Locus of the Points on Circumference of the n-th Circle that Formed by Moving the Center of any Radius Circles on the Outermost Circumference of Preceding set of Circles
This project aimed to study the motion which occurred from the end point on the circumference of the outermost circle by moving the center on the circumference of a preceding circle and the center of an innermost circle at origin. According to the study, when angular velocity was changed, it caused the different of loci. Based on the above information, finding the locus of the point on circumference of n-th circle that formed by moving the center of any radius circles on circumference of preceding set of circles was studied to get general equation. A set of circle and locus were created with GSP program. First, set the same radius circles on the X-axis with the first circle at origin, then found the relationship that occurred from the characteristics of locus. The result showed that if the ratios of angular velocity are 1:1:1, 2:2:2, 3:3:3, ..., …, n:n:n or 1:2:3, 2:4:6, 3:6:9, …,nw1:nw2:nw3, the characteristics of locus will be the same, while the others will be different. Finally, the equation of locus was found as follow: (x,y) = { ..........see in abstract...........} when .........see in abstract........... Where ri is the radius of i-th circle, zeta i is an angle between the radius of i-th circle and X-axis, wi is the angular velocity, t is elapsed time and alpha i is a starting angle between the radius of i-th circle and X-axis.
多邊形的剖分圖形數量之探討
從參考資料[1]可知,將凸n+2邊形利用n-1條不相交的對角線剖分成n個三角形的圖形數量即為卡特蘭數Cn。而我利用不相交的對角線把n+2邊形剖分成數個多邊形和三角形的組合,並從此類的剖分圖形與三角剖分圖形之關聯,進而由卡特蘭數的一般式推導出此類剖分圖形數量的一般式。在本研究中可得,若到把n+2邊形剖分成一個k+2邊形和多個三角形的圖形數量是(2n-k+1 n+1) ;把n+2邊形剖分成一個k+2邊形、一個m+2邊形和多個三角形的圖形數量,當m≠k,數量為n+2/2(2n-k-m+2 n+2) ,當m=k時,數量為n+2/2(2n-2k+2 n+2) ;把n+2邊形剖分成一個k1+2邊形、一個k2+2邊形、一個k3+2邊形、和n-k1-k2-k3 個三角形的剖分圖形,當k1,k2,k3兩兩相異時,數量為(n+2)(n+3)(2n-k1-k2-k3+3 n+3) ;把n+2邊形剖分成一個K1+2邊形、一個K2+2邊形、一個K3+2邊形、一個K4+2邊形和n-K1-k2-k3-k4個三角形的剖分圖形當k1,k2,k3,k4兩兩相異,數量為(n+2)(n+3)(n+4)(2n-k1-k2-k3-k4+4 n=4)。並猜測若k1,k2,...,ki兩兩相異時,把n+2邊形剖分成一個k1+2邊形、一個k2+2邊形、…、一個ki+2邊形、和n-Σkj 個三角形的剖分圖形數量為(n+i)!/(n+1)!(2n-Σkj+i n+i) 。
3進位Kaprekar變換之結構
b進位的n位數字x,數字x各位數字由大到小排列為p,由小到大排列為q,定義Kaprekar變換T(b,n)(x)=p-q,例如T(10,3)(x)= 954-459。當T(b,n)(x)= x,稱x為Kaprekar常數。 Tk(b,n)( x)=T(b,n)( Tk-1(b,n)( x))= x, k >1時,稱x為k階Kaprekar循環數。本文解答了以下問題: 1. b進位的數字不包含數字b-1的Kaprekar常數的形式。 2. 3,4,5,6進位的Kaprekar常數的一般形式。 3. 對於2,3進位的情形,我們引入三元非負整數的形式來討論Kaprekar變換,轉換成Kaprekar數對(p,q),再進一步,由來探討比值p/q,將Kaprekar變換轉成Kaprekar函數g(x),解決 Kaprekar 循環數的所有形式及解。 最後我們得到Tl(b,n)( x)必是Kaprekar循環數。