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topics on calculus

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gmacfadden posted the following as an entry request:

"I am looking for tutor well versed in tensor calculus who would be to provide tutoring on a private, for fee basis. I am not taking any formal classes, I am nearing retirement, and my interest is purely intellectual - ultimate go is to understand special relativity.

"I live in Northern Virginia, US. Please contact "

why don't you put me quizz in the talk page of User:Juan Marquez of wikipedia, or in

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 (

- 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

Let π be a permutation on 14 elements. Define Ni as the minimal number of times the permutation π should be applied to get i back into its original place. What is the remainder, modulo 1299, of the sum over all possible 14! permutations of 4 to the power of the product of (2+the parity of Ni)?

Anbody interested to discuss????

Suppose f:R->R is a smooth function. Let f^(n) represent the nth derivative of f. Suppose that f^(n)(x) = 0 for some x for all n>N. Can you say anything about f^(n) in a small enough neighborhood about x?

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.


Let me be more specific. Let N(x) be the smallest integer for which f^(n)(x) = 0 for all n>N(x). If f is a smooth function, can N(x) be unbounded on a closed interval?

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.


The original problem was
Suppose f:R->R is smooth and for every x, there exists an N=N(x) such that f^(n)(x)=0 for all n>N. Prove that f must be a polynomial. I figured that all I needed to do was show that N(x) is bounded on a closed interval.

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.


Right. If N(x) is bounded on B(x,δ), then f must be a polynomial. What I want to do is show that N(x) must be bounded on B(x,δ) if there is a finite N(x) for each x in B(x,δ).

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.



By contradiction. First of all, the derivatives of all orders are continuous. Suppose that N(x) is unbounded in every closed interval. Start with some x_1 and n_1 such that f^(n_1)(x_1) is not zero. By continuity, there is a closed interval I_1 containing x_1 such that f^(n_1) is not zero in that interval. Because N(x) is not bounded in I_1, there is some x_2 in I_1 and some n_2 > n_1 with f^(n_2)(x_2) not zero, and therefore there is a closed interval I_2 contained in I_1 such that f^(n_2) is not zero in I_2. Proceeding in this fashion we construct a sequence of intervals I_1 \subset I_2 \subset I_3 \subset ... and a sequence n_1 < n_2 < n_3 < ... such that f^(n_j) is not zero in I_j. By a well known property of the real numbers, the intersection of the I_j's is not empty, and for any x in that intersection, f^(n_j)(x) is not zero for all j, contradiction the assumption that there is N(x) such that f^(n)(x) = 0 for all n > N(x). Thus the assumption that N(x) is unbounded in all closed intervals leads to a contradiction and therefore N(x) must be bounded in some closed interval and so it must be a polynomial function.

End of Proof.

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

Please ignore the proof I just gave, as it is fallacious (the interval I construct where f is bounded could be a single point (though I still think the idea can somewhat be salvaged)). Will keep looking :-)



But isn't a single point a closed interval?

Yes, actually there was nothing wrong with the argument except it only shows that there is some non-trivial interval (meaning an interval that is not just a single point) in which the function agrees with a polynomial. In fact, the same argument shows for any a, b with a < b there is a non-trivial interval I_ab contained in [a, b] and a polynomial P_ab(x) such that the function agrees with P_ab on I_ab. But I don't see how one can conclude there is a single polynomial P agreeing with f everywhere.

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