03/26/2015, 11:08 PM

@JmsNxn

I don't yet understand all the analysis behind but I'm tryng to follow the logical implications of the lemmas. If I get it is the Ramanujan's that makes all the works (but we have the exp bound requirmeent).

The other interesting point is the differential equation that holds for the auxiliary function (seems crazy). An this is the really interesting trick imho. But pls, tell me if I get it (started to study calculus just 1 month ago).

We have a power series

and its derivative becomes the following (D distributes over addition, commutes with product and appliyng power rule we get a "shift" of indexes in the series )

Now comes the trick... (Am I right?) Define the coefficents in the following form

that becomes . Replace this in the prevous series and

So if we define

We could assume to have the following

and that

and we hope that

At this point you set (in other notation youre using its xi-based superfunction ) so you can send differentiations by w to iteration by the transfer function (or application of the right-composition operator).

Seeing this I now understand why you told me, long ago, that this can be used for the fractional ranks... but now you talk only about linear operators and the recursion (I mean the b-based superfunction operator) and the antirecursion/subfunction operator aren't linear. Also, just before Lemma 3.1, you say that this is not proved for operators different than the left-composition yet. Why?

Why can't we just repeat the trick using other sequences like using a sequence of values of your bounded hyperoperators indexed by the rank?

We just need to be exp. bounded or I've missed something important somewhere in your paper?

For example we could try to set in it one of these two sequences

Given an invertible function and a in its domain

define the direct antirecursion sequence of and its inverse antirecursion sequence

and

and

Those two sequences of functions give us two sequences of real numbers and for a fixed

btw those sequences satisfie those recurrences

I don't yet understand all the analysis behind but I'm tryng to follow the logical implications of the lemmas. If I get it is the Ramanujan's that makes all the works (but we have the exp bound requirmeent).

The other interesting point is the differential equation that holds for the auxiliary function (seems crazy). An this is the really interesting trick imho. But pls, tell me if I get it (started to study calculus just 1 month ago).

We have a power series

and its derivative becomes the following (D distributes over addition, commutes with product and appliyng power rule we get a "shift" of indexes in the series )

Now comes the trick... (Am I right?) Define the coefficents in the following form

that becomes . Replace this in the prevous series and

So if we define

We could assume to have the following

and that

and we hope that

At this point you set (in other notation youre using its xi-based superfunction ) so you can send differentiations by w to iteration by the transfer function (or application of the right-composition operator).

Seeing this I now understand why you told me, long ago, that this can be used for the fractional ranks... but now you talk only about linear operators and the recursion (I mean the b-based superfunction operator) and the antirecursion/subfunction operator aren't linear. Also, just before Lemma 3.1, you say that this is not proved for operators different than the left-composition yet. Why?

Why can't we just repeat the trick using other sequences like using a sequence of values of your bounded hyperoperators indexed by the rank?

We just need to be exp. bounded or I've missed something important somewhere in your paper?

For example we could try to set in it one of these two sequences

Given an invertible function and a in its domain

define the direct antirecursion sequence of and its inverse antirecursion sequence

and

and

Those two sequences of functions give us two sequences of real numbers and for a fixed

btw those sequences satisfie those recurrences

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