《数学分析原理》设计论文.doc

（此文档为word格式，下载后您可任意编辑修改！）数学分析原理. Walter.Rudin M.北京：机械工业出版社，2004第五章 微 分 法本章除最后一节外，我们集中注意于定义在闭区间或开区间上的实函数，这并不是为了方便，当我们从实函数转到向量函数的时候，就会看到本质的差别，定义在上的函数的微分法将在第九章予以讨论。实函数的导数5.1 定义 设f是定义在a，b上的实值函数，对于任意的xa，b，做（差）商 （1）然后定义 （2）但是这里要由定义4.1可假设等式右端的极限存在。于是与f相联系得到函数f，它的定义域是a，b中所有使极限（2）存在的x点的集，f叫做f的导函数。如果f在x点有定义，就说f在x点可微。如果f在集Ea，b的每一点有定义，就说f在E上可微。可以在（2）中考虑左极限或右极限，从而就可得到左导数和右导数的定义。特别是在端点a，b上，在存在的前提下导数分别是左导数和右导数。但是我们不对单侧导数做详细讨论。如果f定义在（a，b）上，并有a<x<b，这是和前边一样，f(x)依然是以（1）和（2）来定义的，但此时f(a)和f(b)就没有定义了。5.2 定理 设f定义在a，b上，若f在a，b上任意一点x皆可微，则f在x点连续。证 由定理4.4，当时 定理的逆命题不成立。在某些孤立点不可微的连续函数式不难构造的。在第7章我们甚至还能得到一个在整条直线上都连续但处处不可微的函数。5.3 定理 设f和g定义在a，b上，且都在点xa，b可微，那么f+g，fg，f/g也都在x点可微，且：在c中我们自然要假设g（x） 0证 由定理4.4，（a）显然成立 令h=fg，则再两端同除以t-x，再注意当时（定理5.2）。（b）得证。 再令h=f/g，那么 令，再应用定理4.4及定理5.2，就可以得到（c）。 5.4 例 显然任何常数的异数皆为0.若f定义为f（x）=x，则f=1。反复运用（b）和(c)就可以证明xn是可微的，导函数为nxn-1，这里的n是任意整数，且n<0是必须x0.由此，每个多项式都是可微的，所以任一有理函数除掉那些使分母为零的点后也是可微的。 下一条定理通常被称为微分法的“链式法则”，用来求复合函数的导数，它也许是求导数的最重要的定理。在第9章还将看到它的更普遍的表述。 5.5 定理 设f在a，b上连续，f（x）在a，b上的某点x存在，g定义在一个包含f值域的区间I上，又在点f（x）处可微。若 则h在x点可微，且有 （3） 证 设y=f（x），由导数定义，知道 （4） （5）这里ta，b，sI，并且当时，当 时 。现在令s=f（t），先用（5），再用（4）就可以得到 设tx， （6） 由于f的连续性，知道当时，于是（6）右端趋于g(y)f(x)，这就得到了（3）式。 5.6 例 （a）设f定义为 （7） 先承认的导函数是（我们在第8章里讨论三角函数）。当x0时我们可以运用定理5.3及定理5.5，得到 （8） 在x=0点，由于1/x无定义，就不能使用这两个定理了，现在直接按导数定义来计算，对于t0 当t0时，这不能趋于任何极限，所以f(0)不存在。 （b）设f定义为 （9）同上我们可以求得 （10）在x=0，按导数定义计算，得到 令t0，就知道 （11） 所以f在所有点x可微，但是f不是连续函数，这是因为（10） 式右端第二项cos（1/x），当x0时不趋于任何极限。中值定理定义 设f是定义在度量空间X上的实值函数，称f在点pX取得局部极大值，如果存在着>0，当d（p，q）<且qX时有f（q）f（p）。局部极小值可以类似定义。下面的定理是导数的许多应用的基础。5.8 定理 设f定义在a，b上；xa，b，如果f在x点取得局部极大值而且 f（x）存在，那么f(x)=0。 对于局部极小值的类似的命题，自然也是对的。证 按照定义5.7选取，那么 若是x-<t<x，就应该 令tx，便知道f(x)0 若是x<t<x+，就应该 这又将表示f（），所以（）=0。5.9 定理 设 f是a，b上的连续函数，它们在（a，b）中可微，那么便有一点x（a，b），使得 注意：并不要求在区间端点上可微。证 ： 令 那么h在a，b上连续，在（a，b）内可微，且 （12） 要证明本定理，就得证明在某点x（a，b），h（）=0。 若果h是常数，那么不论在哪一点x（a，b），都有h（）=0。如果某个t（a，b）使得h（t）>h（a），设x是使h达到最大值的点（定理4.16），从（12）来看，x（a，b），于是定理5.8说明h（x）=0。如果有某个t（a，b）使得 h（t）<h（a），只要把在a，b内的那个x选的使h达到它的最小值，上述论证仍然成立。 这个定理常常叫做一般中值定理；下面的特殊情形就是通常所说的种植定理。 5.10 定理 设f是定义在a，b上的实连续函数，在（a，b）内可微，那么一定有一点x（a，b），使得 证 在定理5.9中取g（x）=x即得。 5.11 定理 设f在（a，b）内可微，（a）如果对于所有的x（a，b），f（x）0，那么f便是单调递增的。（b）如果对于所有的x（a，b），f（x）=0，那么f便是常数。（c）如果对于所有的x（a，b），f（x）0，那么f便是单调递减的。 证 所有结论都可以从下列等式获得： 这等式对于（a，b）中的任意一对点x1，x2，都成立，而x是x1与x2之间的某个点。 导数的连续性 我们已经看到（例5.6（b）一个函数f可以有处处存在、但在某些点间断的导数f（x）。可是，并不是每个函数都一定可以看做是某个函数的导函数。特别的一点事，在一个闭区间上处处存在的导函数与闭区间上的连续函数之间，却有一个重要的共同性质：任何中间值都能取到。确切的表述是： 5.12 定理 设 f是a，b上的实值可微函数，再设f（a）<<f（b），那么必有一点x（a，b）使f（x）=。 对于f（a）>f(b)的情形，当然也有类似的结果。 证 令g（t）=f（t）-t。于是g（a）<0，从而有某个t1（a，b）使得g（t1）<g(a)；同样，g（b）>0，从而有某个t2（a，b）使得g（t2）<g(b)。 因此，根据定理4.16，g在（a，b）的某点x上达到它在a，b上的最小值。再根据定理5.8，g（x）=0，因而f（x）=。 推论 如果 f 在a，b上可微，那么f在a，b上便不能有一类间断点。 但是f很可能有第二类间断点。 Lospital 法则 下面的定理在求极限时时常用到。 5.13 定理 假设实函数f和g在（a，b）内可微，而且对于所有x（a， b），g（x）0。这里-a<b+,已知 当xa时，A, (13) 如果 当xa时，f（x）0，g（x）0， （14） 或是 当xa时，g（x）+, (15)那么 当xa时，A . (16) 如果是xb,或者(15)中如果是g(x)-,各种类似的叙述自然也都是正确的。