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excel所有公式泰勒公式外文翻译

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2020-09-10 14:00
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泰勒公式外文翻译
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 ,,
f:U,YU,Ux
normed space X into a normed space Y has derivatives up to order n -
1 inclusive in U and has
,,nf,,xan n-th order derivative at the point x, then
1n,,,nn,, (1) f,,,,,,xhfxfxh?f,,xhoh,,,,,,,,,,n!
as. h,0
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.
,f,,xFor it is true by definition of . n,1
Assume formula (1) is true for some . n,1,N
Then by the mean- value theorem, formula (12) of Sect. 10.5, and the
induction hypothesis, we obtain.
1,,,,,nnfx,h,fx,fh,,fxhx?,,,,,,,,,,n!,,
,, 1n,1,,,,,,n,,,,,,,,,,,,,,fx,h,fx,fxh,,fxhh?,,,,
sup ,,,,n,1!,0,,1,,
n,1n,,,,o,hhoh,,,,,,,,,,
as. h,0
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.
f:U,RTheorem 2. Let be a real- valued function defined on an open set
U in a
k,1,1normed space X and having continuous derivatives up to order
inclusive in a
,,k,,fxx,Uneighborhood of a point and a derivative of order k at
the point x itself.
,,,,,k,1k,,,,,,fx,0,?,fx,0fx,0If and , then for x to be
an extremum of the function f it is:
,,kk,,fxhnecessary that k be even and that the form be
semidefinite,
and
1
,,kkf,,xhsufficient that the values of the form on the unit
sphere be bounded away h,1
from zero; moreover, x is a local minimum if the inequalities
,,kkf,,xh,,,0,
hold on that sphere, and a local maximum if
,,kkf,,xh,,,0,
Proof. For the proof we consider the Taylor expansion (1) of f in a
neighborhood of x. The
assumptions enable us to write
1k,,kk f,,,,x,h,fx,f,,,,xh,,hhk!
,,,,where ,h is a real-valued function, and ,h,0 as. h,0
We first prove the necessary conditions.
,,k,,kkf,,x,0Since, there exists a vector on which. Then for
values of the h,0f,,xh,000
real parameter t sufficiently close to zero,
1kk,,kf,,,,x,th,fx,f,,,,,,xth,,thth 0000!k
1,,k,,kkk,,,,,fxh,,thht ,,000!k,,
,,kkand the expression in the outer parentheses has the same sign
as. f,,xh0
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 ,,,, fx,th,fx0
,,kkis the same as that of for sufficiently small t; hence in that
case there cannot be two f,,xh0
,,k,,fxvectors hh, at which the form assumes values with
opposite signs. 01
We now turn to the proof of the sufficiency conditions. For
definiteness we consider the
,,kk,,fxh,,,0h,1case when for. Then
k1,,kk,,fx,h,f,,x,f,,,,xh,,hh k!
k,,,,1h,,k,,k,,,,,,,fx,,hh ,,,,!kh,,,,,,
1,,k,,,,,,hh ,,!k,,
,,,h,0h,0and, since as, the last term in this inequality is
positive for all vectors
h,0sufficiently close to zero. Thus, for all such vectors h,
2
, ,,,,fx,h,fx,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 finite-dimensional, the unit
sphere with center at, ,,Sx;1x,X
being a closed bounded subset of X, is compact. Then the continuous
function
ii,,,,kkk1 (a k-form) has both a maximal and a minimal value
on ,,. If ,,,,Sx;1fxh,,fxh,?,h?ii1k
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
,,kkf,,xhthat the form be either positive- or negative-definite.
It was this form of the condition that we encountered in studying
realvalued functions
non. R
nf:R,RRemark 2. As we have seen in the example of functions, the
semi-definiteness
,,kkf,,xhof the form 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
,,,fx,0differential when seeking an extremum, simply finding the
point x where
3. Some Examples
,,,,131123123,,,,,,,,,,L,CR;Rf,Ca,b; Ru,u,u!Lu,u,uExample
1. Let and . In other words,
3,,x!fxis a continuously differentiable real-valued function
defined in R and a smooth
,,a,b,Rreal-valued function defined on the closed interval .
