User talk:Ikosarakt1/Weak hyper-operators

Wrong beyond \(n\downarrow\downarrow\downarrow\downarrow n\).

Actually, if we make analysis of \(a\downarrow^mb\) instead of \(n\downarrow^mn\), things will be more clear. Now begin with \(a\downarrow\downarrow b\), which is exactly equal to \(a^{a^{b-1}}\).

\begin{eqnarray*} \text{Lower bound} & Notation & \text{Upper bound} \\ a\uparrow\uparrow2 & a\downarrow\downarrow a & a\uparrow\uparrow3 \\ (a\uparrow\uparrow2)\uparrow\uparrow2 & (a\downarrow\downarrow a)\downarrow\downarrow a & (a\uparrow\uparrow3)\uparrow\uparrow3 \\ a\uparrow\uparrow3 & a\downarrow\downarrow\downarrow3 & a\uparrow\uparrow5 \\ a\uparrow\uparrow4 & a\downarrow\downarrow\downarrow4 & a\uparrow\uparrow7 \\ a\uparrow\uparrow b & a\downarrow\downarrow\downarrow b & a\uparrow\uparrow(2b-1) \\ (a\uparrow\uparrow a)^{(a\uparrow\uparrow a)^{a\uparrow\uparrow a-1}} & (a\downarrow\downarrow\downarrow a)\downarrow\downarrow(a\downarrow\downarrow\downarrow a) & (a\uparrow\uparrow(2a-1))^{(a\uparrow\uparrow(2a-1))^{a\uparrow\uparrow(2a-1)-1}} \\ a\uparrow\uparrow(a+1) & (a\downarrow\downarrow\downarrow a)\downarrow\downarrow\downarrow2 & a\uparrow\uparrow(2a+1) \\ a\uparrow\uparrow(a+2) & (a\downarrow\downarrow\downarrow a)\downarrow\downarrow\downarrow3 & a\uparrow\uparrow(2a+3) \\ a\uparrow\uparrow(2a-1) & (a\downarrow\downarrow\downarrow a)\downarrow\downarrow\downarrow a & a\uparrow\uparrow(4a-3) \\ a\uparrow\uparrow(3a-2) & a\downarrow\downarrow\downarrow\downarrow4 & a\uparrow\uparrow(6a-5) \\ a\uparrow\uparrow((a-1)\times(b-1)) & a\downarrow\downarrow\downarrow\downarrow b & a\uparrow\uparrow(2ab) \end{eqnarray*}

Then, something like \((a\downarrow\downarrow\downarrow\downarrow a)\downarrow\downarrow\downarrow\downarrow2=(a\downarrow\downarrow\downarrow\downarrow a)\downarrow\downarrow\downarrow(a\downarrow\downarrow\downarrow\downarrow a)>a\downarrow\downarrow\downarrow(a\downarrow\downarrow\downarrow\downarrow a)>a\uparrow\uparrow(a\uparrow\uparrow((a-1)^2))\). And further, \((a\downarrow\downarrow\downarrow\downarrow a)\downarrow\downarrow\downarrow\downarrow3>a\uparrow\uparrow(a\uparrow\uparrow(a\uparrow\uparrow((a-1)^2)))\), and \(a\downarrow\downarrow\downarrow\downarrow\downarrow3>a\uparrow\uparrow\uparrow(a+1)\), so \(a\downarrow\downarrow\downarrow\downarrow\downarrow b>a\uparrow\uparrow\uparrow(ab-b+2)\). The down-arrow notation is not so weak as you think. hyp$hyp?cos&#38;cos (talk) 09:44, May 19, 2014 (UTC)