泰勒公式外文翻译

1、精品文档1欢迎下载f x -.-h = f x 厂 f x hn!(1)f n x at the point x, thenTaylors Formula and the Study of Extrema1. Taylors Formula for MappingsTheorem 1. If a mapping f：u_；Y from a neighborhood u =u x of a point x ina no rmed space X into a no rmed space Y has derivatives up to order n -1 in elusivein U and h

2、as an n-th order derivativeas hr o.Equality (1) is one of the varieties of Taylors formula, written here for rather gen eral classes of mapp in gs.Proof. We prove Taylors formula by in duct ion.For n =1 it is true by definition ofAssume formula (1) is true for somen _ 1 三 N .The n by the mean-value

3、theorem, formula (12) of Sect. 10.5, and the in duct ion hypothesis, we obta in.as hr o.Weshall not take the time here to discuss other versions of Taylors formula, which are sometimes quite useful. They were discussed earlier in detail for numerical functions. At this point weleave it to the reader

4、 to derive them (see,for example, Problem 1 below).2.Methods of Studying Interior ExtremaUsing Taylors formula, we shall exhibitnecessary conditions and alsosufficie nt con diti ons for an in terior local extremum of real-valued functions defined on an open subset of a normed space. As we shall see,

5、 these conditions are analogous to the differentialconditionsalready known to us for an extremumof a real-valued function of a real variable.Theorem 2. Let f ： u ； R be a real-valued function defined on an open set Uin a normed space X and having continuous derivatives up to order k-1 丄 1 inclusive

6、in a neighborhood of a pointx u and a derivative f k x of order kat the point x itself.丄ff, x f X 十f n x干f , X ：劭-精品文档2欢迎下载of the real parameterf x tho -f x is the same as that off k x hk for sufficiently smallt; hence inthat case there cannot be two vectorsho, h at which the formf k x assumesIf (x)

7、=o,f ( 衣)=o and “卜)工0, then for x to be an extremum of the function f it is:n ecessary that k be eve n and that the formf (k 取 yk be semidefi nite,andsufficient that the values of the formf k xhk on the unit sphere h =1 beboun ded away from zero; moreover, x is a local mi nimum if the in equalitiesf

8、 k Xhk i. .0,hold on that sphere, and a local maximum iff u x hk 会 |kwhere :讪 is a real-valued function, and:逍h = 0 as hr 0.We first prove the n ecessary con diti ons.Since f k x =0, there exists a vectorh =o on which f k x ho 0. Then for valuest sufficiently close to zero,f (x 地)-f (x )=* f & 卜th k

9、 ythg |)h0 k = f(k 权 hk 七(也0 恫k jkand the expressi on in the outer pare ntheses has the same sig n asFor x to be an extremum it is necessary for the left-hand side (and hence also the right-hand side) of this last equality to be of constant sign when t changes sign. But this is possible only ifk is

10、even.This reasoning shows that if x is an extremum, then the sign of the differenee values with opposite sig ns.We now turn to the proof of the sufficie ncy con diti ons. For defi nite ness weconsider the case when f k xhk ：只 o for h =1. Thenf (x +h J-f (x )=丄 f g & hk +a(h ” k!精品文档3欢迎下载and, since :

11、刈h 厂.o as h；o, the last term in this in equalityis positive for allvectorsh =0 sufficie ntly close to zero. Thus, for all such vectors h,f X h -f x .0,that is, x is a strict local minimum.The sufficient condition for a strict local maximum is verified similiarly.Remark 1. If the space X is fin ite-d

12、ime nsion al, the unit spheresx；1 withcenter at x 三x , being a closed bounded subset of X, is compact. Then the continuous functionfk x hk =二 ik f x hhik (a k-form) has both a maximal and amini mal value on Sx;1 . If these values are of opposite sig n, the n f does nothave an extremum at x. If they

