微积分英文版2
微积分常用英文词汇 分章
英汉微积分词汇English-Chinese Calculus Vocabulary第一章函数与极限Chapter 1 Function and Limit高等数学higher mathematics集合set元素element子集subset空集empty set并集union交集intersection差集difference of set基本集basic set补集complement set直积direct product笛卡儿积Cartesian product象限quadrant原点origin坐标coordinate轴axisx 轴x-axis整数integer有理数rational number实数real number开区间open interval闭区间closed interval半开区间half open interval有限区间finite interval区间的长度length of an interval无限区间infinite interval领域neighborhood领域的中心center of a neighborhood领域的半径radius of a neighborhood左领域left neighborhood右领域right neighborhood映射mappingX到Y的映射mapping of X onto Y满射surjection单射injection一一映射one-to-one mapping双射bijection算子operator变化transformation函数function逆映射inverse mapping复合映射composite mapping自变量independent variable因变量dependent variable定义域domain函数值value of function函数关系function relation值域range自然定义域natural domain单值函数single valued function多值函数multiple valued function单值分支one-valued branch函数图形graph of a function绝对值absolute value绝对值函数absolute value function符号函数sigh function整数部分integral part阶梯曲线step curve当且仅当if and only if (iff)分段函数piecewise function上界upper bound下界lower bound有界boundedness最小上界least upper bound无界unbounded函数的单调性monotonicity of a function 单调增加的increasing单调减少的decreasing严格递减strictly decreasing严格递增strictly increasing单调函数monotone function函数的奇偶性parity (odevity) of a function 对称symmetry偶函数even function奇函数odd function函数的周期性periodicity of a function周期period周期函数periodic function反函数inverse function直接函数direct function函数的复合composition of function复合函数composite function中间变量intermediate variable函数的运算operation of function基本初等函数basic elementary function初等函数elementary function线性函数linear function常数函数constant function多项式polynomial分段定义函数piecewise defined function阶梯函数step function幂函数power function指数函数exponential function指数exponent自然指数函数natural exponential function对数logarithm对数函数logarithmic function自然对数函数natural logarithm function三角函数trigonometric function正弦函数sine function余弦函数cosine function正切函数tangent function半角公式half-angle formulas反三角函数inverse trigonometric function常数函数constant function双曲线hyperbola双曲函数hyperbolic function双曲正弦hyperbolic sine双曲余弦hyperbolic cosine双曲正切hyperbolic tangent反双曲正弦inverse hyperbolic sine反双曲余弦inverse hyperbolic cosine反双曲正切inverse hyperbolic tangent最优化问题optimization problems不等式inequality极限limit数列sequence of number复利compound interest收敛convergence收敛的convergent收敛于a converge to a发散divergence发散的divergent极限的唯一性uniqueness of limits收敛数列的有界性boundedness of a convergent sequence子列 subsequence函数的极限 limits of functions函数当x 趋于0x 时的极限 limit of functions as x approaches 0x单侧极限 one-sided limit左极限 left limit右极限 right limit单侧极限 one-sided limits渐近线 asymptote水平渐近线 horizontal asymptote分式 fractions商定律 quotient rule无穷小 infinitesimal无穷大 infinity铅直渐近线 vertical asymptote夹逼准则 squeeze rule (Sandwich theorem)单调数列 monotonic sequence高阶无穷小 infinitesimal of higher order低阶无穷小 infinitesimal of lower order同阶无穷小 infinitesimal of the same order等阶无穷小 equivalent infinitesimal多项式的次数 degree of a polynomial三次函数 cubic function函数的连续性 continuity of a function增量 increment函数在0x 连续 the function is continuous at 0x左连续 left continuous / continuous from the left右连续 right continuous / continuous from the right连续性 continuity不连续性 discontinuity连续函数 continuous function函数在区间上连续 function is continuous on an interval不连续点 discontinuity point第一类间断点 discontinuity point of the first kind第二类间断点 discontinuity point of the second kind初等函数的连续性 continuity of the elementary functions定义区间 defined interval最大值 global maximum value (absolute maximum)最小值 global minimum value (absolute minimum)零点定理 the zero point theorem介值定理 intermediate value theorem第二章 导数与微分Chapter 2 Derivative and Differential 速度velocity速率speed平均速度average velocity瞬时速度instantaneous velocity匀速运动uniform motion平均速度average velocity瞬时速度instantaneous velocity圆的切线tangent line of a circle割线secant line切线tangent line位置函数position function导数derivative求导法differentiation可导的derivable可导函数differentiable function光滑曲线smooth curve变化率rate of change函数的变化率问题problem of the change rate of a function导函数derived function导数定义域domain of derivative左导数left-hand derivative右导数right-hand derivative单侧导数one-sided derivatives在闭区间[a, b]上可导is derivable on the closed interval [a, b] 指数函数的导数derivative of exponential function幂函数的导数derivative of power function切线的斜率slope of the tangent line截距interceptsx 截距x-intercept直线的斜截式slope-intercept equation of a line点斜式point-slope form切线方程tangent equation焦点focus角速度angular velocity成本函数cost function边际成本marginal cost逐项求导法differentiation term by term积之导数derivative of a product商之导数derivative of a quotient链式法则chain rule隐函数implicit function显函数explicit function隐函数求导法implicit differentiation加速度acceleration二阶导数second derivative三阶导数third derivative高阶导数nth derivative / higher derivative莱布尼茨公式Leibniz formula对数求导法log- derivative参数parameter参数方程parametric equation相关变化率correlative change rata微分differential微分学differential可微的differentiable函数的微分differential of function自变量的微分differential of independent variable微商differential quotient逼近法approximation用微分逼近approximation by differentials间接测量误差indirect measurement error绝对误差absolute error相对误差relative error第三章微分中值定理与导数的应用Chapter 3 MeanValue Theorems of Differentials and the Application ofDerivatives均值定理mean value theorem罗尔定理Roll’s theorem费马引理Fermat’s lemma拉格朗日中值定理Lagrange’s mean value theorem驻点stationary point稳定点stable point临界点critical point辅助函数auxiliary function拉格朗日中值公式Lagrange’s mean value formula柯西中值定理Cauchy’s mean value theorem洛必达法则L’Hospital’s Rule不定式indeterminate form“0”型不定式indeterminate form of type “”泰勒中值定理Taylor’s mean value theorem 泰勒公式Taylor formula系数coefficient余项remainder term线性近似linear approximation拉格朗日余项Lagrange remainder term麦克劳林公式Maclaurin’s formula佩亚诺余项Peano remainder term阶乘factorial凹凸性concavity上凹（或下凸）concave upward (concave up)下凹（或向上凸的）concave downward (concave down) 拐点inflection point极值extreme value函数的极值extremum of function极大与极小值maximum and minimum values极大值local (relative) maximum最大值global (absolute) maximum极小值local (relative) minimum最小值global (absolute) minimum目标函数objective function收入函数revenue function斜渐进线slant asymptote曲率curvature弧微分arc differential平均曲率average curvature曲率园circle of curvature曲率中心center of curvature曲率半径radius of curvature渐屈线evolute渐伸线involute根的隔离isolation of root隔离区间isolation interval切线法tangent line method第四章不定积分Chapter 4 Indefinite Integrals 原函数primitive function / antiderivative积分integration积分学integral积分号sign of integration被积函数integrand积分变量integral variable积分常数constant of integration积分曲线integral curve积分表table of integrals换元积分法integration by substitution有理代换法rationalizing substitution三角代换法trigonometric substitutions分部积分法integration by parts分部积分公式formula of integration by parts有理函数rational function真分式proper fraction假分式improper fraction部分分式partial fractions三角积分trigonometric integrals第五章定积分Chapter 5 Definite Integrals曲线下方之面积area under a curve曲边梯形trapezoid with曲边curve edge窄矩形narrow rectangle曲边梯形的面积area of trapezoid with curved edge积分下限lower limit of integral积分上限upper limit of integral积分区间integral interval分割partition黎曼和Riemannian sum积分和integral sum可积的integrableSimpson 法则逼近法approximation by Simpson’s rule梯形法则逼近法approximation by trapezoidal rule矩形法rectangle method曲线之间的面积area between curves积分中值定理mean value theorem of integrals函数在区间上的平均值average value of a function on an interval 牛顿－莱布尼茨公式Newton-Leibniz formula微积分基本公式fundamental formula of calculus微积分基本定理fundamental theorem of calculus变量代换change of variable换元公式formula for integration by substitution递推公式recurrence formula反常积分improper integral反常积分发散the improper integral is divergent反常积分收敛the improper integral is convergent无穷限的反常积分improper integral on an infinite interval无界函数的反常积分improper integral of unbounded functions瑕点flaw绝对收敛absolutely convergent第六章定积分的应用Chapter 6 Applications of the Definite Integrals 元素法the element method面积元素element of area平面plane平面图形的面积area of a plane figure直角坐标（又称“笛卡儿坐标”）Cartesian coordinates / rectangular coordinates x 坐标x-coordinate坐标轴coordinate axes极坐标polar coordinates极轴polar axis极点pole圆circle扇形sector抛物线parabola椭圆ellipse椭圆的轴axes of ellipse蚌线conchoid外摆线epicycloid双纽线lemniscate蚶线limacon旋转体solid of revolution, solid of rotation旋转体的面积volume of a solid of rotation旋转椭球体ellipsoid of revolution, ellipsoid of rotation曲线的弧长arc length of a curve可求长的rectifiable光滑smooth功work水压力water pressure引力gravitation变力variable force第七章空间解析几何与向量代数Chapter 7 Space Analytic Geometry and Vector Algebra纯量（标量）scalar向量vector自由向量free vector单位向量unit vector零向量zero vector相等equal平行parallel平行线parallel lines向量的线性运算linear operation of vector加法addition减法subtraction数乘运算scalar multiplication三角形法则triangle rule平行四边形法则parallelogram rule交换律commutative law结合律associative law分配律distributive law负向量negative vector三角不等式triangle inequality对角线diagonal差difference余弦定理（定律）law of cosines空间space空间直角坐标系space rectangular coordinates坐标平面coordinate plane卦限octant向量的模modulus of vector定比分点definite proportion and separated point中点公式midpoint formula等腰三角形isosceles triangle向量a与b的夹角angle between vector a and b方向余弦direction cosine方向角direction angle投影projection向量在轴上的投影projection of a vector onto an axis向量的分量components of a vector对称点symmetric point数量积（点积，内积）scalar product (dot product, inner product)叉积（向量积，外积）cross product (vector product, exterior product) 混合积mixed product锐角acute angle流体fluid刚体rigid body角速度angular velocity平行六面体parallelepiped平面的点法式方程point-norm form equation of a plane法向量normal vector平面的一般方程general form equation of a plane三元一次方程three-variable linear equation平面的截距式方程intercept form equation of a plane两平面的夹角angle between two planes点到平面的距离distance from a point to a plane空间直线的一般方程general equation of a line in space方向向量direction vector直线的点向式方程point-direction form equations of a line直线的对称式方程symmetric form equation of a line方向数direction number直线的参数方程parametric equations of a line两直线的夹角angle between two lines垂直perpendicular垂直线perpendicular lines直线与平面的夹角angle between a line and a planes 平面束pencil of planes平面束的方程equation of a pencil of planes行列式determinant系数行列式coefficient determinant曲面方程equation for a surface球面sphere球体spheroid球心ball center轨迹方程locus equation旋转轴rotation axis旋转曲面surface of revolution母线generating line圆锥面cone顶点vertex半顶角semi-vertical angle旋转双曲面revolution hyperboloids旋转单叶双曲面revolution hyperboloids of one sheet 旋转双叶双曲面revolution hyperboloids of two sheets 柱面cylindrical surface, cylinder圆柱circular cylinder圆柱面cylindrical surface准线directrix抛物柱面parabolic cylinder二次曲面quadric surface截痕法method of cut-off mark椭圆锥面elliptic cone椭球面ellipsoid双曲面hyperboloid单叶双曲面hyperboloid of one sheet双叶双曲面hyperboloid of two sheets旋转椭球面ellipsoid of revolution抛物面paraboloid椭圆体ellipsoid椭圆抛物面elliptic paraboloid旋转抛物面paraboloid of revolution双曲抛物面hyperbolic paraboloid马鞍面saddle surface椭圆柱面elliptic cylinder双曲柱面hyperbolic cylinder抛物柱面parabolic cylinder空间曲线space curve交线intersection curve空间曲线的一般方程general form equations of a space curve空间曲线的参数方程parametric equations of a space curve螺线spiral / helix螺矩pitch投影柱面projecting cylinder第八章多元函数微分法及其应用Chapter 8 Differentiation of Functions of Several Variables and Its Application 一元函数function of one variable二元函数binary function邻域neighborhood去心邻域noncentral neighborhood方邻域square neighborhood圆邻域circular neighborhood内点interior point外点exterior point边界点frontier point, boundary point聚点point of accumulation导集derived set开集open set闭集closed set连通集connected set开区域open region闭区域closed region有界集bounded set无界集unbounded setn维空间n-dimensional space多元函数function of several variables二重极限double limit多元函数的连续性continuity of function of several variables连续函数continuous function不连续点discontinuity point一致连续uniformly continuous偏导数partial derivative对自变量x的偏导数partial derivative with respect to independent variable x高阶偏导数partial derivative of higher order二阶偏导数second order partial derivative混合偏导数hybrid partial derivative全微分total differential偏增量partial increment偏微分partial differential全增量total increment可微分differentiable必要条件necessary condition充分条件sufficient condition叠加原理superposition principle全导数total derivative中间变量intermediate variable隐函数存在定理theorem of the existence of implicit function 光滑曲面smooth surface曲线的切向量tangent vector of a curve法平面normal plane向量方程vector equation向量值函数vector-valued function切平面tangent plane法线normal line方向导数directional derivative等高线level curve梯度gradient数量场scalar field梯度场gradient field向量场vector field势场potential field引力场gravitational field引力势gravitational potential曲面在一点的切平面tangent plane to a surface at a point曲线在一点的法线normal line to a surface at a point无条件极值unconditional extreme values鞍点saddle point条件极值conditional extreme values拉格朗日乘数法Lagrange multiplier method拉格朗日乘子Lagrange multiplier经验公式empirical formula最小二乘法method of least squares均方误差mean square error第九章重积分Chapter 9 Multiple Integrals重积分multiple integrals二重积分double integral可加性additivity累次（逐次）积分iterated integral体积元素volume element二重积分变量代换法change of variable in double integral极坐标表示的面积area in polar coordinates扇形的面积area of a sector of a circle极坐标二重积分double integral in polar coordinates三重积分triple integral直角坐标系中的体积元素volume element in rectangular coordinate system 柱面坐标cylindrical coordinates柱面坐标系中的体积元素volume element in cylindrical coordinate system 球面坐标spherical coordinates球面坐标系中的体积元素volume element in spherical coordinate system 剥壳法shell method圆盘法disk method反常二重积分improper double integral曲面的面积area of a surface质心center of mass静矩static moment密度density形心centroid转动惯量moment of inertia参变量parametric variable第十章曲线积分与曲面积分Chapter 10 Line (Curve) Integrals and Surface Integrals对弧长的曲线积分line integrals with respect to arc length第一类曲线积分line integrals of the first type对坐标的曲线积分line integrals with respect to x, y, and z第二类曲线积分line integrals of the second type有向曲线弧directed arc单连通区域simple connected region复连通区域complex connected region路径无关path independence格林公式Green formula顺时针方向clockwise逆时针方向counterclockwise区域边界的正向positive direction of region boundary第一类曲面积分surface integrals of the first type旋转曲面的面积area of a surface of a revolution对面积的曲面积分surface integrals with respect to area有向曲面directed surface对坐标的曲面积分surface integrals with respect to coordinate elements第二类曲面积分surface integrals of the second type有向曲面之元素element of directed surface高斯公式gauss formula拉普拉斯算子Laplace operator拉普拉斯变换Laplace transform格林第一公式Green’s first formula通量flux散度divergence斯托克斯公式Stokes formula环流量circulation旋度rotation (curl)第十一章无穷级数Chapter 11 Infinite Series一般项general term部分和partial sum收敛级数convergent series余项remainder term等比级数geometric series几何级数geometric series公比common ratio调和级数harmonic series柯西收敛准则Cauchy convergence criteria, Cauchy criteria for convergence 正项级数series of positive terms达朗贝尔判别法D’Alembert test柯西判别法Cauchy test交错级数alternating series绝对收敛absolutely convergent条件收敛conditionally convergent柯西乘积Cauchy product函数项级数series of functions发散点point of divergence收敛点point of convergence收敛域convergence domain和函数sum function幂级数power series幂级数的系数coefficients of power series阿贝尔定理Abel Theorem收敛半径radius of convergence收敛区间interval of convergence幂级数的导数derivative of power series泰勒级数Taylor series麦克劳林级数Maclaurin series二项展开式binomial expansion近似计算approximate calculation舍入误差round-off error (rounding error)欧拉公式Euler’s formula魏尔斯特拉斯判别法Weierstrass test三角级数trigonometric series振幅amplitude角频率angular frequency初相initial phase矩形波square wave谐波分析harmonic analysis直流分量direct component基波fundamental wave二次谐波second harmonic三角函数系trigonometric function system傅立叶系数Fourier coefficient傅立叶级数Fourier series周期延拓periodic prolongation正弦级数sine series余弦级数cosine series奇延拓odd prolongation偶延拓even prolongation傅立叶级数的复数形式complex form of Fourier series第十二章微分方程Chapter 12 Differential Equation解微分方程solve a differential equation常微分方程ordinary differential equation (ODE)偏微分方程partial differential equation (PDE)微分方程的阶order of a differential equation微分方程的解solution of a differential equation微分方程的通解general solution of a differential equation初始条件initial condition微分方程的特解particular solution of a differential equation初值问题initial value problem微分方程的积分曲线integral curve of a differential equation 可分离变量的微分方程variable separable differential equation 隐式解implicit solution隐式通解implicit general solution衰变系数decay coefficient衰变decay齐次方程homogeneous equation一阶线性方程linear differential equation of first order非齐次non-homogeneous齐次线性方程homogeneous linear equation非齐次线性方程non-homogeneous linear equation常数变易法method of variation of constant暂态电流transient state current稳态电流steady state current伯努利方程Bernoulli equation全微分方程total differential equation积分因子integrating factor高阶微分方程differential equation of higher order悬链线catenary高阶线性微分方程linear differential equation of higher order自由振动的微分方程differential equation of free vibration强迫振动的微分方程differential equation of forced oscillation串联电路的振荡方程oscillation equation of series circuit二阶线性微分方程second order linear differential equation线性相关linearly dependence线性无关linearly independence二阶常系数齐次线性微分方程second order homogeneous linear differential equation with constant coefficient二阶变系数齐次线性微分方程second order homogeneous linear differential equation with variable coefficient特征方程characteristic equation无阻尼自由振动的微分方程differential equation of free vibration with zero damping固有频率natural frequency简谐振动simple harmonic oscillation, simple harmonic vibration微分算子differential operator待定系数法method of undetermined coefficient共振现象resonance phenomenon欧拉方程Euler equation幂级数解法power series solution数值解法numerical solution勒让德方程Legendre equation微分方程组system of differential equations常系数线性微分方程组system of linear differential equations with constant coefficient。
微积分英文版课件
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定理 . 原函数都在函数族
证: 1)
( C 为任意常数 ) 内 .
