# Tetration Forum

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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.
I see another error. I have 10 fingers, not 6.
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 $a_i\in\mathbb C$ for $i\in I$, we can see this sequence as $|I|$ group of fingers where every group $i\in I$ has $a_i$ fingers in it. So we define an "hyper 0" operator $\underset{i\in I}{\rm O}$

$\underset{i\in I}{\rm O}a_i=\frac{\sum_{i\in I}a_i}{|I|}+|I|$

if $\forall i,j\in I: a_i=a_j$ then  $\underset{i=1}{\overset{n}{\rm O}}a=a+n$

for $|I|=2$ (the 2-ary version) we get

$a_1{\rm O}a_2=\frac{a_1+a_2}{2}+2$   and   $a{\rm O}a=a+2$

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 $a_i\in\mathbb C$ for $I=\{1,2,3,..., n\}$. Define $\lambda_i:=a_i/a_1$

$\underset{i=1}{\overset{n}{\rm O}}a_i:=a_1+\sum_{i=1}^n\lambda_i$

if $1\leq\forall i,j\leq n: a_i=a_j$ then  $\underset{i=1}{\overset{n}{\rm O}}a=a+\sum_{i=1}^n1=a+n$

for $n=2$,  $a_1=a\neq 0$ and $a_2=b$

$a{\rm O}b=a+(1+\frac{b}{a})$   and   $a{\rm O}a=a+2$

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.