13、are both of the same sign, then, as was shownin Theorem 2, there is an extremum. In the latter case, a sufficient conditionfor an extremum can obviously be stated as the equivale nt requireme nt that the form f k xhk be either positive- or negative-definite.It was this form of the condition that we

14、encountered in studying realvaluedfunctions on Rn.Remark 2. As we have seen in the example of functions f ：RJR , thesemi-definitenessof the form f k x hk exhibited in the necessary conditions foran extremum is not a sufficient criterion for an extremum.Remark 3. In practice, whe n study ing extrema

15、of differe ntiable functionsone normally uses only the first or second differentials.If the uniqueness andtype of extremum are obvious from the meaning of the problem being studied, one can restrict atte ntio n to the first differe ntial whe n seek ing an extremum,simply finding the point x where f,

16、 x =03. Some ExamplesExample 1.Let L c1 R3;Rand f c1 a,b】R . In other words,J,u2,u3 LJ,u2,u3 is a continuously differentiable real-valued function definedin R3 and xf x a smooth real-valuedfunction defined on the closed intervala,b 二R.Con sider the fun cti onF : C 1 a,b；R R(2)精品文档4欢迎下载defined by the

17、 relationf - C1 a,bjR F fb L x,f x , f, x d RThus, (2) is a real-valued functional defined on the set of functionsC 1 a,b JR .The basic variati onal prin ciples conn ected with moti on are known in physics and mechanics.According to these principles,the actualmotions aredist in guished among all the

18、 con ceivable moti ons in that they proceed along trajectories along which certainfunctionalshave an extremum. Questionsconnected with the extrema of functionalsare central in optimal control theory.Thus, finding and study ing the extrema of function als is a problem of in tri nsic importa nee, and

19、the theory associated with it is the subject of a large area of an alysis - the calculus of variati ons. We have already done a few things to make the transitionfrom the analysis of the extrema of numericalfun cti ons to the problem of finding and study ing extrema of function als seem natural to th

20、e reader. However, we shall not go deeply in to the special problems of variati onal calculus, but rather use the example of the fun cti onal (3) to illustrate only the general ideas of differentiationand study of local extremacon sidered above.Weshall show that the functional(3) is a differentiatem

21、apping and find itsdiffere ntial.We remark that the fun ctio n (3) can be regarded as the compositi on of the mapp ingsFi f Ijx =L x,f x ,f, xdefi ned by the formulaF1 : C 1 a,b；R;C a,b ；R(5)followed by the mapp ingg C a,b JR - F2 g Jg xdx RBy properties of the integral,the mapping F2 is obviously l

22、inear andcontinuous, so that its differentiability is clear.We shall show that the mapp ingF1 is also differe ntiable, and that精品文档5欢迎下载for h wc1 a,b ；R .F1,(f h(x F2L(x, f(x ,f 宰)+&L(x,f (x )f ,(x 加(x)(7)In deed, by the corollary caseto the mean-value theorem, we can writein the prese ntj.:j.:3 32

23、21 1u umaxo空鱼Lx,f x h x ,f , x h x L x, f x , f, x-2L x,f x , f, x h x - IL x,f x ,f, x h, x-sup 乩 u 处 - :：1 L u 1 u .川_迅 u 仁u V3L U .-：0 - 1乞3 maxL u 印 _ ；L u ： max”！(8)0 壬禅1 -,2,3where u =J,u2,u3 and :1, 2，.3.If we now recallthat the norm f c1 of the functionf in C 1 a,b】R ismaxJfL，fj (where | f c

24、 is the maximumabsolute value of the fun cti on on the closed in terval a,b), the n, sett ing u1 二x , u2 =f x , u3 二 f,x, 卫二。，3=hx, an d3=h,x,we obta in from in equality (8), tak ing acco unt of the uniform con ti nuity of the functions :iLu1,u2,u3 ,i =1,2,3, on bounded subsets ofR3 , that= ohc1 as