即
又知
[(x) F(x)] (x) F(x) f (x) f (x) 0
故
(x) F(x) C0 (C0 为某个常数)
即 (x) F(x) C0 属于函数族 F(x) C .
( k 为常数)
(2)
x dx
1
1
x
1
C
( 1)
(3)
dx x
ln
x
C
x 0时 ( ln x ) [ ln(x) ] 1
x
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(4)
1
dx x
2
arctan
x
C
或 arccot x C
(5)
dx arcsin x C 1 x2
或 arccos x C
想到公式
1
d
u u
2
arctan u C
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例. 求 解:
dx a 1 (ax)2
d
(
x a
)
1
(
x a
)2
想到
d u arcsinu C 1u2
f [(x)](x)dx f ((x))d(x)
(直接配元)
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例4. 求 解:
例1. 求
解: 令 u ax b ,则 d u adx , 故
原式 = um 1 d u 1 1 um1 C a a m1
注: 当
时
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微积分双语词汇
MATHEMATICAL TERMS (Part 1)calculus 微积分definition 定义theorem 定理lemma 引理corollary推论prove 证明proof 证明show 证明solution 解formula 公式if and only if ( iff ) 当且仅当x X for all x X x X there exists an x Xsuch that 使得given 已知set集合finite set有限集infinite set 无限集interval区间open interval开区间closed interval 闭区间neighborhood 邻域number 数natural number 自然数integer 整数odd number 奇数even number 偶数real number 实数rational number 有理数irrational number 无理数positive number 正数negative number 负数mapping 映射function 函数monotone function 单调函数increasing function 增函数decreasing function 减函数bounded function 有界函数odd function 奇函数even function 偶函数periodic function 周期函数composite function 复合函数inverse function 反函数domain 定义域range 值域variable 变量independent variable自变量dependent variable因变量sequence 数列convergent sequence收敛数列divergent sequence 发散数列bounded sequence 有界数列decreasing sequence 递减数列increasing sequence 递增数列limit极限one-sided limit 单侧极限left-hand limit 左极限right-hand limit 右极限The Squeeze Theorem 夹逼定理infinity 无穷大infinitesimal 无穷小equivalent infinitesimal 等价无穷小infinitesimal of higher order 高阶无穷小order of infinitesimal 无穷小的阶infinitesimals of the same order 同阶无穷小increment 增量continuous function 连续函数continuity 连续性f(x) is continuous at x 在x连续f(x) is discontinuous at x 在x间断discontinuity 间断点discontinuity of the first (second) kind 第一（二）类间断点removable discontinuity 可去间断点jump discontinuity 跳跃间断点infinite discontinuity 无穷间断点intermediate value 介值The Intermediate Value Theorem 介值定理zero point 零点The Zero Point Theorem 零点定理root 根equation 方程uniform continuity 一致连续derivative 导数rate of change 变化率velocity 速度instantaneous velocity 瞬时速度tangent (line) 切线normal (line) 法线slope 斜率left-hand derivative 左导数right-hand derivative 右导数f(x) is differentiable at x f(x) 在x处可导(可微) differentiation 求导The Chain Rule 链式法则differentiation formulas 求导公式implicit function 隐函数explicit function 显函数implicit differentiation 隐函数求导logarithm 对数Logarithmic differentiation 对数求导法parameter 参数parametric equation 参数方程parametric curve 参数曲线hyperbolic function 双曲函数hyperbolic sine 双曲正弦hyperbolic cosine 双曲余弦hyperbolic tangent 双曲正切hyperbolic cotangent 双曲余切MATHEMATICAL TERMS (Part 2)differential 微分differential quotient 微商approximate value 近似值error 误差relative error 相对误差absolute error 绝对误差invariance of differential form 微分形式不变性higher derivative 高阶导数first derivative 一阶导数second derivative 二阶导数third derivative 三阶导数nth derivative n阶导数twice differentiable 二阶可导acceleration 加速度mean value 中值The Mean Value Theorem 中值定理Rolle’s Theorem 罗尔定理Lagrange’s Mean Value Theorem 拉格朗日中值定理Cauchy’s Mean Value Theorem 柯西中值定理equality 等式inequality 不等式indeterminate form 不定型(未定式) indeterminate form of type ( ) ( )型未定式L’Hospital’s Rule 洛必达法则Taylor’s formula 泰勒公式polynomial 多项式nth-degree polynomial n次多项式remaind 余项Lagrange’s form of remainder 拉格朗日型余项Peano’s form of remainder 皮亚诺型余项Maclaurin formula 麦克劳林公式Taylor polynomial 泰勒多项式Mauclaurin polynomial 麦克劳林多项式polynomial approximation 多项式逼近accuracy 精确度margin 边际marginal cost 边际成本marginal revenue 边际收益elasticity 弹性density 密度mass 质量extreme value 极值local maximum value 极大值local minimum value 极小值(absolute) maximum 最大值(absolute) minimum 最小值stationary point 驻点（稳定点）critical point 临界点The First (Second) Derivative Test （极值的）一（二）阶判别法convex 凸的convex curve 凸曲线concave 凹的concave curve 凹曲线convex function 凸函数point of inflection 拐点asymptote 渐近线horizontal asymptote 水平渐近线vertical asymptote 垂直渐近线slant asymptote 斜渐近线curve sketching 作图sketch a curve 作图curvature 曲率The bisection method 二分法The secant method 弦位法Newton’s method 牛顿（切线）法The tangent method 切线法differential calculus 微分学integral 积分integral calculus 积分学definite integral 定积分indefinite integral 不定积分partition 分割Riemann sum 黎曼和integral sign 积分符号integrand 被积函数upper (lower) limit of integration 积分上（下）限integration 积分（求积）integrable 可积的f(x) is integrable on [a, b] integrable function 可积函数integrability 可积性sufficient condition 充分条件necessary condition 必要条件piecewise continuous 分段连续property 性质The mean value theorem of integral 积分中值定理The fundamental theorem of calculus 微积分基本定理Newton-Leibniz formula primitive function (anti-derivative) 原函数（反导数）The substitution rule for integration 换元积分法The inverse of the chain rule 反链式法（凑微分法）integration by parts 分部积分法rational function 有理函数fraction 分式irreducible fraction 最简分式partial fraction 部分分式partial fraction decomposition 部分分式分解MATHEMATICAL TERMS (Part 3)vector 矢量free vector 自由矢量zero vector 零矢量magnitude of a vector 矢量的模unit vector 单位矢量scalar product 数量积dot product 点积vector product 矢量积cross product 叉积a is perpendicular (orthogonal) tob a与b 垂直coordinate 坐标coordinate system 坐标系coordinate axis 坐标轴x-axis x轴coordinate plane 坐标面direction angle 方向角direction cosine 方向余弦rectangular coordinate system 直角坐标系octant 卦限the first octant 第一卦限variable 变量function of two (three) variables 二（三）元函数function of several variables 多元函数independent variable 自变量dependent variable 因变量domain 定义域range 值域set of points 点集neighborhood 邻域interior point 内点boundary point 边界点bound 边界open set 开集closed set 闭集connected set 连通集region 区域open region 开区域closed region 闭区域bounded region 有界区域unbounded region 无界区域cluster point 聚点double limit 二重极限iterated limit 累次极限continuity 连续性increment 增量total increment 全增量partial increment 偏增量partial derivative 偏导数partial derivative of f(x,y) with respect to x ( y ) f(x,y)关于x（y）的偏导数higher partial derivative 高阶偏导数mixed partial derivative 混合偏导数Laplace equation 拉普拉斯方程total differential 全微分differentiable 可微chain rule 链式法则implicit function 隐函数implicit differentiation 隐函数微分法Jacobian determinant 雅可比行列式curve 曲线space curve 空间曲线tangent vector 切矢tangent line 切线normal plane 法平面surface 曲面normal vector 法矢normal line 法线tangent plane 切平面sphere 球面cylinder 柱面cone 锥面directional derivative 方向导数gradient 梯度gradient vector 梯度矢量Ñf delf level curve 等值线level surface 等值面local extremum 极值local maximum 极大值local minimum 极小值extreme value 最值absolute maximum (minimum) 最大（最小）值stationary point (critical point) 驻点（临界点）conditional extremum 条件极值Lagrange multiplier 拉格朗日乘数method of Lagrange multiplier 拉格朗日乘数法objective function 目标函数constraint 约束条件method of least square 最小二乘法field 场scalar field 数量场vector field 矢量场gradient field 梯度场potential field 势场potential function 势函数conservative field 保守场gravitational field 引力场force field 力场velocity field 速度场MATHEMA TICAL TERMS(Part 4)multiple integral 重积分double integral 二重积分iterated integral 累次积分region 区域region of integration 积分区域type X (Y) region X（Y）型区域order of integration 积分秩序reverse the order of integration 交换积分秩序polar coordinates 极坐标double integrals in polar coordinates 极坐标下的二重积分volume 体积lamina 平面薄片mass 质量density 密度moment about x-axis 关于x轴的（静）力矩center of mass 重心moment of inertia 转动惯量surface 曲面area of a surface 曲面的面积triple integral 三重积分rectangle 矩形rectangular coordinates 直角坐标系cylinder 柱面cylindrical coordinates 柱面坐标系sphere 球面spherical coordinates 球面坐标系change of variables in multiple integrals 重积分的变量替换Jacobian determinant 雅可比行列式line integral 曲线积分line integral with respect to arc length 对弧长的曲线积分（第一型）line integral with respect to x ( y ) 对坐标x（y）曲线积分（第二型）line integral of a vector field 向量场的曲线积分smooth curve 光滑曲线piecewise smooth curve 逐段光滑曲线oriented curve 有向曲线orientation of a curve 曲线的方向work 功the line integral is independent of path 曲线积分与路径无关connected region 连通区域simply-connected region 单连通区域closed curve 闭曲线Green’s theorem 格林定理（公式）positive orientation of a curve 曲线的正向Fundamental theorem for line integrals 曲线积分的基本定理surface integral 曲面积分surface integral of a scalar field 数量场的曲面积分（第一型）surface integral of a vector field 向量场的曲面积分（第二型）orientable surface 可定向曲面oriented surface 有向曲面Möbius strip 莫比乌斯带Klein bottle 克莱因瓶one-sided surface 单侧曲面two-sided surface 双侧曲面closed surface 闭曲面flux 流量、通量electric flux 电通量divergence 散度rotation (curl) 旋度Gauss’theorem 高斯定理（公式）The divergence theorem 散度定理（公式）Stokes’theorem 斯托克斯定理（公式）curl theorem 旋度定理（公式）circulation of v around L v沿L的环流量Hamilton operator 哈密顿算子harmonic field 调和场。
TB_chapter2
1. The point ()2,4P lies on the curve x y =. If Q is the point (x , use your calculator to find the slope of the secant line PQ (correct to six decimal places) for the value .99.3=x2. The displacement (in meters) of an object moving in a straight line is given by 212/4s t t =++, where t ismeasured in seconds. Find the average velocity over the time period [1,3].3.Find the limit.()379lim 25++→x x x4.Find the limit.()51lim 20+-→x x x x5. If ,22)(12++≤≤x x x f for all x find the limit.)(lim 1x f x -→6. Find the limit. 22lim |2|x x x →--7. Evaluate the limit.()xx x 11022lim --→-+8. Use the definition of the derivative to find (2)f '-, where 3()2f x x x =-.9. Find an equation of the tangent line to curve 32y x x =-at the point (2,4).10. Use a graph to find a number N such that 3.031235622<---+x x x whenever N x >.11. If ()g x ()g x '.12. A machinist is required to manufacture a circular metal disk with area 21000cm .a) What radius produces such a disk? b) If the machinist is allowed an error tolerance of 25cm ±in the area of the disk, how close to the idealradius in part (a) must the machinist control the radius?13. Use a graph to find a number δ such that6.0314<-+x whenever .2δ<-x14. For the limit, illustrate the definition by finding values of δ that correspond to .25.0=ε31lim(43)2x x x →+-=15. Determine where f is discontinuous.()20()30333if x f x x if x x if x ⎧<⎪⎪=-≤<⎨⎪->⎪⎩16. For x = 5, determine whether f is continuous from the right, from the left, or neither.17. If a cylindrical tank holds 100,000 gallons of water, which can be drained from the bottom of the tank in an hour,then Torricelli's Law gives the volume of water remaining in the tank after t minutes as2651000,100)(⎪⎭⎫ ⎝⎛-=t t V , 600≤≤tFind the rate at which the water is flowing out of the tank (the instantaneous rate of change of V with respect to t ) as a function of t .18.Find the derivative of the function.25314)(x x x f +-=19. If 2313)(tt f += find )(t f '.20. At what point is the function ()|6|f x x =- not differentiable.ANSWER KEYStewart - Calculus ET 6e Chapter 2 Form A1. 0.2501562. 3/m s3. 2634. -∞5. 16. Limit does not exist7. -1/48. 109. 1016y x =-10. 9≥N11.(),3/5-∞12. cm , 0.0445cm13. 81.0≤δ14. 030.0≤δ15. 03at and16. neither17. ⎪⎭⎫ ⎝⎛--=65165200000t y 18. 310-x19. )3(26t t +- 20. 