泰勒公式外文翻译

Taylor's Formula and the Study of Extrema1. Taylor's Formula for MappingsTheorem 1. If a mapping Y U f →: from a neighborhood ()x U U = of a point x in a normed space X into a normed space Y has derivatives up to order n -1 inclusive in U and has an n-th order derivative ()()x f n at the point x, then()()()()()⎪⎭⎫ ⎝⎛++++=+n n n h o h x f n h x f x f h x f !1,Λ (1)as 0→h .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 1=n it is true by definition of ()x f ,.Assume formula (1) is true for some N n ∈-1.Then by the mean-value theorem, formula (12) of Sect. 10.5, and the induction hypothesis, we obtain.()()()()()()()()()()()()()⎪⎭⎫ ⎝⎛=⎪⎭⎫ ⎝⎛=⎪⎪⎭⎫ ⎝⎛-+++-+≤⎪⎭⎫ ⎝⎛+++-+--<<n n n n n n h o h h o h h x f n h x f x f h x f h x f n h f x f h x f 11,,,,10!11sup !1x θθθθθΛΛ，as 0→h .We shall not take the time here to discuss other versions of Taylor's formula, which are sometimes quite useful. They were discussed 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 extremum 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 R U f →: be a real-valued function defined on an open set U in a normed space X and having continuous derivatives up to order 11≥-k inclusive in a neighborhood of a pointU x ∈ and a derivative ()()x f k of order k at the point x itself. If ()()()0,,01,==-x f x f k Λ and ()()0≠x f k , then for x to be an extremum of the function f it is:necessary that k be even and that the form()()k k h x f be semidefinite, andsufficient that the values of the form ()()k k h x f on the unit sphere 1=h be bounded away from zero; moreover, x is a local minimum if the inequalities()()0>≥δk k h x f ,hold on that sphere, and a local maximum if()()0<≤δk k h x f ,Proof. For the proof we consider the Taylor expansion (1) of f in a neighborhood of x. The assumptions enable us to write()()()()()k k k h h h x f k x f h x f α+=-+!1where ()h α is a real-valued function, and ()0→h α as 0→h .We first prove the necessary conditions.Since ()()0≠x f k , there exists a vector 00≠h on which ()()00≠k k h x f . Then for values of the real parameter t sufficiently close to zero,()()()()()()k k k th th th x f k x f th x f 0000!1α+=-+()()()kk k k t h th h x f k ⎪⎭⎫ ⎝⎛+=000!1αand the expression in the outer parentheses has the same sign as()()k k h x f 0. For 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 if k is even.This reasoning shows that if x is an extremum, then the sign of the difference ()()x f th x f -+0 is the same as that of()()k k h x f 0 for sufficiently small t; hence in that case there cannot be two vectors 0h , 1h at which the form ()()x f k assumes values with opposite signs.We now turn to the proof of the sufficiency conditions. For definiteness we consider the case when ()()0>≥δk k h x f for 1=h . Then()()()()()k k k h h h x f k x f h x f α+=-+!1()()()k k k h h h h x f k ⎪⎪⎪⎭⎫ ⎝⎛+⎪⎪⎭⎫ ⎝⎛=α!1 ()k h h k ⎪⎭⎫ ⎝⎛+≥αδ!1and, since ()0→h α as 0→h , the last term in this inequality is positive for all vectors 0≠h sufficiently close to zero. Thus, for all such vectors h,()()0>-+x f h x f ,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 finite-dimensional, the unit sphere ()1;x S with center at X x ∈, being a closed bounded subset of X, is compact. Then the continuous function ()()()()k k i i i i k k h h x f h x f ⋅⋅∂=ΛΛ11 (a k-form) has both a maximal and a minimal value on ()1;x S . Ifthese values are of opposite sign, then f does not have an extremum at x. 