数学分析pdf
数学分析_复旦_欧阳光中陈传璋第三版3版上下册课后习题答案解析(下)
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《数学分析》知识点整理.pdf
《数学分析（3）》复习资料第十三章 函数列与函数项级数（5%）1．（1）函数列收敛域为(),1,2,nn f x x n == (1,1]-，极限函数为0,1,()1, 1.x f x x ⎧<⎪=⎨=⎪⎩．（2）函数列sin (),1,2,n nxf x n n == 收敛域为(,)-∞+∞，极限函数为()0f x =． 2．（1）函数列在(02(),1,2,nx n f x nxe n -== ,)+∞上不．一致收敛． （2）函数列()1,2,n f x n == 在(1,1)-上一致收敛． （3）函数列22(),1,2,1n xf x n n x ==+ 在(,上一致收敛．)-∞+∞（4）函数列(),1,2,n xf x n n== 在[0上不．一致收敛． ,)+∞（5）函数列()sin,1,2,n xf x n n== 在上不．一致收敛． (,-∞+∞)3．（1）函数项级数nn x∞=∑在(1上不．一致收敛． ,1)-（2）函数项级数2sin nx n ∑，2cos nxn ∑在上一致收敛．(,-∞+∞)（3）函数项级数(1)!nx n -∑在上一致收敛． [,]r r -（4）函数项级数122(1)(1)n nx x --+∑在(,上一致收敛． )-∞+∞（5）函数项级数n n x ∑在11r x r r ∙>⎧⎪>⎨=⎪⎩上一致收敛上不一致收敛．（6）函数项级数2nx n ∑在上一致收敛．[0,1]（7）函数项级数12(1)n x n --+∑在上一致收敛．(,-∞+∞)（8）函数项级数221(1)n x x -+∑在(,上不．一致收敛． )-∞+∞第十四章 幂级数（10%）1．对于幂级数，若0n n n a x ∞=∑lim n ρ=（1limn n na a ρ+→∞=） 则（i ）当0ρ=时，收敛半径R =+∞，收敛域为(,)-∞+∞；（ii ）当ρ=+∞时，收敛半径，仅在0R =0x =处收敛； （iii ）当0ρ<=+∞时，收敛半径1R ρ=，收敛域为(,)R R -，还要进一步讨论区间端点x R =±处的敛散性．2．幂级数展开式： （1）()2(0)(0)(0)()(0)1!2!!n nf f f f x f x x x n '''=+++++（2）011nn x x ∞==-∑，01(1)1n n n x x ∞==-+∑ (1x )<． （3）2(1)(1)(1))12!!m n m m m m m n x mx x x n ---++=+++++ (11)x -<<111]，．1110101m m m ≤--⎧⎪-<<-⎨⎪>-⎩时，收敛域为（，）时，收敛域为（，]时，收敛域为[，(1（4）1110(1)(1)ln(1)(11)1n n n n n n x x x x n n -∞∞+==--+==-<≤+∑∑，1ln(1)nn x x n∞=--=∑ (11)x -≤<． （5）210(1)sin (21)!n n n x x n ∞+=-=+∑，20(1)cos (sin )(2)!n nn x x n ∞=-'==∑ ()x -∞<<+∞．（6）10(1)arctan (11)21n n n x x n ∞+=-=-≤+∑≤（7）0)!nxn x n ∞==-∞<<+∞∑e x3．幂级数的和函数（1）1)(0,1,2,k 1knn kx x x x ∞==<-)∑ = ． （2）()(1)1)1knnn kx x x x ∞=--=<+)∑ ． (0,1,2,k = （3）1ln(1)nn x x n∞==--∑ ．(11)x -≤<（4）121111()1(1)n nn n n n x nxx x x x ∞∞∞-===''⎛⎫⎛⎫'==== ⎪ ⎪--⎝⎭⎝⎭∑∑∑ (1x )<． （5）223)21111(1)()1(1)(1n n n n n n x n n x x x x x x ∞∞∞-==='''''⎛⎫⎛⎫⎛⎫''-===== ⎪ ⎪ ⎪---⎝⎭⎝⎭⎝⎭∑∑∑ (1x <)． 第十五章 傅里叶级数（10%）()f x 是以2π为周期且在[,]ππ-上可积的函数： 1．01()(cos sin )2n n n a f x a nx b nx ∞==++∑，01()a f x πππ-=⎰dx ，1()cos n a f x nx πππ-=⎰dx ，1()sin nbf x nx πππ-=⎰dx 1,2,n ，= ．2．01()cos sin 2n n n a n x n x f x a b l l ππ∞=⎛⎫=++ ⎪⎝⎭∑，01()ll a f x l -=⎰dx ， 1()cos l n l n x a f x dx πl l -=⎰，1()sin l n l n xb f x dx πl l-=⎰，1,2,n = ．3．（1）偶函数的傅里叶级数：01()cos2n n a n x f x a l π∞==+∑，012()cos ()cos l l n l n x n xa f x dx f x dx πl l l l π-==⎰⎰，． 1,2,n = 01()cos 2n n a f x a nx ∞==+∑，012()cos ()cos n a f x nxdx f x nxd πππππ-==⎰⎰x ，1,2,n = ．（2）奇函数的傅里叶级数：1()sinn n n x f x b lπ∞==∑，012()sin ()sin l l n l n x n xf x dx f x dx l l l l πb π-==⎰⎰1,2,，n = ．1()sin n n f x b ∞==∑nx ，012()sin ()sin n b ，f x nxdx f x nxdx πππππ-==⎰⎰1,2,n = ．第十六章 多元函数的极限与连续（5%）1．若累次极限00lim lim (,)x x y y f x y →→，00lim lim (,)y y x x f x y →→和重极限00(,)(,)lim (,)x y x y f x y →都存在，则三者相等．