61.The point P (4, 2) lies on the curve .x y = If Q is the point ()x x ,, use your calculator to find the slope of the secant line PQ (correct to six decimal places) for the value of .01.4=x2.The displacement (in meters) of an object moving in a straight line is given by 212/4s t t =++, where t is measured in seconds. Find the average velocity over the time period [1,1.5].3.Find the limit, if it exists.44lim |4|x x x →-- 4. Find )(a f '.252)(x x x f -+=5. Find an equation of the tangent line to curve 32y x x =-at the point (2,4).6. Evaluate the function 222)(--=x x x f at the given numbers (correct to six decimal places). Use the results to guess the value of the limit ).(lim 2x f x →7. The graph of f is given. State the numbers at which f is not differentiable.⎪⎭⎫ ⎝⎛→x x x 3cos lim 909. If 66)(12++≤≤x x x f for all x find the limit.)(lim 1x f x -→10.Evaluate the limit.()867lim 25++→x x x11. Evaluate the limit.()x x x 11022lim --→-+12.If an arrow is shot upward on the moon, with a velocity of 70 m/s its height (in meters) after t seconds is given by .99.070)(2t t t H -= With what velocity will the arrow hit the moon?13. The cost (in dollars) of producing x units of a certain commodity is .08.013336,4)(2x x x C ++= Find the average rate of change with respect to x when the production level is changed from 101=x to .103=x14.Let ()20()30333if x f x xif x x if x ⎧<⎪⎪=-≤<⎨⎪->⎪⎩Evaluate each limit, if it exists.0lim ()x f x +→ b.) 0lim ()x f x -→15.If f and g are continuous functions with 3)3(=f and []3)()(3lim 3=-→x g x f x , find ).3(g16.Evaluate the limit. 9lim 9+-→x x17.Find a number δsuch that if |2|x δ-<, then |48|x ε-<, where 0.1ε=.then Torricelli's Law gives the volume of water remaining in the tank after t minutes as2651000,100)(⎪⎭⎫ ⎝⎛-=t t V , 600≤≤tFind the rate at which the water is flowing out of the tank (the instantaneous rate of change of V with respect to t ) as a function of t .19. If ()g x ()g x '.20. For the function f whose graph is shown, state the following.)(lim 4x f x -→ANSWER KEYStewart - Calculus ET 6e Chapter 2 Form B1. 0.2498442.2.625 m/s3. Limit does not exist4. a 101-5.1016y x =- 6.(1.6, 0.7465), (1.8, 0.7257), (1.9, 0.7161), (1.99, 0.7079), (1.999, 0.707195), (2.4, 0.674899), (2.2, 0.690261), (2.1, 0.698482), (2.01, 0.706225), (2.001, 0.707018), Limit = 0.707107 7. 1,0,3-8. 09. 110. 21311. -1/412. -7013. 29.3214. a.) 3 b.) 015. 616. 017. 0.025δ=18. ⎪⎭⎫ ⎝⎛--=65165200000t y 19. (),3/5-∞20. -∞1. A cardiac monitor is used to measure the heart rate of a patient after surgery. It compiles the number of heartbeats after t minutes. When the data in the table are graphed, the slope of the tangent line represents the heart rate inbeats per minute. The monitor estimates this value by calculating the slope of a secant line. Use the data to estimate the patient's heart rate after 42 minutes using the secant line between the points with t = 38 and t = 42.Select the correct answer.a. -89b. 180c. 90d. 100e. 89f. 952. If an arrow is shot upward on the moon with a velocity of 55 m/s, its height in meters after t seconds is given by .04.0552t t h -= Find the average velocity over the interval [1, 1.04].Select the correct answer.a. 54.9194b. 55.0284c. 54.8174d. 54.9184e. 54.90843. The displacement (in feet) of a certain particle moving in a straight line is given by 3/8s t = where t is measuredin seconds. Find the average velocity over the interval [1, 1.8].Select the correct answer.a. 0.865b. 0.654c. 0.765d. 0.756e. 0.745f. 0.7554. For the function f whose graph is shown, find the equations of the vertical asymptotes.Select all that apply.a. x = -7b. x = 9c. x = 5d. x = -3e. x = 10f. x = -25. Find the limit, if it exists55lim |5|x x x →--Select the correct answer.a. 5b. 1-c. 1-d. 0e. limit does not exist6. Find the limit.lim x →-∞Select the correct answer.a. -1/2b. 3c. 3-d. 0e. limit does not exist7. Evaluate the limit.()()62lim 231-+→x x xSelect the correct answer.a. 27b. -45c. -135d. 29e. -1258.If 88)(12++≤≤x x x f for all x , find )(lim 1x f x -→. Select the correct answer.a. 1b. 8c. -1/8d. -1/16e. The limit does not exist9. Evaluate the limit.⎪⎭⎫ ⎝⎛→x x x 5cos lim 90Select the correct answer.a. -5b. 1c. 0d. 5e. The limit does not exist10. Use a graph to find a number δ such that 2.021sin <-x whenever δπ<-6x .Round down the answer to the nearest thousandth.Select the correct answer.a. 218.0≤δb. 368.0≤δc. 401.0≤δd. 251.0≤δe. 425.0≤δ11. A machinist is required to manufacture a circular metal disk with area 1000 cm 2. If the machinist is allowed an error tolerance of ±10 cm 2 in the area of the disk, how close to the ideal radius must the machinist control the radius?Round down the answer to the nearest hundred thousandth.Select the correct answer.a. cm 08898.0≤δb. cm 08908.0≤δc. cm 08999.0≤δd. cm 08913.0≤δe. cm 09913.0≤δ12. Consider the function x e x f /121)(+=. Find the value of -→0)(lim x x f . Select the correct answer.a. 1.5b. -0.1c. 0.1d. 0.9e. 0.513.Choose an equation from the following that expresses the fact that a function f is continuous at the number 6.Select the correct answer.a. 6)(lim =∞→x x fb. )6()(lim 6f x f x =→c. )6()(lim f x f x =∞→d. 0)(lim 6=→x x fe. ∞=→6)(lim x x f14. Determine where f is discontinuous.()20()30333if x f x x if x x if x ⎧<⎪⎪=-≤<⎨⎪->⎪⎩Select the correct answer.a. 03andb. 0onlyc. 3onlyd. 03and -e. 3only -15. Use the definition of the derivative for find (2)f '-, where 3()2f x x x =-.Select the correct answer.a. 4b. 10c. -4d. -10e. none of these16.If ()g x ()g x '.Select the correct answer.a. ()(),00,-∞⋃∞b. [3/5,3/5]-c. [,3/5)-∞d. (),3/5-∞e. ()0,∞17. Find an equation of the tangent line to the curve 353+-=x x y at the point (2, 1).Select the correct answer.a. 138+=x yb. 139--=x yc. 137-=x yd. 137+-=x ye. 157-=x yStewart - Calculus ET 6e Chapter 2 Form C18. The cost (in dollars) of producing x units of a certain commodity is 201.019571,4)(x x x C ++=. Find theinstantaneous rate of change with respect to x when x = 103. (This is called the marginal cost .)Select the correct answer.a. 26.06b. 20.06c. 21.06d. 18.06e. 31.0619. If the tangent line to )(x f y = at (8, 4) passes through the point (5, -32), find )8(f '.Select the correct answer.a. 24)8(='fb. 20)8(='fc. 12)8(-='fd. 12)8(='fe. 32)8(='f20. At what point is the function ()|6|f x x =- not differentiable.Select the correct answer.a. 6b. 6-c. 1d. 1-e. 0ANSWER KEYStewart - Calculus ET 6e Chapter 2 Form C1. c2. d3. f4.a, b, c, f5. e6. a7. c8. a9. c10. a11. a12. e13. b14. a15. b16. d17. c18. c19. d20. a1. The position of a car is given by the values in the following table.Find the average velocity for the time period beginning when t = 2 and lasting 2 seconds.Select the correct answer.a. 35.5b. 47.5c. 39d. 37.5e. 33.52. The displacement (in meters) of an object moving in a straight line is given by 212/4s t t =++, where t ismeasured in seconds. Find the average velocity over the time period [1,3].Select the correct answer.a. 3/m sb. 3.5/m sc. 1/m sd. 1.5/m se. none of these3. Find the limit.()71lim 20++→x x x xSelect the correct answer.a. 0b. 71c. 71- d. -∞ e. ∞4. Find the limit.lim x →-∞Select the correct answer.a. -1b. 0c. 1/2d. -∞e. -1/25.The slope of the tangent line to the graph of the exponential function xy 8= at the point (0, 1) is x x x 18lim 0-→. Estimate the slope to three decimal places.Select the correct answer.a. 1.293b. 2c. 2.026d. 1.568e. 2.079f. 2.5566. Find an equation of the tangent line to curve 32y x x =-at the point (2,4).Select the correct answer.a. 1610y x =-b. 108y x =-c. 16y x =-d. 1016y x =+e. none of these7.Find the limit.()10lim tan 1/x x +-→Select the correct answer.a. 0b. ∞c. /2πd. /3πe. π .8. Let |1|1)(2--=x x x F . Find the following limits.),(lim 1x F x +→ )(lim 1x F x -→Select the correct answer.a. both 2b. 2 and 1c. 2 and – 2d. 2 and – 1e. both 19. Use continuity to evaluate the limit.()x x x sin 4sin lim 13+→πSelect the correct answer.a. π13b. - 1c. 0d. ∞e. 110.Let ()20()30333if x f x xif x x if x ⎧<⎪⎪=-≤<⎨⎪->⎪⎩Evaluate the limit, if it exists.0lim ()x f x -→Select the correct answer.e. 3-11. For what value of the constant c is the function f continuous on ()?,∞∞-⎩⎨⎧>-≤+=2527)(2x for cx x for cx x fSelect the correct answer.a. 1=cb. 2=cc. 6=cd. 2-=ce. 7=c12. Find a function g that agrees with f for 25≠x and is continuous on ℜ.xx x f --=255)(Select the correct answer.a. x x g -=51)(b. x x g +=251)(c. x x g +=51)(d. xx g -=51)( e. x x g -=55)(13.Which of the given functions is discontinuous?Select the correct answer.a. 5,5,3121)(<≥⎪⎩⎪⎨⎧-=x x x x fb. 5,5,351)(=≠⎪⎩⎪⎨⎧-=x x x x fc. Both functions are continuous14.Find the limit. 13lim 232-++∞→t t t tSelect the correct answer.a. ∞b. 0c. 3-d. 3e. 215.If ()g x ()g x '.Select the correct answer.a. ()(),00,-∞⋃∞b. [3/5,3/5]-c. [,3/5)-∞d. (),3/5-∞e. ()0,∞16. The cost (in dollars) of producing x units of a certain commodity is .03.013280,4)(2x x x C ++= Find theaverage rate of change with respect to x when the production level is changed from x = 102 to x = 118.Select the correct answer.a. 29.6b. 19.6c. 18.6d. 26.6e. 24.617. Evaluate the limit.|2|lim 2+-→x xSelect the correct answer.a. 2b. 4c. - 2d. 0e. The limit does not exist18. If a ball is thrown into the air with a velocity of 58 ft/s, its height (in feet) after t seconds is given by .11582t t H -=Find the velocity when t = 4.Select the correct answer.a. 27ft/sb. 30ft/sc. 31ft/sd. 25ft/se. 37ft/s19. Is there a number a such that 626lim 223-++++-→x x a ax x x exists? If so, find the value of a and the value of the limit. Select the correct answer.a. a =14, limit equals 1.4b. a =17, limit equals 1.6c. a =28, limit equals 1.4d. a =28, limit equals 1.6e. There is no such number20.If ()g x ()g x '.Select the correct answer.a. ()1/25()352g x x -'=-- b. ()1/21()352g x x '=-- c. ()2()35g x x '=-- d. ()25()352g x x -'=-- e. none of theseANSWER KEYStewart - Calculus ET 6e Chapter 2 Form D1. d2. a3. e4. e5. e6. e7. c8. c9. c10. a11. c12. c13. b14. b15. d16. b17. d18. b19. d20. a1.A tank holds 1000 gallons of water, which drains from the bottom of the tank in half an hour. The values in the table show the volume V of water remaining in the tank (in gallons) after t minutes. If P is the point (15, 263) on thegraph of V, fill the table with the slopes of the secant lines PQ where Q is the point on the graph with the corresponding t .Enter your answer to two decimal places.2. The displacement (in meters) of an object moving in a straight line is given by 212/4s t t =++, where t ismeasured in seconds. Find the average velocity over the time period [1,1.5].3. If an arrow is shot upward on the moon with a velocity of 57 m/s, its height in meters after t seconds is given by 282.057t t h -=. Find the instantaneous velocity after one second.Select the correct answer.a. 55.46b. 55.35c. 55.25d. 55.36e. 55.374. Given that, 3)(lim 7-=→x f x and 9)(lim 7=→xg x . Evaluate the limit.)