User:Ynought/E function

W.I.P. (last edited 10.3.2019)

E function
\(E\) function takes place on a plane with starting point \((0|0)\) at the bottom left corner and the thick ness of the line is \(n^{-n}\) and the lenght is \(n^{-1}\).And the size of the units doesn't matter

Logic and symbols
\(\alpha_n\) is the angle that line nr.\(n\) deviates from the x plane

\(G_1\) is the starting graph.And \(G_i\) is the graph that appears if you repeat the rules for the transformation of \(G_1\) for the \(i\)-th time

\(\star\) denotes homeomorphical embeding lets say i.e. \(G_i\star G_j\) means that \(G_i\) is homeorphically embedable into \(G_j\)

\(\omega=\text{sup}(0,1,2,3...)\)

The definition of line number k:
You start looking for the line with the bottom most point of that line has the least \(y\) value.If that is a tie then you choose the line with the least \(x\) of those lines.call that one line 1.Repeat the rule above.Call that one line 2.Keep repeating that rule until you have every line numbered.And \(L_n\) is the n-th line,and this has to hold \(n\in\mathbb{N}_{>0}\).And \(L_n^{G_i}\) is the \(n\)-th line in \(G_i\)

Termination and the value of E(n)
\(E(n)=\text{max}(k(\nexists i(\omega>k\leqslant i\geqslant 1 \land G_i\star G_k))\) where \(G_1\) has at most \(n\) lines.And in case that case never apears then use this \(E(n)=\text{max}(k(\nexists i(\omega>k\leqslant i\geqslant 1 \land G_i\star G_k\land \nexists a(\nexists b(L_a^{G_n}\| L_b^{G_n}|n\in\mathbb{N}_{\geqslant 1}))))\) \(G_1\) has at most \(n\) lines.if even that doesn't terminate then use this \(E(n)=\text{max}(k(\nexists i(\omega>k\leqslant i\geqslant 1 \land G_i\star G_k\land \nexists a(\nexists b(L_a^{G_n}\nparallel L_b^{G_n}|n\in\mathbb{N}_{\geqslant 1}))))\) \(G_1\) has at most \(n\) lines.If even that doesn't then i give up.