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Topic: Indefinite integral puzzle
Replies: 7   Last Post: Dec 31, 2010 10:50 AM

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Craig Feinstein

Posts: 398
Registered: 12/13/04
Re: Indefinite integral puzzle
Posted: Dec 31, 2010 10:02 AM
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On Dec 31, 3:53 am, William Elliot <> wrote:
> On Thu, 30 Dec 2010, Craig Feinstein wrote:
> > On Dec 30, 10:32 am, 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.


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