泰勒公式外文翻译.doc

1、泰勒公式外文翻译Taylor's Formula and the Study of Extrema1. Taylor's Formula for MappingsTheorem 1. If a mappingf : UY from a neighborhood UU x of a point x in anormed space X into a normed space Y has derivatives up to order1 ninclusive-in U andf nxat the point x, thenhas an n-th order derivativef

2、x hf x f , x h1 f n x hno hnn!(1)ash0 .Equality (1) is one of the varieties of Taylor's formula, written here for rather general classes of mappings.Proof. We prove Taylor's formula by induction.For n1it is true by definition of f , x .Assume formula (1) is true for somen 1N .Then by the mea

3、n-value theorem, formula (12) of Sect. 10.5, and the inductionhypothesis, we obtain.f x hf xf ，x h1 f n x h nn!sup f ,xhf , xf , x h1n x h n 1hf01n1 !on 1honhhash0 .We shall not take the time here to discuss other versions of Taylor's formula, which are sometimes quite useful. They were discusse

4、d earlier in detail for numerical functions. At this point we leave it to the reader to derive them (see, for example, Problem 1 below).2. Methods of Studying Interior ExtremaUsing Taylor's formula, we shall exhibit necessary conditions and also sufficient conditions for an interior local extrem

5、um of real-valued functions defined on an open subset of a normed space. As we shall see, these conditions are analogous to the differential conditions already known to us for an extremum of a real-valued function of a real variable.Theorem 2. Let f : U R be a real-valued function defined on an open

6、 set U in a normed space X and having continuous derivativesup to order k 1 1 inclusive in aneighborhood of a pointx Uand a derivative f k xof order k at the point x itself.Iff , x0 , f k 1 x0and f k x0 , then for x to be an extremum of the function f it2is:necessary that k be even and that the form

7、k x h k be semidefinite,andsufficient that the values of the form f k x hk on the unit sphere h1be boundedaway from zero; moreover, x is a local minimum if the inequalitiesf k x hk0 ,hold on that sphere, and a local maximum iff k x hk0 ,Proof. For the proof we consider the Taylor expansion (1) of f

8、in a neighborhood of x. The assumptions enable us to writef x h f x1 f k x hkh h kk!wherehis a real-valued function, andh0ash0 .We first prove the necessary conditions.Sincefkx 0 , there exists a vector h0 0on which fk x h0k0 .Then for values of thereal parameter t sufficiently close to zero,f x thf

9、 x1fk x thkth th k0k !0001 f k x hkth hkt kk!000and the expression in the outer parentheses has the same signf kasx h0k .For x to be an extremum it is necessaryfor the left-hand side (and hence also the right-hand side) of this last equality to be of constant sign whent changes sign. But this is pos

10、sible only ifk is even.This reasoning shows that if x is an extremum, then the sign of the differencef x th0 f x is the same as that off kx h0k for sufficiently small t; hence in that casethere cannot be two vectors h0 , h1atwhich the form f k x assumes values withopposite signs.We now turn to the p

11、roof of the sufficiency conditions. For definiteness we considerthe case when f k x hk0for h1. Thenf x hf x1 f kx hkkh hk !1kk xhhkfhhk!31h h kk!and, sinceh0ash0 , the last termin this inequality is positive for all vectorsh 0 sufficiently close to zero. Thus, for all such vectors h,f xhf x0 ,that i

12、s, x is a strict local minimum.The sufficient condition for a strict local maximum is verified similiarly.Remark 1. If the space X is finite-dimensional, the unit sphere S x; 1 with center at x X , being a closed bounded subset ofX, is compact. Then the continuous functionf kx h ki1i k f x hi1h ik(a

13、 k-form) has both a maximal and a minimal value on S x;1 .If these values are of opposite sign, then f does not have an extremum atx. If they are both of the same sign, then, as was shown in Theorem 2, there is an extremum. In the latter case, a sufficient condition for an extremum can obviously be

14、stated as theequivalent requirement that the form f k x hk be either positive- or negative-definite.It was this form of the condition that we encountered in studying realvalued functions on Rn .Remark 2.As we have seen inthe example offunctions f : RnR ,thesemi-definitenessof the formf kx hkexhibite

15、d in the necessary conditions for anextremum is not a sufficient criterion for an extremum.Remark 3. In practice, when studying extrema of differentiable functions one normally uses only the first or second differentials. If the uniqueness and type of extremum are obvious from the meaning of the pro

16、blem being studied,one can restrict attention to the first differential when seeking an extremum, simply finding the pointxwhere f , x03. Some ExamplesExample 1.Let LC 1 R3 ; Rand fC 1a ,b ; R . In other words,u1 ,u2 ,u3L u1 ,u 2 ,u 3is a continuously differentiable real-valued function defined inR