()()(2lim 7x f x g x f x -→5. Find an equation of the tangent line to curve 32y x x =-at the point (2,4).Select the correct answer.a. 1610y x =-b. 108y x =-c.16y x =- d. 1016y x =+ e. none of these6. Let ()20()30333if x f x x if x x if x ⎧<⎪⎪=-≤<⎨⎪->⎪⎩Evaluate the limit, if it exists.0lim ()x f x -→Select the correct answer.e. 3-7. For the function f whose graph is shown, find the following.)(lim 7x f x →8.For x = 5, determine whether f is continuous from the right, from the left, or neither.9. Evaluate the limit.()xx x 11077lim --→-+10. Let |1|1)(2--=x x x FFind the following limits.)(lim ),(lim 11x F x F x x -+→→11. Use a graph to find a number δsuch that 3|0.6< whenever |2|x δ-<.Round down the answer to the nearest hundredth.12. Is there a number a such that 6810lim 223-++++-→x x a ax x x exists? If so, find the value of a and the value of the limit.Select the correct answer.a. a =49, limit equals 1.6b. a =13, limit equals 2.2c. a =49, limit equals 2.2d. a =19, limit equals 1.6e. a =49, limit equals 2.713. How close to 2 do we have to take x so that 5x + 3 is within a distance of 0.025 from 13?14. Find a function g that agrees with f for 25≠x and is continuous on .ℜxx x f --=255)( 15. Use the given graph of x x f =)( to find a number δ such that 4.0|2|<-x whenever .|4|δ<-x16.If ()g x ()g x '.17.If ()g x ()g x '.Select the correct answer.a. ()(),00,-∞⋃∞b. [3/5,3/5]-c. [,3/5]-∞d. (),3/5-∞e. ()0,∞18. At what point is the function |6|)(x x f -= not differentiable.19. How close to - 9 do we have to take x so that ()?10000914>+x20.Find the derivative of the function.25314)(x x x f +-=ANSWER KEYStewart - Calculus ET 6e Chapter 2 Form E1. 5, -42.3, 10, -43.6,20, -18.4, 25, -24.1, 30, -17.532. 2.625 m/s3.d 4.-1/2 5. e6. a7.-∞ 8.neither 9. -1/4910. 2, -211. 81.0≤δ12. c13. 005.0|2|<-x14. ()x g +=5115. 44.1≤δ 16. ()1/25()352g x x -'=-- 17. d18. 619. 1.0|9|<+x20. 310-x1. The displacement (in meters) of an object moving in a straight line is given by 212/4s t t =++, where t ismeasured in seconds. Find the average velocity over the time period [1,1.5].2. If a ball is thrown into the air with a velocity of 45 ft/s, its height in feet after t seconds is given by 21545t t y -=. Find the instantaneous velocity when 4=t .3. If 5.4)(lim 3=-→x f x , then if )(lim 3x f x → exists, to what value does it converge?Select the correct answer.a. 6.5b. 4.5c. 1d. 2e. 64. For the function f whose graph is shown, find the limit.)(lim 9x f x +-→5. The function has been evaluated at the given numbers (correct to six decimal places). Use the results to guess the value of the limit.112)(--=x x x f________)(lim 1=→x f xSelect the correct answer.a. 1.255039b. 1.911314c. 1.969944d. 1.473889e. 16.Evaluate the limit.()()104lim 251-+→x x x7. Find the limit.lim x →-∞8.Find the limit.()10lim tan 1/x x +-→9.Evaluate the limit and justify each step by indicating the appropriate properties of limits.393198lim 22-++-∞→x x x x x10. Find an equation of the tangent line to the curve 34x y =at the point ()256,4--.11. Find a number δsuch that if |2|x δ-<, then |48|x ε-<, where 0.1ε=.12. Use a graph to find a number δsuch that 1.021sin <-x whenever δπ<-6x .Round down the answer to the nearest thousandth.13. Use the definition of the limit to find values of δ that correspond to 75.0=ε.Round down the answer to the nearest thousandth.()234lim 31=-+→x x x14. Determine where f is discontinuous.()20()30333if x f x x if x x if x ⎧<⎪⎪=-≤<⎨⎪->⎪⎩15. If f and g are continuous functions with 2)2(=f and [],2)()(2lim 2=-→x g x f x find )2(g .16. Find the limit.)(lim 22bx x ax x x +-+∞→17.State the domain.()sin F x =18.Find the derivative of the function using the definition of derivative.22919)(x x x f +-=19. Find a function g that agrees with f for 4≠x and is continuous on ℜ.xx x f --=42)(20. If an arrow is shot upward on the moon, with a velocity of 70 m/s its height (in meters) after t seconds is given by.99.070)(2t t t H -= With what velocity will the arrow hit the moon?ANSWER KEYStewart - Calculus ET 6e Chapter 2 Form F1. 2.625 m/s2. -753. b4. -∞5. e6. 28125-7. -1/28.2π 9. 38 10. 512192+=x y11. 0.025δ=12. 112.0≤δ13. 085.0≤δ14.03at and 15.2 16. 2b a - 17. ),6[∞18.94-x 19.x g +=2120.-701. If 5.4)(lim 2=-→x f x , then if )(lim 2x f x →exists, to what value does it converge?Select the correct answer.a. 2b. 1c. 5d. 4.5e. 1.52. Consider the following function.()111111)(2≥<≤--<⎪⎩⎪⎨⎧--=x x x x x x x fDetermine the values of a for which )(lim x f ax →exists.3. Evaluate the limit and justify each step by indicating the appropriate properties of limits.443398lim 22-++-∞→x x x x x4. Find )(a f '.233)(x x x f -+=5. Guess the value of the limit.3055tan 3lim xx x x -→Select the correct answer.a. 121b. 135c. 134d. 130e. 1256. Given that 8)(lim 7-=→x f x and 10)(lim 7=→x g x .Evaluate the limit.())()(lim 7x g x f x +→7.Evaluate the limit.()()101lim 231-+→x x x8. Evaluate the limit.⎪⎪⎭⎫⎝⎛--→45lim 233x x x9. Find the derivative of the function using the definition of the derivative.2610)(x x x f +-=10.Let |9|81)(2--=x x x FFind the following limits.)(lim ),(lim 99x F x F x x -+→→Select the correct answer.a. 18 and 9b. 18 and - 18c. both 18d. 18 and – 9e. 81 and 911.Use the given graph of x x f =)(to find a number δsuch that 4.0|2|<-x whenever .|4|δ<-x12. Use a graph to find a number δsuch that 5.0|314|<-+x whenever .|2|δ<-x13. For the limit, illustrate the definition by finding values of δthat correspond to .5.0=ε()234lim 31=-+→x x x14. Find the slope of the tangent line to the curve 35x y = at the point (-4, -320).15. At what point is the function |8|)(x x f -= not differentiable.16.Which of the given functions is discontinuous?a. 5,5,3121)(<≥⎪⎩⎪⎨⎧-=x x x x f b. 5,5,351)(=≠⎪⎩⎪⎨⎧-=x x x x fc. Both functions are continuous17.Select the right number for the following limit and prove the statement using the ,δε definition of the limit. 3183lim 23--+→x x x xSelect the correct answer.a. 6b. 8c. 5d. 9e. 1818.Prove the statement using the ,δε definition of the limit.0|2|lim 2=-→x x19.Prove the statement using the ,δε definition of the limit.()241lim 25=--→x x20.Use continuity to evaluate the limit.()x x x sin 3sin lim 17+-→πSelect the correct answer. a. π17- b. ∞ c. -1 d. 0 e. 1ANSWER KEYStewart - Calculus ET 6e Chapter 2 Form G1. d2. ()()()∞--∞-,11,11,3.38 4.a 61- 5.e 6. 27. -728.22/5 9. 112-x10. b11. 44.1≤δ12. 6875.0≤δ13. 056.0≤δ14. 24015. 816. b17. d18. Given 0>ε, we need 0>δsuch that if | x - 2 | δ< then | | x - 2 | - 0 | ε<. But | | x - 2 | | = | x - 2 |. So this is true ifwe pick .εδ=19. Given 0>ε, we need 0>δsuch that if | x - ( - 5 ) | δ< then | ( x 2 - 1 ) - 24 | ε< or upon simplifying we need | x2 – 25| ε<whenever | x + 5 | δ<. Notice that if | x + 5 | < 1 , then- 1 < x + 5 < 1 - 11 < x - 5 < - 9 | x - 5 | < 11. So take =δmin {ε / 11, 1}. Then | x - 5 | < 11 and | x + 5 | ε</ 11, so | ( x 2 - 1 ) - 24 | = | ( x + 5 ) ( x - 5 ) | = | x + 5 | | x - 5 | < (ε / 11 ) ( 11 ) =ε. Therefore, by the definition of a limit, ().241lim 25=--→x x 20.dStewart - Calculus ET 6e Chapter 2 Form H1. The point P (4, 2) lies on the curve x y =. If Qis the point (,x , use your calculator to find the slope of thesecant line PQ (correct to six decimal places) for the value of 99.3=x .Select the correct answer.a. m PQ = 0.250157b. m PQ = 0.250156c. m PQ = - 0.250154d. m PQ = - 0.250156e. m PQ = 0.2501542. The displacement (in meters) of an object moving in a straight line is given by 212/4s t t =++, where t ismeasured in seconds. Find the average velocity over the time period [1,1.5].3. The displacement (in feet) of a certain particle moving in a straight line is given by 83t s =where t is measured in seconds. Find the instantaneous velocity when t = 3.4. If ,5.7)(lim 2=+→x f x then if )(lim 2x f x →exists, to what value does it converge?Select the correct answer. a. 5 b. 8.5 c. 8 d. 11.5 e. 7.55.If f and g are continuous functions with 3)2(=f and [],5)()(3lim 2=-→x g x f x find ).2(g6. The slope of the tangent line to the graph of the exponential function xy 4=at the point (0, 1) is .14lim 0x x x -→ Estimate the slope to three decimal places. Select the correct answer.a. 1.045b. 1.136c. 0.786d. 1.126e. 1.3867. Find an equation of the tangent line to curve 32y x x =-at the point (2,4).Select the correct answer.a. 1610y x =-b. 108y x =-c. 16y x =-d. 1016y x =+e. none of these8. Find the limit.lim x →-∞9. How close to 2 do we have to take x so that 5x + 3 is within a distance of 0.075 from 13?10. Evaluate the limit and justify each step by indicating the appropriate properties of limits.693958lim 2-++-∞→x x x x x11. Find a number δsuch that if |2|x δ-<, then |48|x ε-<, where 0.01ε=.12. Use the given graph of 2)(x x f =to find a number δsuch that 2112<-x whenever δ<-1x .Round down the answer to the nearest hundredth.13.If ()g x ()g x '.14.If ()g x ()g x '.15.Let ()20()30333if x f x x if x x if x ⎧<⎪⎪=-≤<⎨⎪->⎪⎩Evaluate each limit, if it exists.a.) 0lim ()x f x +→b.) 0lim ()x f x -→16.Which of the given functions is discontinuous?Select the correct answer.a. 5,5,3121)(<≥⎪⎩⎪⎨⎧-=x x x x f b. 5,5,351)(=≠⎪⎩⎪⎨⎧-=x x x x fc. Both functions are continuous17.If a ball is thrown into the air with a velocity of 62 ft/s, its height (in feet) after t seconds is given by21662t t H -=.Find the velocity when t = 5.18.Use continuity to evaluate the limit.()x x x sin 6sin lim 8+→πSelect the correct answer.a. ∞b. - 1c. 1d. 0e. π819. Find a function g that agrees with f for 16≠x and is continuous on ℜ.xx x f --=164)( 20. Consider the function .11)(/1x e x f +=Find the value of )(lim 0x f x +→.Select the correct answer.a. -0.8b. -0.5c. 0.3d. 0e. 0.8ANSWER KEYStewart - Calculus ET 6e Chapter 2 Form H1. b2. 2.625 m/s3. 3.3754. e5. 46. e7. e8. -1/29.015.0|2|<-x 10. 38 11. 0.0025δ=12. 22.0≤δ13. ()1/25()352g x x -'=-- 14. (),3/5-∞15. a.) 3 b.) 016. b17. -9818. d19. x g +=4120. d。
《微积分英文》课件 (2)
Types of Limits
One-sided limits
Limits approached
from one direction
Limits at infinity
Behavior of functions at
infinity
● 02
第2章 Limits and Continuity
01 Definition of a limit
Explanation of what a limit is
02 Properties of limits
Key characteristics of limits
03 Calculating limits algebraically
Graphing functions by analyzing their derivatives and key points
Higher Order Derivatives
Second derivative
Rate of change of the rate of
change
nth derivative
● 03
第3章 Differentiation
Derivatives and Rates of
Change
A derivative is defined as the rate of change of a function at a given point. Notation for derivatives includes symbols such as f'(x) or dy/dx. Derivatives can be interpreted as rates of change in various realworld applications.
