数学及应用数学专业外文翻译多元函数的极值

1、word外文原文EXTREME VALUES OF FUNCTIONS OF SEVERAL REAL VARIABLES1. Stationary PointsDefinition 1.1 Let and . The point a is said to be: (1) a local maximum iffor all points sufficiently close to ;(2) a local minimum iffor all points sufficiently close to ;(3) a global (or absolute) maximum iffor all po

2、ints ;(4) a global (or absolute) minimum iffor all points ;(5) a local or global extremum if it is a local or global maximum or minimum.Definition 1.2 Let and . The point a is said to be critical or stationary point if and a singular point if does not exist at .Fact 1.3 Let and .If has a local or gl

3、obal extremum at the point , then must be either:(1) a critical point of , or(2) a singular point of , or(3) a boundary point of .Fact 1.4 If is a continuous function on a closed bounded set then is bounded and attains its bounds.Definition 1.5 A critical point which is neither a local maximum nor m

4、inimum is called a saddle point.Fact 1.6 A critical point is a saddle point if and only if there are arbitrarily small values of for which takes both positive and negative values.Definition 1.7 If is a function of two variables such that all second order partial derivatives exist at the point , then

5、 the Hessian matrix of at is the matrixwhere the derivatives are evaluated at.If is a function of three variables such that all second order partial derivatives exist at the point , then the Hessian of f at is the matrixwhere the derivatives are evaluated at.Definition 1.8 Let be an matrix and, for

6、each ,let be the matrix formed from the first rows and columns of .The determinants det(),are called the leading minors of Theorem 1.9(The Leading Minor Test). Suppose that is a sufficiently smooth function of two variables with a critical point atand H the Hessian of at.If , then is:(1) a local max

7、imum if 0det(H1) = fxx and 0det(H)=; (2) a local minimum if 0det(H1) = fxx and 0det(H1), 0det(H3);(2) a local minimum if 0det(H1), 0det(H3);(3) a saddle point if neither of the above hold.where the partial derivatives are evaluated at.Key Points.A continuous function on a closed bounded set is bound

8、ed and achieves its bounds.To find the extreme values of a function on a closed bounded set it is necessary to consider the value of the function at stationary points(), singular points (does not exist) and boundary points(points on the edge of the set).Stationary points can be classified as local m

9、axim , local minim or saddle points.If The Leading Minor Test 1.9 is not applicable, the stationary point must be classified by directly applying Definition 1.1 and Fact 1.6. For example in the two variable case, if has a stationary point at ,we consider the sign offor arbitrarily small, positive an

10、d negative values of and (that are not both zero). In each case, if det(H)= 0, then can be either a local extremum or a saddle point.Example. Find and classify the stationary points of the following functions: (1) (2) Solution. (1) ,soijkCritical points occur when ,i.e. when(1) (2) (3) Using equatio

11、ns (2) and (3) to eliminate y and z from (1), we see thator ,giving , and .Hence we have three stationary points: , and . Since, and ,the Hessian matrix is At ,which has leading minors 0,And det .By the Leading Minor Test, then, is a local minimum. At ,which has leading minors 0,And det .By the Lead

12、ing Minor Test, then, is also a local minimum.At , the Hessian isSince det, we can apply the leading minor test which tells us that this is a saddle point since the first leading minor is 0. An alternative method is as follows. In this case we consider the value of the expression，for arbitrarily sma

13、ll values of h, k and l. But for very small h, k and l, cubic terms and above are negligible in comparison to quadratic and linear terms, so that.If h, k and l are all positive, . However, if and and ,then .Hence close to ,both increases and decreases, so is a saddle point.(2) soij.Stationary points

14、 occur when ,i.e. at .Let us classify this stationary point without considering the Leading Minor Test (in this case the Hessian has determinant 0 at so the test is not applicable).LetCompleting the square we see that So for any arbitrarily small values of h and k, that are not both 0, and we see th

15、at f has a local maximum at .2. Constrained Extrema and Lagrange MultipliersDefinition 2.1 Let f and g be functions of n variables. An extreme value of f(x) subject to the condition g(x) = 0, is called a constrained extreme value and g(x) = 0 is called the constraint.Definition 2.2 If is a function

16、of n variables, the Lagrangian function of f subject to the constraint is the function of n+1 variableswhere is known as the Lagrange multiplier. The Lagrangian function of f subject to the k constraints , is the function with k Lagrange multipliers, Key Points.To find the extreme values of f subjec

17、t to the constraint g(x) = 0: (1) calculate, remembering that it is a function of the n+1 variables and (2) find values of such that (you do not have to explicitly find the corresponding values of ): (3) evaluate f at these points to find the required extrema.Note that the equation is equivalent to

18、the equations,and So, in the two variable case, we have Lagranian function and are solving the equations:, , and .With more than one constraint we solve the equation.Theorem 2.3 Let and be a point on the curve C, with equation g(x,y) = 0, at which f restricted to C has a local extremum.Suppose that

19、both and have continuous partial derivatives near to and that is not an end point of and that . Then there is some such that is a critical point of the Lagrangian Function.Proof. Sketch only. Since P is not an end point and ,has a tangent at with normal .If is not parallel to at , then it has non-ze

20、ro projection along this tangent at .But then f increases and decreases away from along ,so is not an extremum. Henceand are parallel and there is somesuch that and the result follows.Example. Find the rectangular box with the largest volume that fits inside the ellipsoid ,given that it sides are pa