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 stated as the equivalent requirement that the form ()()k k h x f be either positive- or negative-definite.It was this form of the condition that we encountered in studying realvalued functions on n R .Remark 2. As we have seen in the example of functionsR R f n →:, the semi-definiteness of the form ()()k k h x f exhibited in the necessary conditions for an extremum 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 problem being studied, one can restrict attention to the first differential when seeking an extremum, simply finding the point x where ()0,=x f3. Some ExamplesExample 1. Let ()()R R C L ;31∈ and ()[]()R b a C f ;,1∈. In other words,()()321321,,,,u u u L u u u α is a continuously differentiable real-valued function defined in3R and ()x f x α a smooth real-valued function defined on the closed interval []R b a ⊂,.Consider the function()[]()R R b a C F →;,:1 (2)defined by the relation()[]()()f F R b a C f α;,1∈()()()R dx x f x f x L b a∈=⎰,,, (3) Thus, (2) is a real-valued functional defined on the set of functions ()[]()R b a C ;,1.The basic variational principles connected 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 central in optimalcontrol 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 analysis 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 ideas 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 mappings()()()()()x f x f x L x f F ,1,,= (4)defined by the formula ()[]()[]()R b a C R b a C F ;,;,:11→(5) followed by the mapping[]()()()R dx x g g F R b a C g ba ∈=∈⎰2;,α (6) By properties of the integral, the mapping2F is obviously linear and continuous, so that its differentiability is clear.We shall show that the mapping 1F is also differentiable, and that()()()()()()()()()()x h x f x f x L x h x f x f x L x h f F ,,3,2,1.,,,∂+∂= (7)for ()[]()R b a C h ;,1∈.Indeed, by the corollary to the mean-value theorem, we can write in the present case()()()i i i u u u L u u u L u u u L ∆∂--∆+∆+∆+∑=32131321332211,,,,,,()()()()()()∆⋅∂-∆+∂∂-∆+∂∂-∆+∂≤<<u L u L u L u L u L u L 3312211110sup θθθθ ()()ii i i u L u u L i ∆⋅∂-+∂≤=≤≤=3,2,110max max 33,2,1θθ (8) where ()321,,u u u u = and ()321,,∆∆∆=∆. If we now recall that the norm()1c f of the function f in ()[]()R b a C ;,1 is ⎭⎬⎫⎩⎨⎧c c f f ,,max (where c f is the maximum absolute value of the function on the closed interval []b a ,), then,setting x u =1,()x f u =2, ()x f u ,3=, 01=∆, ()x h =∆2, and ()x h ,3=∆, we obtain from inequality (8), taking account of the uniform continuity of the functions ()3,2,1,,,321=∂i u u u L i , on boundedsubsets of 3R , that()()()()()()()()()()()()()()()()x h x f x f x L x h x f x f x L x f x f x L x h x f x h x f x L b x ,,3,2,,,0,,,,,,,,max ∂-∂--++≤≤ ()()1c h o = as ()01→c hBut 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, and()()()()()()()()()()⎰∂+∂=ba dx x h x f x f x L x h x f x f x L h f F ,,3,2,,,,, (9) We often consider the restriction of the functional (3) to the affine space consisting of the functions ()[]()R b a C f ;,1∈ that assume fixed values ()A a f =, ()B b f = at the endpoints of the closed interval []b a ,. In this case, the functions h in the tangent space ()1f TC , must have the value zero at the endpoints of the closed interval []b a ,. Taking this fact into account, we may integrate by parts in (9) and bring it into the form()()()()()()()()⎰⎪⎭⎫ ⎝⎛∂-∂=b a dx x h x f x f x L dx d x f x f x L h f F ,3,2,,,,, (10)of course under the assumption that L and f belong to the corresponding class ()2C .In particular, if f is an extremum (extremal) of such a functional, then by Theorem 2 we have ()0,=h f F for every function ()[]()R b a C h ;,1∈ such that ()()0==b h a h . From this and relation (10) one can easily conclude (see Problem 3 below) that the function f must satisfy the equation()()()()()()0,,,,,3,2=∂-∂x f x f x L dx d x f x f x L (11)This is a frequently-encountered form of the equation known in the calculus of variations as the Euler-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 curve 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, ()0,0 and ()0,1 . We confine ourselves to just the curves that are the graphs of functions()[]()R C f ;1,01∈ assuming the value zero