泰勒公式外文翻译

1 Taylor s Formula and the Study of Extrema 1 Taylor s Formula for Mappings Theorem 1 If a mapping from a neighborhood of a point x in a normed YUf xUU space X into a normed space Y has derivatives up to order n 1 inclusive in U and has an n th order derivative at the point x then xf n 1 n nn hohxf n hxfxfhxf 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 it is true by definition of 1 n xf Assume formula 1 is true for some Nn 1 Then by the mean value theorem formula 12 of Sect 10 5 and the induction hypothesis we obtain nn nn nn hohho hhxf n hxfxfhxf hxf n hfxfhxf 1 1 10 1 1 sup 1 x 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 Extrema Using 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 be a real valued function defined on an open set U in a normed RUf space X and having continuous derivatives up to order inclusive in a neighborhood of 11 k a point and a derivative of order k at the point x itself Ux xf k If and then for x to be an extremum of the function f it is 0 0 1 xfxf k 0 xf k necessary that k be even and that the form be semidefinite kk hxf and 2 sufficient that the values of the form on the unit sphere be bounded away kk hxf1 h from zero moreover x is a local minimum if the inequalities 0 kk hxf hold on that sphere and a local maximum if 0 kk hxf Proof For the proof we consider the Taylor expansion 1 of f in a neighborhood of x The assumptions enable us to write k kk hhhxf k xfhxf 1 where is a real valued function and as h 0 h 0 h We first prove the necessary conditions Since there exists a vector on which Then for values of the 0 xf k 0 0 h 0 0 kk hxf real parameter t sufficiently close to zero k kk thththxf k xfthxf 0000 1 k k kk ththhxf k 000 1 and the expression in the outer parentheses has the same sign as kk hxf 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 xfthxf 0 is the same as that of for sufficiently small t hence in that case there cannot be two kk hxf 0 vectors at which the form assumes values with opposite signs 0 h 1 h xf k We now turn to the proof of the sufficiency conditions For definiteness we consider the case when for Then 0 kk hxf1 h k kk hhhxf k xfhxf 1 k k k hh h h xf k 1 k hh k 1 and since as the last term in this inequality is positive for all vectors 0 h 0 h sufficiently close to zero Thus for all such vectors h 0 h 3 0 xfhxf 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 with center at 1 xSXx being a closed bounded subset of X is compact Then the continuous function a k form has both a maximal and a minimal value on If k k ii ii kk hhxfhxf 1 1 1 xS these 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 be either positive or negative definite kk hxf 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 functions the semi definiteness RRf n of the form exhibited in the necessary conditions for an extremum is not a sufficient kk hxf 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 xf 3 Some Examples Example 1 Let and In other words is a RRCL 31 RbaCf 1 321321 uuuLuuu continuously differentiable real valued function defined in and a smooth real 3 R xfx valued function defined on the closed interval Rba Consider the function 2 RRbaCF 1 defined by the relation fFRbaCf 1 3 RdxxfxfxL b a Thus 2 is a real valued functional defined on the set of functions RbaC 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 optimal control theory Thus finding and studying the extrema of functionals is a problem 4 of 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 4 xfxfxLxfF 1 defined by the formula 5 RbaCRbaCF 1 1 followed by the mapping 6 RdxxggFRbaCg b a 2 By properties of the integral the mapping is obviously linear and continuous so that its 2 F differentiability is clear We shall show that the mapping is also differentiable and that 1 F 7 xhxfxfxLxhxfxfxLxhfF 3 2 1 for RbaCh 