0
$\begingroup$

Consider the set $i_0,i_1,...$ defined as

$i_0$ is the smallest element and $i_0 = 0$.

If $ (i_a - i_b)^2 < 1$ then $i_c$ is Also in the set and given by

$$ i_c = (i_a + i_b + 1)/2$$ [*]

Let $T(x,y)$ be the cardinality of $i_j$ Numbers in the interval $[x,y]$. Notice $1/2,3/4,7/8,...$ are all in the set hence $T(x,y)$ can be infinite and therefore I can not " simply say " $T(x,y) = T(0,y) - T(0,x) $ Because of the case $ \infty - \infty$.

However the set is not dense like the rationals and we can Find a nonzero gap between any pair of elements in the set.

Apparantly the gap between the real $x$ and the smallest next element $i_k$ is given by

for $ x < 0 : $ $$ f(x) = -x$$

for $x > 0 :$ $$ f(x) = f(x - f(x-1))/2 $$ [ ** ]

It turns out

$f(0) = 2^{-1} $

$f(1) = 2^{-3} $

$f(2) = 2^{-10}$

$f(3) = 2^{- 1541023937}$

And this sequence must Go to zero very fast !

Tommy1729 claimed for large $H$ , $ - ln(f(K)) > exp^{[K]}(1)$ where ^[ .. ] means iteration. Hence going to 0 faster Then tetration ( base e).

( I looked at the tetration forum and Some ackermann type papers but could not Find this )

Now I assume the function $f$ is hard to compute and can not be computed differently from Its definition above.

I assume the function $T(x,y) $ is Also hard to compute. I assume $T(x,y)$ does not help much for computing $f $, Unless perhaps $T(x,y)$ is small.

In fact I think computing $T(x,y)$ can only be done by multiple computations of $f$ 's !!?

And I think we need a way to test for $T(x,y) =$ oo.

Many questions.

But the main 2 questions for here are the following.

[ I have to restrict to 2 , since the answer would be too long ; better asking others in a seperate post ]

How does one show [ * ] implies [ ** ] ? How does one show [ ** ] implies [ * ] ?

Btw What field of math is this ? Does not feel like combinatorics to me ...

References are very Welcome too !!

Thanks in advance.

$\endgroup$
  • $\begingroup$ Is this order theory ? $\endgroup$ – mick Nov 6 '17 at 7:29
  • $\begingroup$ I was thinking about using this set for a 2 player game ... $\endgroup$ – mick Nov 6 '17 at 7:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.