注意，我们现在是按照定义4.33推广了的意义来使用极限概念的。 证 先考虑-A+的情形。选择一个实数q使A<q，再选一个r使A<r<q。由（13）知道有一点c（a，b），使得当a<x<c有 <r， （17） 如果a<x<y<c，那么定理5.9说明有一点t（x，y）使得 =<r. （18） 先看（14）成立的情形.在（18）中令xa，便看到 r<q （a<y<c）. （19） 再看（15）成立的情形。在（18）中让y固定，我们可以选一点c1（a，y），使a<x<c1能够保证g（x）>g(y)及g（y）>0.将（18）两端乘以g(x)-g(y)/g(x),便得到<r-r+ (a<x<c1) (20)如果在（20）式中令xa，（15）式说明必有一点c2（a，c1）使 <q (a<x<c2). (21) 总之，（19）与（21）式都说明对于任意的q，只要A<q,便有一点c2，使得a<x<c2足以保证f（x）/g（x）<q。 同理，若是-<A+,选择p<A,便可找到一点c3，使得 P< (a<x<c3). (22) 结合起这两方面就得到了（16）式。 高阶导数 定义 如果在一个区间上有导函数f，而自身又是可微的，把的导函数记做，就叫做的二阶导数。照这样继续下去就得到，（），（），这许多函数，其中每一个是前一个的导函数。（）叫做f的n阶导函数。 为了要（）（x）在x点存在f（n-1）（x）必须在x点的某个邻域例存在（当x是定义f的区间的端点时，f（n-1）（x）必须在它有意义的那个单侧邻域例存在），而且f（n-1）必须在x点可微。因为在f（n-1）必须在x的邻域例存在，那么，f（n-2）又必须在x的邻域里可微。文献原文附录：Walter.Rudin. principles of mathematical analysisM.北京：机械工业出版社，2004 5 DIFFERNTIATIONIn this chapter we shall(except in the final section)confine our attention to real functions dfine on intervals or segements.This is not just a matter of convenience,since genuine differences appear when we pass from real functions to vector-valued ones.Differentiation of functions defined on Rk will be discussed in Chap 9.THE DERIVATIVE OF A REAL FUNCTION5.1 Definition Let f be defined (and real-valued) on a,b. Fof any xa,b from the quotient (1) and define(2)provided this limit exists in accordance with Definition 4.1 We thius associante with the function f a function f whose domain is the set of points x at which the limit (2) exists; f is called the derivative of f. If f is defined at a point x, we say that f is differentiable at x. If f is defined at every point of a set Ea,b,we say that f is differentiable on E. It is possible to consider right-hand and left-hand derivative,respectively.We shall not,however,discuss one-side derivatives in any detail. If f is defined on a segment(a,b) and if a<x<b,then f(x)is defined by (1) and (2),ad above. But f(a) and f(b) are not defined in this case.5.2 Theorem Let be defined on a,b.If it is differentiable at a point x a,b,then f is continuous at x. Proof As tx,we have,by Theorem 4.4 The converse of this theorem is not true. It is easy to construct continuous functions which fail to be differentiable at isolated points.In Chap 6 we shall even become acquainted with a function which is continuous on the whole line without being differentiable at any point!5.3 Theorem Suppose f and g are defined on a,b and are differentiable at a point xa,b.Then f+g,fg,and f/g are differentiable at x,and(a) (f+g)(x)=f(x)+g(x);(b) (fg)(x)=f(x)g(x)+f(x)g(x);(c) (f/g)(x)=g(x)f(x)-g(x)f(x)/g2(x)In (c),we assume of course that g(x)0. Proof (a) is clear,by Theorem 4.4.Let h=fg.Then If we divide this by t-x and note that f(t)f(x) as tx(Theorem5.2),(b) follows. Next, let h=f/g. Then Letting tx,and applying Teorems 4.4 and 5.2, we obtain(c).5.4 Examples The derivative of any