数学分析1-1(修改)
数学分析I,II,III
中国海洋大学本科生课程大纲课程属性：学科基础课程性质：必修一、课程介绍1.课程描述：数学分析是以极限为工具研究函数的学科，是数学专业的一门重要基础课，共分三个学期讲授。
数学分析针对数学类专业一、二年级学生开设，它一方面为后继课程提供所需的基础知识，同时又为培养学生利用数学工具进行独立工作的能力提供必需的训练。
学生学好这门课程的基本内容和方法，对后继课程的学习具有关键性的作用。
通过本课程的学习，要求学生掌握一元函数微积分学、多元函数微积分学与级数理论中的基本概念、基本理论和基本运算，并培养学生对数学问题的思维能力、论证能力、运算技能和独立分析、解决问题的能力。
本课程主要内容包括：数学分析I——函数、极限和连续、实数基本定理、导数与微分、微分学基本定理及应用、不定积分。
数学分析II——定积分、定积分的应用和近似计算、数项级数、广义积分、函数项级数、幂级数、Fourier级数和Fourier变换、多元函数的极限与连续。
数学分析III——多元函数的偏导数和全微分，极值和条件极值，隐函数存在定理，含参量的积分和含参量的反常积分，多元函数各种积分的定义、性质和运算，场论初步。
2.设计思路：本课程是专业基础课，为数学专业一二年级新生设置，教学历时3个学期，教学内- 9 -容为学生专业发展的后继学习奠定必要的理论基础。
课程内容的选取基于该课程作为分析类课程的基础性地位。
课程内容主要包括三大模块：单变量微积分学、多变量微积分学、级数理论；三大模块相互联系，体现了数学分析研究的基本内容和方法。
单变量微积分学是数学分析中最基础的部分，内容是研究函数的微分、积分及其应用，重用极限与连续的工具。
主要包括函数、极限和连续、实数基本定理、导数与微分、微分学基本定理及应用、不定积分、定积分、定积分的应用和近似计算、广义积分。
多变量微积分学是在单变量微积分学的基础上，将研究的一元函数推广为更为广泛的多元函数上去。
内容包括多元函数的极限与连续、多元函数的偏导数和全微分、极值和条件极值、隐函数存在定理、含参量的积分和含参量的反常积分、多元函数各种积分的定义、性质和运算、场论初步。
人B版数学选修1-1讲义：第1章 1-1-1 命题
1．1 命题与量词1．1.1 命题1．了解命题、真命题、假命题的概念及命题的构成．(重点)2．会判断所给语句是不是命题，并判断命题的真假性．(难点、易错点) 3．理解命题的结构形式，并能把命题改写成“若p，则q”的形式．[基础·初探]教材整理命题的概念及结构阅读教材P3～P4，完成下列问题．1．命题的定义在数学中，把用语言、符号或式子表达的，可以判断真假的陈述句叫做命题．2．命题的分类(1)真命题：判断为真的语句叫做真命题；(2)假命题：判断为假的语句叫做假命题．3．命题的结构(1)结构形式：若p，则q.(2)命题的条件是：命题中的p；命题的结论是：命题中的q.判断(正确的打“√”，错误的打“×”)(1)“指数函数的图象真漂亮”是命题．()(2)语句“陈述句都是命题”不是命题．()(3)命题“实数的平方是非负数”是真命题．()(4)“mx2＋2x－1＝0是一元二次方程”是真命题．()(5)“一个素数的平方仍是素数”的条件是“一个数是素数”．() 【解析】(1)×.因为漂亮没有明确的标准，无法判断对错，故(1)错．(2)×.这个句子无法判断真假，故(2)错．(3)√.(4)×.m＝0时2x－1＝0是一元一次方程，故(4)错．(5)√.【答案】(1)×(2)×(3)√(4)×(5)√[质疑·手记]预习完成后，请将你的疑问记录，并与“小伙伴们”探讨交流：疑问1：_____________________________________________________ 解惑：______________________________________________________ 疑问2：_____________________________________________________ 解惑：______________________________________________________ 疑问3：_____________________________________________________ 解惑：_______________________________________________________[小组合作型](1)求证3是无理数；(2)若x∈R，则x2＋4x＋4≥0；(3)你是高一的学生吗？。
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。
数学分析第一章
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设
k
是满足
k n
a
的最大的正整数，即
k +1 n
> a.
