You may be interested in the following books on non-Newtonian calculus and related matters. They are all available for reading at Google Book Search (http://books.google.com/books?q=%22Non-Newtonian+Calculus%22&btnG=Search...).

- Michael Grossman and Robert Katz. "Non-Newtonian Calculus", ISBN 0912938013, 1972

- Michael Grossman. "The First Nonlinear System of Differential and Integral Calculus", ISBN 0977117006, 1979

- Jane Grossman, Michael Grossman, Robert Katz. "The First Systems of Weighted Differential and Integral Calculus", ISBN 0977117014, 1980

- Jane Grossman. "Meta-Calculus: Differential and Integral", ISBN 0977117022, 1981

- Michael Grossman. "Bigeometric Calculus: A System with a Scale-Free Derivative", ISBN 0977117030, 1983

- Jane Grossman, Michael Grossman, Robert Katz. "Averages: A New Approach", ISBN 0977117049, 1983

Anbody interested to discuss????

And what would you like to know? I don't see how value in one point can determine values in neighbourhood. For example every polynomial function satisfies your conditions.

joking

It can. Let me give you an example. Consider function f:R->R which is smooth and there is c such that

f(x)=0 for x<=c
f(x) is nonzero for x>c in a small neighbourhood of c

It can be shown that such function exists. Then obviously we have

f^(n)(x) = 0 for x<=c

but in a small neighbourhood of c the derivative f^(n)(x) cannot be zero. Because if f^(n)(x)=0 in an open interval, then f(x) is a polynomial in this interval. But there is no polynomial satisfying conditions I gave at the begining.

To give an explicit example, consider function

F(x) = e^{-1/(1-x^2)} for -1 < x < 1
F(x) = 0 otherwise

This function satisfies my conditions for c = -1. Note that such functions are important in constructions of so called smooth partitions of unity.

joking

Hi Steve,
let me show you my idea.
Let B(x,\delta), \delta > 0, an arbitrary neighborhood about x. Let f be a smooth function in B(x,\delta). Consider an arbitrary point u such that 0 < |u - x| < \delta. Assume that for all n > N, f^{(n)}(u) = 0, where N is some positive integer (your claim N = N(x), is wrong). Since, by hypothesis, f(u) is smooth in B(x,\delta), it is continuous in there and, therefore, it is Riemann-integrable in that neighborhood. Thus, for n = N, we assume
f^{(N)}(u) = c_N (a constant different from zero).
By iterative integration we get,
f^{(N-1)}(u) = (c_N/1!)u + c_{N-1}/0!,
f^{(N-2)}(u) = (c_N/2!)u^2 + c_{N-1}u/1! + c_{N-2}/0!,
.................................................
f^{(N-k)}(u) = (c_N/k!)u^k + c_{N-1}u^{k-1}/(k-1)1! + ... + c_{N-k}/0!.
So, for k = N
f^{(0)}(u) := f(u) = (c_N/N!)u^N + c_{N-1}u^{N-1}/(N-1)1! + ... + c_0/0!.
We set c_k/k! = a_k, so that
f(u) = a_N.u^N + a_{N-1}.u^{N-1} + ... + a_1.u + a_0.
Finally, since f is continuous in B(x,\delta)
\lim_{u \to x} f(u) = f(x) = a_N.x^N + a_{N-1}.x^{N-1} + ... + a_1.x + a_0.
This is the polynomial we need to find.

perucho

Yes. Unfortunetly I don't see any reason for N(x) to be bounded on an interval. I tried to prove this, but I couldn't. Also I didn't find any counterexample. But my intuition tells me that there might be a pathological smooth map with those properties which is not a polynomial.

joking

Correction: I meant I_1 \superset I_2 \superset I_3 \superset ...

But isn't a single point a closed interval?