

Store








CTK Exchange
Ramsey_KJ
Member since Sep2304

Sep2404, 06:37 AM (EST) 

""Weighted" Multiplication"

Has anyone seen this before? Define the operation A @ B as equal to A*B^2 + (B^21)/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 + (31)/8) * 3 + (31)/8 = 10 = 1*9 + (91)/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 + (31)/2 and 25 = 8*3 + (31)/2. 1 @ 3 = 10, thus (1*9)^2 = T(13)10 and (6*9)^2 = T(76)10. note that 13 = 1*9 + (91)/2 and 76 = 8*9 + (91)/2. 1 @ 5 = 28, thus (1*15)^2= T(22)28 and (6*15)^2 = T(127)28 note that 22 = 1*15 +(151)/2 and 127 = 8*15 + (151)/2 Have a Good Day KJ Ramsey 

Alert  IP 
Printerfriendly page 
Reply 
Reply With Quote  Top 


JJ
guest

Sep2504, 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^21)/8 = 2*(3^2)+(3^21)/8 = 2*9+1 = 19 (A @ B) @ C = D @ C = D*C^2 + (C^21)/8 = 19(4^2)+(4^21)/8 = 305 E = B*C = 3*4 = 12 A @ (B*C)= A @ E = A*E^2 + (E^21)/8 = 2(12^2)+(12^21)/8 = 256+143/8 = 273.875 Hence (A @ B) @ C NOT equals A @ (B*C) 

Alert  IP 
Printerfriendly page 
Reply 
Reply With Quote  Top 


You may be curious to have a look at the old CTK Exchange archive. Please do not post there.
Front page
Contents
Copyright © 19962018 Alexander Bogomolny
66596653