Consider the function
,,1,,,,F:Ca,b;R,R (2)
defined by the relation
,,1,,,,,,f,Ca,b;R!Ff
b,,,,,,,,Lx,fx,fxdx,R (3) ,a
,,1,,,,Ca,b;RThus, (2) is a real-valued functional defined on
the set of functions .
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
3
control theory. Thus, finding and studying the extrema of
functionals is a problem 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) ,,F,,,,,,,,fx,Lx,fx,fx1
defined by the formula
,,1 (5) ,,,,F:C,,a,b;R,C,,a,b;R1
followed by the mapping
b,,g,C,,,,,,a,b;R!Fg,gxdx,R (6) 2,a
By properties of the integral, the mapping is obviously linear and
continuous, so that F2
its differentiability is clear.
We shall show that the mapping is also differentiable, and that F1
,,,, (7) ,,,,F,,,,,,,,fhx,,Lx,fx,fxh,,,,,,
x,,Lx,,,x231
,,1,,h,C,,a,b;Rfor .
Indeed, by the corollary to the mean-value theorem, we can write in
the present case
33i,,,,,,Lu,,,u,,,u,,,Lu,u,u,,Lu,u,u, i,,1i ,,,,,,,,,,,,,sup,Lu,,,,,Lu,Lu,,,,,Lu,Lu,,,,,Lu,,
11223311,,,01
i,,,,,3max,Lu,,u,,Lu,max, (8) ii0,,,1,1,2,3ii,1,2,3
123123,,,,u,u,u,u,,,,,,,where and .
,,,,1,,,,,Ca,b;Rfmaxf,fIf we now recall that the norm of the
function f in is ,,1,,ccc,,
,,a,bf(where is the maximum absolute value of the function on the
closed interval), then, c
23,23,11,,,,,,,,u,fx,,hx,,hxu,fxu,x,,0setting, , , , , and,
we obtain from inequality (8),
123taking account of the uniform continuity of the functions,,, on
bounded ,Lu,u,u,i,1,2,3i
4
3subsets of , that R
,,,,,,,,,,,,,, ,,,,,,,,,,,,,,,,,,,,,,
,,maxLx,fx,hx,fx,hx,Lx,f x,fx,,Lx,fx,fxhx,,Lx,fx,fxhx230,x,b
,, as ,ohh,0,,1,,1cc
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
b,,,,,,,,,,F,,,,,,fh,,Lx,fx,fxh,,,,,,
x,,Lx,fx, fxh,,xdx (9) 23,a
We often consider the restriction of the functional (3) to the
affine space consisting of the
,,1,,f,C,,a,b;R,,,,functions that assume fixed values fa,A,
fb,B at the endpoints of the
,,1,,closed interval a,b. In this case, the functions h in the
tangent spaceTC, must have the f
,,a,bvalue zero at the endpoints of the closed interval . Taking
this fact into account, we may
integrate by parts in (9) and bring it into the form
bd,,,,,,,,,,,,,,,,,,,,,FfhLx,fx,fxLx,fx,fxhxdx
(10) ,,,,,,23,dxa,,
,,2Cof course under the assumption that L and f belong to the
corresponding class .
In particular, if f is an extremum (extremal) of such a functional,
then by Theorem 2 we have
,,,1,,,,Ffh,0h,C,,a,b;R,,,,ha,hb,0 for every function such
that . From this and relation (10)
one can easily conclude (see Problem 3 below) that the function f
must satisfy the equation
d,,,,,,,Lx,f,,,,x,fx,,Lx,f,,,,x,fx,0 (11) 23dx
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
,,0,0,,1,0which the two points are, for example, and . We
confine ourselves to just the
,,1,,,,f,C0,1;Rcurves that are the graphs of functions assuming
the value zero at both ends of
,,0,1the closed interval . The length of such a curve
12,,,,,,,Ff,1,fxdx (12) ,0
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
21233,,,,Lu,u,u,1,u
5
and therefore the necessary condition (11) for an extremal here
reduces to the equation
,,,,,,,dfx ,0,,2dx,,,,,,,1,fx,,
from which it follows that
,fx,, (13) ,常数2,,,,,1,fx
,,on the closed interval 0,1
uSince the function is not constant on any interval, Eq. (13) is
possible only if 21,u
,f,,x,,,const on a,b. Thus a smooth extremal of this problem must
be a linear function whose
,,,,1,0,,graph passes through the points 0,0fx,0 and . It
follows that , and we arrive at the
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 P0
at a lower level under the action of gravity in the shortest time.
another fixed point P1
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 PP01
coordinate system such that is at the origin, the x-axis is directed
vertically downward, P0
,,and the point P has positive coordinates x, shall find the
shape of the track among 111
,,the graphs of smooth functions defined on the closed interval 0,x
and satisfying the 1
f,,0,0,,condition fx,y,. At the moment we shall not take time to
discuss this by no means 11
uncontroversial assumption (see Problem 4 below).
