12/04/2010, 12:16 AM

another brainstorm ...

exp^[1/2](x) probably isnt an entire function.

right ?

i mean exp^[1/2](x) = sexp(slog(x)+1/2) and since both sexp and slog are not entire ...

furthermore an entire function is of the form exp(f(x))*(x-a_0)...(x-a_n) where f(x) is entire too.

but such a product identity is unlikely for exp^[1/2](x) , especially the exp(f(x)) part since an entire function grows faster than polynomial but f(x) grows slower hence a contradiction.

but what if f(x) = constant ??

we still have the argument that sexp and slog are not entire.

maybe there is a simple proof that exp^[1/2](x) is not entire.

if it miraculously turns out to be entire it has to be of the form

e^C *(x-a_0)...(x-a_n).

eitherhow , already the zero's of exp^[1/2](x) interest me.

in fact toying around a bit we find :

exp^[1/2](x) = sexp(slog(x)+1/2) = 0

hence bye reversing : x = sexp(slog(0)-1/2) = exp^[-1/2](0)

i know slog(x) has period 2pi i.

hence the zero's of exp^[1/2](x) appear to be "exp^[-1/2](n*2pi i)" what reduces to exp^[-1/2](0).

but things dont make sense , i must have made a mistake.

first , there must be other zeros or the zeros are wrong :

if exp^[-1/2] is entire then having only zero's exp^[-1/2](n*2pi i) or even just exp^[-1/2](0) cannot hold.

now assume exp^[-1/2] is not entire.

well log isnt , sexp isnt and slog isnt , so that makes more sense.

i dont believe it only has one zero ?

exp^[-1/2](0) = exp[1/2](log(0) + 2 n pi i) = sexp(slog(log(0))+1/2)

hmm

so ok , it seems exp^[-1/2](0) is indeed the only solution ?

and it also seems 2 pi i is a period of exp^[1/2](x) , which we ofcourse actually already knew earlier.

but wait a sec , if exp^[1/2] is periodic then we have as solutions :

exp^[1/2](x) = 0

x = exp^[-1/2](0) + 2pi i.

is that correct or are there others as well ?

does this constitute a proof that exp^[1/2] is not entire ?

exp^[1/2](x) probably isnt an entire function.

right ?

i mean exp^[1/2](x) = sexp(slog(x)+1/2) and since both sexp and slog are not entire ...

furthermore an entire function is of the form exp(f(x))*(x-a_0)...(x-a_n) where f(x) is entire too.

but such a product identity is unlikely for exp^[1/2](x) , especially the exp(f(x)) part since an entire function grows faster than polynomial but f(x) grows slower hence a contradiction.

but what if f(x) = constant ??

we still have the argument that sexp and slog are not entire.

maybe there is a simple proof that exp^[1/2](x) is not entire.

if it miraculously turns out to be entire it has to be of the form

e^C *(x-a_0)...(x-a_n).

eitherhow , already the zero's of exp^[1/2](x) interest me.

in fact toying around a bit we find :

exp^[1/2](x) = sexp(slog(x)+1/2) = 0

hence bye reversing : x = sexp(slog(0)-1/2) = exp^[-1/2](0)

i know slog(x) has period 2pi i.

hence the zero's of exp^[1/2](x) appear to be "exp^[-1/2](n*2pi i)" what reduces to exp^[-1/2](0).

but things dont make sense , i must have made a mistake.

first , there must be other zeros or the zeros are wrong :

if exp^[-1/2] is entire then having only zero's exp^[-1/2](n*2pi i) or even just exp^[-1/2](0) cannot hold.

now assume exp^[-1/2] is not entire.

well log isnt , sexp isnt and slog isnt , so that makes more sense.

i dont believe it only has one zero ?

exp^[-1/2](0) = exp[1/2](log(0) + 2 n pi i) = sexp(slog(log(0))+1/2)

hmm

so ok , it seems exp^[-1/2](0) is indeed the only solution ?

and it also seems 2 pi i is a period of exp^[1/2](x) , which we ofcourse actually already knew earlier.

but wait a sec , if exp^[1/2] is periodic then we have as solutions :

exp^[1/2](x) = 0

x = exp^[-1/2](0) + 2pi i.

is that correct or are there others as well ?

does this constitute a proof that exp^[1/2] is not entire ?