2．若累次极限00lim lim (,)x x y y f x y →→与00lim lim (,)y y x x f x y →→存在但不相等，则重极限00(,)(,)lim (,)x y x y f x y →必不存在．3．2222(,)(0,0)lim 0x y x y x y →=+，2222(,)(0,0)1lim x y x y x y →++=+∞+，22(,)lim 2x y →=，22(,)(0,0)1lim ()sin 0x y x y x y →+=+，2222(,)(0,0)sin()lim 1x y x y x y →+=+． 第十七章 多元函数微分学（20%）1．全微分：z zdz dx dy x y ∂∂=+∂∂． 2．zzz x y x yx x y yt t∂∂s t s sts∂∂∂∂∂∂∂∂∂∂z z x z y s y t∂∂∂∂∂=+s x s y z z x z t x t y ∂∂∂∂∂∂∂∂∂∂=+∂∂∂∂∂． 3．若函数f 在点可微，则0P f 在点沿任一方向的方向导数都存在，且0P 000(,,)l x y z 0000()()cos ()cos ()cos l x y z f P f P f P f P αβγ=++，其中cos α，cos β，cos γ为方向l x 的方向余弦，000(,,)y z即cos α=cos β=，cos γ=4．若(,,)f x y z 在点存在对所有自变量的偏导数，则称向量0000(,,)P x y z 000((),(),())x y z f P f P f P 为函数f 在点的梯度，记作0P 000(),()ad )z ((),x y gr f P f =P f P f ．向量grad f 的长度（或模）为gra d f =．5．设，(,z f x y xy =+)f 有二阶连续偏导数，则有1211z 212()z f yf z x x y y y ∂⎛⎫∂ ⎪''∂+∂∂⎝⎭==∂∂∂∂2f f y f yf x∂'''=⋅+⋅=+∂'，11122212221112221(1)()f f x f y f f x f f x y f xyf ''''''''''''''''=⋅+⋅++⋅+⋅=++++．6．设，令00()()0x y f P f P ==0()xx f P A =，0()xy f P B =，0()yy f P C =，则（i ）当，时，20AC B ->0A >f 在点取得极小值； 0P （ii ）当，20AC B ->0A <时，f 在点取得极大值； 0P （iii ）当时，20AC B -<f 在点不能取得极值； 0P （iv ）当时，不能肯定20AC B -=f 在点是否取得极值．0P 第十八章 隐函数定理及其应用（10%）1．隐函数，则有(,)0F x y =x yF dydx F =-． 2．隐函数，则有(,,)0F x y z =x z F zx F ∂=-∂，y zF z y F ∂=-∂(,,,)0(,,,)0F x y u v G x y u v ． =⎧⎨3．隐函数方程组：=⎩，有x yu v xyuv F F F F F F F F x y u v G G G G GG G G x yuv ∂∂∂∂∂∂∂∂∂∂∂∂∂∂∂∂⎛⎫ ⎪⎛⎫ ⎪= ⎪ ⎪⎝⎭ ⎪⎝⎭， 则uv uv uv F F J G G =，xv xv xv F F J G G =，uxux u x F F J G G =，y v yv y v F F J G G =，uyuy uyF F JG G =， xv uv J u x J ∂=-∂ ，ux uv J vx J ∂=-∂，yv uv J u y J ∂=-∂，uy uvJ v y J ∂=-∂． 4．平面曲线在点的切线．．方程为(,)0F x y =000(,)P x y 000000(,)()(,)()0x y F x y x x F x y y y -+-=， 法线．．方程为000000(,)()(,)()0y x F x y x x F x y y y -+-=． 5．空间曲线：在点处的L (,,)0(,,)0F x y z G x y z =⎧⎨=⎩0000(,,)P x y z切线．．方程为00z x yz x y z x y z x y 0x x y y z z F F F F F F G G G G G G ---==⎛⎫⎛⎫⎛ ⎪ ⎪ ⎝⎭⎝⎭⎝⎫⎪⎭00000()()()0x y z F x x F y y F z z ， 法线．．方程为． 00()()()yz xy zx yz xy zx F F F F F F x x y y z z G G G G G G ⎛⎫⎛⎫⎛⎫-+-+-= ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭6．曲面在点处的切平面．．．方程为(,,)0F x y z =0000(,,)P x y z -+-+-=， 法线．．方程为00x y 0zx x y y z z F F F ---==． 7．条件极值例题：求函数在约束条件22u x y z =++222z x y =+与4x y z ++=下的最大值和最小值．解：令，22222(,,,,)()(4)L x y z x y z z x y x y z λμλμ=+++--+++-则由，得稳定点22220222040x yz L x x L y y L z L z x y L x y z λμλμλμλμ=-+=⎧⎪=-+=⎪⎪=++=⎨⎪=--=⎪=++-=⎪⎩00112x y z =⎧⎪=⎨⎪=⎩及228x y z =-⎧⎪=-⎨⎪=⎩，故当1x y ==，时函数在约束条件下取得最小值， 2z =22u x y z =++28z =26当，时函数在约束条件下取得最大值．2x y ==-22u x y z =++72第十九章 含参量积分（5%）1．，；10()s xs x e +∞--Γ=⎰dx 0s >(1)(s s )s Γ+=Γ；1(2Γ=；1()2n Γ+=，1()2n Γ-=． 