于是, a < k + 1 < k + 2 < b, 则 k + 1, k + 2 是
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5.实数集R具有稠密性.即任何两个不相等的实数之间必 有另一个实数,且既有有理数,也有无理数.
6.实数集R与数轴上的点具有一一对应关系.即任一实数 都对应数轴上唯一的一点,反之,数轴上的每一点也都唯 一的代表一个实数.
证 倘若a > b，设 a b > 0, 则 a b + ,
与 a < b + 矛盾.
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（6）实数与数轴上的点一一对应
实数集 R与数轴上的点可建立一一对应关系.
1. 这种对应关系，粗略地可这样描述： 设 P 是数轴上的一点 (不妨设在 0的右边), 若 P 在 整数 n与 n + 1之间，则 a0 n. 把(n, n + 1]十等分, 若点 P 在第 i 个区间，则 a1 i. 类似可得到 an, n 2, 3, L . 这时, 令点 p 对应于 a0 .a1a2 L an L .
大一数学分析1知识点总结
大一数学分析1知识点总结数学分析是学习高等数学的基础课程之一，它以极限理论为中心，研究函数的性质和变化规律。
作为大一学习数学的入门课程，数学分析1包含了许多重要的知识点。
在本文中，我将对这些知识点进行总结和归纳，以便更好地帮助大家巩固和理解这些概念。
1. 极限的定义和性质数学分析的核心概念之一就是极限。
极限的定义是指当自变量趋近于某一值时，函数的输出趋于一个确定的值。
对于给定的数列或函数，我们可以通过求取极限来确定其收敛性和发散性。
在学习极限的时候，我们需要掌握极限的基本性质，如唯一性、有界性、保号性等。
2. 函数的连续性连续性是数学分析中的一个重要概念。
如果一个函数在某一点的极限等于该点的函数值，则称该函数在该点连续。
函数的连续性可以根据极限的定义来判断。
一般来说，多项式函数、指数函数、对数函数等都是连续函数。
而分段函数、有理函数则可能存在不连续点。
3. 导数的定义和计算方法导数是函数变化率的度量，描述了函数在某一点的瞬时增量。
导数的定义是函数在该点的极限，也可以理解为函数的斜率。
通过导数，我们可以判断函数的增减性、凹凸性以及切线的斜率等。
常见的函数求导方法有常数法、幂函数法、指数函数法、对数函数法等。
4. 高阶导数和泰勒展开高阶导数是导数的导数，描述了函数变化的二阶、三阶……n阶性质。
高阶导数可以通过逐次求导来计算。
泰勒展开是将函数在某一点展开为无穷级数的形式，利用这种形式可以近似计算函数的值。
泰勒展开在物理、工程等领域常常被广泛应用。
5. 积分的概念和计算方法积分是函数的面积与变量的乘积，描述了函数的累积效应。
积分的概念可以通过极限来定义，常见的积分符号为∫。
对于给定的函数，我们可以通过不同的积分方法来计算其积分值，如定积分、不定积分、换元法、分部积分法等。
6. 微分方程和其应用微分方程是数学分析的重要分支，研究了函数与其导数之间的关系。
微分方程广泛应用于自然科学、工程技术等领域。
在大一数学分析1中，我们主要学习了常微分方程的基本概念和解法，如可分离变量法、线性齐次方程法、齐次线性方程法等。
1.数学分析1
《数学分析1》课程实践教学大纲课程编号：0501B01课程名称：数学分析课程类别：必修本课程适用专业：数学教育专业本课程总学时： 56 本课程实践学时： 10执笔人：朱海燕审核人：王习娟一、本实践课程简要介绍本课程是数学教育专业的主干专业基础课程，主要讲授数列及函数的极限与一元微分学，旨在对学生进行系统而严格的近现代数学思想和方法的教育与训练，使学生牢固掌握数学分析的理论和方法，为学生学习一元积分学及物理、工程技术、计算机等学科的有关科目打下坚实的基础，也为学生今后从事中学数学教育提供系统而全面的分析知识。
二、本课程实践教学目标通过教学，应使学生正确理解和掌握数学分析的基本概念、基本理论，基本掌握数学分析中的论证方法；较熟练地获得本课程所要求的基本演算能力，为进一步学习数学专业课程打下必要的基础；同时，深刻理解本课程与初中数学中有关的内在联系，以指导初中数学教学。
三、本课程实践学时分配四、本课程实践教学成绩评定方法本课程实践任务是根据数学分析课程的章节而定，重在培养学生对各章知识点的掌握，每次实践课后会布置相应的讨论和测试题，采用百分制，或开卷或闭卷，以总成绩的10% 记入平时成绩。
五、本课程实践教学参考资料[1] 刘玉链编.数学分析[M].北京：高等教育出版社.[2] 吉米多维奇.数学分析习题集[M]. 北京：高等教育出版社.六、本课程实践教学建议及说明执行本大纲应注意：1．要加强实践性，让学生多动手、动脑思考问题；2．指导教师要精心设计实践内容，让学生都能进行主动思考问题，避免实践课成为教师一言堂。
附实践内容及要求实践一【1】实践目的：使学生掌握实数的基本性质和确界原理，并使学生理解函数的概念，熟悉与函数性态有关的一些常用术语。
