Thread Rating:
  • 0 Vote(s) - 0 Average
  • 1
  • 2
  • 3
  • 4
  • 5
hyper 0
#1
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.
Reply
#2
I see another error. I have 10 fingers, not 6.
Reply
#3
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
You have 5 fingers total.That is 1 and 2/3s groups of 3. So the answer is 5/3+3 = 14/3

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.

MathStackExchange account:MphLee

Fundamental Law
Reply


Possibly Related Threads...
Thread Author Replies Views Last Post
  On my old fractional calculus approach to hyper-operations JmsNxn 13 529 05/29/2021, 01:13 AM
Last Post: MphLee
  On to C^\infty--and attempts at C^\infty hyper-operations JmsNxn 11 1,987 03/02/2021, 09:55 PM
Last Post: JmsNxn
  Thoughts on hyper-operations of rational but non-integer orders? VSO 2 3,046 09/09/2019, 10:38 PM
Last Post: tommy1729
  Hyper-volume by integration Xorter 0 2,640 04/08/2017, 01:52 PM
Last Post: Xorter
  Hyper operators in computability theory JmsNxn 5 8,326 02/15/2017, 10:07 PM
Last Post: MphLee
  Recursive formula generating bounded hyper-operators JmsNxn 0 2,865 01/17/2017, 05:10 AM
Last Post: JmsNxn
  holomorphic binary operators over naturals; generalized hyper operators JmsNxn 15 25,621 08/22/2016, 12:19 AM
Last Post: JmsNxn
  on constructing hyper operations for bases > eta JmsNxn 1 4,634 04/08/2015, 09:18 PM
Last Post: marraco
  Bounded Analytic Hyper operators JmsNxn 25 34,994 04/01/2015, 06:09 PM
Last Post: MphLee
  Incredible reduction for Hyper operators JmsNxn 0 3,532 02/13/2014, 06:20 PM
Last Post: JmsNxn



Users browsing this thread: 1 Guest(s)