數學

有二系統A和B，A中一開始有2k個物體，，B中有0個物體。在一個單位時間內，兩系統可以互相轉移最多一個物體。當B中物體的個數為 i-1，i∈{1,2,...,k+1}，我們稱其為狀態 i，從狀態1﹝初態﹞開始計時，到達狀態 k+1﹝相同態﹞便即刻停止實驗，經過之時間為一隨機變數T，稱之為實驗時間。問當兩個系統的物體數剛好相等時，經過的實驗時間之分佈為何？本文將以上述問題為核心，分別探討不同條件下系統的實驗時間所反映出來的現象，如機率、期望值、變異數等等。 Define two systems, A includes 2k objects, and B has none. They can transfer at most one object from one system to another in a time unit. When the number of objects in B is i-1, i∈{1,2,...,k+1} , we say the system is at state i. As soon as system transfer form state 1 ( initial state ) to state k+1 ( the same state ), the experiment stop. Random variable T, called the experiment time, is the time before stop. What would be the distribution of the experiment time if all systems have the same amount of objects within? This article will focus on the described question and discuss what property the experiment time of the system under various conditions has, such as probability, mean, and variance.

在此研究中，我們用類似反演變換的方法，以一個定圓創立並證明了一種新的幾何變換，稱為 「調和變換」 · 我們得到點、直線、圓與圓錐曲線經過變換的關係 ·。1 .直線可以映射成原直線或一圓錐曲線 · 2.圓可以映射成一種特殊曲線。 3 .圓錐曲線可以映射成兩條圓錐曲線或一條圓錐曲線和一直線。此外我們還發現調和變換和反演變換的特殊關係 · 最後，由於調和變換可以簡化圓錐曲線的關係，我們將調和變換應用在行星輾些的證明上，並得到了良好的結果。In this research, we use a method similar to the inversion to establish a new geometric transformation, called harmonic transformation, by a fixed circle O, we prove some of its properties. We have gotten the relationship among points. lines, circs, conies and their images: 1 .The image of a line is a conic or a line itself. 2.Thc image of a circle is a special category of curve. 3.The image of a conic with its focus at the center of O is two conies or a line and a conic. Further mote, we also find the special connection between harmonic transformation and inversion. Finally, since the harmonic transformation can simplify the conic, we apply the harmonic transformation to identify the orbit of a planet, and obtain a nice conclusion.

This study was to explore the nature of two basic constitutes of the regular pentagon,With these two constitutes, the regular pentagon could be multiplied into any times in size. We used four multiplication methodsto show how the regular pentagon enlarge and to verify that the enlarged regular pentagons derived from computer did exist. By integrating these four multiplication rules, we were able to arrange regular pentagon of any length of side, and evidenced the equation was ( If m,n is the number of A,B of a regular pentagon respectively ) When we tried to verify if any regular pentagon could be constituted by other smaller regular pentagons, we found that it was un-dividable only if the length of pentagon side were (the number of A, B were the 2n and 2n-1 item of Lucas Sequence), otherwise, any regular pentagon is able to be constituted by other smaller regular pentagons. The divided forms could be multiple. We also found that any pentagon could be divided by two successive un-dividable pentagons, which is called “standard division rule”. We expected to derive all kinds of division by analysis of two successive un-dividable pentagons in standard division rule. 這個研究起源於一個拼圖玩具：利用兩種黃金三角形排出指定大小的正五邊形。我們的研究動機是：一、 假如無限量供應A 和B，能夠拼出哪些邊長的正五邊形？二、 哪些拼好的正五邊形不能拆成一些較小的正五邊形？我們將研究的主要結果分述如下：

