )()(x g e x f = )()()(x g e x f x g '=' )(ln )(x g x f = )()()(x g x g x f '=' )(ln )()(ln )()()()(x u x v x u x v e e x u x f x v ===,])()()()(ln )([)()(ln )(x u x u x v x u x v ex f x u x v '+'='6. Inverse function, Parametric function and Implicit function Inverse function:dy dx dx dy 1=, ])([1)(1'='-x f x f , i.e., x y arcsin =, y x sin =Parametric function:dtdx dtdy dx dy =, i.e., )(t y ϕ=,)(t x ψ=→)(1x t -=ψ, )]([1x y -=ψϕ)()(t t dt dx dt dy dx dt dt dy dx dy ψϕ''=== Implicit function: 0))(,(=x y x F , 0))(,(=x f x F .0-222=+a y x ,ta y t a x sin cos ==, t ]2,0[π∈t ta t a dx dy x y cot sin cos )(-=-=='7. High derivativexx f x x f dx y d x f x ∆'-∆+'==''→∆)()(lim )(022 ta t a t dt dx dt y d dx y d x y x y x 32sin 1sin csc ])([)(-=-='='=''='' xx f x x f x f n n x n ∆-∆+=--→∆)()(lim )()1()1(0)( y=sinx )2sin(cos π+=='x x y , )22sin()2cos(ππ⨯+=+=''x x y )2sin()(π⨯+=n x ynB APPLICATIONS OF DIFFERENTIAL CALCULUS1. Monotonicitya) If S is an interval of real numbers and f(x) is defined for all x in S, then ：f(x) is increasing on S ⇔ 0)(≥'x f for all x in S, and f(x) is decreasing on S ⇔0)(≤'x f for all x in S. b) Find the monotone interval ● Find domain of the function,● Find )(x f ', and x which make 0)(='x f , ● Draw sign diagram, find the monotone interval. 2. Maxima/Minima, Horizontal inflection, Stationary pointC INTEGRAL1. The idea of definite integralWe define the unique number between all lower and upper sums as⎰badx x f )( and call it “the definite integral of )(x f from a to b ”,i.e., ∑∑⎰=-=∆〈〈∆ni i n i ba i x x f dx x f x x f 110)()()( where nab x -=∆.We note that as ∞→n ,∑⎰-=→∆10)()(n i ba idx x f x x f and⎰∑→∆=ba ni i dx x f x x f )()(1We write ⎰∑=∆=∞→ba ni i n dx x f x x f )()(lim 1. If 0)(≥x f for all x on [a,b] then⎰badx x f )( is the shaded area.2. Properties of definite integrals⎰⎰-=-bab adx x f dx x f )()]([⎰⎰=ba b a dx x f c dx x cf )()(, c is any constant ⎰⎰⎰=+ca ba cb dx x f dx x f dx x f )()()( ⎰⎰⎰+=+bababadx x g dx x f dx x g x f )()()]()([)()()()(a F b F x F dx x f bab a-==⎰, where ⎰=dx x f x F )()(⎰-=a adx x f 0)(（f(x) odd ）,⎰⎰-=a aadx x f dx x f 0)(2)((f(x)even)If 0)(≥x f on b x a ≤≤ then⎰≥badx x f 0)(If )()(x g x f ≥ on b x a ≤≤ then⎰⎰≥b a ba dx x g dx x f )()(The average value of a function on an interval [a,b]⎰-=ba avedx x f ab f )(13. The infinite integral If )()(x f x F =', then ⎰+=C x F dx x f )()(Formulas:⎰++=+C x n dx x n n111, C a a dx a xx +=⎰ln 1 ⎰+-=C x inxdx cos s ,⎰+=Cx xdx sin cos ,C x xdx +-=⎰cos ln tan ,⎰+=C x xdx sin ln cotC x xdx +=-⎰arcsin 12(12<x ), C x x dx +=+⎰arctan 12 U Substitution⎰'dx x g x g f )())(( substitution u=g(x) ⎰du u f )(Integration by Parts⎰⎰-=vdu uv udv。
AP Calculus Series——Infinite Series 2021.03.10【AP
AP Calculus Series ——Infinite Series2021.03.101. Definition ：An infinite series is an expression of the form 123n a a a a +++++or1n n a ∞=∑. Thenumbers 12,,a a are the terms of the series; n a is the nth term.2. Convergence of an infinite series An infinite series1nn a∞=∑ converges to the sum S if its partialsum converges to S : lim N N S S →∞=.In this case, we write1nn aS ∞==∑. Otherwise, we say that the infinite series diverges.3. Linearity of infinite series If1nn a∞=∑and1nn b∞=∑converges ，then()1n n n a b ∞=±∑ and 1nn ca∞=∑(c is anconstant) also converge, and()111nnn n n n na b a b ∞∞∞===±=±∑∑∑,11nnn n ca ac ∞∞===∑∑.4. THEOREM If the infinite series1nn a∞=∑converges, then lim 0n n a →∞=.Diverge Test If the nth term n a does not converges to 0, then the infinite series 1nn a∞=∑diverges.Exercise1. Calculate 345S S S ，，and find the sum of the telescope series ()11112n n n ∞=-++∑2. Find a formula for the partial sum of N S of()11nn ∞=-∑and show the series diverges.3. Use the theorem to prove that the following series diverge.1) 11012n nn ∞=+∑2)211n n ∞=+3)()211nn n ∞=-∑4)01231234-+-+5) 111cos cos cos 234+++ 6))2141n n n ∞=+∑4. Calculate the sume of the geometric series or state the series diverges. 1)1311nn -∞=⎛⎫ ⎪⎝⎭∑ 2) 1nn e π∞=⎛⎫ ⎪⎝⎭∑3) ()13258nnn n ∞=--∑5. Prove 1935n nnn ∞=+∑ diverges.AP Calculus Series ——Infinite Series Answer2021.03.10Exercise1. 3453151103142S S S S ====，，，2. 1 is odd0 is even N N S N -⎧=⎨⎩，，3. Use the theorem to prove that the following series diverge.1) 1 1012110n n n ∞=+∑2) 2111n n ∞=+3)()211 nn n ∞=∞-±∑4) 012313124-+-+±5) 111cos cos cos 2341+++6))2141 n n n ∞=∞+∑4. Calculate the sume of the geometric series or state the series diverges. 1) 1Di 3 ve 1r e 1g nn -∞=⎛⎫ ⎪⎝⎭∑2) 1Diver e g nn e π∞=⎛⎫⎪⎝⎭∑3) ()132583415nn n n ∞=---=∑5. Prove 1935n nnn ∞=+∑ diverges. Proof: Assume 1935n nnn ∞=+∑ converges. Let 1935n nnn S ∞=+=∑,then 11933552n n n n n n S S ∞∞===-=-∑∑,that is 195nn n ∞=∑. But we know that 195nn n ∞=∑ diverges. Therefore, 1935n nnn ∞=+∑ diverges.。