17、3and xf xasmooth real-valued function defined on the closed interval ,bR .Consider the functionF : C 1a ,b ; RR(2)defined by the relationfC 1a ,b ; RF fbL x , f x , f , x dxRa4(3)Thus, (2) is a real-valued functional defined on the set of functionsC 1 a,b ; R .The basic variational principles connec

18、ted with motion are known in physics and mechanics. According to these principles, the actual motions are distinguished among all the conceivable motions in that they proceed along trajectories along which certain functionals have an extremum. Questions connected with the extrema of functionals are

19、central in optimal control theory. Thus, finding and studying the extrema of functionals is a problemof intrinsic importance, and the theory associated with it is the subject of a large area of analysis - the calculus of variations. We have already done a few things to make the transition from the a

20、nalysis of the extrema of numerical functions to the problem of finding and studying extrema of functionals seem natural to the reader. However, we shall not go deeply into the special problems of variational calculus, but rather use the example of the functional (3) to illustrate only the general i

21、deas of differentiation and study of local extrema considered above.We shall show that the functional (3) is a differentiate mapping and find its differential.We remark that the function (3) can be regarded as the composition of the mappingsF1 fxL x, f x , f , x(4)defined by the formulaF1 : C 1a,b ;

22、 RC a,b ; R(5)followed by the mappingbgCa,b ; RF2 gg x dxRa(6)By properties of the integral, the mapping F2 is obviously linear and continuous, sothat its differentiability is clear.We shall show that the mapping F1 is also differentiable, and thatF1, f h x2L x, f x , f , x h x3 L x , f x . f , x h

23、, x(7)for hC 1a,b ; R .Indeed, by the corollary to the mean-value theorem, we can write in the present case3L u11,u22 ,u33 L u1 ,u2 ,u3i L u1, u2 ,u3 ii15sup1L u1L u 1 2 L u2 L u 1 3 L u3L u013 maxi L uui L umaxi01i1,2,3i1,2,3(8)where uu1,u2 ,u3and1 ,2 ,3 .If we now recall that the normf c 1of the f

24、unction f inC 1 a, b ; R is max f c, f ,c(wheref c is the maximum absolute value of the function on the closed interval a ,b ),then, settingu1x , u 2fx ,u3f ,x ,1 0,2 h x , and3h , x , we obtainfrominequality (8), taking account of the uniform continuity of the functionsL u1 ,u2 ,u3 ,i1,2 ,3 ,R3, th

25、aton bounded subsets ofmax L x, f xh x , f , xh, xL x , f x , f ,x2L x , f x , f , xh x3 L x, f x , f , x h, x0 x bo h c 1ash c 10But this means that Eq. (7) holds.By the chain rule for differentiating a composite function, we now conclude that the functional (3) is indeed differentiable, andF , f h

26、b, x h x3 L x , f x , f , x h , x dx2L x, f x , fa(9)We often consider the restriction of the functional (3) to the affine space consisting ofthe functions fC 1a,b ; Rthat assumefixed values f aA ,f bBat the endpointsof the closed interval a ,b . In this case, the functions h in the tangent spaceTC

27、f1 , musthave the value zero at the endpoints of the closed interval a ,b . Taking this fact into account, we may integrate by parts in (9) and bring it into the formF , f hb, xd, x h x dx2 L x, f x , f3 L x, f x , fadx(10)of course under the assumption thatL and f belong to the corresponding classC

28、 2 .In particular, if f is an extremum (extremal) of such a functional, then by Theorem 2 wehave F , f h 0 for every function hC 1 a,b ; R such that h a h b 0 . From this andrelation (10) one can easily conclude(see Problem 3 below) that the function f mustsatisfy the equationd3L x , f x , f , x 02L

29、 x , f x , f , xdx60 , and we(11)This is a frequently-encountered form of the equation known in the calculus of variations as theEuler-Lagrange equation.Let us now consider some specific examples.Example 2.The shortest-path problemAmong all the curves in a plane joining two fixed points, find the cu

30、rve that has minimal length.The answer in this case is obvious, and it rather serves as a check on the formal computations we will be doing later.We shall assume that a fixed Cartesian coordinate system has been chosen in the plane,in which the two points are, for example,0,0and 1,0. We confine ours

31、elves to justfC 10,1 ;Rassuming the value zero at boththe curves that are the graphs of functionsends of the closed interval 0,1 . The length of such a curveFf1f , 2x dx10(12)depends on the function f and is a functional of the type considered in Example 1. In this case the functionL has the formL u

32、1,u 2 ,u31u 3 2and therefore the necessary condition (11) for an extremal here reduces to the equationdf , x0dxf ,21xfrom which it follows that,fx常数(13)on the closed intervalSince the function0,1uis not constant on any interval, Eq. (13) is possible only if1u2f , xconst on a ,b . Thus a smooth extre

