信科专业英语论文.doc

中北大学 理学院英语论文 专业：信息与计算科学班级：07070242学号：0707024231姓名：朱胜明 2010年5月13日 Not successful breakthroughFrankly, in my personal problem-solving experience, not successful and even failures, they are much more than success of exciting. However, not successful is not only to the author left a negative result, the smooth face occasionally I always look for them to continue to the problem solving stupid , I do not know the personal experience to say whether the laughing stock of the readers a little help, but I can confirm that this is my already meager wealth of the main assets of problem solving, and my views (including Articles 1998 problem solving analysis of serial and Introduction to mathematical problem solving learning book) has caused some concern and sympathy counterparts, need to apologize for that, sometimes, on the problem solving and problem-solving analysis of a large number of readers have written I can not reply to everyone, todays topic is largely a deliberate cover. Next, I want to be three cases of problem solving to demonstrate how the failure to success, how to control the shallow success.1. Case 1 - to obtain useful information from the failureExample 1 If a, b, c are real and unequal, and x / (a-b) = y / (b-c) = z / (c-a), seeking x + y + z.Solution: get the geometric theorems x / (a-b) = y / (b-c) = z / (c-a)= (X + y + z) / (a-b) + (b-c) + (c-a).However, the denominator is zero-based(a-b) + (b-c) + (c-a) = 0,Our problem-solving efforts failed.Comment: This is a case of failed problem solving, 3 referred to in the solution after the direction of some deal, actually used style. Therefore, the failure of the process was exactly the subject of an implied condition, which is a positive gain, as we remove the unsuccessful style , to look at the type and style at the same time, style make us see the coincidence of two lines:xX + yY + z = 0,(A-b) X + (b-c) Y + (c-a) = 0.And style has enabled us to see through the point lineX = 1,Y = 1. for further inference, linear also through point (1.1), thenx + y + z = 0.Compared with 3, it is an innovative solution ,its birth depends on two things:1, to obtain a useful problem-solving information from failure that is style.2, type on , style is to focus shift from the perspective of analytic geometry to see them.With this step, the remaining work at best can be completed within 30 seconds.2. Case 2 - to have not yet succeeded does not mean failureIf f (n) is the positive items on the n increment series, M is greater than f (1) the normal number, to prove inequality when using mathematical kownledgef (n) M (n N), Its step 2 will appear this situation: Suppose f (k) 0) M + a,Can not launch f (k +1) M.Accordingly, many people suggested to use the approach to strengthen the proposition,someone also comes to such a proposition (see article 4 P.32 and paper 5 P.12):Proposition if (f (n) is increasing on the n series of positive items, M is the normal number, then the inequality f (n) M (n N) can not be directly proved by mathematical induction.Comment: is not denmonstratedby , there are two possibilities, one is the lack of skill of mathematical induction, the other is the improper use of mathematical induction. The no use as can not, its loss is irreparable.We analyze the deal not successful, the key is recursive , which prompted us to think about: f (k +1) and f (k) Is it only a recursive relationship between you?Indeed, some functional of its f (k +1) and f (k) the relationship between the complex and can not use mathematical induction to prove directly; while some relationships are more simple and can only be expressed in Math. But whether it is very complex or simple and its expression are not unique necessarily paper 6 P. 278 had given a counter example to show that the proposition is not true:Example 2 Using mathematical induction to provef (n) = 1 + (1 / 2) + (1 / 22) + . + (1 / 2n-1) 2.Explain: When n = 1, the proposition clearly established.It is