2．1110(,)(1)p q p q x x ---⎰)dx (0,0p q >>B =；(,)(,)p q q p B =B ；1(,)(,1)1q p q p q p q -B =B -+- ；(0,1p q >>)1(,)(1,)1p p q p q -p q B =B -+-) ；(1,0p q >>(1)(1)(,)(1,1)(1)(2)p q p q p q p q p q --B =B --+-+- ．(1,1p q >>)3．()()(,)()p q p q p q ΓΓB =Γ+ ．(0,0p q >>)第二十章 曲线积分（5%）1．设有光滑曲线：L (),(),x t y t ϕψ=⎧⎨=⎩t [,]αβ∈，函数(,)f x y 为定义在L 上的连续函数，则(,)((),(Lf x y ds f t t βαϕψ=⎰⎰；当曲线由方程L ()y x ψ=，[,]x a b ∈表示时，(,)(,(bLaf x y ds f x x ψ=⎰⎰．2．设平面曲线：L (),(),x t y t ϕψ=⎧⎨=⎩t [,]αβ∈，其中()t ϕ，在[,]αβ上具有一阶连续导函数，且((),())A ϕαψα，((),())B ϕβψβ． 又设与为上的连续函数，则沿L 从A 到(,)P x y (,)Q x y L B 的第二型曲线积分(,)(,)[((),())()((),())()]LP x y dx Q x y dy P t t t Q t t t dt βαϕψϕϕψψ''+=+⎰⎰．第二十一章 重积分（20%）1．若(,)f x y 在平面点集}{12(,)()(),D x y y x y y x a x b =≤≤≤≤（x 型区域）上连续，其中1()y x ，2()y x 在[,上连续，则]a b 21()()(,)(,)b y x ay x Df x y d dx f x y dy σ=⎰⎰⎰⎰，即二重积分可化为先对y ，后对x 的累次积分．若}{12(,)()(),D x y x y x x y c y d =≤≤≤≤，其中1()x y ，2()x y 在]上连续，则二重积分可化为先对[,c d x ，后对y 的累次积分21()()(,)(,dx y cx y D)f x y d dy f x y σ=⎰⎰⎰⎰dx ．在二重积分中，每次积分的上、下限一定要遵循“上限大，下限小”的原则，且一般来说，第一次（先）积分的上、下限一般为第二次（后）积分的积分变量的函数或常数，而第二次（后）积分的上、下限均为常数． 2．格林公式：若函数，在闭区域上连续，且有一阶偏导数，则有(,)P x y (,)Q x y D ()L DQ Pd Pdx Qdy x yσ∂∂-=+∂∂⎰⎰⎰ （或L Dx y d Pdx Q +dy P Qσ∂∂∂∂=⎰⎰⎰ D ），这里为区域的边界曲线，并取正方向． L 3．设是单连通闭区域．若函数，在内连续，且具有一阶连续偏导数，则以下四个条件等价：D (,)P x y (,)Q x y D （i ）沿内任一按段光滑封闭曲线，有D L 0LPdx Qdy +=⎰；（ii ）对中任一按段光滑曲线，曲线积分与路线无关，只与的起点及终点有关；D L LPdx Qdy +⎰L （iii ）是内某一函数的全微分，即在内有Pdx Qdy +D (,)u x y D du Pdx Qdy =+；（iv ）在内处处成立D P Qy x∂∂=∂∂． (,)4．设f x y 在极坐标变换cos ,:sin ,x r T y r θθ=⎧⎨=⎩0r ≤<+∞，02θπ≤≤下，xy 平面上有界闭区域与D r θ平面上区域∆对应，则成立(,D)(cos ,sin )f x y dxdy f r r rdrd θθθ∆=⎰⎰⎰⎰．通常积分区域为圆形、扇形、环形或为其一部分，或积分区域的边界线用极坐标方程表示较简单，且被积函数为22()f x y +，(y f x ，(xf y，()f x y +等形式时常选用在极坐标系下计算二重积分．。
2024考研数学满分笔记pdf
2024考研数学满分笔记pdf一、数学分析1.极限与连续性极限的定义：对于数列的极限，若对于任意的ε>0，存在正整数N，当n>N时，|an - a| < ε，则称数列{an}收敛于a，记作lim(an) = a。
连续性的定义：若函数f在点x0处连续，则对于任意ε>0，存在δ>0，使得当|x - x0| < δ时，有|f(x) - f(x0)| < ε成立。
2.微分与积分微分的定义：函数f在点x0处可导，则存在常数A，使得当x→x0时，有Δf = f(x) - f(x0) ≈ A(x - x0)成立。
积分的定义：对于定积分∫[a,b]f(x)dx，若存在分点ξk∈[xk-1,xk]，使得S = ∑(i=1)^n f(ξi)Δxi = limn→∞ Σ(i=1)^nf(ξi)Δxi成立，则称f在[a,b]上可积。
二、线性代数1.向量空间向量空间的定义：对于域F上的n维数组空间Vn(F)，若满足以下条件，则称Vn(F)为F上的n维向量空间：（1）对于任意u、v∈Vn(F)，有u+v∈Vn(F)；（2）对于任意k∈F、u∈Vn(F)，有ku∈Vn(F)；（3）存在零向量0∈Vn(F)使得对于任意u∈Vn(F)，有u+0=u；（4）对于任意u∈Vn(F)，存在-u∈Vn(F)，使得u+(-u)=0。
2.矩阵与行列式矩阵的定义：对于m×n矩阵A=(aij)，其中aij∈F，则称A为m×n矩阵。
对于n×n矩阵A，若存在n阶单位矩阵En，使得EA=AE=A 成立，则称A为可逆矩阵。
行列式的定义：对于n阶行列式Det(A)，其定义为Det(A)=Σα(i1i2...in)Ai1i1Ai2i2...Ainin，其中α(i1i2...in)为排列的符号，Ai1i1Ai2i2...Ainin为n个元素所组成的乘积。
三、概率论与数理统计1.随机变量与概率分布随机变量的定义：对于样本空间Ω上的实函数X(ω)，若X(ω)是Ω上的一个实数值函数，则称X(ω)为随机变量。
数学分析讲义(第一章)
Ⅱ 典型例题与方法
1. 利用极限定义验证极限
前提：知道数列（函数）的极限值；
关键：寻找 N (δ ) .
