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inspired by an equation - tommy1729 - 07/27/2010
i was inspired by f(z) = A^(Lz) = L^z. which looks neat to me. i havent really thought about it deep but here is an attempt : A^(Lz) = L^z => A^L = L ; L^p = 1 p = 2pi i / (ln(A) L) ( the period ) => A^L = L ; L^p = L^ 2pi i / (ln(A) L) = 1 => L ln(A) = ln(L) => ln(A) = ln(L)/L => L^ 2pi i / ( ln(L) * L/L ) = 1 => L^ 2pi i / ln(L) = 1 and now ? ln on both sides ? => 2 pi i / ln(L) * ln(L) = 0 => 2pi i = 0 ?? certainly not all A and L satisfy A^(Lz) = L^z. so where did the variables go to ? is there no solution ? i assume taking ln on both sides is the error => solve for L : L^ 2pi i / ln(L) = 1 and now ... Lambert W function ? or is that equation already wrong because of branches and we need to return to => A^L = L ; L^p = L^ 2pi i / (ln(A) L) = 1 but thats again an equation in 2 complex variables :s of course f(z) = A^(Lz) = L^z has the trivial solution f(z) = 1 or f(z) = 0. but im looking for others. maybe => 2 pi i / ln(L) * ln(L) = 0 means that all solutions for A must be integer and hence only the trivial solutions f(z) = 1 or f(z) = 0 exist. i cant find or imagine any other. also the generalizations of this equation inspire me. analogues with double periodic functions , finding solutions in terms of other numbers ( 3d complex or other ) , replacing exp with another function and multiplication with its inv superfunction of that other function etc etc this may be trivial sorry ... ( should have paid attention in class as a teenager :p ... if that was in class ... ) tommy1729 note to myself : post this kind of stuff in the general section tommy ! bo always puts it there so thats where it belongs ------------- edit : Ok sorry this is nonsense. A^(Lz) = L^z => take ln on both sides : ln(A) * L * z = ln(L) * z. take z = 1 : ln(A) = ln(L)/L This is a relationship between A and L , not an equation to be solved for. ------------- |