一 Motivation and Purpose: In this study, we want to completely know about “The number abc…de, which times m/n, 1≦n≦m≦9?N can get ed… cba?”, and also expect to find out “The good rule within them”. 二 Procedure:Using method of enumeration, induction to collect sample of all and beginning from two digits to get information “good rule”. When get some useful idea, put them into the following research for the step easy go on, the method try and error is a very tiresome works, especially when we deal higher digits. till enough information is obtained, we solve problem and find new one, then likewise again research steps, just the basic science research ways, we are glad have the key of these problem. 三 Result and conclusion :Those number we named “converse No.” There are two groups: S=m+n=10 and 11 S=11, then Q=m/n=9/2,8/3,7/4,6/5=4.5,2.6,1.75,1.2 S=10, then Q=m/n=9/1,8/2,7/3,6/4=9,4,2.3,1.5 Each group have four type. When S=11,Q=7/4=1.75,if converse No.each digit is a multiple of 3, then can cancellation or extension of fraction to get another 3 or 4. Growth up rule: Converse No. = type factor x heritable factor x growth factor=rx hx g S=11,r=2~5,h=9, s=10, r=1~4, h=99 一 研究目的：盼能找出”顛倒一族”的族譜。二 研究過程：確定研究題目為ab…cde×m/n=edc...ba,0≦n≦m≦9?N 求ab… cde？以窮舉法收集觀察資料，歸納演繹尋求規律。1.先觀察兩位數，分析共有顛倒對36對。2.建立乘數Q=m/n一覽表，共有27個3.設計顛倒對大/小及其商一覽表，以利觀察、歸納獲得規律。4.接著觀察三位數，共有360對，綜合二、三位數規律，找出選擇式窮舉法：9之倍數法。5.再接著找出四位數，再綜合而知另有 全調法 重現法 半調法 GCD遺傳基因法等來繁衍高位數顛倒數。6.於是依諸法找得六位數資料，得知GCD遺傳基因法為繁衍通則，完成族譜建立模式。7.研究顛倒數位數與其個數間關係式，完成研究。研究結論：1.顛倒一族有兩大類：S=10與S=11 S=m+n。2.每一大類有四型： S=10中，Q =9/1,8/2,7/3,6/4(9,4,2.3,1.5)S=11中，Q =9/2,8/3,7/4,6/5(4.5,2.6,1.75,1.2)3.每一型均有一個顛倒數，除了S=11中，Q=7/4=1.75者可約、擴分而得3or4個。4.顛倒數原則上均為9之倍數，除了Q=7/4經約、擴分可能得非9倍數者。

想像一下，原來固定的一個黑色大正方形(以下我們簡稱S1)四個角經過較小之正方形(以下簡稱S2)遮蓋後旋轉，為什麼會有放大縮小的感覺呢?(圖一) 我們猜測這是因為未被遮蓋部分(以下簡稱S3)的面積改變了，所造成的結果。令S1的邊長為 r，S2的一個小正方形邊長為n，k=r-n，θ為旋轉角度，我們實際算出了旋轉角度θ對S3的面積函數，

In this work I make the analysis of the possibility of the existence of the number system with non-natural base & its investigation. The question examined in my work is totally opened:\r ‧ making the list of new characteristics, rules of the translation of the numbers, and also rules of the simple calculating operations, checked the operations of subtraction & division;\r ‧ checked the Euclidean algorithm, its characteristics by means of estimating the coefficients;\r ‧ found the practical appliance of new method in compiling & solving of the tasks.\r Investigation I’ve suggested stipulates for independence of new system & its appliance in type of tasks, that is beyond the course of school program & also beyond the whole system of school education.