基本方法：
（1）求最小的 N ：从不等式 an − a < ε 直接解出 n ；
（2）适当放大法：不等式 an − a < ε 较为复杂，无法直接解出，或求解的过程较繁，
为此先将表达式 an − a 进行化简，并适当放大，使之成为关于 n 的简单函数 H (n) （仍为无
(5). lim f (x) = A ⇔ ∀ε > 0, ∃M > 0, 当 x > M 时，有 f (x) − A < ε . x→+∞
(6) lim f (x) = A ⇔ ∀ε > 0, ∃M > 0, 当 x < −M 时，有 f (x) − A < ε . x→−∞ 2
特别地，若函数以零为极限，则称之为该情形下的无穷小量.理解无穷小量阶的比较的定
义及其意义，掌握等价无穷小量在极限计算中的应用，熟记常用的等价无穷小量：当 x → 0
时，
x ~ sin x ~ tan x ~ arcsin x ~ arctan x ~ ln(1 + x) ~ e x −1，
1 − cos x ~ x2 , (1 + x)α ~ αx, a x − 1 ~ x ln a . 2
n →∞
yn xn
= ⎪⎨+ ∞， ⎪⎩− ∞.
二 函数极限
1 定义 函数极限的六种形式：
(1)
lim f (x) = A ⇔ ∀ε > 0, ∃δ > 0, 当 0 <
x → x0
x − x0
< δ 时，有
数学分析第一章
Chapter1.Metric Spaces§1.Metric SpacesA metric space is a set X endowed with a metricρ:X×X→[0,∞)that satisfies the following properties for all x,y,and z in X:1.ρ(x,y)=0if and only if x=y,2.ρ(x,y)=ρ(y,x),and3.ρ(x,z)≤ρ(x,y)+ρ(y,z).The third property is called the triangle inequality.We will write(X,ρ)to denote the metric space X endowed with a metricρ.If Y is a subset of X,then the metric space(Y,ρ|Y×Y)is called a subspace of(X,ρ).Example1.Letρ(x,y):=|x−y|for x,y∈I R.Then(I R,ρ)is a metric space.The set I R equipped with this metric is called the real line.Example2.Let I R2:=I R×I R.For x=(x1,x2)∈I R2and y=(y1,y2)∈I R2,defineρ(x,y):=(x1−y1)+(x2−y2).Thenρis a metric on I R2.The set I R2equipped with this metric is called the Euclidean plane.More generally,for k∈I N,the Euclidean k space I R k is the Cartesian product of k copies of I R equipped with the metricρgiven byρ(x,y):=kj=1(x j−y j)21/2,x=(x1,...,x k)and y=(y1,...,y k)∈I R k.Example3.Let X be a nonempty set.For x,y∈X,defineρ(x,y):=1if x=y, 0if x=y.In this case,ρis called the discrete metric on X.Let(X,ρ)be a metric space.For x∈X and r>0,the open ball centered at x∈X with radius r is defined asB r(x):={y∈X:ρ(x,y)<r}.A subset A of X is called an open set if for every x∈A,there exists some r>0 such thatB r(x)⊆A.1Theorem1.1.For a metric space(X,ρ)the following statements are true.1.X and∅are open sets.2.Arbitrary unions of open sets are open sets.3.Finite intersections of open sets are open sets.Proof.Thefirst statement is obviously true.For the second statement,we let(A i)i∈I be a family of open subsets of X and wish to prove that∪i∈I A i is an open set.Suppose x∈∪i∈I A i.Then x∈A ifor some i0∈I.Since A i0is an open set,there exists some r>0such that B r(x)⊆A i.Consequently,B r(x)⊆∪i∈I A i.This shows that∪i∈I A i is an open set.For the third statement,we let{A1,...,A n}be afinite collection of open subsets of X and wish to prove that∩n i=1A i is an open set.Suppose x∈∩n i=1A i.Then x∈A i for every i∈{1,...,n}.For each i∈{1,...,n},there exists r i>0such that B ri(x)⊆A i. Set r:=min{r1,...,r n}.Then r>0and B r(x)⊆∩n i=1A i.This shows that∩n i=1A i is an open set.Let(X,ρ)be a metric space.A subset B of X is called an closed set if its complement B c:=X\B is an open set.The following theorem is an immediate consequence of Theorem1.1.Theorem1.2.For a metric space(X,ρ)the following statements are true.1.X and∅are closed sets.2.Arbitrary intersections of closed sets are closed sets.3.Finite unions of closed sets are closed sets.Let(X,ρ)be a metric space.Given a subset A of X and a point x in X,there are three possibilities:1.There exists some r>0such that B r(x)⊆A.In this case,x is called an interiorpoint of A.2.For any r>0,B r(x)intersects both A and A c.In this case,x is called a boundarypoint of A.3.There exists some r>0such that B r(x)⊆A c.In this case,x is called an exteriorpoint of A.For example,if A is a subset of the real line I R bounded above,then sup A is a boundary point of A.Also,if A is bounded below,then inf A is a boundary point of A.A point x is called a closure point of A if x is either an interior point or a boundary point of A.We denote by A the set of closure points of A.Then A⊆A.The set A is called the closure of A.2Theorem1.3.If A is a subset of a metric space(X,ρ),then A is the smallest closed set that includes A.Proof.Let A be a subset of a metric space.Wefirst show that A is closed.Suppose x/∈A. Then x is an exterior point of A;hence there exists some r>0such that B r(x)⊆A c.If y∈B r(x),thenρ(x,y)<r.Forδ:=r−ρ(x,y)>0,by the triangle inequality we have Bδ(y)⊆B r(x).It follows that Bδ(y)⊆A c.This shows y/∈A.Consequently,B r(x)⊆A c. Therefore,A c is open.In other words,A is closed.Now assume that B is a closed subset of X such that A⊆B.Let x∈B c.Then there exists r>0such that B r(x)⊆B c⊆A c.This shows x∈A c.Hence,B c⊆A c.It follows that A⊆B.Therefore,A is the smallest closed set that includes A.A subset A of a metric space(X,ρ)is said to be dense in X if A=X.