《微积分英文版》课件
Limits and continuity
Definition: A limit is the value that a function approaches as the input approaches a certain point Continuity means that the function doesn't have any breaks or jumps at any point
Course structure
03
The course is divided into several modules, each focusing on a specific topic in calculus Learners can complete the course at their own pace and in any order of the modules
Properties: One side limits, absolute continuity, uniform continuity, etc
Differentiation
Definition: The derivative of a function at a point is the slope of the tangent line to the graph of the function at that point It can be used to find the rate of change of a function
Integral definition: The integral of a function is a measure of the area under its curve It is calculated by finding the limit of the sum of areas of rectangles under the curve as the width of the rectangles approaches zero
微积分(下)英文教材
Chapter 1 Infinite SeriesGenerally, for the given sequence,.......,......,3,21n a a a a the expressionformed by the sequence ,.......,......,3,21n a a a a.......,.....321+++++n a a a ais called the infinite series of the constants term, denoted by ∑∞=1n n a , that is∑∞=1n n a =......., (32)1+++++n a a a aWhere the nth term is said to be the general term of the series, moreover, the nth partial sum of the series is given by=n S......321n a a a a ++++1.1 Determine whether the infinite series converges or diverges.Whil e it’s possible to add two numbers, three numbers, a hundred numbers, or even a million numbers, it’s impossible to add an infinite number of numbers.To form an infinite series we begin with an infinite sequence of real numbers:.....,,,3210a a a a , we can not form the sum of all the ka (there isan infinite number of the term), but we can form the partial sums∑===000k ka a S∑==+=1101k ka a a S∑==++=22102k ka a a a S∑==+++=332103k ka a a a a S……………….∑==+++++=nk kn n a a a a a a S 03210.......Definition 1.1.1If the sequence {n S } of partial sums has a finite limit L, We write∑∞==k ka Land say that the series ∑∞=0k k a converges to L. we call L the sum ofthe series.If the limit of the sequence {n S } of partial sums don’t exists, we say that the series ∑∞=0k k a diverges.Remark it is important to note that the sum of a series is not a sum in the ordering sense. It is a limit.EX 1.1.1 prove the following proposition: Proposition1.1.1: (1) If 1<xthen the ∑∞=0k k a converges, and ;110xx k k -=∑∞=(2)If,1≥xthen the ∑∞=0k kx diverges.Proof: the nth partial sum of the geometric series ∑∞=0k k a takes theform1321.......1-+++++=n n xx x x S ① Multiplication by x gives).......1(1321-+++++=n n xx x x x xS =nn xxx x x +++++-1321.......Subtracting the second equation from the first, we find thatnn xS x -=-1)1(. For ,1≠x this givesxxS nn --=11 ③If,1<x then→nx ,and this by equation ③.xxxS nn n n -=--=→→1111limlimThis proves (1).Now let us prove (2). For x=1, we use equation ① and device that,n S n =Obviously, ∞=∞→n n S lim, ∑∞=0k k a diverges.For x=-1 we use equation ① and we deduce If n is odd, then 0=n S ,If n is even, then.1-=n SThe sequence of partial sum nS like this 0,-1,0,-1,0,-1………..Because the limit of sequence }{n S of partial sum does not exist. By definition 1.1.1, we have the series ∑∞=0k K x diverges. (x=-1).For 1≠x with,1>x we use equation ③. Since in this instance, wehave -∞=--=∞→∞→xxS nn n n 11limlim . The limit of sequence of partial sum not exist,the series ∑∞=0k kxdiverges.Remark the above series is called the geometric series. It arises in so many different contexts that it merits special attention.A geometric series is one of the few series where we can actually give an explicit formula for n S ; a collapsing series is another.Ex.1.1.2 Determine whether or not the series converges ∑∞=++0)2)(1(1k k kSolution in order to determine whether or not this series converges we must examine the partial sum. Since2111)2)(1(1+-+=++k k k kWe use partial fraction decomposition to write2111111 (4)1313121211)2111()111(..............)4131()3121()2111()2)(1(1)1(1 (3).212.11+-+++-++-+-+-=+-+++-++-+-+-=++++++⨯+⨯=n n n n n n n n n n n n S nSince all but the first and last occur in pairs with opposite signs, the sum collapses to give211+-=n S nObviously, as.1,→∞→n S n this means that the series converges to 1.1)211(lim lim =+-=∞→∞→n S n n ntherefore 1)2)(1(10=++∑∞=n k kEX.1.1.3 proves the following theorem:Theorem 1.1.1 the kth term of a convergent series tends to 0; namely if∞=0k ka Converges, by definition we have the limit of the sequence}{n S ofpartial sums exists. NamelylaS nk kn n n ==∑=∞→∞→0limlimObviously.limlim 01l aS nk kn n n ==∑=∞→-∞→since1--=n n s s a n , we have0lim lim )(lim lim 11=-=-=-=-∞→∞→-∞→∞→l l S S S S a n n n n n n n n nA change in notation gives 0lim =∞→n k a .The next result is an obviously, but important, consequence of theorem1.1.1. Theorem 1.1.2 (A diverges test) iflim ≠∞→k k a , or ifn k a ∞→lim does not exist, then the series ∑∞=0k ka diverges.Caution, theorem 1.1.1 does not say that iflim =∞→k k a , and then∑∞=0k ka converge. In fact, there are divergent series for which 0lim=∞→k k a . Forexample, the series .....1 (2)11111++++=∑∞=nkk . Since it issequence }{n S of partial sum nn n n S n =>+++=1 (2)111}{ is unbounded. So∞===∞→∞→n S n n n limlim , therefore the series diverges.But01limlim ===∞→∞→k a k k kEX.1.1.3 determine whether or not the series:..........5443322101+++++=+∑∞=k k kConverges.Solution since 01111lim1limlim≠=+=+==∞→∞→∞→kk k a k k k k , this series diverges.EX.1.1.4 Determine whether or not the series ∑∞=021k kSolution1 thegivenseries is a geometric series.121,)21(0<==∑∑∞=∞=x and xk k k k,by proposition 1.1.1 we know that seriesconverges. Solution 2 ,21 (412)111-++++=n n S ① ,2121 (2)1212121132nn n S +++++=-②①-② (1-21))211(2,211nn nnS S-=-=.2)211(2lim lim =-=∞→∞→nn n n SBy definition of converges of series, this series converges.EX.1.1.5 proofs the following theorem:Theorem 1.1.2 If the series ∑∑∞=∞=0k k k k b and a converges, then (1))(0∑∞=+k k kb aalso converges, and is equal the sum of the two series.(2) If C is a real number, then ∑∞=0k k Ca also converges. Moreover iflak k=∑∞=0then ClCa k k=∑∞=0.Proof let∑∑====nk knnk knbS aS20)1(,∑∑===+=nk knnk k knCa Sb aS40)3(,)(Note that )1()4()2()1()3(nnnnnCSS andS S S =+=Since (),lim ,lim )2(1m S l S nn n n ==∞→∞→Thenm l S S S S S nn n n n n n nn +=+=+=∞→∞→∞→∞→)2()1()2()1()3(lim lim )(lim lim.lim lim lim )1()1()4(Cl S C CSS nn nn nn ===∞→∞→∞→Theorem 1.1.4 (squeeze theorem) Suppose that}{}{n n c and a both converge to l and thatnn n c b a ≤≤ for,k n ≥(k is a fixed integer), then }{n b also converges to l .Ex.1.1.6 show that 0sinlim3=∞→n nn .Solution For ,1≥n ,1)sin(13nnnn≤≤-since,0)1(lim ,0)1(lim ==-∞→∞→n and nn nthe result follows by the squeeze theorem.For sequence of variable sign, it is helpful to have the following result.EX1.1.7 prove that the following theorem holds.Theorem 1.1.5 If 0lim ,0lim==∞→∞→n n n n a then a ,Proof since,n n n a a a ≤≤-from the theorem 1.1.4Namely the squeeze theorem, we know the result is true.Exercise 1.1(1) An expression of the form 123a a a +++…is called(2) A series123a a a +++…is said to converge if the sequence {}S n converges, whereS n=1. The geometric series 2a ar ar +++…converges if; in this casethe sum of the series is 2. Iflim 0n n a →∞≠, we can be sure that the series1nn a∞==∑3. Evaluate 0(1),02k k r r r ∞=-<<∑.4. Evaluate 0(1),11k k k x x ∞=--<<∑.5. Show that 1ln 1k k k ∞=+∑diverges.Find the sums of the series 6-11 6. 31(1)(2)k k k ∞=++∑7.112(1)k k k ∞=+∑8.11(3)k k k ∞=+∑9.0310kk ∞=∑10.0345kkkk ∞=+∑11.3023k kk +∞=∑12. Derivethefollowingresultsfromthegeometricseries 221(1),||11k kk x x x∞=-=<+∑.Test the following series for convergence: 13. 11n nn∞=+∑ 14.312k k ∞+=∑1.2 Series With Positive T erms1.2.1 The comparison TestThroughout this section, we shall assume that our numbers n a are x≥,then the partial sum12n nS a a a =+++… are increasing, i.e.1231n n S S S S S +≤≤≤≤≤≤……If they are to approach a limit at all, they cannot become arbitrarily large. Thus in that case there is a number Bsuch thatn S B≤ for all n.Such a number Bis called an upper bound. By a least upper bound we mean a number Swhich is an upper bound, and such that every upperboundBis S ≥. We take for granted that a least upper bound exists. Thecollection of numbers {}n S has therefore a least upper bound, i.e., there is a smallest numbers such that n S S≤ for all n. In that case, the partialsumsn SapproachSas a limit. In other words, given any positivenumber 0ε>, we have n S S Sε-≤≤ for all n sufficiently large.This simply expresses the factSis the least of all upper bounds forour collection of numbers n S . We express this as a theorem.Theorem 1.2.1 Let {}(1,2,n a n =…) be a sequence of numbers≥and let 12n n S a a a =+++…. If the sequence of numbers {}n S is bounded,then it approaches a limit S , which is its least upper bound.Theorem 1.2.2 A series with nonnegative terms converges if and only if the sequence of partial sums is bounded above.Theorem 1.2.1 and 1.2.2 give us a very useful criterion to determine when a series with positive terms converges.The convergence or divergence of a series with nonnegative terms isusually deduced by comparison with a series of known behavior.S 1 S 2 S n STheorem 1.2.3(The Ordinary Comparison Test) Let 1n n a ∞=∑and1nn b ∞=∑be two series, with0n a ≥for all n and0n b ≥for all n. Assume thatthere is a numbers 0c >, such thatn na cb ≤ for all n, and that 1n n b ∞=∑converges, then 1n n a ∞=∑converges, and 11nnn n a c b ∞∞==≤∑∑.Proof: We have1212121()n n n nn a a a cb cb cb c b b b c b ∞=+++≤+++=+++≤∑……….This means that 1n n c b ∞=∑ is a bound for the partial sums 12n a a a +++….The least upper bound of these sums is therefore 1n n c b ∞=≤∑, thus proving ourtheorem.Theorem 1.2.3 has an analogue to show that a series does not converge.Theorem 1.2.4(Ordinary Comparison Test) Let 1n n a ∞=∑ and 1n n b ∞=∑ betwo series, withna and 0nb ≥ for all n. Assume that there is a number0c >such that n n a cb ≥for all n sufficiently large, and 1n n b ∞=∑ does notconverge, then 1n n a ∞=∑ diverges.Proof. Assume n na cb ≥forn n ≥, since 1n n b ∞=∑diverges, we canmake the partial sum0001Nn n n Nn n b b b b +==+++∑…arbitrarily large as N becomes arbitrarily large. But 0NNNn n nn n n n n n a cb c b ===≥=∑∑∑.Hence the partial sum 121NnNn a a a a ==+++∑… are arbitrarily large as Nbecomes arbitrarily large, are hence 1n n a ∞=∑ diverges, as was to be shown.Remark on notation you have easily seen that for each 0j ≥, 0k k a ∞=∑converges iff 1kk j a ∞=+∑converges. This tells us that, in determining whetheror not a series converges, it does not matter where we begin the summation, where detailed indexing would contribute nothing, we will omit it and write ∑without specifying where the summation begins. For instance, it makes sense to you that 21k∑converges and 1k∑diverges without specifying where we begin the summation. But in the convergent case it does, however, affect the sum. Thus for example122kk ∞==∑,1112kk ∞==∑,21122kk ∞==∑, and so forth.Ex 1.2.1 Prove that the series 211n n∞=∑converges.Solution Let us look at the series:22222222211111111112345781516+++++++++++………We look at the groups of terms as indicated. In each group of terms, if we decrease the denominator in each term, then we increase the fraction. We replace 3 by 2 , then 4,5,6,7 by 4, then we replace the numbers from 8 to 15 by 8, and so forth. Our partial sums therefore less than or equal to222222221111111112244488++++++++++………and we note that 2 occurstwice, 4 occurs four times, 8 occurs eight times, and so forth. Our partialsum are therefore less than or equal to222222221111111112244488++++++++++………and we note that 2 occurs twice, 4 occurs four times, 8 occurs eight times, and so forth. Hence the partial sums are less than or equal to2222124811124848+++++++1 (1)2Thus our partial sums are less than or equal to those of the geometric series and are bounded. Hence our series converges.Generally we have the following result: The series 1111111234pppppn nn∞==++++++∑……, wherepis a constant,is called a p-series.Proposition1.2.1. If 1p >, the p-series converges; and if 1p ≤, thenthe p-series diverges.Ex 1.2.2 Determine whether the series 2311n nn ∞=+∑converges.Solution We write2323111(1)1111nn nn nn==++++. Then we see that23111122nn nn ≥=+. Since 11n n∞=∑ does not converge, it follows that the series 2311n nn∞=+∑ does not converge either. Namely this series diverges.Ex 1.2.3 Prove the series 241723n n n n ∞=+-+∑converges.Proof :Indeed we can write2222424334477(1)171331123(2())2()n n nnn n n n n n n n+++==-+-+-+For n sufficiently large, the factor23471312()nn n+-+ is certainly bounded,and in fact is near 1/2. Hence we can compare our series with 21n∑ tosee converges, because ∑21nconverges and the factor is bounded.Ex.1.2.5 Show that 1ln()k b +∑diverges.Solution 1 We know that ask →∞,ln 0k k→. It follows that ln()0k b k b+→+, and thus thatln()ln()0k b k b k b kk bk+++=→+. Thus forksufficiently large, ln()k b k+< and11ln()kk b <+. Since 1k∑diverges,we can conclude that 1ln()k b +∑diverges.Solution 2 Another way to show that ln()k b k+< for sufficiently largekis to examine the function()ln()f x x x b =-+. At3x = thefunction is positive:(3)3ln 93 2.1970f =-=->Since '1()10f x x b=->+ for all 0x >, ()0f x > for all3x >. It followsthatln()x b x+< for all 3x ≥.We come now to a somewhat more comparison theorem. Our proof relies on the basic comparison theorem.Theorem 1.2.5(The Limit Comparison Test) Let k a ∑ and k b ∑ beseries with positive terms. If lim()k k ka lb →∞=, where l is some positivenumber, then k a ∑ and k b ∑converge or diverge together.Proof Choose ε between 0 and l , sincekka lb →, we know for allksufficiently large (for all k greater than some 0k ) ||k ka lb ε-<.For such k we have k ka l lb εε-<<+, and thus()()k k kl b a l b εε-<<+this last inequality is what we needed.(1) If k a ∑converges, then ()k l b ε-∑converges, and thuskb ∑converges.(2) If k b ∑converges, then ()k l b ε+∑converges, and thuska ∑converges.To apply the limit comparison theorem to a series k a ∑, we must first find a series k b ∑of known behavior for which k ka b converges to apositive number.Ex 1.2.6 Determine whether the series sin kπ∑converges ordiverges.Solution Recall that as sin 0,1x x x→→.As,k kπ→∞→ and thussin 1k kππ→. Sincekπ∑diverges, so sin()kπ∑diverges.Ex 1.2.7Determine whether the series 100∑converges ordiverges.Solution For large value of k,dominates the numeratorand22kthe denominator, thus, for such k,differs252k=. Since2210051020012k kk÷==→And 2255122kk=∑∑converges, this series converges.Theorem 1.2.6 Let k a ∑ and k b ∑ be series with positive terms and suppose thus0k ka b →, then(1) If k b ∑converges, then k a ∑converges. (2) If k a ∑diverges, then k b ∑diverges.