數學

We investigate the machinery producing successive Simpson’s paradoxical reverse. Taking advantage of algebraic and geometric techniques, we obtain the following results. Take playing baseball for example. In our study, we find that Simpson’s paradox only occurs when the hitter’s hits over 3 times in one game. Set n equal to the times I will hit in one game. If my batting average in each game is at least(n ?1)/2 times higher than the others’; then I am sure that my total batting average would not be invert by the others. In order to find how many the lattice points in the triangle, we use Pick’s formula. But sometimes, the Pick’s formula is not appropriate to triangles whose vertex are not all lattice points. So we develop New Pick’s formula to estimate the number of lattice points in such kind of triangles. Besides, we also find an iterative algorithm to produce successive “Simpson reverse” phenomenon by using C++ language, and we can therefore produce as many “Simpson’s set of four sequences” terms as we like(not beyond the computers’ upper limit).Moreover, if both sequences of ratios converge, then they must have the same limit.我們探討了一般人乍看之下顯得頗弔詭的辛普森詭論。我們配合GSP 作圖，用解析幾何、設立直角座標系和C++ 程式的運算，找出在特殊情況下或一般情況下所產生的辛普森數列組和特殊的性質，並且以棒球場上的打擊率為例子來做印證。通常一場棒球賽中，每個人平均上場3 次～4 次，經過我們的討論，發現要發生逆轉的機會只有在上場達到4 次或以上時才會發生。?了求出在直角座標系中可以滿足的格子點個數，我們用了Pick公式，但?了更準確的估計，我們引進了虛點的概念，重新推導出了新Pick 公式。另外，我們還發現，假設兩個人上場比賽，若打了2 場，且每場最多上場打擊K 次，其中的一個人的打擊率只要是另一個人的(k-1)/2倍以上就保證不會被逆轉。我們又找到了連續產生辛普森逆轉的演算法，利用C++ 寫出程式，經由演算法和遞迴式，製造出項數可任意多(只要電腦能夠承受)的辛普森數列組，且我們發現若兩個比值數列接收斂，則極限趨近於同一個數值。

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，

Sicherman 已經找出與兩顆六面的正常骰子有相同機率分布的Sicherman 骰子，並進一步獲得與三顆六面的正常骰子有相同機率分布的骰子必為一對Sicherman 骰子與一顆六面的正常骰子之結果，我們試圖由已知的Sicherman 六面骰子的處理方法出發，透過對割圓多項式的分析來累積足夠的相關資料，以處理由兩顆四面骰子至兩顆三十面骰子，處理由三顆四面骰子至三顆三十面骰子的各種Sicherman 骰子的答案，來探索兩顆與三顆的n 面Sicherman骰子存在的充要條件與求法，並進一步將所得之結果分類，得到 ”有相同標準分解式的類型的數n，會具有相同組數的Sicherman 骰子”之猜測結果與特殊情形下的證明。 Sicherman has found out the Sicherman dice which have the same probability distribution as the normal two six-sides dice. Furthermore , he also found out a pair of Sicherman dice and a normal six-sides dice has the same result as 3 normal six-sides dice . We try to begin with the given algorithm of six-sides Sicherman dice , through the analysis of Cyclotomic Polynomials to accumulate sufficient related information then to come up with the solution from discussion of 2 four-sides dice to 2 thirty-sides dice , from 3 four-sides dice to 3 thirty-sides dice to explore the existence of necessary and sufficient condition and solution of 2 n-sides Sicherman dice and 3- sides Sicherman dice , and even to classify the results to come to a conclusion of the guessing results and proofs under special cases about “the numbers n which have the same Canonical Prime Factorization will have the same numbers of n-sides Sicherman dice.”

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.