§pletenessLet(x n)n=1,2,...be a sequence of elements in a metric space(X,ρ).We say that (x n)n=1,2,...converges to x in X and write lim n→∞x n=x,ifρ(x n,x)=0.limn→∞From the triangle inequality it follows that a sequence in a metric space has at most one limit.Theorem2.1.Let A be a subset of a metric space(X,ρ).Then a point x∈X belongs to A if and only if there exists a sequence(x n)n=1,2,...in A such that lim n→∞x n=x. Proof.If x∈A,then B1/n(x)∩A=∅for every n∈I N.Choose x n∈B1/n(x)∩A for each n∈I N.Thenρ(x n,x)<1/n,and hence lim n→∞x n=x.Suppose x/∈A.Then there exists some r>0such that B r(x)∩A=∅.Consequently, for any sequence(x n)n=1,2,...in A,we haveρ(x n,x)≥r for all n∈I N.Thus,there is no sequence of elements in A that converges to x.A sequence(x n)n=1,2,...in a metric space(X,ρ)is said to be a Cauchy sequence if for any givenε>0there exists a positive integer N such thatm,n>N impliesρ(x m,x n)<ε.Clearly,every convergent sequence is a Cauchy sequence.If a metric space has the property that every Cauchy sequence converges,then the metric space is said to be complete.For example,the real line is a complete metric space.3The diameter of a set A is defined byd(A):=sup{ρ(x,y):x,y∈A}.If d(A)<∞,then A is called a bounded set.Theorem2.2.Let(X,ρ)be a complete metric space.Suppose that(A n)n=1,2,...is a sequence of closed and nonempty subsets of X such that A n+1⊆A n for every n∈I N and lim n→∞d(A n)=0.Then∩∞n=1A n consists of precisely one element.Proof.If x,y∈∩∞n=1A n,then x,y∈A n for every n∈I N.Hence,ρ(x,y)≤d(A n)for all n∈I N.Since lim n→∞ρ(A n)=0,it follows thatρ(x,y)=0,i.e.,x=y.To show∩∞n=1A n=∅,we proceed as follows.Choose x n∈A n for each n∈I N.Since A m⊆A n for m≥n,we haveρ(x m,x n)≤d(A n)for m≥n.This in connection with the assumption lim n→∞d(A n)=0shows that(x n)n=1,2,...is a Cauchy sequence.Since (X,ρ)is complete,there exists x∈X such that lim n→∞x n=x.We have x m∈A n for all=A n.This is true for all n∈I N.Therefore,x∈∩∞n=1A n.m≥n.Hence,x∈A§pactnessLet(X,ρ)be a metric space.A subset A of X is said to be sequentially compact if every sequence in A has a subsequence that converges to a point in A.For example,afinite subset of a metric space is sequentially compact.The real line I R is not sequentially compact.But a bounded closed interval in the real line is sequentially compact.A subset A of a metric space is called totally bounded if,for every r>0,A can be covered byfinitely many open balls of radius r.For example,a bounded subset of the real line is totally bounded.On the other hand, ifρis the discrete metric on an infinite set X,then X is bounded but not totally bounded. Theorem3.1.Let A be a subset of a metric space(X,ρ).Then A is sequentially compact if and only if A is complete and totally bounded.Proof.Suppose that A is sequentially compact.Wefirst show that A is complete.Let (x n)n=1,2,...be a Cauchy sequence in A.Since A is sequentially compact,there exists a )k=1,2,...that converges to a point x in A.For anyε>0,there exists subsequence(x nka positive integer N such thatρ(x m,x n)<ε/2whenever m,n>N.Moreover,there exists some k∈I N such that n k>N andρ(x n,x)<ε/2.Thus,for n>N we havek4ρ(x n,x)≤ρ(x n,x nk )+ρ(x nk,x)<ε.Hence,lim n→∞x n=x.This shows that A iscomplete.Next,if A is not totally bounded,then there exists some r>0such that A cannot be covered byfinitely many open balls of radius r.Choose x1∈A.Suppose x1,...,x n∈A have been chosen.Let x n+1be a point in the nonempty set A\∪n i=1B