Purpose of the Research:\r 1) To determine whether a poor understanding and inability of Grade 7 and 8 learners to apply the BODMAS principle in mathematics, influences scores obtained in a mathematics test.\r 2) To determine whether scores obtained in the given mathematics test can be improved with a BODMAS learning tool.\r Procedures:\r 1. Get the educators opinion on mathematics in schools. Send a total of 50 questionnaires to four schools.\r 2. Determine what percentage of a mathematical test/examination requires the application of BODMAS\r 3. Do a pre-test at two schools, a total of 370 grade 7 and 8 learners.\r 4. Design a BODMAS learning tool and verify it with three educators.\r 5. Implement the tool at the two schools.\r 6. Do a post-test at the two schools.\r 7. Get all the educators who were at the implementation session to evaluate the session.\r 8. Investigate two other schools, by sending 270 pre-tests to those two schools, to determine whether applying the BODMAS principle correctly is also a problem for learners in those schools.\r 9. Implement the BODMAS learning tool into the intermediate phase syllabus.\r Data:\r 1. Of the 41 educators in the sample, 52% think the standard of maths in their schools is average.\r 2. 38.9% of a grade 8 mathematics examination paper and 46% of grade 8 mathematics tests contains questions that are BODMAS related.\r 3. The learners achieved an overall average of 22.57% in the pre-test\r 4. The educators evaluated the BODMAS learning tool as very good as it is.\r 5. Learners and educators enjoyed the implementation session of the BODMAS learning tool.\r 6. In the post test learners did much better, the overall average increased by 21.00% to 43.57%.\r 7. Educators were positive about the way in which the tool was explained.\r 8. The learners in the other two schools also struggled with applying the BODMAS principle.\r 9. A second pilot study is being done in four primary schools by the Department of Education for the implementation in the Free State mathematics 2013 syllabus. \r Conclusion:\r My hypothesis is supported. \r 1) A poor understanding and inability of Grade 7 and 8 learners to apply the BODMAS principle in mathematics, influenced scores obtained in a mathematics test.\r 2) Scores obtained in the given mathematics test were improved with a BODMAS learning tool.

Motivated by Napoleon theorem, we study the properties of the triangles obtained by moving the midpoint of each side of a given trianle along the perpendicular bisector of corresponding sides, and extend the results to the case of quadrilaterals. On the other hand ,we consider the method of erecting a regular M-gon to each side of a given N-gon and joint the N centers of these M-gons to form a new N-gon. (abbreviated as CRG method),and get the following results. 1. We characterize some kinds of N-gons that can be transformed to regular N-gons via CRG method. 2. Of M,N are nature numbers with M|N, then it is possible to find a N-gon that can be transformed to a regular N-gon by CRG method. \r 3. If a polygon P is symmetric with respect to a fixed point or a fixed line, then P can be transformed by CRG to a polygon with similar symmetries. 4. If a polygon P is transformed by CRG to ′P,there exists a commonpoint G such that ΣGA=0 andΣGB=0, where A and B runs through vertices of and P′P, respectively. 本研究將拿破崙定理加以延伸。先探討由各邊中點沿中垂線延伸得出之三角形的性質並推廣至四邊形之情形條列式報告成果。另一個推廣是將給定的多邊形的每邊外接一個正多邊形，再以這些外接的正多邊形的中心為頂點造出一個新的多邊形。我們發現此幾何變換具有以下性質：（1） 「哪些多邊形能被變換成正多邊形呢?」，我們觀察出能被變換成正多邊形的多邊形其限制條件隨邊數增加而增多，並進一步區分了哪些多邊形可以被變換成正多邊形。 （2） 在將非正N邊形做變換時，不一定須外接正N邊形才能得到正N邊形，我們區分出可外接哪些正多邊形而得到正多邊形。 （3） 對給定的多邊形作此變換時，若原多邊形有點對稱或線對稱等性質，則新多邊形也將具有相同的性質。 （4） 此變換得到的新多邊形會與原多邊形共重心，亦即新舊兩多邊形內到各自的頂點向量和為0的點會是同一點。