【2】实践要求：理解实数的有序性、稠密性、封闭性及确界原理等概念，做出本章的小结，并在课上和课后完成一定的实践作业。
【3】实践方式：先由学生讨论并小结，再由教师提出问题并共同探讨和解决问题。
人教A版数学选修1-1-1
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(1)“x>2, y>3” “x+ y>5”； (2) “a是菱形”____ “ a是正方形” (3) “x2 -9=0”____ “x=3”; (4)“A= ｛x丨2＜x＜5｝” “B=｛x丨0＜x＜6｝”； (5)“x=y=0”____ “x2 + y2 =0 ” ； (6)“三角形ABC是等腰三角形”____ “三角形ABC是等腰直角三角形” ;
南昌航空大学 往年 数学分析(1)试卷 A卷参考答案
南昌航空工业学院2006—2007学年第一学期期末考试课程名称：数学分析（1）A 卷参考答案及评分标准 一、计算下列极限：（每小题5分，共20分）1.1132lim()2n nnn →∞+解：1111322lim n()322lim ()x x x x x xx l xe→∞+→∞+=1132ln 21lim x x x xe⎛⎫ ⎪⎪ ⎪⎪⎪⎪⎝⎭+→∞=()1111322lim exp 3ln32ln 22x x x x x →∞+⎧⎫⎪⎪⎪⎪=⎨⎬⎛⎫⎪⎪ ⎪⎪⎪⎝⎭⎩⎭+=分由归结原则11322lim ()n n n n →∞+= --------5分2.ln sin 5lim ln sin 3x xx +→解： 原式5cos5/sin 5lim 3sin 3/cos3x x xx x +∞∞→= ---------3分05sin 5lim 3sin 3x x x +→=001= ---------5分3. 21lim(2)[]12x x x →-=-解： 2x >当时知11(2)(2)[]12x x x --<-≤-，则21lim (2)[]12x x x +→-=-；2x <当时知1(2)[]1(2)2x x x ≤-<---1，则21lim(2)[]12x x x -→-=-.----4分 故21l i m (2)[]12x x x →-=-. --------5分4. 1lim1ln xx x x x x →--+解： 原式011(1ln )lim1x x x x x x →-+=-+1(1ln )lim 1x x x x x x x x →-+=- --------3分()21lim 1(1ln )(1ln )x x x x x x x x x x →=--+-+-2= ---------5分二、求下列导数：（每小题5分，共20分）1. 22log 237y x x =--， 求dydx解： 243ln 2(237)dy x dx x x -=⋅-- --------5分2.2216x y x x +=--，求)(n y 解： 由735(3)5(2)y x x =+-+ ---------3分 所以()1173(1)!5(3)5(2)n n n n y n x x ++⎛⎫=-+ ⎪-+⎝⎭ ---------5分3. tan 2t x t t y =+⎧⎨=⎩, 求22dx yd解： 2/2l n 2/s e c 1td y d y d t d x d x d t t ==+ --------3分221/d y dy dy dx dt dx dx dt ⎛⎫= ⎪⎝⎭=22122(sec 1)(ln 2)2ln 2sec tan 2t t t t t ++---------5分4.30()0(,)x e x f x ax b x a b ⎧>=⎨+≤⎩为常数，)0(f '存在，求)(x f 的导数. 解： '(0)f 存在知()f x 在x=0连续得300(0)lim()lim 1xx x f ax b e -+→→=+==故1b =又''(0)(0)f f -+=(0)f '=得 30lim 33xx a e +→== -------3分故33,0()3,0x e x f x x ⎧>=⎨≤⎩ . ------5分三、设0()f x ''存在,试证: 00020()()2()limh f x h f x h f x h →++--=)(0x f ''.（8分）证明： 由0''(),f x 存在则0'()()f x f x x x =和在连续，于是左式000'()'()lim2h f x h f x h h →+--= -------4分=000000'()'()'()'()1lim lim 2h h f x h f x f x f x h h h →→+---⎛⎫+ ⎪⎝⎭ -------6分 0''()f x ==右式. 证毕 -------8分四、设0)(,),0(,],0[=a f a a f 且内可导在上连续在,证明:存在一点),0(a ∈ξ,使 ()'()0f f ξξξ+=.