r(x i).If m,n∈I N and m=n,thenρ(x m,x n)≥r.Therefore,the sequence(x n)n=1,2,...has no convergent subsequence.Thus,if A is sequentially compact,then A is totally bounded.Conversely,suppose that A is complete and totally bounded.Let(x n)n=1,2,...be a sequence of points in A.We shall construct a subsequence of(x n)n=1,2,...that is a Cauchy sequence,so that the subsequence converges to a point in A,by the completeness of A.For this purpose,we construct open balls B k of radius1/k and corresponding infinite subsets I k of I N for k∈I N recursively.Since A is totally bounded,A can be covered byfinitely many balls of radius1.Hence,we can choose a ball B1of radius1such that the set I1:={n∈I N:x n∈B1}is infinite.Suppose that a ball B k of radius1/k and an infinite subset I k of I N have been constructed.Since A is totally bounded,A can be covered by finitely many balls of radius1/(k+1).Hence,we can choose a ball B k+1of radius1/(k+1) such that the set I k+1:={n∈I k:x n∈B k+1}is infinite.Choose n1∈I1.Given n k,choose n k+1∈I k+1such that n k+1>n k.By our construction,I k+1⊆I k for all k∈I N.Therefore,for all i,j≥k,the points x niandx nj are contained in the ball B k of radius1/k.It follows that(x nk)k=1,2,...is a Cauchysequence,as desired.Theorem3.2.A subset of a Euclidean space is sequentially compact if and only if it is closed and bounded.Proof.Let A be a subset of I R k.If A is sequentially compact,then A is totally bounded and complete.In particular,A is bounded.Moreover,as a complete subset of I R k,A is closed.Conversely,suppose A is bounded and closed in I R k.Since I R k is complete and A is closed,A is complete.It is easily seen that a bounded subset of I R k is totally bounded.Let(A i)i∈I be a family of subsets of X.We say that(A i)i∈I is a cover of a subset A of X,if A⊆∪i∈I A i.If a subfamily of(A i)i∈I also covers A,then it is called a subcover. If,in addition,(X,ρ)is a metric space and each A i is an open set,then(A i)i∈I is said to be an open cover.Let(G i)i∈I be an open cover of A.A real numberδ>0is called a Lebesgue number for the cover(G i)i∈I if,for each subset E of A having diameter less thanδ,E⊆G i for5some i∈I.Theorem3.3.Let A be a subset of a metric space(X,ρ).If A is sequentially compact, then there exists a Lebesgue numberδ>0for any open cover of A.Proof.Let(G i)i∈I be an open cover of A.Suppose that there is no Lebesgue number for the cover(G i)i∈I.Then for each n∈I N there exists a subset E n of A having diameter less than1/n such that E n∩G c i=∅for all i∈I.Choose x n∈E n for n∈I N.Since A is sequentially compact,there exists a subsequence(x nk)k=1,2,...which converges to a point x in A.Since(G i)i∈I is a cover of A,x∈G i for some i∈I.But G i is an open set.Hence, there exists some r>0such that B r(x)⊆G i.We canfind a positive integer k such that1/n k<r/2andρ(x nk ,x)<r/2.Let y be a point in E nk.Since x nkalso lies in the setE nk with diameter less than1/n k,we haveρ(x nk,y)<1/n k.Consequently,ρ(x,y)≤ρ(x,x nk)+ρ(x nk,y)<r2+1n k<r.This shows E nk ⊆B r(x)⊆G i.However,E nkwas so chosen that E nk∩G c i=∅.Thiscontradiction proves the existence of a Lebesgue number for the open cover(Gi)i∈I.A subset A of(X,ρ)is said to be compact if each open cover of A possesses afinite subcover of A.If X itself is compact,then(X,ρ)is called a compact metric space. Theorem3.4.Let A be a subset of a metric space(X,ρ).Then A is compact if and only if it is sequentially compact.Proof.If A is not sequentially compact,then A is an infinite set.Moreover,there exists a sequence(x n)n=1,2,...in A having no convergent subsequence.Consequently,for each x∈A,there exists an open ball B x centered at x such that{n∈I N:x n∈B x}is afinite set.Then(B x)x∈A is an open cover of A which does not possess afinite subcover of A. Thus,A is not compact.Now suppose A is sequentially compact.Let(G i)i∈I be an open cover of A.By Theorem3.3,there exists a Lebesgue numberδ>0for the open cover(G i)i∈I.By Theorem 3.1,A is totally bounded.Hence,A is covered by afinite collection{B1,...,B m}of