constant is clearly zero.If f is defined byf(x)=x,then f(x)=1. Repeated application of (b) and (c) then shows that xn is differentiable,and that its derivative is nxn-1,for any integer n (if n<0, we have to restrict ourselves to x 0).Thus every polynomial is differentiable,and so is every rational function,except at the points where the denominator is zero. The following theorem is known as the “chain rule”for differentiation.It deals with differentiation of composite functions and is probably the most important theorem about derivateves.Ws shall meet more general versions of it in Chap 9.5.5 Theorem Suppose f is continuous on a,b,f(x) exists at some point xa,b,g is defined on an interval I which contains the range of f,and g is differentiable at the point f(x).If h(t)=g(f(t) (atb),then h is differentiable at x,and(3) h(x)=g(f(x)f(x). Proof Let y=f(x).By the definition of the derivative,we have(4) (5) where ta,b,sI,and u(t)0 as tx,v(s)0 as sy.Let s=f(t). Using first (5) and then (4),we botain or,if tx, (6) Letting tx,we see that sy,by the continuity of f,so that the right side of (6) tends to g(y)f(x),which gives (3).5.6 Examples (a) Let f be defined by (7) Taking for granted that the derivative of sinx is cos x (we shall discuss the trigonometric functions in Chap 8),we can apply Theorems 5.3 and 5.5 whenever x0,and obtain (8) At x=0,these theorems do not apply any longer,since 1/x is not denfined there,and we appeal directly to the definition:for t0, As to,this does not tend to any limit,so that f(0) does not exist. (b) Let f be defined by (9) As above,we obtain (10) At x=0,we appeal to the definition,and obtain Letting t0,we see that (11) Thus f is differentiable at all points x,but f is not a continuous function,since cos(1/x) in (10) does not tend to a limit as x0. MEAN VALUE THEOREMS 5.7 Definition Let f be a real function defined on a metric space X.We sayThat f has a local maximum at a point pX if there exists >0 such that f(q)f(p)for all qX with d(p,q)< . Local minima are defined likewise. Our next theorem is the basis of many applications of differentiation.5.8 Theorem Let f be defined on a,b;if f has a local maximum at a point x(a,b),and if f(x) exisets,then f(x)=0. The analogous statement for local minima is of course also ture. Proof choose in accordance with Definition 5.7,so that If x- <t<x,then Letting tx,we see that f(x)0. If x<t<x+,then Which shows that f(x)0.Hence f(x)=0. 