assumed that f (k) 2, thenf (k +1) = f (k) + (1 / 2k) 2 + (1 / 2k),As the 2 + (1 / 2k) Evergrande at 2, so mathematical induction has not succeeded.However, it is only used inappropriately. For a recursive manner, that is not difficult.f (k +1) = 1 + (1 / 2) f (k) 1 + (1 / 2) 2 = 2.The following is taken directly from a counter-examples in 4, Example 2.Example 3 verify (1 / 1!) + (1 / 2!) + (1 / 3!) + . + (1 / n!) 2.Proof: When n = 1, the proposition is clearly established.Suppose n = k when the propositions are true, then(1 / 1!) + (1 / 2!) + . + (1 / k!) + 1 / (k +1)! = 1 + (1 / 2) + (1 / 3) (1 / 2!) + . + (1 / k) 1 / (k-1)! + 1 / (k +1) (1 / k!) 1 + (1 / 2) (1 + (1 / 2!) + . + 1 / (k-1)! + (1 / k!) 0),Or f (x) = (x2+1) / 2 + (1-) x (0 1). Under normal circumstances should be the x of positive function (8 default constant is incomplete; Similarly, the 2000 College Entrance Question 20 (2), on cn= An+ BnSet upan= Cncos2,bn= Cnsin2Is wrong), but f (x) is a quadratic function, can only constant. in order to calculate , the f (-1) = 0 substituted into can be obtained = 1 (or in the = 1 / 2). type of different type and, reflecting the difference between special and general, reflecting the verification and demonstrated the difference between. In fact, the original method 1 came out, can be drawn immediately type, and whether the application of fundamental inequality has nothing to do. Similarly, the original method 1 thought about the author of reasoning is tight in the Solving China is still a problem. All these circumstances that we should not only reflect on the problem-solving activities, but also on the Reflection for further reflection. Thinking of a solution of the following is wrong with the reader?Solution: There are known condition is equivalent to k 0, so thatF (x)-x f (x) - (x2+1) / 2 = k 0,To x =- 1 时, f (x) = 0 Substituting get k =- 1,So f (x)-x f (x) - (x2+1) / 2 =- 1,That f2(X) - (x +1)2/ 2 f (x) + (x3+ X +2) / 2 = 0.This solution out of f (x) is an irrational function, not a quadratic function, so no solution for this question.As a re-reflection is another reflection of the new example, we pointed out in 9 Example 2 (the 1997 college entrance examination problems) 1 Q, you can take = a (x2-X) (0,1) ( is a function of x), thenf (x) = a (x1-X) (x2-X) + x= 1（1）,Will score points according to the nature,we have the result of x f (x) x1.译文：尚未成功的坦率坦率说，在我个人的解题经历中，“尚未成功”乃至失败，实在是比激动人心的成功多得多但是，“尚未成功”并非只给笔者留下消极的结果，而面对偶尔的顺利笔者也总是要继续寻找当中的“解题愚蠢”（见文1、2），我不知道这些说来见笑的个人体验是否对广大读者有点帮助，但我能肯定地说，这是我本来就少得可怜的解题财富中的主要资产，并且我的看法（包括本刊1998年开始的解题分析连载以及数学解题学引论一书）已引起了一部分同行的关注与共鸣，需要致歉的是，二三年来，关于解题与解题分析的大批读者来信我不能一一作复，今天的话题很大程度上是一种有意的弥补下面，笔者要进行3个解题个案的分析，以展示如何由失败走向成功，又如何对浅层的成功进行深层的调控1个案1由失败中获取有用的信息例1若、为互不相等的实数，且（）（）（），求解：由等比定理得 （）（）（）（）（）（）（）但是，式的分母为零（）（）（）0，我们的解题努力失败了评析：这是一个失败的解题案例，文3谈到了调整解题方向后的一些处理，其实都用到式所以，失败的过程恰好显化了题目的一个隐含条件，这是一个积极的收获，当我们将不成功的式去掉，把目光同时注视式与式时，式使我们看到了两条直线重合：0，（）（）（）0而式又使我们看到了直线通过点1，1作一步推理，直线也通过点（1，1），于是0与文3相比，这是一个不无新意的解法，其诞生有赖于两点：第1，从失败的解题中获取一条有用的信息，即式第2，对式、式都作“着眼点的转移”，从解析几何的角度去看它们有了这两步，剩下来的工作充其量在30秒以内就可以完成2个案2尚未成功不等于失败设（）为关于的正项递增数列，为大于（1）的正常数，当用数学归纳法来证不等式（）（）时，其第2步会出现这样的情况：假设（），则（1）（）（1）（）0），无法推出（1）据此，许多人建议，用加强命题的办法来处理，还有人得出这样的命题（见文432及文512）：命题设（）为关于的正项递增数列，为正常数，则不等式（）（）不能直接用数学归纳法证明评析：不等式没能用递推式证出来，有两种可能，其一是数学归纳法的功力不足，其二是数学归纳法的使用不当把“不会用”当作“不能用”，其损失是无法弥补的我们分析上述处理的“尚未成功”，关键在于递推式，这促使我们思考：（1）与（）之间难道只有一种递推关系吗？确实，有的函数式其（1）与（）之间的关系很复杂，无法用数学归纳法来直接证明；而有的关系则较简单，仅用加减乘除就可以表达出来但无论是“很复杂”还是“较简单”，其表达式都未必惟一，文6278给出过一个反例，说明上述“命题”不真：例2用数学归纳法证明（）1（12）（122）（121）2讲解：当1时，命题显然成立现假设（）2，则（1）（）（12）2（12），由于2（12）恒大于2，所以数学归纳法证题尚未成功然而，这仅是“方法使用不当”换一种递推方式，证明并不困难（1）1（12）（）1（12）22下面一个反例直接取自文4的例2例3求证（11！）（12！）（13！）（1！）2证明：当1时，命题显然成立假设时命题成立，则（11！）（12！）（1！）1（1）！1（12）（13）（12！）（1）1（1）！1（1）（1！）1（12）1（12！）1（1）！（1！）1（12）22这表明1时命题成立由数学归纳法知，不等式已获证3个案3对尚未成功的环节继续反思文7有很好的立意也有很好的标题，叫做“反思通解引出简解创造巧解”，它赞成反思“失败”并显示了下面一道二次函数题目的调控过程：例4二次函数（）2的图象经过点（1，0），是否存在常数、使不等式（）（21）2对一切实数都成立？若存在，求出、；若不存在，说明理由讲解：作者从解两个二次不等式（21）2（）0，（）0开始（解法1），经过数形结合的思考（解法2）等过程，最后“经学生相互讨论后得到巧解”（解法4）