Date: Dec 31, 2010 10:02 AM Author: Craig Feinstein Subject: Re: Indefinite integral puzzle On Dec 31, 3:53 am, William Elliot <ma...@rdrop.remove.com> wrote:

> On Thu, 30 Dec 2010, Craig Feinstein wrote:

> > On Dec 30, 10:32 am, Pfss...@aol.com wrote:

> >> On Wed, 29 Dec 2010 10:35:22 -0800 (PST), Craig Feinstein

> >>> Let I(f(x))dx be the indefinite integral of a function f(x). Then what

> >>> is:

> >>> I(I(I(I(...)dx)dx)dx)dx ?

>

> >> well, first of all, what does this mean???

> >> Give a precise definition!!!

>

> > I(I(I(I(...)dx)dx)dx)dx means that we want the (indefinite) integral

> > of the integral of the integral of the integral ..... ad infinitum.

>

> That's not a precise definition. It's not even a definition.

> It's the same intuitive description of your notion, as was

> depicted with your notation.

>

> Let f(x) = 0. Then integral 0 dx = c1

> integral^2 0 dx^2 = c1.x + c2

> ...

> integral^j 0 dx^j is any polynomial in R[x] of degree at most j.

>

> Thus in the limit one may opine that

> integral^oo 0 dx^j is any polynomial p in R[[x]], ie

> all formal infinite series of the form sum(j=0,oo) aj.x^j.

>

> See my other post where I indicate how the concept

> is ludicrous by sketching out how

> integral^oo f(x) dx^oo = integral^oo g(x) dx^oo

> for all f,g with Taylor series.

>

> You can't even define an operator I:R^R -> R^R

> and take the limit lim(n->oo) I^n(f) which you

> could do if I(f) = integral(0,x) f(x) dx.

>

> You have to define the operator I:P(R^R) -> P(R^R)

> and then create a topology for P(R^R) so that

> lim(n->oo) I^n({f}) is definable.

>

> Though there are several topologies for R^R,

> which you may want to review and pick on the

> most likely to use, there are none for P(R^R)

> that I know, which are in use.

>

> In other words, you've got a heap of work to do

> to give a precise definition of integral^oo f(x) dx^oo

> for indefinite integration. In addition, my intuitive

> exploration given in my other post, indicates that the

> result of integral^oo f(x) dx^oo, for a wide range of

> functions, lacks significance.

>

> For example

> integral^oo 0 dx^oo = R[[x]]

> is every function with a Taylor series and a lot

> of other infinite series in powers of x that have

> a variety of convergent and divergent properties.

>

> That's because you've a infinite number of constants of

> integration. Happily, it's merely a countable infinite.

>

> My claim is

> integral^oo f(x) dx^oo = R[[x]]

> for all f with a Taylor series.

>

> By default, the functions are taken to have domain R.

> R[[x]] however, does include partial functions, ie

> functions with domains other than R.

You don't need a precise definition to find the answer:

Clue: Let f(x) = I(I(I(I(...)dx)dx)dx)dx. Then f(x)= I(f(x))dx.

Craig