1 Indeed by the corollary to the mean value theorem we can write in the present case i i i uuuLuuuLuuuL 321 3 1 321332211 uLuLuLuLuLuL 33122111 10 sup 8 i i ii uLuuL i 3 2 110 maxmax3 3 2 1 where and 321 uuuu 321 If we now recall that the norm of the function f in is where 1 c f RbaC 1 c c ff max is the maximum absolute value of the function on the closed interval then setting c f ba and we obtain from inequality 8 taking xu 1 xfu 2 xfu 3 0 1 xh 2 xh 3 account of the uniform continuity of the functions on bounded subsets of 3 2 1 321 iuuuL i 5 that 3 R xhxfxfxLxhxfxfxLxfxfxLxhxfxhxfxL bx 3 2 0 max as 1 c ho 0 1 c h But 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 9 b a dxxhxfxfxLxhxfxfxLhfF 3 2 We often consider the restriction of the functional 3 to the affine space consisting of the functions that assume fixed values at the endpoints of the RbaCf 1 Aaf Bbf closed interval In this case the functions h in the tangent space must have the value ba 1 f TC zero at the endpoints of the closed interval Taking this fact into account we may ba integrate by parts in 9 and bring it into the form 10 b a dxxhxfxfxL dx d xfxfxLhfF 3 2 of course under the assumption that L and f belong to the corresponding class 2 C In particular if f is an extremum extremal of such a functional then by Theorem 2 we have for every function such that From this and relation 10 0 hfF RbaCh 1 0 bhah one can easily conclude see Problem 3 below that the function f must satisfy the equation 11 0 3 2 xfxfxL dx d xfxfxL 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 problem Among 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 and We confine ourselves to just the curves 0 0 0 1 that are the graphs of functions assuming the value zero at both ends of the RCf 1 0 1 closed interval The length of such a curve 1 0 12 1 0 2 1dxxffF 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 2 3321 1 uuuuL 6 and therefore the necessary condition 11 for an extremal here reduces to the equation 0 1 2 xf xf dx d from which it follows that 13 常数 xf xf 2 1 on the closed interval 1 0 Since the function is not constant on any interval Eq 13 is possible only if 2 1u u const on Thus a smooth extremal of this problem must be a linear function whose xf ba graph passes through the points and It follows that and we arrive at the 0 0 0 1 0 xf closed interval of the line joining the two given points Example 3 The brachistochrone problem The 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 to 0 P another fixed point at a lower level under the action of gravity in the shortest time 1 P 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 points and we introduce a rectangular 0 P 1 P coordinate system such that is at the origin the x axis is directed vertically downward and 0 P the point has positive coordinates We shall find the shape of the track among the 1 P 11 y x graphs of smooth functions defined on the closed interval and satisfying the condition 1 0 x At the moment we shall not take time to discuss this by no means 00 f 11 yxf uncontroversial assumption see Problem 4 below If the particle began its descent from the point with zero velocity the law of variation of 0 P its velocity in these coordinates can be written as 14 gxv2 Recalling that the differential of the arc length is computed by the formula 15 dxxfdydxds 2 22 1 we find the time of descent 7 16 1 0 2 1 2 1 x dx x xf g fF along the trajectory defined by the graph of the function on the closed interval xfy 1 0 x For the functional 16 1 2 3 321 1 u u uuuL and therefore the condition 11 for an extremum reduces in this case to the equation 0 1 2 xfx xf dx d from which it follows that 17 xc xf xf 2 1 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 form 18 xc ds dy However from the geometric point of view 19 cos ds dx sin ds dy 