21、rallel to the axes.Solution. Clearly the box will have the greatest volume if each of its corners touch the ellipse. Let one corner of the box be corner (x, y, z) in the positive octant, then the box has corners (x,y,z) and its volume is V= 8xyz.We want to maximize V given that . (Note that since th

22、e constraint surface is bounded a max/min does exist). The Lagrangian isand this has critical points when , i.e. when (Note that will always be the constraint equation.) As we want to maximize V we can assume that so that .)Hence, eliminating , we getso that and But then so or ,which implies that an

23、d (they are all positive by assumption). So L has only one stationary point (for some value of , which we could work out if we wanted to). Since it is the only stationary point it must the required max and the max volume is.中文译文 多元函数的极值1. 稳定点定义1.1 使并且. 对于任意一点有以下定义: (1)如果对于所有充分地接近时，那么是一个局部极大值;(2)如果对于

24、所有充分地接近时，那么是一个局部极小值;(3)如果对于所有点成立，那么是一个全局极大值或绝对极大值;(4) 如果对于所有点成立，那么是一个全局极小值或绝对极小值; (5) 局部极大小值统称为局部极值；全局极大小值统称为全局极值.定义 1.2 使并且.对于任意一点，如果，并且对于任意奇异点都不存在，那么称是一个关键点或稳定点.结论 1.3 使并且.如果有局部极值或全局极值对于一点, 那么 一定是:(1)函数的一个关键点, 或者(2)函数的一个奇异点, 或者 (3)定义域的一个边界点.结论 1.4 如果函数是一个在闭区间上的连续函数，那么在区间上有边界并且可以取到边界值.定义 1.5 对于任一个关

25、键点，当既不是局部极大值也不是局部极小值时，叫做函数的鞍点.结论 1.6 对于一个关键点是鞍点当且仅当任意小时，对于函数取正值和负值.定义 1.7 如果 是二元函数，并且在点处所有二阶偏导数都存在，那么那么根据函数在点处导数，有在点处的Hessian矩阵为：.推广：如果 是三元函数，并且在点处所有二阶偏导数都存在，那么根据函数在点处导数，有在点处的Hessian矩阵为：.定义 1.8 矩阵是 阶矩阵，并且对于每一个都有,从矩阵中选取左上端的行和列，令其为阶的矩阵.那么行列式det(),叫做矩阵的顺序主子式. 定理 1.9 假设是一个充分光滑的二元函数，且在点处稳定，其Hessian 矩阵为H

26、.如果，那么根据偏导数判定点是：(1) 一个局部极大值点， 如果0det(H1) = fxx并且0det(H)=; (2) 一个局部极小值点， 如果0det(H1) = fxx并且0det(H1), 0det(H3)时;(2) 一个局部极小值点， 如果当0det(H1), 0det(H3)时;(3) 一个鞍点，如果点既不是局部极大值点也不是局部极小值点. 在不同的情况下 ,当det(H)= 0时, 点是一个局部极值点，或者是一个鞍点.关键点.在有界闭集上的连续函数有边界，并且可以取到其边界值.当确定函数在有界闭集上的极值时，必须考虑函数在稳定点(即 时), 奇异点 (当 不存在时) 和边界点(

27、点在集合的边缘)处的函数值.稳定点可以分为局部极大值点、局部极小值点或鞍点. 对于稳定点，当应用定理 1.9 不能分类时，可依据定义1.1和结论1.6对稳定点进行直接分类.例如，在二元情况下，如果 在点 处的点是稳定点，我们可以考虑函数的符号，当 和 任意小( 和 可为正值和负值,但不同时为0)时. 例. 确定以下函数的稳定点并说明是哪一类点: (1) (2) 解. (1) ,soijk当时有稳定点,也就是说, 当 (1) (2) (3) 时,将方程(2)和方程(3)带入到方程(1)可以消去变量y和z, 由此可以得到即,得,和.因此我们可以得到函数的三个稳定点:,和. 又因为,和,那么Hess

28、ian矩阵为 在点处, 那么顺序主子式 0,并且行列式.根据主子式判定方法，那么点是一个局部极小值点. 在点处, 那么顺序主子式 0,并且行列式.根据主子式判定方法，那么点也是一个极小值点. 在点处，Hessian矩阵为因此det,根据主子式判定方法，第一主子式为0，由此我们可以知道该点是一个鞍点. 下面是另一种计算方法，在这种情况下，我们考虑现在下面函数表达式，的值，对于任意h, k和l无限小时. 担当h, k和l非常小时, 三次及三次以上方程相对线性二次方程时可忽略不计,那么原方程可为.当h, k和l 都为正时,.然而， 当、和,那么.因此当接近时,同时增加或者同时减少, 所以 是一个鞍点.(2) soij.当时有稳定点,也就是说, 当在时. 现在我们在不考虑主子式判定方法的情况下为该稳定点进行分类因为在时Hessian矩阵的行列式为0，所以该判定方法在此刻无法应用.令配成完全平方的形式为所以对h和k为任意小时h和k都不为0，有，因此我们可以确定函数f 在点处有局部极大值.2. 条件极值和Lagrange乘数法定义 2.1 函数f和函数g都是n元函数.对于限制在条件g(x) = 0下的函数f (x)的极值叫做