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1、泰勒公式外文翻譯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常數(shù)(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泰勒公式和極值的研究1. 映射的泰勒公式定理 1如果從賦范空間 X 的點 x 的鄰域 U U x 到賦范空間 Y 的映射 f : UY 在 U中有直到 n-1階 ( 包括 n-1 在內(nèi) ) 的導(dǎo)數(shù),而在點 x 處有 n 階導(dǎo)數(shù)。 f
44、n x ,那么當(dāng) h 0時有f x hf x f , x h1 f n x hno h nn!(1)10等式 (1) 是各種形式的泰勒公式中的一種,這一次它確實是對非常一般的函數(shù)類寫出來的公式了。我們用歸納法證明泰勒公式(1) 。當(dāng) n 1 時,由 f ,x 的定義, (1) 式成立。假設(shè) (1) 對 n 1N 成立。于是根據(jù)有限增量定理,5 章中公式 (12) 和所作的歸納假設(shè),我們得到,當(dāng)h0 時成立。f x hf xf ,x h1 f n x hnn!sup f ,xhf , xf , , x h1f n xh n 1h01n1 !on 1hnho h這里我們不再繼續(xù)討論其他的,有時甚至
45、是十分有用的泰勒公式形式。當(dāng)時,在研究數(shù)值函數(shù)時,曾詳細地討論過它們?,F(xiàn)在我們把它們的結(jié)論提供給讀者 ( 例如,可參看練習(xí) 1) 。2. 內(nèi)部極值的研究我們將利用泰勒公式指出定義在賦范空間的開集上的實值函數(shù)在定義域內(nèi)部取得局部極值的必要微分條件和充分微分條件。 我們將看到, 這些條件類似于我們熟知的實變量的實值函數(shù)的極值的微分條件。定理 2設(shè) f : UR 是定義在賦范空間 X 的開集 U上的實值函數(shù),且 f 在某個點 x U的鄰域有直到 k11階.(包括 k-1階在內(nèi)的 ) 導(dǎo)映射,在點 x 本身有 k 階導(dǎo)映射 f k x 。如果 f ,x0, fk 1 x0 且 f kx 0 ,那么為使
46、 x 是函數(shù) f 的極值點必要條件是 : k是偶數(shù), f k x hk 是半定的。充分條件是 :f kx hk 在單位球面h1 上的值不為零;這時,如果在這個球面上fkx hk0 ,那么 x 是嚴格局部極小點;如果fkx hk0 ,那么 x 是嚴格局部極大點f在點 x 鄰域內(nèi)的泰勒展開式。由所作的假設(shè)可得為了證明定理,我們考察函數(shù)f x hf x1f k x hkh h k ,k!其中h 是實值函數(shù),而且當(dāng) h 0時。h0我們先證必要條件。因為 f k x 0 ,所以有向量 h00 ,使 fk x h0k0 。于是,對于充分接近于零的實參11量 t ,f x th0f x1fk x thkk!0th0 th0k1 fk x hkk!0括號內(nèi)的表達式與f kx h0k 同號。為使 x 是極值點,當(dāng) t 變號時最后一個等式的左邊只有當(dāng) k 為偶數(shù)時才可能。th0 h0 k t k( 從而右邊 ) 必須不改變符號。 這上述討論表明,如果x 是極值點,那么對于充分小的t ,差 f x th0 f x 的符號與fk x
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