本研究中，我們將提出一些新穎結果，著重討論其在三角中的應用；同時，找出其遞迴關係式，得出三角展開式與其所對應之多項式分解式，進而討論出多種的規律性及所涵蓋的內容及推廣性質，我得到很多高中數學公式無法推導出在【4】和【8】中的漂亮公式及創新的結果，且這些等式都是由我們不太瞭解的無理數所構成的。 主要是討論我們在【7】中所得到的收穫與經驗；複數是三角、幾何、代數互動的橋樑，我是以不同的角度及嶄新的方法來綜合探討在【6】中相關的應用。提出關於正整數平方的倒數和公式更為精簡且基本的證明，將 sin−2 x 表示成級數形式的部分分式，進而應用在(a,b) = 1的機率問題上；並研究相關的等式，直接透過三角與代數來研究關於 2p 次方的倒數之求和問題，得出級數 之和的有用遞迴公式，並與最重要的常數扯上關係。 For one thing, we present diverse methods to evaluate finite trigonometric summation and related sums. Trigonometric summations over the angles equally divided on the upper half plane are investigated systematically. Several related trigonometric identities are also exhibited. What is more, we use methods of calculus, and make several surprising and unexpected transformations. A useful recursive formula for obtaining the infinite sums of even order harmonic series, infinite sums of a few even order harmonic series, which are calculated using the recursive formulas, are tabulated for easy references. Furthermore, is there any interesting results and applications？ Finally, the purpose of this paper is to develop a new proof of and related identities, but their derivations are more complicated. The following studies are completed under the instruction of the professor.

Hamiltonian和Hiroimono路徑問題都有百年以上的歷史，且沒有直接解決方法，可能因為是題目本身的條件在討論時尚不足解決問題，現階段只能用電腦去跑。

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.

1. 遊戲規則：將1～ 2m × 2n的連續正整數，由上而下、由左而右依序填入 2m × 2n的方格內。操作規則允許將2m × 2n做往右或往左或往上或往下的完全對摺，直到操作至所有單位方格均疊成一行，此同時有數字也由上而下形成一數列。2. 本研究即是探討操作完成的數列之數量與數字間的關連性。3. 我們發現：(1) 數列之數量與巴斯卡三角形有關。(2) 形成的數列必符合內文的 [ R(L) 性質]、 [ D(U) 性質]、[ R&D 性質]、[D&R 性質]。 1. Rules of thegame: Fill in order the continuous positive integers 1~ 2m × 2n, from top to bottom and from left to right in the 2m × 2n check. The operational rule allows a complete fold of 2m × 2n either rightward or leftward, or upward or downward, until all the check units pile up in a line. At the same time, all the integers form a series from top to bottom. 2. This study explores the relationship between the number of the series and the integers after the operation. 3. Our findings are: (1) The number of the series is related to Pascal triangles. (2) The series formed meet the properties mentioned in the study: [the property of R(L)], [the property of D(U)], [the property of R & D], and [the property of D & R].

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。

本研究從理想城鎮(Ideal City)街道開始，討論平面上相異n 點到某一點的最短距離和。經研究後發現：當n 為偶數時，則到相異n 點的最短距離和所形成的區域可能是一個點、一個線段或是一個矩形；當n 為奇數時，則相異n 點的最短距離和所形成的區域將會退化成一個點。此外，本研究將理想城鎮的街道換成正三角形的街道幾何平面，同樣是討論平面上相異n 點到某一點的最短距離和。經研究後發現：當n 為偶數時，則相異n 點的最短距離和所形成的區域可能為一個點、一個線段、一個四邊形、一個五邊形及一個六邊形；當n 為奇數時，相異n 點的最短距離和所形成的區域則可能為點、三角形的情況。假使考量各點重要性的比重，分別加權後再求最小點。研究發現無論在理想城鎮或正三角形幾何平面上，皆可將各點視為多個權數相同之點重疊於此點上，便可利用先前的方式求得最小點區域。透過這次的研究，可以利用n 個相異點到某一點的最短距離和實際應用在貨物運送的問題或是消防設施配置等問題。The present study was intended to start with the Ideal City and proceed to discuss the sum of the shortest distance between a point and n different points on a plane. After the discussion, it was found that if n is even, the formed region could be a point, a line segment, or a rectangle. If n is odd, then the formed region must be a mere point. Further, the current study transformed the Ideal City into the geometric plane of an equilateral triangle. Similar to the previous discussion, if n is even, the formed region could be a point, a line segment, a quadrangle, a pentagon, or a hexagon. On the other hand, if n is odd, then the formed region could be a point, or a triangle. The result of this study, which investigated the sum of the shortest distance of a certain point to n different points can be applied to the real life situation, such as transporting goods or distributing fire control facilities.