open balls with radius less thanδ/2.For each k∈{1,...,m},the diameter of B k is less thanδ.Hence,B k⊆G ik for some i k∈I.Thus,{G ik:k=1,...,m}is afinite subcover of A.This shows that A is compact.6§4.Continuous FunctionsLet(X,ρ)and(Y,τ)be two metric spaces.A function f from X to Y is said to be continuous at a point a∈X if for everyε>0there existsδ>0(depending onε)such thatτ(f(x),f(a))<εwheneverρ(x,a)<δ.The function f is said to be continuous on X if f is continuous at every point of X.Theorem4.1.For a function f from a metric space(X,ρ)to a metric space(Y,τ),the following statements are equivalent:1.f is continuous on X.2.f−1(G)is an open subset of X whenever G is an open subset of Y.3.If lim n→∞x n=x holds in X,then lim n→∞f(x n)=f(x)holds in Y.4.f(A)⊆f(A)holds for every subset A of X.5.f−1(F)is a closed subset of X whenever F is a closed subset of Y.Proof.1⇒2:Let G be an open subset of Y and a∈f−1(G).Since f(a)∈G and G is open,there exists someε>0such that Bε(f(a))⊆G.By the continuity of f,there exists someδ>0such thatτ(f(x),f(a))<εwheneverρ(x,a)<δ.This shows Bδ(a)⊆f−1(G). Therefore,f−1(G)is an open set.2⇒3:Assume lim n→∞x n=x in X.Forε>0,let V:=Bε(f(x)).In light of statement2,f−1(V)is an open subset of X.Since x∈f−1(V),there exists someδ>0 such that Bδ(x)⊆f−1(V).Then there exists a positive integer N such that x n∈Bδ(x) for all n>N.It follows that f(x n)∈V=Bε(f(x))for all n>N.Consequently, lim n→∞f(x n)=f(x).3⇒4:Let A be a subset of X.If y∈f(A),then there exists x∈A such that y=f(x).Since x∈A,there exists a sequence(x n)n=1,2,...of A such that lim n→∞x n=x. By statement3we have lim n→∞f(x n)=f(x).It follows that y=f(x)∈f(A).This shows f(A)⊆f(A).4⇒5:Let F be a closed subset of Y,and let A:=f−1(F).By statement4we have f(A)⊆⊆F=F.It follows that A⊆f−1(F)=A.Hence,A is a closed subset of X.5⇒1:Let a∈X andε>0.Consider the closed set F:=Y\Bε(f(a)).By statement5,f−1(F)is a closed subset of X.Since a/∈f−1(F),there exists someδ>0 such that Bδ(a)⊆X\f−1(F).Consequently,ρ(x,a)<δimpliesτ(f(x),f(a))<ε.So f is continuous at a.This is true for every point a in X.Hence,f is continuous on X.As an application of Theorem4.1,we prove the Intermediate Value Theorem for continuous functions.7Theorem 4.2.Suppose that a,b ∈I R and a <b .If f is a continuous function from [a,b ]to I R ,then f has the intermediate value property,that is,for any real number d between f (a )and f (b ),there exists c ∈[a,b ]such that f (c )=d .Proof.Without loss of any generality,we may assume that f (a )<d <f (b ).Since the interval (−∞,d ]is a closed set,the set F :=f −1((−∞,d ])={x ∈[a,b ]:f (x )≤d }is closed,by Theorem 4.1.Let c :=sup F .Then c lies in F and hence f (c )≤d .It follows that a ≤c <b .We claim f (c )=d .Indeed,if f (c )<d ,then by the continuity of f we could find r >0such that c <c +r <b and f (c +r )<d .Thus,we would have c +r ∈F and c +r >sup F .This contradiction shows f (c )=d .The following theorem shows that a continuous function maps compact sets to compact sets.Theorem 4.3.Let f be a continuous function from a metric space (X,ρ)to a metric space (Y,τ).If A is a compact subset of X ,then f (A )is compact.Proof.Suppose that (G i )i ∈I is an open cover of f (A ).Since f is continuous,f −1(G i )is open for every i ∈I ,by Theorem 4.1.Hence,(f −1(G i ))i ∈I is an open cover of A .By thecompactness of A ,there exists a finite subset {i 1,...,i m }of I such that A ⊆∪m k =1f−1(G i k ).Consequently,f (A )⊆∪mk =1G i k .This shows that f (A )is compact.Theorem 4.4.Let A be a nonempty compact subset of a metric space (X,ρ).If f is a continuous function from A to the real line I R ,then f is bounded and assumes its maximum and minimum.Proof.By Theorem 4.3,f (A )is a compact set,and so it is bounded and closed.Let t :=inf f (A ).Then t ∈f (A )=f (A ).Hence,t =min f (A )and t =f (a )for some a ∈A .Similarly,Let s :=sup f (A ).Then s ∈f (A )=f (A ).Hence,s =max f (A )and s =f (b )for some b ∈A .A function f from a metric space (X,ρ)to a metric space (Y,τ)is said to be uni-formly continuous on