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Re: Indefinite integral puzzle
Posted:
Dec 31, 2010 10:02 AM


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