(3) If k a ∑converges, then k b ∑may converge or diverge. (4) If k b ∑diverges, then k a ∑may converge or diverge. [Parts (3) and (4) explain why we stipulated 0l >in theorem 1.2.5]1.2.2 The root test and the ratio testTheorem 1.2.7 (the root test, Cauchy test) let ∑k a be a series with nonnegative terms and suppose thatρ==∞→∞→k k k kk k a a 1lim lim, ifρ<1, ∑k a converges, ifρ>1,∑ka diverges, ifρ=1, the test is inconclusive.Proof we suppose firstρ<1 and chooseμso that 1<<u ρ. Sinceρ→k k a 1)(, we have μ<kk a 1, for all k sufficiently large thus kk a μ< for allk sufficiently large since ∑k μ converges (a geometric series with 0<1<μ), we know by theorem 1.2.5 that∑ka converges.We suppose now that1>ρand chooseμso that 1>>u ρ. sinceρ→k k a 1)(, we haveμ>k k a 1)( for all k sufficiently large. Thuskk a μ>for all k sufficiently large.Since ∑k μ diverges (a geometric series with 1>μ ) the theorem1.2.6 tell us that ∑k a diverges.To see the inconclusiveness of the root test when 1=ρ, note that1)(1→k k a for both:112∑∑kandk,11)1()1()(221121=→==kk k k kka 11)1()(11→==k k k k kk aThe first series converges, but the second diverges. EX.1.2.7 Determine whether the series ∑kk )(ln 1converges ordiverges.Solution For the series ∑kk )(ln 1, applying the root test we have0ln 1lim)(lim 1==∞→∞→ka k kk k , the series converges.EX.1.2.8 Determine whether series ∑3)(2k k converges or diverges.Solution For the series ∑kk )3(2, applying the root test, we have1212]1[2)1(.2)(3331>=⨯→==kk k k kka . So the series diverges.EX1.2.9 Determines whether the series kk∑-)11(converges ordiverges.Solution in the case of kk∑-)11(, we have 111)(1→-=ka kk . Ifapplying the root test, it is inconclusive. But since kkka )11(-=convergestoe1 and not to 0, the series diverges.We continue to consider only series with terms≥. To comparesuch a series with a geometric series, the simplest test is given by the ratio test theoremTheorem 1.2.8 (The ratio test, DAlembert test) let ∑k a be a series with positive terms and suppose thatλ=+∞→kk k a a 1lim,If ,1<λ∑k a converges, if,1>λ∑ka diverges.If the,1=λthe test is inconclusive.Proof we suppose first that,1<λsince1lim1<=+∞→λkk k a aSo there exists some integer N such that if n ≥NCa a nn ≤+1 ThenNN N N N a C Caa Caa 212,1≤≤≤+++ and in general byinduction,N kk N a C a ≤+Thusca c c c c a a c a c ca a aNkN N kN N N kN Nn n-≤++++≤++++≤∑+=11)........1( (3)22Thus in effect, we have compared our series with a geometric series, and we know that the partial sums are bounded. This implies that our series converges.The ratio test is usually used in the case of a series with positive terms nasuch that.1lim1<=+∞→λnn n a aEX.1.2.10 show that the series ∑∞=13n n nconverges.Solution we let ,3nn n a =then31.13.3111n n n n a a nn nn +=+=++,this ratioapproaches∞→n as 31, and hence the ratio test is applicable: the seriesconverges.EX1.2.11 show that the series ∑!k kkdiverges.Solution we have kkkk nn kkk kk k k a a )11()1(!)!1()1(11+=+=++=++Soe ka a kk nn n =+=∞→+∞→)11(lim lim1Since 1>e , the series diverges. EX.1.2.12 proves the series ∑+121k diverges.Solution sincekk k k k k a a k k 32123212112.1)1(211++=++=+++=+ 13212limlim1=++=∞→+∞→kk a a k kk k .Therefore the ratio test is inconclusive. We have to look further. Comparison with the harmonic series shows that the series diverges:∑++=+>+)1(21,11.21)1(21121k k k k dverges.Exercise 1.21. The ordinary comparison test says that if ____ and if ∑i b converges. Then ∑k a also converges.2. Assume that 00>≥k kb and a . Thelimit comparison Test says that if0<____<+∞ then ∑k a and ∑k b converges or diverge together. 3. Let nn n a a 1lim+∞→=ρ. The ratio Test says that a series ∑k a of positive termsconverges if ___, diverges if ____and may do either if ___. Determine whether the series converges or diverges 4. ∑+13kk 5. ∑+2)12(1k 6. ∑+11k 7. ∑-kk2218. ∑+-1tan 21kk9. ∑321k10. ∑-k)43( 11. ∑k kln 12. ∑!10k k13. ∑kk1 14. ∑kk 100! 15. ∑++kk k623216. kk ∑)32( 17.∑+k11.18. ∑kk 410!19. Let}{n a be a sequence of positive number and assume thatna a nn 111-≥+ for all n. show that the series ∑n a diverges.1.3 Alternating series, Absolute convergence and conditional convergenceIn this section we consider series that have both positive and negative terms.1.3.1 Alternating series and the tests for convergence The series of the form .......4321+-+-u u u u iscalled the alternatingseries, where 0>n ufor all n, here two example:∑∞=--=+-+-+-11)1( (6)1514131211n n n,11)1( (6)5544332211+-=+-+-+-∑∞=n nnWe see from these examples that the nth term of an alternating series is the form n n n n n nu a or u a )1()1(1-=-=-,wherenu is a positive number (infact nna u =.)The following test says that if the terms of an alternating series decrease toward 0 in absolute value, then the series converges. Theorem 1.3.1 (Leibniz Theorem) If the alternating series n n nu ∑∞=-1)1(satisfy:(1)1+≥n n u u(n=1,2………); (2)lim =∞→n n u ,then the series converges. Moreover, it is sum 1u s ≤, and the error nr makeby usingn s of the first n terms to approximate the sum s of the series isnot more than 1+n u , that is, 1+≤n n u r namely 1+≤-=n n n u s s r .Before giving the proof let us look at figure 1.3.1 which gives a picture of the idea behind the proof. We first plot 11u s =on a number line.To find2swe subtract 2u , so 2s is the left of 1s . Then to find3s weadd 3u , so 3s is to the right of 2s . But, since3u <2u ,3s is to the left of 1s .Continuing in this manner, we see that the partial sums oscillate back and forth. Since 0→nu , the successive steps are becoming smaller and smaller.The even partial sums ,........,,642s s s areincreasing and the odd partialsums,........,,531s s s aredecreasing. Thus it seems plausible that both areconverging to some number s, which is the sum of the series. Therefore, in the following proof we consider the even and odd partial sums separatelyWe give the following proof of the alternating series test. We first consider the even partial sums:,0212≥-=u u s Since 12u u ≤,)(24324s u u s s ≥-+= since uu ≤4 In general, 22212222)(---≥-+=n n n n n s u u s s since122-≤n n u uThus.........................02642≤≤≤≤≤≤n s s s sBut we can also writenn n n u u u u u u u u s 21222543212)(....)()(--------=--Every term in brackets is positive, so 12u s n ≤for all n. therefore, thesequence }{2n s of even partial sums is increasing and bounded above. It is therefore convergent by the monotonic sequence theor em. Let’s call it is limit s, that is, s s n n =∞→2limNow we compute the limit of the odd partialsums:sconditionby s u s u s s n n n n n n n n n =+=+=+=+∞→∞→+∞→+∞→))2((0lim lim )(lim lim 12212212Since both the even and odd partial sums converge to s, wehave s s n n =∞→lim , and so the series is convergent.EX.1.3.1 shows that the following alternating harmonic series is convergent:.)1(..........413121111∑∞=--=+-+-n n nSolution the alternating harmonic series satisfies (1)nu n u n n 1111=<+=+; (2)1limlim ==∞→∞→nu n n nSo the series is convergent by alternating series Test. Ex. 1.3.2 Test the series ∑∞=--1143)1(n nn n for convergence and divergence.Solution the given series is alternating but043143lim143limlim ≠=-=-=∞→∞→∞→nn n u n n n nSo condition (2) is not satisfied. Instead, we look at the limit of the nthterm of the series:143)1(limlim --=∞→∞→n n a n n n This limit does not exist, so the series diverges bythe test for divergence. EX.1.3.3 Test the series ∑∞=+-121)1(n nn for convergence or divergence.Solution the given series is alternating so we try to verify conditions (1) and (2) of the alternating series test.Unlike the situation in example 1.3.1, it is not obvious the sequence given by12+=n n u n is decreasing. If we consider the related function1)(2+=x x x f ，we easily find that10)1(1)1(21)(22222222'><+-=+-+=xwhenver x xx x x x f .Thus f is decreasing on [1,∞) and so )1()(+>n f n f .Therefore, }{n u isdecreasingWe may also show directly that nn u u <+1, that is11)1(122+<+++n n n nThis inequality it equivalent to the one we get by cross multiplication:nn n n n n n n n n n n n n n n +<⇔++<+++⇔++<++⇔+<+++2232322221221]1)1[()1)(1(11)1(1Since 1≥n , we know that the inequality12>+n n istrue. Therefore,n n u u <+1and }{n u is decreasing.Condition (2) is readily verified:。
微积分英文版课件
极限和连续性的关系：极限是连续 的必要条件，但不是充分条件
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
普林斯顿微积分中英文读本
普林斯顿微积分中英文读本
普林斯顿微积分中英文读本是一本微积分教材，旨在帮助学生掌握微积分的基本知识和解题技巧。
这本书的英文原版名称为“The Calculus Lifesaver”，中文版则将其翻译为《普林斯顿微积分读本》。
这本书涵盖了微积分的基础知识、极限、连续、微分、导数的应用、积分、无穷级数、泰勒级数与幂级数等内容，通过详细讲解和丰富的例题，帮助学生深入理解微积分的概念和解题方法。
同时，这本书也着重训练学生的解题能力，教会学生如何思考问题并找到解题所需的知识点。
微积分英文教材
Chapter 1 Infinite SeriesGenerally, for the given sequence,.......,......,3,21n a a a a theexpression 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 n a , that is∑∞=1n n a =.......,.....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 k a converges to L. we call L thesum of the series.If the limit of the sequence {n S } of partial sums don’t exists, we say that the series ∑∞=0k k a diverges.Remark it is important to note that the sum of a series is not a sum in the ordering sense. It is a limit.EX 1.1.1 prove the following proposition: Proposition1.1.1:(1) If 1<x then the ∑∞=0k k a converges, and ;110xx k k -=∑∞= (2)If ,1≥x then the ∑∞=0k k x diverges.Proof: the nth partial sum of the geometric series ∑∞=0k k atakes the form 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 thatn 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, ∞=∞→n n S lim , ∑∞=0k k a diverges.For x=-1 we use equation ① and we deduce If n is odd, then 0=n S , If n is even, then .1-=n S ThesequenceofpartialsumnS like this0,-1,0,-1,0,-1………..Because the limit of sequence }{n S of partial sum does not exist.By definition 1.1.1, we have the series ∑∞=0k K x diverges. (x=-1).For 1≠x with ,1>x we use equation ③. Since in thisinstance, we have -∞=--=∞→∞→xx S nn n n 11lim lim . The limit of sequence of partial sum not exist, the series ∑∞=0k k x diverges.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 k Solution 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 nSince 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 convergesto 1. 1)211(lim lim =+-=∞→∞→n S n n n therefore 1)2)(1(10=++∑∞=n k k 1.1.3 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 of partial sums exists. Namelyl a S nk k n n n ==∑=∞→∞→0lim limObviously .lim lim1l a S nk k n n n ==∑=∞→-∞→since 1--=n n s s a n, we have 0lim 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 ifn k a ∞→lim does not 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 forwhich 0lim =∞→k k a .For example, theseries.....1. (2)11111++++=∑∞=nkk . Since it is sequence}{n S of partial sumnnn 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 kSolution 1 the given series is a geometric series.121,)21(00<==∑∑∞=∞=x and x k k k k , by proposition 1.1.1 we know that series converges.Solution 2 ,21 (412111)-++++=n n S ① ,2121.........21212121132n n n S +++++=-②①-② (1-21))211(2,211n n n n S S -=-=.2)211(2lim lim =-=∞→∞→nn n n S By definition of converges of series, this series converges.1.1.5 p the following theorem: Theorem 1.1.2 If the series ∑∑∞=∞=00k kk k banda converges, then(1) )(0∑∞=+k k k b a also converges, and is equal the sum of the twoseries.(2) If C is a real number, then ∑∞=0k k Ca also converges.Moreover if l a k k =∑∞=0then Cl Ca k k =∑∞=0.Proof let ∑∑====nk k nn k k nb S a S20)1(,∑∑===+=nk k nn k 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 thatn n n c b a ≤≤ for ,k n ≥(k is a fixed integer), then }{n b alsoconverges to l .Ex.1.1.6 show that 0sin lim3=∞→nnn .Solution For,1≥n ,1)sin (13nn 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 series 1n n 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. Derive the following results from the geometric series 2201(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 sum 12n n S a a a =+++… are increasing, .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, givenany 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 isbounded, 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 terms is usually deduced by comparison with a series of known behavior.Theorem 1.2.3(The Ordinary Comparison Test) Let 1n n a ∞=∑and1nn b∞=∑be two series, with 0n a ≥ for all n and 0n b ≥for all n.S 1 S 2 S n SAssume that there is a numbers 0c >, such that n n a cb ≤ for all n, and that1nn 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 partialsums 12n a a a +++….The least upper bound of these sums is therefore 1n n c b ∞=≤∑, thusproving our theorem.Theorem 1.2.3 has an analogue to show that a series does not converge.Theorem 1.2.4(Ordinary Comparison Test) Let 1n n a ∞=∑ and 1n n b ∞=∑be two series, with n a and 0n b ≥ for all n. Assume that there is a number 0c > such that n n a cb ≥ for all n sufficiently large, and 1n n b ∞=∑ does not converge, then 1n n a ∞=∑ diverges.Proof. Assume n n a cb ≥for 0n n ≥, since 1n n b ∞=∑diverges, we canmake the partial sum0001Nnn n N n n bb b b +==+++∑…arbitrarily large as N becomes arbitrarily large. But 0N N N n n n n n n n n n a cb c b ===≥=∑∑∑. Hence the partial sum 121Nn N n a a a a ==+++∑…are arbitrarily large as N becomes arbitrarily large, are hence1nn a∞=∑ diverges, as was to be shown.Remark on notation you have easily seen that for each 0j ≥,kk a∞=∑ converges iff1kk j a∞=+∑ converges. This tells us that, indetermining whether 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 summation begins. For instance, it makes sense to you that 21k ∑converges and 1k∑ diverges without specifying where we begin the summation. But in the convergent case it does, however, affect the sum. Thus for example0122k k ∞==∑, 1112k k ∞==∑, 21122kk ∞==∑, and so forth. Ex 1.2.1 Prove that the series 211n n ∞=∑converges. Solution Let us look at the series:22222222211111111112345781516+++++++++++………We look at the groups of terms as indicated. In each group ofterms, 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, and so 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 series1111111234p p p p pn nn ∞==++++++∑……, where p is a constant, is called a p-series.Proposition1.2.1. If 1p >, the p-series converges; and if1p ≤, 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 that23111122n n n n ≥=+. Since 11n n ∞=∑ does not converge, it follows that the series 2311n n n ∞=+∑ does not converge either. Namely this seriesdiverges.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 ∑to see converges, because ∑21n converges and the factor is bounded.Ex.1.2.5 Show that 1ln()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 that 1ln()k b +∑diverges.Solution 2 Another way to show that ln()k b k +< forsufficiently large k is to examine the function ()ln()f x x x b =-+. At 3x = the function 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) Let k a ∑ and k b ∑ be series with positive terms. If lim()k k ka lb →∞=, where l is somepositive number, then ka∑ andkb∑converge or divergetogether.