X if for every ε>0there exists δ>0(depending on ε)such that τ(f (x ),f (y ))<εwhenever ρ(x,y )<δ.Clearly,a uniformly continuous function is continuous.A function from (X,ρ)to (Y,τ)is said to be a Lipschitz function if there exists a constant C f such that τ(f (x ),f (y ))≤C f ρ(x,y )for all x,y ∈X .Clearly,a Lipschitz function is uniformly continuous.8Example.Let f and g be the functions from the interval(0,1]to the real line I R given by f(x)=x2and g(x)=1/x,x∈(0,1],respectively.Then f is uniformly continuous, while g is continuous but not uniformly continuous.Theorem4.5.Let f be a continuous function from a metric space(X,ρ)to a metric space(Y,τ).If X is compact,then f is uniformly continuous on X.Proof.Letε>0be given.Since f is continuous,for each x∈X there exists r x>0suchthatτ(f(x),f(y))<ε/2for all y∈B rx (x).Then(B rx(x))x∈X is an open cover of X.Since X is compact,Theorem3.3tells us that there exists a Lebesgue numberδ>0for this open cover.Suppose y,z∈X andρ(y,z)<δ.Then{y,z}⊆B rx(x)for some x∈X. Consequently,τ(f(y),f(z))≤τ(f(y),f(x))+τ(f(x),f(z))<ε/2+ε/2=ε.This shows that f is uniformly continuous on X.9。
数学分析习题课讲义
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Lecture Notes at Fudan
Hint. a) f (0) ∈ {0, 1}, and f (0) = 0 ⇒ f ≡ 0. b) Now, we assume f (0) = 1. Then, f (−n) = f (n)∀n ∈ Z, and f (2m) = 2f (m)2 − 1, ∀m ∈ Z. If there is an integer k such that |f (k )| > 1, then |f (2k )| = |2f (k )2 − 1| = (f (k )2 − 1) + f (k )2 > |f (k )| > 1, and so we have an increasing sequence of integer |f (k )| < |f (2k )| < |f (4k )| < · · · < |f (2l k )| < · · · . But it contradict that f is a bounded function with all values in Z. Therefore f (Z) ⊂ {−1, 0, 1}. c) Let f (1) = cos θ with θ ∈ {π, by induction and the formula 2 cos α cos β = cos(α + β ) + cos(α − β ). Example 1.4. Define the function G : N ∪ {0} → Z by G(0) = 0 ; G(n) = n − G(G(n − 1)), n ∈ N. To show that G(n) = [ Proof. 1. Actually, we have 1 ≤ G(n) ≤ n and G(n − 1) ≤ G(n) ∀n ≥ 1. At first G(1) = 1, G(2) = 1.By induction, we assume 1 ≤ G(k ) ≤ k and G(k − 1) ≤ G(k ) ∀1 ≤ k ≤ n − 1. Then, we have 1 ≤ G(n − 1) ≤ n − 1, and so G(G(n − 1)) ≤ G(n − 1) ≤ n − 1, 1 = n − (n − 1) ≤ G(n) = n − G(G(n)) ≤ n − 1 < n, G(n) − G(n − 1) = 1 − [G(G(n − 1)) − G(G(n − 2))] ≥ 0 since 1 ≤ G(n − 2) ≤ G(n − 1) ≤ n − 1. 2. By induction, we can show G(n + 1) − G(n) = 1 or 0 ∀n. 3. Define F (n) = [α(n + 1)] where α =
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pdf高等数学习题精选张天德高等数学作为理工科学生必须学习的一门课程，难度较大，需要大量的练习才能够掌握。
而在众多的高等数学教材中，张天德教授的《数学分析习题集》可以说是经典之一。
因为其讲解深入浅出，提供了大量的例题和习题，能够帮助学生更好地掌握高等数学的理论和应用。
首先，张天德教授的这本《数学分析习题集》是一个PDF版本，非常方便学生进行下载和使用。
PDF文档格式可以在各种设备上进行浏览和打印，同时还可以进行笔记和注释，非常适合学生进行自主学习。
其次，张天德教授在讲解数学分析的理论知识时，采用了深入浅出的方式，让学生更加容易理解。
其讲解过程简单明了，将抽象的数学公式具体化，减轻了学生的学习负担。
因此，这本习题集不仅适合初学者，也适合有一定数学基础的学生进行深入学习。
最重要的一点是，这本习题集提供了大量的例题和习题，从易到难，从基础到应用，涵盖了高等数学各个方面的知识点。
这些例题和习题不仅有助于学生巩固理论知识，更能够帮助学生掌握数学的实际应用，提高解题能力。
在学习这本习题集时，建议学生注意以下几点：第一，不要只看习题的答案，一定要自己动手尝试解题。
这样才能够真正掌握数学的解题方法和思路，并且能够更好地应对考试中的各种复杂题目。
第二，适当地进行错题集的整理和总结。
这将帮助学生更好地理解自己的薄弱环节，并能够更好地弥补自己的不足。
当然，也可以找老师进行讲解和指导，从而更好地提高自己的数学水平。
第三，不断地进行练习和总结。
数学是一门基础性非常强的学科，只有不断地练习，不断地总结，才能够真正掌握其中的应用技巧和解题方法。
总之，张天德教授的《数学分析习题集》提供了大量的例题和习题，能够帮助学生更好地掌握高等数学的理论和应用。
这本习题集既适合初学者，也适合有一定数学基础的学生深入学习。
在学习过程中，学生需要注重练习和总结，才能够真正掌握高等数学的精髓。
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数学分析pdf
数学分析是一种应用于数学研究的技术。
它使用精密的数学语言对外部客观世界和内部抽象世界的大量杂乱的事实、规律、关系、性质、过程和结果进行深入地描述、解释和预测。
数学分析技术围绕着许多学科展开，如概率数学、统计数学、动态系统分析、矩阵分析、拓扑学等。
一、数学分析的定义
数学分析是一种专门研究函数、极限、积分、微分方程以及复杂几何体的数学技术。
它主要关注该学科的理论基础，并研究在特定条件下的函数的行为以及它们之间的关系。
二、数学分析的用途
数学分析有着应用于各行各业的广泛，它可以被运用在物理学和工程学中，以解决各类实际问题，如拟计划优化、精确测量、力学和热学等。
它还是建立数学模型的基础，可用于研究现实世界的有限变量的不确定性。
三、数学分析的内容
数学分析含有诸多概念、定义和定理，主要包括下列几部分：
（1）实数与有理数：实数和有理数的定义，以及它们的性质。
（2）函数：定义、基本概念，多项式、参数方程和曲线的性质，例如
局部极值、凹凸性等。
（3）微积分：求导数、积分、初等函数，定义和求证坐标系下函数的
最大值、最小值等内容。
（4）复数分析：复数的定义及其在极坐标、相位表达式和极角表示中
的性质，以及与微积分相关的定理。
（5）线性代数：向量、向量空间、矩阵、特殊形式、行列式、线性等
式组、变换和子空间等，还包括齐次线性方程组和线性方程组的解法。
四、数学分析的应用
数学分析也是物理学、工程学中数学运用的基础。
数学分析在许多领
域都得到了广泛应用，如品质管理、计算机科学、金融学、经济学、
生命科学、机械工程等。
它的理论和方法在许多实用领域得到了广泛，如建模仿真、最优化解决方案、计算解析和数值计算等。