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hyper 0 - dantheman163 - 10/25/2009
I have a couple of ideas about the hyper 0 operator. I have see a couple of different definitions of it that don't agree with each other and nobody seems to have a definite solution. Some people say that a[0]b= b+1. Others have sort of a piece wise approach that is discontinuous and frankly doesn't make much sense. I have found a different approach (that may be completely wrong). The solution that i have found is that a[0]b= (a+b)/a + a The initial conditions that i worked with were as follows: 1. a[0]a=a+2 2. 2[0]4=5 (2[x]4 = 2[x+1]3) 3. This operation should be something that is fundamental (this is not really an initial condition but rather something i figured to be true) Now what is operation really means is something that may not be inherently obvious from the the definition a[0]b= (a+b)/a + a. I am going to explain using fingers. To start you would put a fingers in one hand and b fingers in the other. Next you would figure out how many groups of a fingers you had total. Then you would perform the sum (# of groups) + (# in each group ) which is the same as (# of groups) + a. For example 3[0]3 You have 6 fingers total. So you have 2 groups of 3 fingers. So the answer is 2+3 = 5 3[0]2 You have 5 fingers total.That is 1 and 2/3s groups of 3. So the answer is 5/3+3 = 14/3 An error that some people may see is that a[0](a[0]a) DOES NOT= a+3 however, I feel that a[0]a[0]a = a+3. a[0]b[0]c using "the finger method" would equal (a+b+c)/a + a. I hope that you guys will not have to struggle too hard to understand what I am saying and I also hope that all of this is not completely wrong Thanks. RE: hyper 0 - andydude - 10/26/2009
I see another error. I have 10 fingers, not 6. RE: hyper 0 - MphLee - 03/09/2021
Ok, at the beginning I was convinced that the grouping operation on fingers was just the arithmetic mean. Quote:To start you would put a fingers in one hand and b fingers in the other. Next you would figure out how many groups of a fingers you had total. Then you would perform the sum (# of groups) + (# in each group ) which is the same as (# of groups) + a. In that interpretation we have that given a sequence for , we can see this sequence as group of fingers where every group has fingers in it. So we define an "hyper 0" operator if then
for (the 2-ary version) we get and
But then an example of computation proposed is Quote:3[0]2 So the groups are meant to be weighted and the operation is clearly not commutative anymore. In fact the operation proposed is the following. Let for . Define if then
for , and and
It is clear that the solutions work in some way for preaddition. It is not clear to me how these two solutions can meet the requirment of fundamentality Quote:3. This operation should be something that is fundamental since they require summation and ratios to be defined. |