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...to evaluate the infinitely growing superfunction of exponential, \(\mathrm{SuExp}_{\sqrt{2},5}\).

...рис.11.4 || \(y\!=\!\mathrm{tet}_\eta(x)\!=\!F_{3}(x)\) || \(y\!=\!\mathrm{SuExp}_{\eta,3}(x)\!=\!F_{3}(x)\). | стр.138, рис.11.4 || \(y\!=\!\mathrm{SuExp}_{\eta,3}(x)\!=\!F_{1}(x)\) || \(y\!=\!\mathrm{tet}_\eta(x)\!=\!F_{1}(x)\).