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 find 20 2 2 sin 1 c x But it follows from 19 and 20 that dx dy d dy 2 2 2 2 sin 2 sin cc d d tg d dx tg d dx from which we find 21 b c y 2sin2 2 1 2 Setting and we write relations 20 and 21 asa c 2 2 1 t 2 22 bttay tax sin cos1 Since it follows that only for It follows from the form of the function 0 a0 x kt2 Zk 22 that we may assume without loss of generality that the parameter value corresponds 0 t to the point In this case Eq 21 implies and we arrive at the simpler form 0 0 0 P0 b 8 23 ttay tax sin cos1 for the parametric definition of this curve Thus the brachistochrone is a cycloid having a cusp at the initial point where the tangent 0 P is vertical The constant a which is a scaling coefficient must be chosen so that the curve 23 also passes through the point Such a choice as one can see by sketching the curve 23 1 P 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 9 泰勒公式和极值的研究泰勒公式和极值的研究 1 1 映射的泰勒公式映射的泰勒公式 定理定理 1 1 如果从赋范空间 X 的点 x 的邻域到赋范空间 Y 的映射在 U xUU YUf 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 那么当 xf n 时有0 h 1 n nn hohxf n hxfxfhxf 1 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 当时 由的定义 1 式成立 1 n xf 假设 1 对成立 Nn 1 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 时成立 0 h nn nn nn hohho hhxf n hxfxfhxf hxf n hfxfhxf 1 1 10 1 1 sup 1 x 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 2 内部极值的研究内部极值的研究 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 定理定理 2 2 设是定义在赋范空间 X 的开集 U 上的实值函数 且 f 在某个点RUf 的邻域有直到阶 包括 k 1 阶在内的 导映射 在点 x 本身有 k 阶导映射Ux 11 k xf k 如果且 那么为使 x 是函数 f 的极值点 0 0 1 xfxf k 0 xf k 必要条件是 k 是偶数 是半定的 kk hxf 充分条件是 在单位球面上的值不为零 这时 如果在这个球面上 kk hxf1 h 0 kk hxf 10 那么 x 是严格局部极小点 如果 0 kk hxf 那么 x 是严格局部极大点 为了证明定理 我们考察函数 f 在点 x 邻域内的泰勒展开式 由所作的假设可得 k kk hhhxf k xfhxf 1 其中是实值函数 而且当时 h 0 h 0 h 我们先证必要条件 因为 所以有向量 使 于是 对于充分接近于零的实参 0 xf k 0 0 h 0 0 kk hxf 量 t k kk thththxf k xfthxf 0000 1 k k kk ththhxf k 000 1 括号内的表达式与同号 kk hxf 0 为使 x 是极值点 当 t 变号时最后一个等式的左边 从而右边 必须不改变符号 这只有当 k 为偶数时才可能 上述讨论表明 如果 x 是极值点 那么对于充分小的 t 差的符号与 xfthxf 0 相同 因而在这种情况下不可能有两个向量 使在它们上的取值有 kk hxf 00 h 1 h xf k 不同的符号 我们转到极值充分条件的证明 为了确定起见 我们研究 当 0 kk hxf1 h 的情况 这时 k kk hhhxf k xfhxf 1 k k k hh h h xf k 1 k hh k 1 又因时 所以不等式的右端对于所有充分接近于零的向量均为正 因0 h 0 h 0 h 而对所有这些向量 h 0 xfhxf 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 注注 1 1 如果空间 X 是有限维的 那么以点为中心的单位球面是 X 中的有Xx 1 xS 界闭集 因而是紧集 这时 连续函数 k 形式 在上 k k ii ii kk hhxfhxf 1 1 1 xS 有最大值和最小值 如果最大值和最小值异号 那么函数 f 在点 x 没有极值 如果它 11 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 形式是定的 正定的或负定的 kk hxf 我们在研究中的实值函数时所遇到的正是这种形式的极值条件 n R 注注 2 2 像我们在函数的例子所看到的那样 在极值的必要条件中所说的形RRf n 式的半定性还不是极值的充分条件 kk hxf 注注 3 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 求满足的点 x 0 xf 3 3 一些例子一些例子 例例 1 1 设 而 换句话说 RRCL 31 RbaCf 1 321321 uuuLuuu 是定义在中的连续可微的实值函数 而是定义在区间上的光滑实值函 3 R xfx Rba 数 我们研究函数 RRbaCF 1 2 它由以下关系式给出 fFRbaCf 1 RdxxfxfxL b a 3 因此 2 是定义在函数集上的实泛函 RbaC 1 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 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 4 xfxfxLxfF 1 给出的映射 5 RbaCRbaCF 1 1 12 和映射 6 RdxxggFRbaCg b a 2 的复合 由积分的性质 映射显然是线性连续映射 因而它的可微性问题是明显的 2 F 我们来证明也是可微的 而且 1 F 7 xhxfxfxLxhxfxfxLxhfF 3 2 1 其中 RbaCh 1 事实上 由有限增量定理的推论 在我们的情况下可得 i i i uuuLuuuLuuuL 321 3 1 321332211 uLuLuLuLuLuL 33122111 10 sup 8 i i ii uLuuL i 3 2 110 maxmax3 3 2 1 其中 321 uuuu 321 如果记起中函数 f 的范数是 其中是函数在区间上 RbaC 1 1 c f c c ff max c f ba 的最大模 那么设 和 考虑到函数xu 1 xfu 2 xfu 3 0 1 xh 2 xh 3 在的有界子集上的一致连续性 从不等式 8 得到 3 2 1 321 iuuuL i 3 R xhxfxfxLxhxfxfxLxfxfxLxhxfxhxfxL bx 3 2 0 max 当时 1 c ho 0 1 c h 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 9 b a dxxhxfxfxLxhxfxfxLhfF 3 2 经常把泛函 3 限制在那样一些函数的仿射空间上 它们在区间的 RbaCf 1 ba 端点取固定的值 在这种情况下 切空间中的函数 h 在区间的 Aaf Bbf 1 f TC ba 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 10 b a dxxhxfxfxL dx d xfxfxLhfF 3 2 当然要预先假设 L 和 f 属于相应的函数类 2 C 特别地 如果 f 是这个泛函的极值点 极值曲线