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Subject: ""Weighted" Multiplication"     Previous Topic | Next Topic
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Ramsey_KJ
Member since Sep-23-04
Sep-24-04, 06:37 AM (EST)
Click to EMail Ramsey_KJ Click to view user profileClick to add this user to your buddy list  
""Weighted" Multiplication"
 
   Has anyone seen this before?
Define the operation A @ B as equal to A*B^2 + (B^2-1)/8 and note that (A @ B) @ C equals A @ (B*C).
For instance if A= 1 and B=C = sqrt(3) then we have
(1 @ sqrt(3)) @ sqrt(3) = (3 + (3-1)/8) * 3 + (3-1)/8
= 10
= 1*9 + (9-1)/8
= 1 @ 3
Also (N @ B) @ 1/B = N @ 1 = N.

This came to me while considering triangular numbers that are squared. For instance 1 = T(1) - 0 and 6^2 = T(8) - 0.
0 @ 3 = 1, thus (1*3)^2 = T(4) - 1 and (6*3)^2 = T(25) - 1
note that 4 = 1*3 + (3-1)/2 and 25 = 8*3 + (3-1)/2.
1 @ 3 = 10, thus (1*9)^2 = T(13)-10 and (6*9)^2 = T(76)-10.
note that 13 = 1*9 + (9-1)/2 and 76 = 8*9 + (9-1)/2.
1 @ 5 = 28, thus (1*15)^2= T(22)-28 and (6*15)^2 = T(127)-28
note that 22 = 1*15 +(15-1)/2 and 127 = 8*15 + (15-1)/2

Have a Good Day
KJ Ramsey


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JJ
guest
Sep-25-04, 00:23 AM (EST)
 
1. "RE: "Weighted" Multiplication"
In response to message #0
 
   < note that (A @ B) @ C equals A @ (B*C) >.
is FALSE.
Example :
A=2 ; B=3 ; C=4
D = A*B^2 + (B^2-1)/8 = 2*(3^2)+(3^2-1)/8 = 2*9+1 = 19
(A @ B) @ C = D @ C = D*C^2 + (C^2-1)/8 = 19(4^2)+(4^2-1)/8 = 305
E = B*C = 3*4 = 12
A @ (B*C)= A @ E = A*E^2 + (E^2-1)/8 = 2(12^2)+(12^2-1)/8 = 256+143/8 = 273.875
Hence (A @ B) @ C NOT equals A @ (B*C)


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Ramsey_KJ
guest
Sep-26-04, 11:13 AM (EST)
 
2. "RE: "Weighted" Multiplication"
In response to message #1
 
   Check your math. In each case the answer should be 305.875


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