25、hc1L；But this means that Eq. (7) holds.By the cha in rule for differe ntiati ng a composite fun cti on, we now con clude that the fun cti onal (3) is in deed differe ntiable, andbF, f h2Lx,f x,fx hx ；：3Lx,f x ,f, x h,x dx(9)aWe ofte n con sider the restrictio n of the fun cti onal (3) to the affine

26、spaceconsistingof the functions f C 1 a,b；R that assume fixed values f a =A ,f b =B at the endpoints of the closed interval a,bl. In this case, the functionsh in the tangent space TC： , must have the value zero at the endpoints of the closed interval a,b . Taking this fact into account, we may integ

27、rate by parts in (9) and bring it into the form2U2U3 3 u u精品文档6欢迎下载0,0 and 10 . We confinef,x.1 f Xon the closed in terval0,11(10)of course un der the assumpti on that L and f bel ong to the corresp onding class C 2 .In particular, if f is an extremum (extremal) of such a function al, the n byTheore

28、m 2 we have Ff h = for every functionh =C 1 a,bR such that ha 二hb=0.From this and relation(10) one can easily conclude (see Problem 3 below) thatthe fun cti on f must satisfy the equati on-2Lx,f x,f,x -d- .3Lx,f x,f,x =0(11)This is a freque ntly-e ncoun tered form of the equati on known in the calcu

29、lusof variati ons as theEuler-Lagra nge equati on.Let us now con sider some specific examples.Example 2. The shortest-path problemAmong all the curves in a pla ne joining two fixed poin ts, findthe curve thathas mini mal le ngth.The an swer in this case is obvious, and it rather serves as a check on

30、 the formal computations we will be doing later.Weshall assume that a fixed Cartesia n coord in ate system has bee n chose n in theplane, in which the two points are, for example, ourselves to just the curves that are the graphs of functions m1 0,1； Rassu ming the value zero at both ends of the clos

31、ed in terval0,11 . The len gthof such a curveF(f讥1+(f (x dx(12)depe nds on the functionf and isa fun cti onalof the type con sidered in Example1. In this case the functionL has the formLu1,u2,u-J_u32and therefore the n ecessary con diti on (11) for an extremal here reduces to the equati ond 丨f Yx)=0

32、叫1+(叹)from which it follows that二常数(13):2LXf精品文档7欢迎下载0,0 and 10 . It followsv = _ 2gx(15)y = f x on the closedSince the function一uis not constant on any interval,Eq. (13) is possible.1 u2only if f x =const ona,b . Thus a smooth extremal of this problem must be alin ear fun cti on whose graph passes

33、through the points that f x 三0 , and we arrive at the closed interval of the line joining the two give n poin ts.Example 3. The brachistochr one problemThe classical brachistochr one problem, posed by Joha nn Berno ulli I in 1696,was to find the shape of a track along which a point mass would pass f

34、rom a prescribed point P to another fixed point P at a lower level under the action of gravity in the shortest time.Weneglect friction,of course. In addition, we shall assume that the trivialcase in which both points lie on the same vertical li ne is excluded.In the vertical plane passing through th

35、e pointsP0 and R we introduce arecta ngular coord in ate system such thatP is at the origi n, the x-axis isdirected vertically dow nward, and the point R has positive coord in ates 百小.We shall find the shape of the track among the graphs of smooth functions defined on the closed interval0,百 1 and sa

36、tisfyingthe conditionf0 =0, f 为二At the momentwe shall not take time to discuss this by no meansuncontroversial assumpti on (see Problem 4 below).If the particle bega n its desce nt from the pointP with zero velocity, thelaw of variati on of its velocity in these coord in ates can be writte n as(14)R

37、ecalling that the differential of the arc length is computed by the formulawe find the time of descentalong the trajectory defined by the graph of the function(16)ds精品文档8欢迎下载in terval0,x1精品文档9欢迎下载For the functional (16)ddxL u1,u2 ,u3and therefore the condition (11) for an extremum reduces in this ca

