# topics on calculus

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### Looking for calculus tutor

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.

### Re: Looking for calculus tutor

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

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

### June Ponder This!!!!!!!!!!

Let Ï€ be a permutation on 14 elements. Define Ni as the minimal number of times the permutation Ï€ 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)?

### Re: June Ponder This!!!!!!!!!!

Anbody interested to discuss????

### smooth functions

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?

### Re: smooth functions

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

### Re: smooth functions

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?

### Re: smooth functions

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

### Re: smooth functions

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.

### Re: smooth functions

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

### Re: smooth functions

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

### Re: smooth functions

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

### Re: smooth functions

Proof:

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.

### Re: smooth functions

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

### Withdraw proof

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 :-)

Regards,

csguy

### Re: Withdraw proof

But isn't a single point a closed interval?

### Re: Withdraw proof

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.