(7分)证明：()()F x xf x =令，则()[0,](0,)F x a a 在连续，在内可导，而(0)()0F F a == 由罗尔定理,知(0,),'()0.a F ξξ∈=存在使得证毕. ------7分五、求函数1)(23+--=x x x x f 的极值点,曲线的拐点及单调与凹凸区间.(15分)解：易知x ∈ ，'()(31)(1)f x x x =+-，''()21f x x =-（3）.()0f x '=1211,3x x ==-()0f x ''=13x =由上表得：()y f x =的单增区间：(,],[1,)3-∞-+∞；单减区间：[,1]3-；极大值：132()327f -=；极小值：(1)0f =； 凹区间：1[,)3+∞；凸区间：1(,]3-∞；拐点：116(,)327. --------15分六、1.什么是区间套?试叙述区间套定理；2.试叙述数列极限Cauchy(柯西)收敛准则. （每小题5分，共10分）解： 1. 闭区间列n n {[a ,b ]}具有如下性质：11()[,][,],1,2,;n n n n i a b a b n ++⊂= ()lim()0n n n ii b a →∞-=则称n n {[a ,b ]}为闭区间套，简称区间套. -----3分区间套定理：n n 若{[a ,b ]}是一个闭区间套，则存在唯一的实数ξ，使得 ξ[,]1,2,,n n a b n ∈= ， ,1,2,.n n a b n ξ≤≤= 即2.{}n a 数列收敛的充要条件是： 0,N ε+∀∃∈> ，使当,n m N >时，有 .n m a a ε-<七、 试叙述：(每小题5分，共20分)1.f 在某区间I 上连续的定义；2.f 在某区间I 上一致连续的定义；3.f 在某区间I 上非一致连续的定义；4.并利用定义证明:2)(x x f =在(,)-∞+∞上非一致连续.解：1.,f I 设在上有定义0,x I ∈对任意00lim ()().x x f x f x →=若f I 则称在上连续.----5分2.,f I 设在上有定义,0,()0εδδε∀>∃=>对∀对','',x x I ∈,'''x x δ-<当时有(')('').f x f x ε-< ----10分 3.,f I 设在上有定义00,ε∃>0,δ∀>对,',''x x I ∃∈''',x x δ-<虽但0(')('').f x f x ε-≥ -----15分4.102,ε=取0,δ∀>对1142',,''x x δδδδ=+=+取 ---18分 虽'''x x δ-<，但321442(')('').()f x f x δδδ-=+> 所以,f ∞+∞（x ）在(-)上非一致连续. --20分。
2019-2020人教A版数学选修1-1 第1章 1.1 1.1.1 命题课件PPT
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2．下列语句是命题的是( )
①三角形内角和等于 180°；②2>3；③一个数不是正数就是负数；
④x>2；⑤019 央视猪年春晚真精彩啊！
A．①②③ C．①②⑤
B．①③④ D．②③⑤
A [①、②、③是陈述句，且能判断真假，因此是命题，④不 能判断真假，⑤是感叹句，故④、⑤不是命题．]
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2．下列命题： ①若 xy＝1，则 x，y 互为倒数； ②同一平面内四条边相等的四边形是正方形； ③平行四边形是梯形； ④若 ac2>bc2，则 a>b. 其中真命题的序号是________． ①④ [①④是真命题，②同一平面内四条边相等的四边形是菱 形，但不一定是正方形，③平行四边形不是梯形．]
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2．命题的结构 思考 2：命题“实数的平方是非负数”的条件与结论分别是什 么？ [提示] 条件是：“一个数是实数”，结论是：“它的平方是非 负数”．
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1．语句“若 a>b，则 a＋c>b＋c”( )
A．不是命题
B．是真命题
C．是假命题
D．不能判断真假
B [结合不等式的性质可知，若 a>b，则 a＋c>b＋c，是真命题．]
(2)①中 x 有范围，可以判断真假，因此是命题；②是疑问句， 不是命题；③是陈述句，但“大”的标准不确定，无法判断真假，因 此不是命题；④是陈述句且能判断真假，因此是命题；⑤是祈使句， 不是命题．]
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判断一个语句是否是命题的两个关键点 1命题是可以判断真假的陈述句，因此，疑问句、祈使句、感 叹句等都不是命题. 2对于含变量的语句，要注意根据变量的取值范围，看能否判 断其真假，若能，就是命题；若不能，就不是命题. 提醒：若语句中含有变量，但变量没有给出范围，则该语句不是 命题.