38、se to the equati on=0,from which it follows thatfx.c.x(17)“+(f，2(x)where c is a non zero con sta nt, since the points are not both on the line.Taking account of (15), we can rewrite (17) in the form空=c. xdsHowever, from the geometric point of view空=costp=sin Cpds dssamevertical(19)(18)where is the a

39、n gle betwee n the tangent to the trajectory and the positivex-axis.By comparing Eq. (18) with the second equation in (19), we findx =4 sin2 :c(20)But it follows from (19) and (20) thatsin2 :cdy _dy dx _tg ,dx d齐芯，芥旳頁旳乔from which we findsin2 :c(21)1Setting 2 =a and 2 =t, we write relations (20) and

40、(21) as2 c2x = a 1 - cost y =at -sint 亠bSi nee a=o, it follows that x =o on ly for t=2k：，kwz. It follows of the fun ctio n (22) that we may assume without loss of gen erality that the(22)from the formparameter value t =o corresp onds to the pointF0 =o,o . In this case Eq. (21)精品文档10欢迎下载精品文档(23)11欢迎下

41、载implies b =o, and we arrive at the simpler formx =a 1 -cost y =a t sint for the parametric definition of this curve.Thus the brachistochr one is a cycloid hav ing a cusp at the in itial point where the tangent is vertical. The con sta nt a, which is a scali ng coefficie nt,must be chosen so that th

42、e curve (23) also passes through the pointR . Such a choice, as one can see by sketching the curve (23), is by no meansalways unique, and this shows that the n ecessary con diti on )for an extremum is in gen eral not sufficient. However, from physical considerations it is clear which of the possible

43、 values of the parameter a should be preferred (and this, of course, can be con firmed by direct computati on).精品文档12欢迎下载-sup f X + 0 1=o，x - f，x诂泰勒公式和极值的研究1.映射的泰勒公式定理1如果从赋范空间X的点x的邻域U二U x到赋范空间Y的映射f ：U Y在U中有直到n-1阶(包括n-1在内)的导数，而在点x处有n阶导数。f n x，那么当h； o时有f(x卄冃心口卡+；卫汀+0 ；等式(1)是各种形式的泰勒公式中的一种，这一次它确实是对非常一般的

44、函数 类写出来的公式了。我们用归纳法证明泰勒公式(1)。当n=1时，由f,x的定义，式成立。假设(1)对n 一1 EN成立。于是根据有限增量定理，5章中公式(12)和所作的归纳假设，我们得到，当h 0时 成立。丄 f (n Jx hn ； n! *这里我们不再继续讨论其他的，有时甚至是十分有用的泰勒公式形式。当时，在研究数值函数时，曾详细地讨论过它们。现在我们把它们的结论提供给读者(例如，可参看练习1)。2.内部极值的研究我们将利用泰勒公式指出定义在赋范空间的开集上的实值函数在定义域内部取得 局部极值的必要微分条件和充分微分条件。我们将看到，这些条件类似于我们熟知的实 变量的实值函数的极值的微

45、分条件。定理2设f :U R是定义在赋范空间X的开集U上的实值函数，且f在某个点Xu的邻域有直到k -1 _1阶.(包括k-1阶在内的)导映射，在点x本身有k阶导映射f k x 0时h 0，所以不等式的右端对于所有充分接近于零的向量h=0均为正。因而对所有这些向量hf x h _f X ,即X是严格局部极小点。严格局部极大点的充分条件可类似地验证。注1如果空间X是有限维的，那么以点X. X为中心的单位球面Sx;1是X中的有界 闭集，因而是紧集。这时，连续函数f k X hk =hikf X hhik （k-形式）在S x;1上有最 精品文档14欢迎下载大值和最小值。如果最大值和最小值异号，那么