PIf the particle began its descent from the point with zero velocity,
the law of variation 0
of its velocity in these coordinates can be written as
(14) v,2gx
Recalling that the differential of the arc length is computed by the
formula
222,,,,,,,,,ds,dx,dy,1,fxdx (15)
we find the time of descent
6
2,x1,,fx,,11, (16) ,,Ffdx,,x0g2
along the trajectory defined by the graph of the function on the
closed interval. ,,y,fx,,0,x1
For the functional (16)
23,,1,u123,,Lu,u,u,, 1u
and therefore the condition (11) for an extremum reduces in this
case to the equation
,,,,,,,,,dfx,0, ,,dx2,,,,,,,,,x1,fx,,,,,,,,
from which it follows that
,fx,, (17) ,cx2,,,,,1,fx
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
dy,cx (18) ds
However, from the geometric point of view
dydx,cos,,sin,, (19) dsds
,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
12 (20) x,sin,2c
But it follows from (19) and (20) that
22,,dydydxdxdsin,sin,,,,, ,,,,tgtg2,,22,,d,dxddd,,,cc,,
from which we find
1,,y,2,,sin2,,b (21) 22c
12,,t,aSetting and , we write relations (20) and (21) as 2c2
x,a,,1,cost (22) ,,y,at,sint,b
a,0x,0t,2k,k,ZSince , it follows that only for ,. It follows from
the form of the
t,0function (22) that we may assume without loss of generality that
the parameter value
,,P,0,0b,0corresponds to the point . In this case Eq. (21) implies,
and we arrive at the 0
7
simpler form
x,a1,cost,, (23) ,,y,at,sint
for the parametric definition of this curve.
Thus the brachistochrone is a cycloid having a cusp at the initial
point where the P0
tangent 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 P1
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).
8
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,,到赋范空间Y的映射在
UU,Uxf:U,Y
,,n,, fx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。,那么当时h,0有
1n,,,nn,, (1) f,,,,,,xhfxfxh?f,,xhoh,,,,,,,,,,n!
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)。
,f,,x 当时,由的定义,(1)式成立。 n,1
假设(1)对成立。 n,1,N
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
h,0成立。
1,,,,,nnfx,h,fx,fh,,fxhx?,,,,,,,,,,n!,,
,, 1n,1,,,,,,n,,,,,,,,,,,,,,fx,h,fx,fxh,,fxhh?,,,,
sup ,,,,n,1!,0,,1,,
n,1n,,,,o,hhoh,,,,,,,,,,
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分条件。
f:U,Rx,U定理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

,,k,,fxk,1,1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。
,,,,,k,1k,,,,,,fx,0,?,fx,0fx,0 如果且,那么为使x是函数f
的极值点
,,kk,,fxh 必要条件是: k是偶数,是半定的。
,,kkf,,xhh,1 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
9
,,kkf,,xh,,,0, 那么x是严格局部极小点;如果
,,kkf,,xh,,,0, 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

1k,,kk, f,,,,x,h,fx,f,,,,xh,,hhk!