本文由‘‘分數7/17是否能表示成兩個相異的埃及分數之和’’這個問題出發，藉由簡單數論的性質以及反證法，得到一個真分數可表示成兩個相異埃及分數之和的定理檢驗法(定理1)。有了這個基礎，我們進ㄧ步推廣定理1 的結果，做出了嶄新的結果(定理2、定理3) 。此定理分別可以用來檢驗真分數表示成三個、四個相異埃及分數之和的存在性; 至於將真分數表示為5 項、6 項….k 項相異埃及分數之和的部分尚在嘗試。利用定理1、2，我們寫了兩個Matlab 軟體工具的電腦程式，使得我們可以檢驗任意真分數是否可以表示成兩項及三項的和，並可把所有的解列出來; 最後我們研究的是一個有關埃及分數的猜想(Erdos-Strauss Conjecture)問題，當分子為4，且分母為4k、4k+2、4k+3 時，猜想皆成立。對於分母為4k+1 而言，當k 為3r+1、3r+2 猜想亦成立，k=3r 且r 為奇數時也是成立的，因此目前需解決的問題只剩分母為24t+1 的情況了。值得一提的是，我們用Matlab 的程式檢驗出當分母為1014 至1014 +240000 之內的正整數時，猜想都是成立的，這已經超越了已知文獻的結果。This paper begins with the question: ‘‘Is 7/17 able to be the sum of two different Egyptian fractions?’’ to discuss the problem of Egyptian fractions. According to the complete division properties and the counter-evidence method, we get a back-check theorem which is about a true fraction can be the sum of two different Egyptian fractions (see theorem 1). Using the same method we obtain a new back-check theorem that is a fraction can be the sum of three or four different Egyptian fractions (thereom2, thereom3). Similarly, we can follow the same procedure to get the rule that a fraction can be the sum of five or six …or even more different Egyptian fractions. By the theorem1 and 2, we propose two programs written vie the Matlab software to examine that any true fraction can be the sum of two items and three items or not. Finally we focus on the Erdos-Straus Conjecture, which related about true fractions can be divided by three different Egyptian fractions. The conjecture is when the denominator is 4k, 4k+2, or 4k+3, the problem mentioned above can be solved. As for the denominator is 4k+1, then the conjecture also can be solved, as k equals to 3r+1 or 3r+2. Also, k being 3r and r is an odd number, the conjecture is satisfied. As for the case of r equals to even number, the problem has not been solved. But it is worth to mention here that we use Matlab software to examine the conjecture is agreeable as the denominator is between 1014to 1014+ 240000. This is beyond the results from the literatures.

將一個集合{1,2,...,n}分成數個非空的集合（組，區塊），稱為此集合的一個分割。如果可以找到1 ≦ a 已知無孤立點無交錯分割以Riordan 數{rn}n≥0 =1,0,1,1,3,6,15,36,... 來計數。在這篇文章中我們研究無孤立點無交錯分割的一些性質。 首先我們考慮無孤立點的無交錯分割按區塊的細分。我們得出：集合{1,2,...,n}恰含k個區塊的無孤立點的無交錯分割的個數為： 其次，我們證明bn,k和多邊形的剖分有令人訝異的關連。令dn,k是用不相交對角線將凸n 邊形分成k 塊的方法。我們用代數方法證出 bn,k = dn+2−k ,k，也給了一個新的組合證明。 最後，透過對應的方法，我們找出了七個嶄新的組合結構，這些結構都是以Riordan 數來計數。 Partition the set {1,2,...,n} into several nonempty sets (blocks) and call it a partition. If there exists 1 ≦ a It is known that the nonsingleton noncrossing partitions are counted by Riordan numbers {rn}n≥0 =1,0,1,1,3,6,15,36,... In this paper we study the properties of them. First we consider the enumeration of nonsingleton noncrossing partitions in respect to the blocks. We prove that the number of nonsingleton noncrossing partitions of {1,2,...,n} with k blocks is Then we give a connection between nonsingleton noncrossing partitions and polygon dissections. Let dn,k be the ways to dissect an n –gon with noncrossing diagonals. We prove that bn,k = dn+2−k ,k We also give a combinatorial proof. Furthermore, by way of the technic of bijection, we find 7 new combinatorial structures counted by Riordan numbers.