For those who are confused , let me explain the next big step about this.

If we want a superfunction sexp(z) that is analytic for Re(z) > C >> 0 for some C then we need tommysexp(z) or sheldontommysexp(z) to have the same property. Lets call this C property.

tommysexp(z) and sheldontommysexp(z) agree on the reals (converge speed may differ though ).

Now tommysexp(z) seems to be only defined for the real z so it requires analytic continuation ( assuming it is possible ).

However sheldontommysexp(z) IS periodic and might be analytic in that period. However to go beyond that period we will need analytic continuation because the C property forbids nonreal periodicity. ( again assuming it is possible ).

The analytic continuation of tommysexp(z) and sheldontommysexp(z) are identical function.

Hence either/both tommysexp and sheldontommysexp have property C.

To prove property C of tommysexp(z) = sheldontommysexp(z) one can choose one of those 2 and investigate its analytic continuation/singularities/discontinu on the complex plane within the periodic strip for Re(z) > C.

If it turns out there are no singularities in the periodic semistrip near the reals , then both functions have property C and are equal.

If there are singularities , property C is broken however it might still have regions of analyticity. A natural boundary proves nowhere analytic ( because exp has fixpoints ).

The reason why tommysexp and sheldontommysexp are equivalent is because of a base change like trick explained by sheldon in this thread. The periodicity is also explained here by him.

Notice that a log(f(z)) with f(z) periodic remains periodic.

Also notice that the C property is a uniqueness criterion !

To see this think of (my proof of) TPID 4.

Or the 1 period function wobble of sexp(z).

regards

tommy1729

If we want a superfunction sexp(z) that is analytic for Re(z) > C >> 0 for some C then we need tommysexp(z) or sheldontommysexp(z) to have the same property. Lets call this C property.

tommysexp(z) and sheldontommysexp(z) agree on the reals (converge speed may differ though ).

Now tommysexp(z) seems to be only defined for the real z so it requires analytic continuation ( assuming it is possible ).

However sheldontommysexp(z) IS periodic and might be analytic in that period. However to go beyond that period we will need analytic continuation because the C property forbids nonreal periodicity. ( again assuming it is possible ).

The analytic continuation of tommysexp(z) and sheldontommysexp(z) are identical function.

Hence either/both tommysexp and sheldontommysexp have property C.

To prove property C of tommysexp(z) = sheldontommysexp(z) one can choose one of those 2 and investigate its analytic continuation/singularities/discontinu on the complex plane within the periodic strip for Re(z) > C.

If it turns out there are no singularities in the periodic semistrip near the reals , then both functions have property C and are equal.

If there are singularities , property C is broken however it might still have regions of analyticity. A natural boundary proves nowhere analytic ( because exp has fixpoints ).

The reason why tommysexp and sheldontommysexp are equivalent is because of a base change like trick explained by sheldon in this thread. The periodicity is also explained here by him.

Notice that a log(f(z)) with f(z) periodic remains periodic.

Also notice that the C property is a uniqueness criterion !

To see this think of (my proof of) TPID 4.

Or the 1 period function wobble of sexp(z).

regards

tommy1729