46、函数f在点X没有极值。如果它们同号， 那么像定理2所指出的，f在点X有极值。在后一种情况下，显然极值的充分条件可叙 述为与它等价的形式：形式f k Xhk是定的（正定的或负定的）。我们在研究Rn中的实值函数时所遇到的正是这种形式的极值条件。注2像我们在函数f ：Rn R的例子所看到的那样，在极值的必要条件中所说的形式f f卜hk的半定性还不是极值的充分条件。注3实际上，在研究可微函数的极值时，通常只利用一阶微分或一阶和二阶微分。 如果根据所研究问题的意义，极值点的唯一性及极值的特性是显然的，那么在求极值点 时就可只用一阶微分：求满足f，X =0的点X。3. 一些例子例1设L C1 R3;R，而

47、f C 1 a,b；R，换句话说，uf3 LJ,u3是定义在R3中的连续可微的实值函数，而xfx是定义在区间a,b! R上的光滑实值函 数。我们研究函数F : C 1 a,b JR L R（2）它由以下关系式给出f C 1 a,b JR F fbL x, f x , f, x d R（ 3）a因此，（2）是定义在函数集C1 a,b JR上的实泛函。在物理学和力学中，与运动密切相关的基本变分原理是众所周知的。根据这些原理， 在所有可能的运动中真实运动的特点是，它们总是沿着使某些泛函有极值的轨道进行。 与泛函的极值有关的问题是最优控制理论中的中心问题。因此，寻求和研究泛函的极值 是重要的独立课题，

48、分析中以大量篇幅讨论这个课题的理论，这就是变分学，为使读者 对从数值函数的极值分析到寻求和研究泛函的极值的转变不感到突然的，我们己做了某精品文档(9)15欢迎下载L u-J,ui2 ,u- i3-Lu13 3口2 2U UJ J. .: :3 3口2 2U U其中 u =J,u2,u3max0空迪些工作。但是我们不准备深入讨论变分法的专门问题，仅以泛函（3）为例说明上面讲过的微分法和局部极值研究的一般思想。我们要证明泛函（3）是可微映射并求出它的微分。首先指出，函数（3）可以看作由公式F1 f X ALx,f x,f, x(4)给出的映射F1 : C 1 a,b；R .C a,b；R(5)和映

49、射g .二C a,b；R F2 gg xdx .二R11_a的复合。由积分的性质，映射F2显然是线性连续映射，因而它的可微性问题是明显的我们来证明F1也是可微的，而且Ff f h X = ：2L x,f x ,f , x h x ；：3Lx,f x .f , x h, x其中 h C1 a,bh。事实上，由有限增量定理的推论，在我们的情况下可得_ sup| iL U 川 -:：1 L u ! ：2L u 八 -：2L u ! ：3L u 八：-:L u : .： 0 13 max 亂 fu卜占 L(u |max 应(8)0 至空f i =1,2,3i 3,2,3如果记起cfja,bR ）中函数

50、f的范数|f|c（i是 maxj f , f, （其中ifc是函数在区间a,b上的最大模），那么设u1 =x , u2 =f x , u3 =f，x , . =0 , . =h x和3 =h，x，考虑到函数 :iLu1,u2,u3 ,i =12,3在R3的有界子集上的一致连续性，从不等式（8）得到Lx,fx 广 hx,f,x - h, x _Lx,fx,f,x _；2L x,f x , f, x h _；3L x, f x , f, x h, x而这意味着等式（7）成立。现在，根据复合映射的微分定理断定，泛函（3）确实可微，并且bF, f h 2L x,f x ,f , x h x fL x,f x , f x hx dxa精品文档16欢迎下载F，fh = a -2Lx, f x,f, x -dX BL x,f x,f, x h x dx(10)(13)经常把泛函限制在那样一些函数f c1 a,b；R的仿射空间上，它们在区间la,b的端点取固定的值f a =A, f b =B o在这种情况下，切空间TCf1中的函数h在区间a,b的端点应该有零值。考虑到这一点，在这种情况下，利用分部积分，显然可把等式(9)化为当然要预先假设L和f属于相应的函数类c2特别地，如果f