33、mal of this problem must be a linear functionwhose graph passesthrough the points 0 ,0 and 1,0 . It follows that f x arrive at the closed interval of the line joining the two given points.Example 3.The brachistochrone problemThe classical brachistochrone problem, posed by Johann Bernoulli I in 1696,

34、 was to find the shape of a track along which a point mass would pass from a prescribed pointP0to another fixed point P1 at a lower level under the action of gravity in the shortesttime.We neglect friction, of course. In addition, we shall assume that the trivial case in7which both points lie on the

35、 same vertical line is excluded.Inthe vertical plane passing through the points P0andP1we introduce arectangular coordinate system such that P0 is at the origin, the x-axis is directedvertically downward, and the point 1has positive coordinates1,y1 .We shall find thePxshape of the track among the gr

36、aphs of smooth functions defined on the closed interval 0,x1 and satisfying the condition f 0 0 , f x1 y1 . At the moment we shall not take time to discuss this by no means uncontroversial assumption (see Problem 4 below).If the particle began its descent from the point P0 with zero velocity, the la

37、w ofvariation of its velocity in these coordinates can be written asv2 gx(14)Recalling that the differential of the arc length is computed by the formuladsdx2dy21 f,2x dx(15)we find the time of descentFf1x1 1f , 2xdx2 g0x(16)yf xon the closedalong the trajectory defined by the graph of the functioni

38、nterval 0 ,x1 .For the functional (16)L u1 ,u 2 ,u31 u3 2u1,and therefore the condition (11) for an extremum reduces in this case to the equationdf , x0 ,dxf ,2x 1xfrom which it follows thatf , xcx1f , 2 x(17)where c is a nonzero constant, since the points are not both on the same vertical line.8Tak

39、ing account of (15), we can rewrite (17) in the formdyxcds(18)However, from the geometric point of viewdxcos , dysindsds(19)where is the angle between the tangent to the trajectory and the positive x-axis. By comparing Eq. (18) with the second equation in (19), we find12xc2 sin(20)But it follows fro

40、m (19) and (20) thatdydydxtgdxtgdsin2sin2ddx ,dddc222,cfrom which we find12sin 2b(21)y22cSetting1aand2t, we write relations (20) and (21) as2c2xa 1costya tsin t b(22)Since a 0 , it follows that x 0 only for t 2k , k Z . It follows from the form of the function (22) that we may assumewithout loss of

41、generality that the parameter valuet 0corresponds to the point00 ,0. In this case Eq. (21) impliesb0, and we arrivePat the simpler formxa 1costya tsin t(23)for the parametric definition of this curve.Thus the brachistochrone is a cycloid having a cusp at the initial pointP0 where thetangent is verti

42、cal. The constant a, which is a scaling coefficient, must be chosen so thatthe curve (23) also passesthrough the point P1 .Such a choice, as one can see bysketching the curve (23), is by no means always unique, and this shows that the necessarycondition (11) for an extremum is in general not suffici

43、ent. However, from physical considerations it is clear which of the possible values of the parametershould be preferred (and this, of course, can be confirmed by direct computation).9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 U x 到赋范空间 Y 的映射 f : UY 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

44、n x ，那么当 h 0时有f x hf x f , x h1 f n x hno h nn!(1)10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， (1) 式成立。假设 (1) 对 n 1N 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h0 时成立。f x hf xf ，x h1 f n x hnn!sup f ,xhf , xf , , x h1f n xh n 1h01n1 !on 1hnho h这里我们不再继续讨论其他的，有时甚至

45、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分条件。定理 2设 f : UR 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 x U的邻域有直到 k11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 。如果 f ,x0, fk 1 x0 且 f kx 0 ，那么为使

46、 x 是函数 f 的极值点必要条件是 : k是偶数， f k x hk 是半定的。充分条件是 :f kx hk 在单位球面h1 上的值不为零；这时，如果在这个球面上fkx hk0 ，那么 x 是严格局部极小点；如果fkx hk0 ，那么 x 是严格局部极大点f在点 x 邻域内的泰勒展开式。由所作的假设可得为了证明定理，我们考察函数f x hf x1f k x hkh h k ，k!其中h 是实值函数，而且当 h 0时。h0我们先证必要条件。因为 f k x 0 ，所以有向量 h00 ，使 fk x h0k0 。于是，对于充分接近于零的实参11量 t ，f x th0f x1fk x thkk!0th0 th0k1 fk x hkk!0括号内的表达式与f kx h0k 同号。为使 x 是极值点，当 t 变号时最后一个等式的左边只有当 k 为偶数时才可能。th0 h0 k t k( 从而右边 ) 必须不改变符号。 这上述讨论表明，如果x 是极值点，那么对于充分小的t ，差 f x th0 f x 的符号与fk x