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)If k a ∑converges, then ()k l b ε-∑converges, and thuskb∑converges.(2)If k b ∑converges, then ()k l b ε+∑converges, and thuska∑converges.To apply the limit comparison theorem to a series k a ∑, we must first find a series k b ∑of known behavior for whichkka b converges to a positive number. Ex 1.2.6 Determine whether the series sin kπ∑converges ordiverges.Solution Recall that as sin 0,1x x x →→. As ,0k kπ→∞→ and thussin 1k kππ→. Since k π∑diverges, so sin()k π∑diverges.Ex 1.2.7 Determine whether theseriesconverges or diverges.Solution For large value of k, dominates the numeratorand 2k dominates the denominator, thus, for such k,252k=. Since22512k ÷==→And 2255122k k =∑∑converges, this series converges. Theorem 1.2.6 Let k a ∑ and k b ∑ be series with positive terms and suppose thus0kka b →, then (1) If k b ∑converges, then k a ∑converges. (2) If k a ∑diverges, then k b ∑diverges.(3)If k a ∑converges, then k b ∑may converge or diverge.(4)If k b ∑diverges, then k a ∑may converge or diverge.[Parts (3) and (4) explain why we stipulated 0l >in theorem 1.2.5]1.2.2 The root test and the ratio testTheorem 1.2.7 (the root test, Cauchy test) let ∑k a be a series with nonnegative terms and suppose thatρ==∞→∞→k 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.Since ρ→kk a 1)(, we have μ<k ka 1, for all k sufficiently large thus k k a μ< for all k sufficiently large since ∑k μ converges (a geometric series with 0<1<μ), we know by theorem 1.2.5 that∑kaconverges.We suppose now that 1>ρand choose μso that 1>>u ρ. sinceρ→k k a 1)(, we have μ>kk a 1)( for all k sufficiently large. Thusk k a μ> for all k sufficiently large.Since ∑k μ diverges (a geometric series with 1>μ ) the theorem 1.2.6 tell us that ∑k a diverges.To see the inconclusiveness of the root test when 1=ρ, note that 1)(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 or diverges.Solution For the series ∑kk )(ln 1, applying the root test we have0ln 1lim)(lim 1==∞→∞→ka k kk k , the series converges. EX.1.2.8 Determine whether series ∑3)(2k kconverges ordiverges.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 sincek k k a )11(-=converges to e1and not to 0, the series diverges.We continue to consider only series with terms 0≥. Tocompare 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 ∑k a be a series with positive terms and suppose thatλ=+∞→kk k a a 1lim, If ,1<λ ∑k a converges, if ,1>λ ∑k a diverges. 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 by induction ,N k k N a C a ≤+Thusca c c c c a a c a c ca a aNk N Nk 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 terms n a such that .1lim1<=+∞→λnn n a a EX.1.2.10 show that the series ∑∞=13n nnconverges.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 kk 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 lim1 Since 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 ∑k a and ∑k b converges or divergetogether. 3. Let nn n a a 1lim+∞→=ρ. The ratio Test says that a series ∑k a of positive terms converges if ___, diverges if ____and may do either if ___.Determine whether the series converges or diverges 4. ∑+13k k5. ∑+2)12(1k6. ∑+11k7. ∑-kk 221 8. ∑+-1tan 21k k 9. ∑321k 10. ∑-k)43( 11. ∑k k ln 12. ∑!10k k13. ∑k k 1 14. ∑k k 100!15. ∑++k k k 6232 16. kk ∑)32( 17.∑+k11.18. ∑kk 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 ∑n a diverges.Alternating series, Absolute convergence and conditional convergenceIn this section we consider series that have both positive and negative terms.1.3.1 Alternating series and the tests for convergence The series of the form .......4321+-+-u u u u 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 series n n 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 error n r make by using n s of the first n terms to approximate the sum s of the series 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 find3s we add 3u , so 3s is to the right of 2s . But, since 3u <2u , 3s isto the left of 1s . Continuing in this manner, we see that thepartial 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 areconverging to some number s, which is the sum of the series. Therefore, in the following proof we consider the even and odd partial sums separatelyWe give the following proof of the alternating series test. We first consider the even partial sums: ,0212≥-=u u s Since 12u u ≤,)(24324s u u s s ≥-+= since 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 =∞→2limNow we compute the limit of the odd partial 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 anddivergence.Solution the given series is alternating but 043143lim 143limlim ≠=-=-=∞→∞→∞→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 ordivergence.Solution the given series is alternating so we try to verifyconditions (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 function1)(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 n n 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 by the Alternating series Test. 1.3.2 Absolute and conditional convergenceIn this section we consider series that have both positive andnegative terms. Absolute and conditional convergence. Definition 1.3.1 suppose that the series ∑∞=1k k a is not serieswith positive terms, if the series∑∞=1k kaformed with theabsolute value of the terms n a converges, the series ∑∞=1k k a iscalled absolutely convergent. The series∑∞=1k kais calledconditionally convergent, if the series ∑∞=1k k a converges but∑∞=1k kadiverges.Theorem 1.3.2 if ∑k a converges, the ∑k a converges. Proofforeachk,kk k a a a ≤≤-, and thereforekk k a a a 20≤+≤.if∑kaconverges,then∑∑=k ka a22converges, and therefore, by theorem 1.2.3 (theordinary comparison theorem),∑+)(k ka aconverges. Sincek k k k a a a a -+=)(by the theorem (1), we can 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 (41)31211222++++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,。
微积分英文课件PPT (2)
x 1 0 1/3 1 2
f (x) 3 4 0 3 4 / 3 0 3 2
4.2 The Mean Value Theorem
Rolle’ s Theorem: Let f be a function that satisfy the following three hypotheses:
f (c) 0
c
Caution: The conditions cannot be weakened.
Y
Example
f (x) x x [1,1]
-1
0 1X
Example
f (x) x2 x [0,1]
Y 0 1X
Caution:
The conditions is sufficient, but not necessary.
f (x) x3 at x 0
2)There may be an extreme value even when f (c) does not exist.
For example: f (x) x at x 0
Fermat’s Theorem does suggest that we should at least start looking for extreme values of f at the
1) f is continuous on the closed interval [a,b]. 2) f is differentiable on the open interval (a,b). 3) f (a) = f (b)
Then there is a number c in (a,b) such that f (c) 0
The maximum and minimum values of f are called the extreme values of f.
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 。
普林斯顿微积分读本英文名(3篇)
第1篇Introduction:Princeton Calculus: A Comprehensive Guide is a comprehensive读本 that aims to provide a thorough understanding of calculus, as taught at Princeton University. This读本 covers all the essential topics and concepts of calculus, making it an invaluable resource for students, educators, and anyone interested in delving into the fascinating world of mathematics. With its clear explanations, abundant examples, and challenging problems, this读本将帮助读者掌握微积分的核心原理，并在实际应用中取得成功。
Chapter 1: Introduction to CalculusIn this chapter, we introduce the fundamental concepts of calculus, including limits, derivatives, and integrals. We explore the historical development of calculus and its significance in various fields of science, engineering, and economics. Additionally, we discuss the notations and conventions used in calculus, providing a solid foundation for the rest of the读本。
微积分英文专业词汇
微积分词汇第一章函数与极限Chapter1 Function and Limit集合set元素element子集subset空集empty set并集union交集intersection差集difference of set基本集basic set补集complement set直积direct product笛卡儿积Cartesian product开区间open interval闭区间closed interval半开区间half open interval有限区间finite interval区间的长度length of an interval无限区间infinite interval领域neighborhood领域的中心centre of a neighborhood 领域的半径radius of a neighborhood 左领域left neighborhood右领域right neighborhood映射mappingX到Y的映射mapping of X ontoY 满射surjection单射injection一一映射one-to-one mapping双射bijection算子operator变化transformation函数function逆映射inverse mapping复合映射composite mapping自变量independent variable因变量dependent variable定义域domain函数值value of function函数关系function relation值域range自然定义域natural domain 单值函数single valued function多值函数multiple valued function单值分支one-valued branch函数图形graph of a function绝对值函数absolute value符号函数sigh function整数部分integral part阶梯曲线step curve当且仅当if and only if(iff)分段函数piecewise function上界upper bound下界lower bound有界boundedness无界unbounded函数的单调性monotonicity of a function 单调增加的increasing单调减少的decreasing单调函数monotone function函数的奇偶性parity(odevity) of a function 对称symmetry偶函数even function奇函数odd function函数的周期性periodicity of a function周期period反函数inverse function直接函数direct function复合函数composite function中间变量intermediate variable函数的运算operation of function基本初等函数basic elementary function 初等函数elementary function幂函数power function指数函数exponential function对数函数logarithmic function三角函数trigonometric function反三角函数inverse trigonometric function 常数函数constant function双曲函数hyperbolic function双曲正弦hyperbolic sine双曲余弦hyperbolic cosine双曲正切hyperbolic tangent反双曲正弦inverse hyperbolic sine反双曲余弦inverse hyperbolic cosine反双曲正切inverse hyperbolic tangent极限limit数列sequence of number收敛convergence收敛于a converge to a发散divergent极限的唯一性uniqueness of limits收敛数列的有界性boundedness of a convergent sequence子列subsequence函数的极限limits of functions函数当x趋于x0时的极限limit of functions as x approaches x0左极限left limit右极限right limit单侧极限one-sided limits水平渐近线horizontal asymptote无穷小infinitesimal无穷大infinity铅直渐近线vertical asymptote夹逼准则squeeze rule单调数列monotonic sequence高阶无穷小infinitesimal of higher order低阶无穷小infinitesimal of lower order同阶无穷小infinitesimal of the same order作者：新少年特工2007-10-8 18:37 回复此发言--------------------------------------------------------------------------------2 高等数学-翻译等阶无穷小equivalent infinitesimal函数的连续性continuity of a function增量increment函数在x0连续the function is continuous at x0左连续left continuous右连续right continuous区间上的连续函数continuous function函数在该区间上连续function is continuous on an interval 不连续点discontinuity point第一类间断点discontinuity point of the first kind第二类间断点discontinuity point of the second kind初等函数的连续性continuity of the elementary functions定义区间defined interval最大值global maximum value (absolute maximum)最小值global minimum value (absolute minimum)零点定理the zero point theorem介值定理intermediate value theorem第二章导数与微分Chapter2 Derivative and Differential速度velocity匀速运动uniform motion平均速度average velocity瞬时速度instantaneous velocity圆的切线tangent line of a circle切线tangent line切线的斜率slope of the tangent line位置函数position function导数derivative可导derivable函数的变化率问题problem of the change rate of a function导函数derived function左导数left-hand derivative右导数right-hand derivative单侧导数one-sided derivatives在闭区间【a,b】上可导is derivable on the closed interval [a,b]切线方程tangent equation角速度angular velocity成本函数cost function边际成本marginal cost链式法则chain rule隐函数implicit function显函数explicit function二阶函数second derivative三阶导数third derivative高阶导数nth derivative莱布尼茨公式Leibniz formula对数求导法log- derivative参数方程parametric equation相关变化率correlative change rata微分differential可微的differentiable函数的微分differential of function自变量的微分differential of independent variable微商differential quotient间接测量误差indirect measurement error绝对误差absolute error相对误差relative error第三章微分中值定理与导数的应用Chapter3 MeanValue Theorem of Differentials and the Application of Derivatives罗马定理Rolle’s theorem费马引理Fermat’s lemma拉格朗日中值定理Lagrange’s mean value theorem驻点stationary point稳定点stable point临界点critical point辅助函数auxiliary function拉格朗日中值公式Lagrange’s mean value formula柯西中值定理Cauchy’s mean value theorem洛必达法则L’Hospital’s Rule0/0型不定式indeterminate form of type 0/0 不定式indeterminate form泰勒中值定理Taylor’s mean value theorem 泰勒公式Taylor formula余项remainder term拉格朗日余项Lagrange remainder term麦克劳林公式Maclaurin’s formula佩亚诺公式Peano remainder term凹凸性concavity凹向上的concave upward, cancave up凹向下的，向上凸的concave downward’concave down 拐点inflection point函数的极值extremum of function极大值local(relative) maximum最大值global(absolute) mximum极小值local(relative) minimum最小值global(absolute) minimum目标函数objective function曲率curvature弧微分arc differential平均曲率average curvature曲率园circle of curvature曲率中心center of curvature曲率半径radius of curvature渐屈线evolute渐伸线involute根的隔离isolation of root隔离区间isolation interval切线法tangent line method第四章不定积分Chapter4 Indefinite Integrals原函数primitive function(antiderivative)积分号sign of integration被积函数integrand积分变量integral variable积分曲线integral curve积分表table of integrals换元积分法integration by substitution分部积分法integration by parts分部积分公式formula of integration by parts 有理函数rational function真分式proper fraction假分式improper fraction第五章定积分Chapter5 Definite Integrals曲边梯形trapezoid with曲边curve edge窄矩形narrow rectangle曲边梯形的面积area of trapezoid with curved edge积分下限lower limit of integral积分上限upper limit of integral积分区间integral interval分割partition积分和integral sum可积integrable矩形法rectangle method积分中值定理mean value theorem of integrals函数在区间上的平均值average value of a function on an integvals牛顿－莱布尼茨公式Newton-Leibniz formula微积分基本公式fundamental formula of calculus换元公式formula for integration by substitution递推公式recurrence formula反常积分improper integral反常积分发散the improper integral is divergent反常积分收敛the improper integral is convergent无穷限的反常积分improper integral on an infinite interval无界函数的反常积分improper integral of unbounded functions绝对收敛absolutely convergent第六章定积分的应用Chapter6 Applications of the Definite Integrals元素法the element method面积元素element of area平面图形的面积area of a luane figure直角坐标又称“笛卡儿坐标(Cartesian coordinates)”极坐标polar coordinates抛物线parabola椭圆ellipse旋转体的面积volume of a solid of rotation 旋转椭球体ellipsoid of revolution, ellipsoid of rotation曲线的弧长arc length of acurve可求长的rectifiable光滑smooth功work 水压力water pressure引力gravitation变力variable force第七章空间解析几何与向量代数Chapter7 Space Analytic Geometry and Vector Algebra向量vector自由向量free vector单位向量unit vector零向量zero vector相等equal平行parallel向量的线性运算linear poeration of vector三角法则triangle rule平行四边形法则parallelogram rule交换律commutative law结合律associative law负向量negative vector差difference分配律distributive law空间直角坐标系space rectangular coordinates坐标面coordinate plane卦限octant向量的模modulus of vector向量a与b的夹角angle between vector a and b方向余弦direction cosine方向角direction angle向量在轴上的投影projection of a vector onto an axis数量积，外积，叉积scalar product,dot product,inner product曲面方程equation for a surface球面sphere旋转曲面surface of revolution母线generating line轴axis圆锥面cone顶点vertex旋转单叶双曲面revolution hyperboloids of one sheet旋转双叶双曲面revolution hyperboloids oftwo sheets柱面cylindrical surface ,cylinder圆柱面cylindrical surface准线directrix抛物柱面parabolic cylinder二次曲面quadric surface椭圆锥面dlliptic