inspired by an equation
#1
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 Smile

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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.

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Messages In This Thread
inspired by an equation - by tommy1729 - 07/27/2010, 11:18 PM

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