苏教版高中数学选修1-1第1章§1.11.1.2.docx
1.1.2 充分条件和必要条件 课时目标 1.理解充分条件、必要条件、充要条件的意义.2.会求(判定)某些简单命题的条件关系．1．如果已知“若p ，则q ”为真，即p ⇒q ，那么我们说p 是q 的____________，q 是p的______________．2．如果p ⇒q ，且q ⇒p ，就记作__________．这时p 是q 的______________条件，简称________条件，实际上p 与q 互为________条件．如果p ⇒q 且q ⇒p ，则p 是q 的________________________条件．一、填空题1．“x >0”是“x ≠0”的____________条件．2．对于三个集合A ，B ，C ，条件A ⊆B ，B ⊆C ，C ⊆A 是A ＝B ＝C 的________条件．3．设集合M ＝{x |0<x ≤3}，N ＝{x |0<x ≤2}，那么“a ∈M ”是“a ∈N ”的____________条件．4．“k ＝1”是“直线x －y ＋k ＝0与圆x 2＋y 2＝1相交”的____________条件．5．a <0是方程ax 2＋2x ＋1＝0至少有一个负数根的____________条件．6．α＝π6＋2k π(k ∈Z )”是“cos 2α＝12”的____________条件． 7．不等式(a ＋x )(1＋x )<0成立的一个充分而不必要条件是－2<x <－1，则a 的取值范围是________．8．函数y ＝ax 2＋bx ＋c (a >0)在[1，＋∞)上单调递增的充要条件是__________．二、解答题9．下列命题中，判断条件p 是条件q 的什么条件：(1)p ：|x |＝|y |，q ：x ＝y .(2)p ：△ABC 是直角三角形，q ：△ABC 是等腰三角形；(3)p ：四边形的对角线互相平分，q ：四边形是矩形．10.已知P ＝{x |a －4<x <a ＋4}，Q ＝{x |x 2－4x ＋3<0}，若x ∈P 是x ∈Q 的必要条件，求实数a 的取值范围．能力提升11．记实数x 1，x 2，…，x n 中的最大数为max{x 1，x 2，…，x n }，最小数为min {}x 1，x 2，…，x n .已知△ABC 的三边边长为a ，b ，c (a ≤b ≤c )，定义它的倾斜度为l ＝max ⎩⎨⎧⎭⎬⎫a b ，b c ，c a ·min ⎩⎨⎧⎭⎬⎫a b ，b c ，c a ， 则“l ＝1”是“△ABC 为等边三角形”的____________条件．12．已知数列{a n }的前n 项和为S n ＝(n ＋1)2＋c ，探究{a n }是等差数列的充要条件．1．判断p 是q 的什么条件，常用的方法是验证由p 能否推出q ，由q 能否推出p ，对于否定性命题，注意利用等价命题来判断．2．证明充要条件时，既要证明充分性，又要证明必要性，即证明原命题和逆命题都成立，但要分清必要性、充分性是证明怎样的一个式子成立．“A 的充要条件为B ”的命题的证明：A ⇒B 证明了必要性；B ⇒A 证明了充分性．“A 是B 的充要条件”的命题的证明：A ⇒B 证明了充分性；B ⇒A 证明了必要性．1.1.2 充分条件和必要条件知识梳理1．充分条件 必要条件2．p ⇔q 充分必要 充要 充要 既不充分又不必要作业设计1．充分不必要解析 对于“x>0”⇒“x ≠0”，反之不一定成立．因此“x>0”是“x ≠0”的充分不必要条件．2．充要解析 由A ⊆B ，B ⊆C ，得A ⊆C ；又因C ⊆A ，所以A ＝C ，同理得A ＝B.