cone椭球面ellipsoid单叶双曲面hyperboloid of one sheet双叶双曲面hyperboloid of two sheets旋转椭球面ellipsoid of revolution椭圆抛物面elliptic paraboloid旋转抛物面paraboloid of revolution双曲抛物面hyperbolic paraboloid马鞍面saddle surface椭圆柱面elliptic cylinder双曲柱面hyperbolic cylinder抛物柱面parabolic cylinder空间曲线space curve空间曲线的一般方程general form equations of a space curve空间曲线的参数方程parametric equations of a space curve螺转线spiral螺矩pitch投影柱面projecting cylinder投影projection平面的点法式方程pointnorm form eqyation of a plane法向量normal vector平面的一般方程general form equation of a plane两平面的夹角angle between two planes点到平面的距离distance from a point to a plane空间直线的一般方程general equation of a line in space方向向量direction vector直线的点向式方程pointdirection form equations of a line方向数direction number直线的参数方程parametric equations of a line两直线的夹角angle between two lines 垂直perpendicular直线与平面的夹角angle between a line and a planes平面束pencil of planes平面束的方程equation of a pencil of planes行列式determinant系数行列式coefficient determinant第八章多元函数微分法及其应用Chapter8 Differentiation of Functions of Several Variables and Its Application一元函数function of one variable多元函数function of several variables内点interior point外点exterior point边界点frontier point,boundary point聚点point of accumulation开集openset闭集closed set连通集connected set开区域open region闭区域closed region有界集bounded set无界集unbounded setn维空间n-dimentional space二重极限double limit多元函数的连续性continuity of function of seveal连续函数continuous function不连续点discontinuity point一致连续uniformly continuous偏导数partial derivative对自变量x的偏导数partial derivative with respect to independent variable x高阶偏导数partial derivative of higher order 二阶偏导数second order partial derivative 混合偏导数hybrid partial derivative全微分total differential偏增量oartial increment偏微分partial differential全增量total increment可微分differentiable必要条件necessary condition充分条件sufficient condition叠加原理superpostition principle全导数total derivative中间变量intermediate variable隐函数存在定理theorem of the existence of implicit function曲线的切向量tangent vector of a curve法平面normal plane向量方程vector equation向量值函数vector-valued function切平面tangent plane法线normal line方向导数directional derivative梯度gradient数量场scalar field梯度场gradient field向量场vector field势场potential field引力场gravitational field引力势gravitational potential曲面在一点的切平面tangent plane to a surface at a point曲线在一点的法线normal line to a surface at a point无条件极值unconditional extreme values条件极值conditional extreme values拉格朗日乘数法Lagrange multiplier method 拉格朗日乘子Lagrange multiplier经验公式empirical formula最小二乘法method of least squares均方误差mean square error第九章重积分Chapter9 Multiple Integrals二重积分double integral可加性additivity累次积分iterated integral体积元素volume element三重积分triple integral直角坐标系中的体积元素volume element in rectangular coordinate system柱面坐标cylindrical coordinates柱面坐标系中的体积元素volume element in cylindrical coordinate system 球面坐标spherical coordinates球面坐标系中的体积元素volume element in spherical coordinate system反常二重积分improper double integral曲面的面积area of a surface质心centre of mass静矩static moment密度density形心centroid转动惯量moment of inertia参变量parametric variable第十章曲线积分与曲面积分Chapter10 Line(Curve)Integrals and Surface Integrals对弧长的曲线积分line integrals with respect to arc hength第一类曲线积分line integrals of the first type对坐标的曲线积分line integrals with respect to x,y,and z第二类曲线积分line integrals of the second type有向曲线弧directed arc单连通区域simple connected region复连通区域complex connected region格林公式Green formula第一类曲面积分surface integrals of the first type对面的曲面积分surface integrals with respect to area有向曲面directed surface对坐标的曲面积分surface integrals with respect to coordinate elements第二类曲面积分surface integrals of the second type有向曲面元element of directed surface高斯公式gauss formula拉普拉斯算子Laplace operator格林第一公式Green’s first formula通量flux散度divergence斯托克斯公式Stokes formula环流量circulation旋度rotation,curl第十一章无穷级数Chapter11 Infinite Series一般项general term部分和partial sum余项remainder term等比级数geometric series几何级数geometric series公比common ratio调和级数harmonic series柯西收敛准则Cauchy convergence criteria, Cauchy criteria for convergence正项级数series of positive terms达朗贝尔判别法D’Alembert test柯西判别法Cauchy test交错级数alternating series绝对收敛absolutely convergent条件收敛conditionally convergent柯西乘积Cauchy product函数项级数series of functions发散点point of divergence收敛点point of convergence收敛域convergence domain和函数sum function幂级数power series幂级数的系数coeffcients of power series阿贝尔定理Abel Theorem收敛半径radius of convergence收敛区间interval of convergence泰勒级数Taylor series麦克劳林级数Maclaurin series二项展开式binomial expansion近似计算approximate calculation舍入误差round-off error,rounding error欧拉公式Euler’s formula魏尔斯特拉丝判别法Weierstrass test三角级数trigonometric series振幅amplitude角频率angular frequency初相initial phase矩形波square wave谐波分析harmonic analysis直流分量direct component 基波fundamental wave二次谐波second harmonic三角函数系trigonometric function system傅立叶系数Fourier coefficient傅立叶级数Forrier series周期延拓periodic prolongation正弦级数sine series余弦级数cosine series奇延拓odd prolongation偶延拓even prolongation傅立叶级数的复数形式complex form of Fourier series第十二章微分方程Chapter12 Differential Equation解微分方程solve a dirrerential equation常微分方程ordinary differential equation偏微分方程partial differential equation,PDE 微分方程的阶order of a differential equation 微分方程的解solution of a differential equation微分方程的通解general solution of a differential equation初始条件initial condition微分方程的特解particular solution of a differential equation初值问题initial value problem微分方程的积分曲线integral curve of a differential equation可分离变量的微分方程variable separable differential equation隐式解implicit solution隐式通解inplicit general solution衰变系数decay coefficient衰变decay齐次方程homogeneous equation一阶线性方程linear differential equation of first order非齐次non-homogeneous齐次线性方程homogeneous linear equation 非齐次线性方程non-homogeneous linear equation常数变易法method of variation of constant 暂态电流transient stata current稳态电流steady state current伯努利方程Bernoulli equation全微分方程total differential equation积分因子integrating factor高阶微分方程differential equation of higher order悬链线catenary高阶线性微分方程linera differential equation of higher order自由振动的微分方程differential equation of free vibration强迫振动的微分方程differential equation of forced oscillation串联电路的振荡方程oscillation equation of series circuit二阶线性微分方程second order linera differential equation线性相关linearly dependence线性无关linearly independce二阶常系数齐次线性微分方程second order homogeneour linear differential equation with constant coefficient二阶变系数齐次线性微分方程second order homogeneous linear differential equation with variable coefficient特征方程characteristic equation无阻尼自由振动的微分方程differential equation of free vibration with zero damping 固有频率natural frequency简谐振动simple harmonic oscillation,simple harmonic vibration微分算子differential operator待定系数法method of undetermined coefficient共振现象resonance phenomenon欧拉方程Euler equation幂级数解法power series solution数值解法numerial solution勒让德方程Legendre equation微分方程组system of differential equations 常系数线性微分方程组system of linera differential equations with constant coefficient。
大一微积分知识点英文
大一微积分知识点英文Calculus Knowledge Points for FreshmenCalculus, a fundamental branch of mathematics, is essential for students majoring in science, engineering, and mathematics. Mastering the core concepts and principles of calculus is crucial for a successful academic journey in these fields. In this article, we will explore some key calculus knowledge points for freshmen.1. Limits:Limits are fundamental to the study of calculus. A limit represents the value that a function or sequence approaches as its input or index approaches a certain point. Limits are extensively used to define derivatives and integrals.2. Derivatives:Derivatives measure the rate at which a function changes. It represents the slope of the tangent line to a curve at a particular point. Derivatives allow us to analyze the behavior of functions, determine critical points, and solve optimization problems. Notation for derivatives includes the prime symbol (') and the differential operator d/dx.3. Differentiation Rules:Differentiation rules provide shortcuts for computing derivatives. Some of the important rules include the power rule, product rule, quotient rule, chain rule, and trigonometric derivatives. Understanding these rules simplifies the process of finding derivatives of functions.4. Applications of Derivatives:Derivatives have various applications in real-life scenarios. They can be used to determine velocity and acceleration, solve related rates problems, find maximum and minimum values, and analyze the behavior of functions. Application areas include physics, economics, engineering, and biology.5. Integrals:Integrals, also known as antiderivatives, are the reverse process of derivatives. They represent the accumulation of quantities over an interval. Integrals are used to find areas, volumes, average values, and solve differential equations. Notation for integrals includes the integral symbol (∫) and the differential operator dx.6. Integration Techniques:Integration techniques provide methods for computing integrals. These techniques include u-substitution, integration by parts, trigonometric substitutions, and partial fractions. Mastery of these techniques enables students to evaluate a wide range of integrals efficiently.7. Applications of Integrals:Integrals have numerous applications, particularly in calculating areas and volumes. They can be used to find the area between curves, volumes of solids of revolution, work done by a force, and average values of functions. Integration is a powerful tool in physics, engineering, and economics.8. Fundamental Theorem of Calculus:The Fundamental Theorem of Calculus establishes the relationship between differentiation and integration. It states that the derivative of an integral of a function is equal to the original function. This theorem allows for the evaluation of definite integrals using antiderivatives.9. Sequences and Series:Sequences and series involve the summation of infinite terms. Convergence and divergence of sequences and series are crucialconcepts in calculus. Tests such as the ratio test, comparison test, and integral test can determine the convergence or divergence of a series.10. Multivariable Calculus:Multivariable calculus extends the concepts of calculus to functions of multiple variables. It involves partial derivatives, gradient vectors, multiple integrals, line integrals, and surface integrals. Multivariable calculus is essential for fields such as physics, computer science, and engineering.In summary, these calculus knowledge points provide a foundation for freshmen to embark on their study in calculus. Understanding and applying these concepts will enable students to solve complex problems and analyze real-world phenomena. By building a solid understanding of calculus, students can pave the way for success in their academic and professional pursuits.。
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Factoring the numerator creates: (x -1)(x2 +x+1), thus the
(x – 1) cancels, leaving
limx2 x 1 3 x1
This indicates that the function values are getting very, very, very close to 1 as the x values approach 3.
时，
BD
1x
oC
A
△AOB 的面积＜ 圆扇形AOB的面积＜△AOD的面积
即
1 2
sin
x
1 2
tan
x
亦故即有 显然有
sin x x tan x
(0
x
2
)
cos x sin x 1 x
(0
x
2
)
注
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example.
calculate
解:
lim
x0
tan x
x
lim (1 1 )x
x
x
2.7
1.2
Rigorous Study of Limits
Definition of Limit: L is the limit of f(x) as x approaches c if for any value of epsilon (no matter how small) around f(x)=A, there’s a value of sigma around x=c, such that all function values fall into the “target area” of epsilon about L.
故
1 a 0
因此
a 1
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Squeeze Theorem
A) g(x) is bounded above and below by f(x) and h(x)
B) the limit of f(x) as x approaches a = L and the limit of h(x) as x approaches a=L
THEN C) the limit of g(x) as x approaches a = L. (WHY is this useful? Sometimes we can’t
calculate a limit of a specific function, but we CAN calculate the limit of other functions right above & below it!)
xx0
lim f (x) lim f (x) A
xx0
xx0
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Function
f
(
x)
x 0
1, ,
x 1 ,
x0 x0 x0
y
y x 1
1
o 1
x
y x 1
讨论 x 0 时 f (x) 的极限是否存在 .
lim f (x) lim (x 1) 1
x0
有界
因此 为无穷小, f (x) A
g(x) 1 B
由极限与无穷小关系定B 理 ,
得
1 g(x)
2 B
x (x0 )
(详见P44)
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若
lim f (x) A, lim g(x) B , 则有
说明: 可推广到有限个函数相乘的情形 .
推论 1 . lim[C f (x)] C lim f (x) ( C 为常数 )
x
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3. 左极限与右极限
左极限 :
f
(x0 )
lim
xx0
f
(x)
A
0, 0, 当 x ( x0 , x0 )
时, 有
右极限 :
f
(x0 )
lim
xx0
f
(x)
A
0, 0, 当 x ( x0 , x0 )
时, 有
定理 3 .
lim f (x) A
Find
lim x3 1
x1 x 1
The function is not defined at x = 1, but since we’re only interested in what happens as we get VERY close to that
point, it does not matter.
How do you prove a limit exists?
For any given epsilon, you must find sigma, in terms of that epsilon such that it would always be true.
Pr ove: lim 2x 1 7 x3
x1
x 1 u2 1 u 1 x 1 u 1
∴ 原式 lim(u 1) 2
u 1
方法 2
lim (x 1)( x 1) lim( x 1)
x1 x 1
x1
2
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试确定常数 a 使
解: 令t1,则 x
0
lim
t0
3
1
1 t3
a t
3
lim
t0
t3
1 t
a
lim 3 t3 1 a 0 t0
0 x 3 (2x 1) 7
2x 6
2x3
x 3 , 2
2
Prove
Proof: f (x) A
0, To any 0, 当
时,
总有 So
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Prove
Proof: f (x) A
So 0, 取 , 当 x2 1 2
x0
lim f (x) lim (x 1) 1
x0
x0
So f (0 ) f (0 ) ,
lim f (x) NOT Exist .
x0
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2.6 Limit Theorems
1.lim k k xc
2.lim x c xc
3.lim k f ( x) k lim f ( x)
(1 n11)n (1 1x)x (1 1n)n1
lim (1
n
n11)n
lim
n
(1 n11)n1 e
1
1 n1
lim (1
n
1 n
)n1
lim [(1
n
1n)n（1
1n）]
e
lim (1
x
1x) x
e
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Calculate
lim (1 x)1/ x
x0
Calculate
such that f ( x) A , provided that x M .
lim f ( x) A
x
Similarly,
lim f (x) A
x
lim f (x) A
x
0,M 0, s.t x M f ( x) A 0,M 0, s.t x M f ( x) A 0,M 0, s.tx M f ( x) A
lim x0
sin x
x
1 cos
x
lim sin x lim 1 1 x0 x x0 cos x
example.
解: 令 t arcsin x, 则 x sin t , 因此
原式 lim t t0 sin t
sin t 1
t
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2.
证: 当 x 0 时, 设 n x n 1, 则
推论 2 . lim[ f (x)]n [ lim f (x) ] n ( n 为正整数 )
设 n 次多项式
试证
lim
xx0
Pn
(
x)
Pn
(
x0
).
证:
lim
x x0
Pn
(x)
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定理. 若 lim f (x) A, lim g(x) B , 且 B≠0 , 则有
x = 3 时分母为 0 !
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example
解: x = 1 时分母 = 0 , 分子≠0 , 但因
lim x2 5x 4 12 51 4 0
x1 2x 3
21 3
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例. 求
解: 方法 1 令 u x , 则 limu 1,
证: 因 lim f (x) A, lim g(x) B , 则有
f (x) A , g(x) B (其中 , 为无穷小) 于是 f (x) g(x) (A ) (B )
(A B) ( ) 由定理 1 可知 也是无穷小, 再利用极限与无穷小
的关系定理 , 知定理结论成立 .
CHAPTER 2
LIMITS
2.1
Introduction to Limits (Calculus is ALL about limits!)
An Intuitive Understanding
A limit implies what happens to our function as we get closer and closer to a specific x value. We may never get to that point, and that does not matter, we are only interested in the behavior of the function (y-values) as we get VERY close to a specific x value.