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

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William Elliot

Posts: 1,948
Registered: 5/30/08
Re: Indefinite integral puzzle
Posted: Dec 31, 2010 3:53 AM
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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.




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