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.中文譯文 多元函數的極值1. 穩定點定義1.1 使并且. 對于任意一點有以下定義: (1)如果對于所有充分地接近時，那么是一個局部極大值;(2)如果對于

24、所有充分地接近時，那么是一個局部極小值;(3)如果對于所有點成立，那么是一個全局極大值或絕對極大值;(4) 如果對于所有點成立，那么是一個全局極小值或絕對極小值; (5) 局部極大小值統稱為局部極值；全局極大小值統稱為全局極值.定義 1.2 使并且.對于任意一點，如果，并且對于任意奇異點都不存在，那么稱是一個關鍵點或穩定點.結論 1.3 使并且.如果有局部極值或全局極值對于一點, 那么 一定是:(1)函數的一個關鍵點, 或者(2)函數的一個奇異點, 或者 (3)定義域的一個邊界點.結論 1.4 如果函數是一個在閉區間上的連續函數，那么在區間上有邊界并且可以取到邊界值.定義 1.5 對于任一個關

25、鍵點，當既不是局部極大值也不是局部極小值時，叫做函數的鞍點.結論 1.6 對于一個關鍵點是鞍點當且僅當任意小時，對于函數取正值和負值.定義 1.7 如果 是二元函數，并且在點處所有二階偏導數都存在，那么那么根據函數在點處導數，有在點處的Hessian矩陣為：.推廣：如果 是三元函數，并且在點處所有二階偏導數都存在，那么根據函數在點處導數，有在點處的Hessian矩陣為：.定義 1.8 矩陣是 階矩陣，并且對于每一個都有,從矩陣中選取左上端的行和列，令其為階的矩陣.那么行列式det(),叫做矩陣的順序主子式. 定理 1.9 假設是一個充分光滑的二元函數，且在點處穩定，其Hessian 矩陣為H

26、.如果，那么根據偏導數判定點是：(1) 一個局部極大值點， 如果0det(H1) = fxx并且0det(H)=; (2) 一個局部極小值點， 如果0det(H1) = fxx并且0det(H1), 0det(H3)時;(2) 一個局部極小值點， 如果當0det(H1), 0det(H3)時;(3) 一個鞍點，如果點既不是局部極大值點也不是局部極小值點. 在不同的情況下 ,當det(H)= 0時, 點是一個局部極值點，或者是一個鞍點.關鍵點.在有界閉集上的連續函數有邊界，并且可以取到其邊界值.當確定函數在有界閉集上的極值時，必須考慮函數在穩定點(即 時), 奇異點 (當 不存在時) 和邊界點(

27、點在集合的邊緣)處的函數值.穩定點可以分為局部極大值點、局部極小值點或鞍點. 對于穩定點，當應用定理 1.9 不能分類時，可依據定義1.1和結論1.6對穩定點進行直接分類.例如，在二元情況下，如果 在點 處的點是穩定點，我們可以考慮函數的符號，當 和 任意小( 和 可為正值和負值,但不同時為0)時. 例. 確定以下函數的穩定點并說明是哪一類點: (1) (2) 解. (1) ,soijk當時有穩定點,也就是說, 當 (1) (2) (3) 時,將方程(2)和方程(3)帶入到方程(1)可以消去變量y和z, 由此可以得到即,得,和.因此我們可以得到函數的三個穩定點:,和. 又因為,和,那么Hess

28、ian矩陣為 在點處, 那么順序主子式 0,并且行列式.根據主子式判定方法，那么點是一個局部極小值點. 在點處, 那么順序主子式 0,并且行列式.根據主子式判定方法，那么點也是一個極小值點. 在點處，Hessian矩陣為因此det,根據主子式判定方法，第一主子式為0，由此我們可以知道該點是一個鞍點. 下面是另一種計算方法，在這種情況下，我們考慮現在下面函數表達式，的值，對于任意h, k和l無限小時. 擔當h, k和l非常小時, 三次及三次以上方程相對線性二次方程時可忽略不計,那么原方程可為.當h, k和l 都為正時,.然而， 當、和,那么.因此當接近時,同時增加或者同時減少, 所以 是一個鞍點.(2) soij.當時有穩定點,也就是說, 當在時. 現在我們在不考慮主子式判定方法的情況下為該穩定點進行分類因為在時Hessian矩陣的行列式為0，所以該判定方法在此刻無法應用.令配成完全平方的形式為所以對h和k為任意小時h和k都不為0，有，因此我們可以確定函數f 在點處有局部極大值.2. 條件極值和Lagrange乘數法定義 2.1 函數f和函數g都是n元函數.對于限制在條件g(x) = 0下的函數f (x)的極值叫做