User:Sligocki/up-arrow properties


 * This article is now available on my blog: https://www.sligocki.com/2009/10/07/up-arrow-properties.html

Some useful definitions and properties for Knuth's up-arrow notation

Definition
$$a \uparrow^n b$$ is defined for a, b and n are integers $$n \ge 1$$ and $$b \ge 0$$.


 * 1) $$ a \uparrow^1 b = a^b $$
 * 2) $$ a \uparrow^n 0 = 1 $$
 * 3) $$ a \uparrow^n b = a \uparrow^{n-1} (a \uparrow^n (b-1)) $$

Therefore $$a\uparrow^n b = a \uparrow^{n-1} a \uparrow^{n-1} \cdots \uparrow^{n-1} a$$ (with b copies of a, where $$\uparrow$$ is right associative) and so it is seen as an extension of the series of operations $$+, \times, \uparrow, \uparrow^2, \dots$$ where $$\uparrow$$ is basic exponentiation

Basic Properties

 * 1) $$ a \uparrow^n 1 = a $$
 * 2) $$ 2 \uparrow^n 2 = 2^2 = 2 \times 2 = 2 + 2 = 4 $$

Extension
We can extend the uparrows to include multiplication and addition as the hyper operator.

This system may be consistently expanded to include multiplication, addition and incrementing:


 * 1) $$ a \uparrow^{-2} b = b+1 $$
 * 2) $$ a \uparrow^{-1} b = a+b $$
 * 3) $$ a \uparrow^0 b = ab $$
 * 4) $$ a \uparrow^1 b = a^b $$
 * 5) $$ a \uparrow^n 0 = 1 $$ (for $$ n \ge 1 $$)
 * 6) $$ a \uparrow^n b = a \uparrow^{n-1} (a \uparrow^n (b-1)) $$

We will show that Rules 3, 5 and 6 imply rule 4 Assume that $$ a \uparrow^1 \beta = a^\beta $$ for any $$ \beta < b $$, then
 * Proof of consistency by induction.
 * $$ a \uparrow^1 b = a \uparrow^0 (a \uparrow^1 (b-1)) = a \cdot a^{b-1} = a^b $$ by rule 6, rule 3 and assumption

Furthermore, $$ a \uparrow^1 0 = 1 = a^0 $$ by rule 5

Thus the assumption is true for all $$\beta \ge 0 $$

Likewise we can show that Rules 2, 5, 6 imply Rule 3 and that Rules 1, 5, 6 imply Rule 2. Therefore, Rules 1, 5, 6 imply Rules 4, 5, 6 and so consistently extend the system.
 * QED

Clearly some of the properties do not extend.

Changing Bases
Todo: How do you change bases.

Example:
 * $$ 3 \uparrow^k n \approx 10 \uparrow^k n^\prime $$ what is n'?

For k = 1:
 * $$ a \uparrow n = a^n = b^{n\log_b(a)} = b \uparrow (\log_b(a) n) $$

For k = 2. For all $$a < b$$ there is a unique $$\delta$$ such that
 * $$ b \uparrow \uparrow (n - \delta - 1) < a \uparrow \uparrow n < b \uparrow \uparrow (n - \delta) $$ for all sufficiently large n

Examples: Thus the base of a tetration is not very important, they all grow at approximately the same rate eventually.
 * $$ 10 \uparrow \uparrow (n-1) < 3 \uparrow \uparrow n < 10 \uparrow \uparrow n $$ for all $$ n \ge 3 $$
 * $$ 10 \uparrow \uparrow (n-3) < 2 \uparrow \uparrow n < 10 \uparrow \uparrow (n-2) $$ for all $$ n \ge 4 $$

In fact these numbers $$\delta$$ grow very slowly.

Claim:
 * $$ a \uparrow \uparrow (n+k-1) < (a \uparrow \uparrow k) \uparrow \uparrow n < a \uparrow \uparrow (n+k) \,$$

Note, the left inequality is easy to prove:
 * $$ (a \uparrow \uparrow k) \uparrow \uparrow n = ((a \uparrow \uparrow k) \uparrow)^{n-1} (a \uparrow \uparrow k) > (a \uparrow)^{n-1} (a \uparrow \uparrow k) = a \uparrow \uparrow (n+k-1) \,$$

Claim:
 * $$ n \uparrow^n n < (3 \uparrow^n 3) \uparrow^n (3 \uparrow^n 3 - 3) < 3 \uparrow^n (3 \uparrow^n 3) = 3 \uparrow^{n+1} 3 \,$$