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Mathematics Dictionary
Dr. K. G. Shih
P/q rule of floral functions
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Q01. Floral functions

Floral functions 
   1. R = a + b*sin(p*A/q)^m\M
   2. R = a + b*cos(p*A/q)^M
   3. Where a, b, p, q are constant parameters
   4. A is cycle domain from 0 to 2*q*pi

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Q02. P/q rule of floral function : R = sin(p*A/q)^M

Graphic Theory 
   1. Case M = 1
      a. It is p petals if p = odd integer and q is odd integer
         and A = (0, pi)
      b. It is 2*p petals if p = odd integer and q is even integer
         and A = (0, 2*q*pi)
      c. It is 2*p petals if p = even integer and q is odd integer
         and A = (0, 2*q*pi)
   2. Case M = 2
      See Pattern Mathematics published by Dr. Shih
Example 
   1. R = sin(0.6*A). How many petals
      Since 0.6 = 3/5. Hence p = 3 and q = 5
      Hence it has 3 petals with A = (0, 5*pi)
   2. How to prove ?
      Grphical and numerical proof are given in Q03 and Q04
   3. How to find cycle domain A ?
      Use pattern mathematics program
      a. Click command Domain
      b. Give data : a,b,p,q,M
      c. Continue to click Domain 
   4. How to find number petals ?
      Same question 3 to click command Petal continuously to see demo
Pattern Math programs
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Q03 Floral function : R = sin(p*A) 

   1. This is the case a = 0, b = 1, q = 1, M = 1
   2. Find number of petals
   3. R = sin(3*A) has 3 petals with A = (0, pi)
   4. R = sin(4*A) has 4 petals with A = (0, 2*pi)
   5. 
      
      R = sin(p*A) Graphic and numerical proof

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Q04 Floral function : R = sin(p*A/2) 

   1. This is the case a = 0, b = 1, q = 2, M = 1
   2. Find number of petals
   3. R = sin(3*A/2) has 6 petals with A = (0, 2*2*pi)
   4. R = sin(4*A/3) has 8 petals with A = (0, 2*3*pi)
   5. 
      
      R = sin(p*A/2) Graphic and numerical proof

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