其中,,,,,h是实值函数,而且当时。,h,0 h,0
我们先证必要条件。
,,k,,kkf,,x,0因为,所以有向量,使。于是,对于充分接近于零的实
参h,0f,,x h,000
量t,
1kk,,kf,,,,x,th,fx,f,,,,,,xth,,thth 0000!k
1,,k,,kkk ,,,,,fxh,,thht ,,000!k,,
,,kk同号。 括号内的表达式与f,,xh0
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
fx,th,fx0
,,k,,kk,,fx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hhf,,xh010
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
,,kk,,fxh,,,0h,1,当 的情况。这时
k1,,kk,,fx,h,f,,x,f,,,,xh,,hh k!
k,,,,1h,,k,,k,,,,,,,fx,,hh ,,,,!kh,,,,,,
1,,k,,,,,,hh , ,,!k,,
,,,h,0h,0h,0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
而对所有这些向量h
,,,,fx,h,fx,0,
10
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,那么以点为中心的单位球面是X中的有界,,
Sx;1x,X
ii, ,,,kkk1闭集,因而是紧集。这时,连续函数(k-形式)在上有
最,,,,,,Sx;1fxh ,,fxh,?,h?ii1k
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
,,kk,,fxh述为与它等价的形式:形式是定的(正定的或负定的)。
n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。 R
nf:R,R注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

,,kkf,,xh式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
,f,,x,0时就可只用一阶微分:求满足的点x。
3.一些例子
,,,,131,,,,,,L,CR;Rf,Ca,b;R,而,换句话说, 例1 设
123123,,,,u,u,u!Lu,u,u
3,,,,a,b,Rx!fx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R
数。
我们研究函数
,,1,,,,F:Ca,b;R,R (2) 它由以下关系式给出
,,1,,,,,,f,Ca,b;R!Ff
b,,,,,,,,Lx,fx,fxdx,R (3) ,a
,,1,,,,Ca,b;R因此,(2)是定义在函数集上的实泛函。
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,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。
11
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) ,,F,,,,,,,,fx,Lx,fx,fx1
给出的映射
,,1 (5) ,,,,F:C,,a,b;R,C,,a,b;R1
和映射
bg,C,,,,,,,,a,b;R!Fg,gxdx,R (6) 2,a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。
F2
我们来证明也是可微的,而且 F1
,,,, (7) ,,,,F,,,,,,,,fhx,,Lx,fx,fxh,,,,,,
x,,Lx,,,x231
,,1,,h,C,,a,b;R其中。
事实上,由有限增量定理的推论,在我们的情况下可得
33i,,,,,,Lu,,,u,,,u,,,Lu,u,u,,Lu,u,u, i,,1i ,,,,,,,,,,,,,sup,Lu,,,,,Lu,Lu,,,,,Lu,Lu,,,,,Lu,,
11223311,,,01
i,,,,,3max,Lu,,u,,Lu,max, (8) ii0,,,1,1,2,3ii,1,2,3
123123,,,,u,u,u,u,,,,,,,其中,。
,,,,1,,,,,Ca,b;R,,a,bffmaxf,f 如果记起中函数f的范数是(其中是函
数在区间上,,1,,cccc,,
23,23,11 ,,,,,,,,u,fx,,hx,,hxu,fxu,x,,0的最大模),那么设,,,,
和,考 虑到函数
3123R在的有界子集上的一致连续性,从不等式(8)得
到 ,,,Lu,u,u,i,1,2,3i
,,,,,,,,,,,,,,,,,,,,,,,,,,, ,,,,,,,,,,
,maxLx,fx,hx,fx,hx,Lx,fx,fx,,Lx,fx,f xhx,,Lx,fx,fxhx 230,x,b
,,,ohh,0 当时 ,,1,,1cc
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
b,,,,,,,,,, F,,,,,,fh,,Lx,fx,fxh,,,,,,