由A ＝B ＝C ，得A ⊆B ，B ⊆C ，C ⊆A.3．必要不充分解析 因为N M.所以a ∈M 是a ∈N 的必要而不充分条件．4．充分不必要解析 把k ＝1代入x －y ＋k ＝0，推得“直线x －y ＋1＝0与圆x 2＋y 2＝1相交”；但“直线x －y ＋k ＝0与圆x 2＋y 2＝1相交”不一定推得“k ＝1”．故“k ＝1”是“直线x －y ＋k ＝0与圆x 2＋y 2＝1相交”的充分不必要条件．5．充分不必要解析 当a<0时，由韦达定理知x 1x 2＝1a <0，故此一元二次方程有一正根和一负根，符合题意；当ax 2＋2x ＋1＝0至少有一个负数根时，a 可以为0，因为当a ＝0时，该方程仅有一根为－12，所以a 不一定小于0.由上述推理可知，a<0是方程ax 2＋2x ＋1＝0至少有一个负数根的充分不必要条件．6．充分不必要解析 ∵当α＝π6＋2k π(k ∈Z )时， cos 2α＝cos ⎝⎛⎭⎫π3＋4k π＝12， ∴“α＝π6＋2k π(k ∈Z )”是“cos 2α＝12”的充分条件．而当α＝－π6时，cos 2α＝12， 但不存在k ∈Z 使得－π6＝π6＋2k π， ∴“α＝π6＋2k π(k ∈Z )”不是“cos 2α＝12”的必要条件． 7．a >2解析 不等式变形为(x ＋1)(x ＋a )<0，因当－2<x <－1时不等式成立，所以不等式的解为－a <x <－1.由题意有(－2，－1)(－a ，－1)，∴－2>－a ，即a >2.8．b ≥－2a 解析 由二次函数的图象可知当－b 2a≤1，即b ≥－2a 时，函数y ＝ax 2＋bx ＋c 在[1，＋∞)上单调递增．9．解 (1)∵|x |＝|y |⇒x ＝y ，但x ＝y ⇒|x |＝|y |，∴p 是q 的必要不充分条件．(2)△ABC 是直角三角形⇒△ABC 是等腰三角形．△ABC 是等腰三角形⇒△ABC 是直角三角形．∴p 既不是q 的充分条件，也不是q 的必要条件．(3)四边形的对角线互相平分⇒四边形是矩形．四边形是矩形⇒四边形的对角线互相平分．∴p 是q 的必要不充分条件．10．解 由题意知，Q ＝{x |1<x <3}，Q ⇒P ，∴⎩⎪⎨⎪⎧a －4≤1a ＋4≥3，解得－1≤a ≤5. ∴实数a 的取值范围是[－1,5]．11．必要不充分解析 当△ABC 是等边三角形时，a ＝b ＝c ，∴l ＝max ⎩⎨⎧⎭⎬⎫a b ，b c ，c a ·min ⎩⎨⎧⎭⎬⎫a b ，b c ，c a ＝1×1＝1. ∴“l ＝1”是“△ABC 为等边三角形”的必要条件．∵a ≤b ≤c ，∴max ⎩⎨⎧⎭⎬⎫a b ，b c ，c a ＝c a. 又∵l ＝1，∴min ⎩⎨⎧⎭⎬⎫a b ，b c ，c a ＝a c， 即a b ＝a c 或b c ＝a c， 得b ＝c 或b ＝a ，可知△ABC 为等腰三角形，而不能推出△ABC 为等边三角形． ∴“l ＝1”不是“△ABC 为等边三角形”的充分条件．12．解 当{a n }是等差数列时，∵S n ＝(n ＋1)2＋c ，∴当n ≥2时，S n －1＝n 2＋c ，∴a n ＝S n －S n －1＝2n ＋1，∴a n ＋1－a n ＝2为常数．又a 1＝S 1＝4＋c ，∴a 2－a 1＝5－(4＋c )＝1－c ，∵{a n }是等差数列，∴a 2－a 1＝2，∴1－c ＝2.∴c ＝－1.反之，当c ＝－1时，S n ＝n 2＋2n ，可得a n ＝2n ＋1 (n ≥1)为等差数列，∴{a n }为等差数列的充要条件是c ＝－1.。