5.9 Theorem If f and g are continuous real functions on a,b which aredifferentiable in (a,b), then there is a point x(a,b)at which Note that differentiability is not required at the endpointds. Proof Put Then h is continuous on a,b,h is differentiable in (a,b),and (12) To prove the theorem,we have to show that h(x)=0 for some x(a,b). If h is constant,this holds for every x(a,b).If h(t)>h(a) for some t(a,b),let x be a point on a,b at which h attains its maximum(Theorem 4.16). By(12),x(a,b),and Theorem 5.8 shows that h(x)=0.If h(t)<h(a) for some t(a,b),the same argument applies if we choose for x a point on a,b where h attains its minimum.This theorem is often called a generalized mean value theorem;the Following special case is usually referred to as “the”mean value theorem:5.10 Theorem If f is a real continuous function on a,b which is differetiable in (a,b),then there is a point x(a,b) at which Proof Take g(x)=x in Theorem 5.9.5.11 Theorem Suppose f is differentiable in (a,b). (a) If f(x)0 for all x(a,b),then f is monotonically increasing. (b) If f(x)=0 for all x(a,b),then f is constant. (c) If f(x)0 for all x(a,b),then f is monotonically decreasing. Proof All conclusions can be read off from the equation Which is valid,for each pair of numbers x1,x2in (a,b),for some x between x1 and x2. THE CONTINUITY OF DERIVATIVES We have already seen Example 5.6(b)that a function f my have a derivative fwhich exisets at every point,but is discontinuous at some point.Howere,not every function is a derivative.In particular,derivatives Which exist at every point of an interval have one important property incommon with functions which are continuous on an interval:Intermediate values are assumed(compare Theorem 4.23).The pecise statement follows.5.12 Theorem Suppose f is a real differentiable function on a,b andSuppose f(a)<<f(b).Then there is a point x(a,b) such that f(x)=A similar result holds of course if f(a)>f(b).Proof Put g(t)=f(t)- t.Then g(a)<0,so that g(t1)<g(a) for some t1(a,b),and g(b)>0,so that g(t2)<g(b) for some t2(a,b).Hence g attains its minimum on a,b(Theorem 4.16)at some point x such that a<x<b.By Theorem 5.8,g(x)=0.Hence f(x)= . Corollary If f is a differentiable on a,b,then fcannot have any simple discontinuities on a,b. But f my very well have discontinuities of the second kind. LHOSPITALS RULE The following theorem is frequently useful in evaluation of limits. 5.13 Theorem Suppose f and g are real and differentiable in (a,b),and g(x)0 for all x(a,b),where -ab+.Suppose (13) If (14) ), Or if (15) then (16) The analogous statement is of course also true if xb,or if g(x)- in (15). Let us note that we now use the limit concept in the extend sense of Definition 4.33 Proof We first consider the case in which -A<+.Choose a real number q such that A<q,and then choose r such that A<r<q.By(13)there is a point c(a,b) such that a<x<c implies (17) If a<x<y<c,then Theorem 5.9 shows that there is a point t(x,y)such that (18) Suppose (14) holds.Letting xa in (18),we see that (19) Next,suppose (15) holds. Keeping y fixed in (18),we can choose a point c1(a,y) such that g(x)>g(y) and g(x)>0 if a<x<c1.Multiplying (18) by g(x)>g(x),we obtain (20) If we let xa in (20),(15)shows that there is a point c2(a,c1) such that (21)