x,,Lx,fx,fxh,,xdx (9) 23,a
,,1,,f,C,,a,b;R,,a,b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
12
,,1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
端,,,,,,fa,Afb,Ba,bTCf
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
bd,,,,, (10) ,,,,,,,,,,,,,,,,
FfhLx,fx,fxLx ,fx,fxhxdx,,,,,,23,dxa,,
,,2当然要预先假设L和f属于相应的函数类C。
特别地,如果f是这个泛函的极值点(极值曲线),那么根据定理2,对于任意
使得
,,1,,,h,C,,a,b;RF,,fh,0,,,,的函数均有。由此,由(10)不难推
出( 见练习3),函数fha,hb,0
应该满足欧拉-拉格朗日方程。
d,,,,,, (11) ,Lx,f,,,,x,fx,,Lx,f,,,,x,fx,023dx
这是在变分学中被称为欧拉-拉格朗日方程的特殊的形式,现在研究具体的例
子。
例2 短程线问题
在平面内连接两个固定点的曲线中,求长度最小的那条曲线。
在这种情况下,答案是显然的,宁愿把它作为对以下推理的一个检验。
,,,,1,00, 0我们将认为,在平面上给出了笛卡儿坐标系,在该坐标系中不
妨认为点和是
,,1,,f, C,,0,1;R,,0,1给定的点。我们只限于研究那些曲线,它们是在区
间的端点取零值的函数的 图像,这种曲线的长度
12,,,,,,,Ff,1,fxdx (12) ,0
依赖于函数f且是例1中所研究的那种类型的泛函。在所给的情况下,函数L
有形式
21233,,,,Lu,u,u,1,u,
因此,在这里极值的必要条件(11)归结为方程
,,,,,,,dfx, ,0,,2dx,,,,,,,1,fx,,
,,0,1由它推出,在区间上
,fx,,,常数 (13) 2,,,,,1,fx
u,,,fx,,,a,b因为函数是 严格单调增的增函数,所以(13)只有在上常数时才
能成立。21,u
,,0,0,,1, 0f,,x,0这样一来,要求的光滑极值函数应是线性函数,其图
形通过点,。由此推出,于是我们得 到,连接两个已知点的直线段为所求的曲线。
例3 最速降线(或捷线)问题
这个于1696年由约翰?伯努利首先提出的关于捷线的经典问题乃是寻求那样的

P槽形式,质点沿着该沟槽在重力作用下在最短的时间内从已知点滑落到另一
更低的固0
13
定点。 P1
当然,我们忽略摩擦力。此外,在以后的研究中我们不考虑两个点位于同一铅

线上的平凡情况。
在通过点,的铅垂平面内,引进直角坐标系,使得点是它的原点,横轴垂 直
PPP010向下,而点有正的坐标。我们将只在定义在区间上满足条件,,,
f0,0,, ,,,,Px,y0,xfx,y111111
的光滑函数的图形中寻求沟槽的形式。我们暂且不讨论这个绝非无可争议的假
定(见习
题4)。
如果质点从点以零速度开始自己的运动,那么在所选择的坐标系中它的速度的
变P0
化规律为
(14) v,2gx
回忆弧长微分的计算公式,
222,,, (15) ,,,,,,ds,dx,dy,1,fxdx
,,y,fx求出沿着由定义在区间上的函数的图形确定的轨道运动的时
间 ,,0,x1
2,x1,,fx,,11, (16) F,,fdx,,x0g2
对于泛函(16),
23,,1,u123,,Lu,u,u,, 1u
因此,极值的必要条件(11)在所给情况下归结为方程
,,,,,,,,,dfx,0, ,,dx2,,,,,,,,,x1,fx,,,,,,,,从它推出
,fx,,,cx (17) 2,,,,,1,fx
PP其中c是不为零的常数。(因为,不在同一铅垂线上!) 01
考虑到(15),方程(17)可以改写为
dy,cx (18) ds
但从几何的观点有
dydx,cos,,sin,,, (19) dsds
14
其中是曲线的切线与横轴正方向间的夹角。 ,
比较方程(18)和(19)的第二个方程,我们求出
12 (20) x,sin,2c
但从(19)和(20)得到
22,,dydydxdxdsin,sin,,,,, ,,,,tgtg2,,22,,d,dxddd,,,cc,,由此求出
1 (21) ,,y,2,,sin2,,b22c
12,,t设和,可把关系式(20)和(21)改写为 ,a2c2
x,a1,cost,, (22) ,,y,at,sint,b
因为,所以 只在,时。从函数(22)的形式推出,不失一般性,可
a,0t,2k,k,Zx,0以认为点对应于 参量。这时,我们得到所求曲线的更简单的参数
形式 ,,P,0,0t,0b,00
x,a,,1,cost (23) ,,y,at,sint
这样,捷线是摆线,它在起始点是具有铅垂切线的尖点。常数a,即同位相似
系P0
数,应当这样选择,它使曲线(23)也通过点。画出曲线(23)后,可以发现,这
样的选P1择不总 是唯一的,这也证实了极值的必要条件(11)一般来说不是充分
的。然而从物理意
义考虑,在参数a的可能的值中应该选取怎样的值是显然的(其实也可以用直
接计算来
证实)。
15

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