at both ends of the closed interval []1,0 . The length of such a curve ()()()⎰+=102,1dx x f f F (12) depends on the function f and is a functional of the type considered in Example 1. In this case the function L has the form ()()233211,,u u u u L +=and therefore the necessary condition (11) for an extremal here reduces to the equation ()()()012,,=⎪⎪⎪⎭⎫ ⎝⎛+x f x f dx d from which it follows that ()()()常数≡+x f x f 2,,1 (13)on the closed interval []1,0Since the function 21u u+ is not constant on any interval, Eq. (13) is possible only if()≡x f ,const on []b a ,. Thus a smooth extremal of this problem must be a linear function whose graph passes through the points ()0,0 and ()0,1. It follows that ()0≡x f , and we 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, was to find the shape of a track along which a point mass would pass from a prescribed point 0P to another fixed point 1P at a lower level under the action of gravity in the shortest time.We neglect friction, of course. In addition, we shall assume that the trivial case in which both points lie on the same vertical line is excluded.In the vertical plane passing through the points0P and 1P we introduce a rectangular coordinate system such that0P is at the origin, the x-axis is directed vertically downward, and the point 1P has positive coordinates ()11,y x .We shall find the shape of the track among the graphs of smooth functions defined on the closed interval []1,0x and satisfying the condition ()00=f ,()11y x f =. 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 0P with zero velocity, the law of variation of its velocity in these coordinates can be written asgx v 2= (14)Recalling that the differential of the arc length is computed by the formula()()()()dx x f dy dx ds 2,221+=+= (15) we find the time of descent()()()⎰+=102,121x dx x x f g f F (16) along the trajectory defined by the graph of the function()x f y = on the closed interval []1,0x . For the functional (16) ()()1233211,,u u u u u L +=,and therefore the condition (11) for an extremum reduces in this case to the equation ()()()012,,=⎪⎪⎪⎪⎪⎭⎫⎝⎛⎪⎭⎫ ⎝⎛+x f x x f dx d , from which it follows that ()()()x c x f x f =+2,,1 (17)where c is a nonzero constant, since the points are not both on the same vertical line.Taking account of (15), we can rewrite (17) in the formx c ds dy = (18)However, from the geometric point of viewϕcos =ds dx ,ϕsin =ds dy (19)where ϕ 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 findϕ22sin 1c x = (20) But it follows from (19) and (20) that dx dy d dy =ϕ,2222sin 2sin c c d d tg d dx tg d dx ϕϕϕϕϕϕϕ=⎪⎪⎭⎫ ⎝⎛==,from which we find()b c y +-=ϕϕ2sin 2212 (21)Setting a c =221and t =ϕ2, we write relations (20) and (21) as()()b t t a y t a x +-=-=sin cos 1 (22)Since 0≠a , it follows that 0=x only for πk t 2=,Z k ∈. It follows from the form of the function (22) that we may assume without loss of generality that the parameter value 0=t corresponds to the point ()0,00=P . In this case Eq. (21) implies 0=b , and we arrive at thesimpler form()()t t a y t a x sin cos 1-=-= (23)for the parametric definition of this curve.Thus the brachistochrone is a cycloid having a cusp at the initial point 0P where the tangent is vertical. The constant a, which is a scaling coefficient, must be chosen so that the curve (23) also passes through the point 1P . Such a choice, as one can see by sketching the curve (23), is by no means always unique, and this shows that the necessary condition (11) for an extremum is in general not sufficient. However, from physical considerations it is clear which of the possible values of the parameter a should be preferred (and this, of course, can be confirmed by direct computation).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()x U U =到赋范空间Y 的映射Y U f →: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。