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Hometopics on calculus
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topics on calculus
This entry is an overview of many calculus related entries which can be found here, at PlanetMath.org. By calculus we mean real analysis at the highschool level or college level, and the entries in this page should be at either level. If an entry is written at a higher level, it will be indicated with a “GL” tag.
1 Expositions
The following are books or notes on calculus:
2 $\mathbb{R}$ : The real line
3 Functions of one real variable

Concept of function and real function
4 Computing limits
5 Continuity
6 Differentiation in one variable
7 Integration of functions of one real variable

A lecture on integration by substitution

A lecture on integration by parts

A lecture on trigonometric integrals and trigonometric substitutions

A lecture on the partial fraction decomposition method
8 Definite integral

Approximate integration:
Left Hand Rule Right Hand Rule Midpoint Rule Trapezoidal Rule Simpson’s Rule
9 Integral Transforms
10 Multivariable Calculus
10.1 Differentiation
10.2 Integration

Stokes’ theorem
11 Differential Equations
12 Infinite Series
12.1 Series of Numbers

(GL?) Nonexistence of universal series convergence criterion

(GL?) Multiplication of series
12.2 Function Sequences and Series

Limit of function sequence

Weierstrass’ criterion of uniform convergence

(GL) Fourier series
12.3 Power Series and Taylor Series

Taylor’s theorem

Example of Taylor polynomials for $\sin x$

Examples on how to find Taylor series from other known series
13 Additional Topic

NonNewtonian calculus
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Comments
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.
"I live in Northern Virginia, US. Please contact gmacfadden@aol.com "
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
Additional References
You may be interested in the following books on nonNewtonian calculus and related matters. They are all available for reading at Google Book Search (http://books.google.com/books?q=%22NonNewtonian+Calculus%22&btnG=Search...).
 Michael Grossman and Robert Katz. "NonNewtonian 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. "MetaCalculus: Differential and Integral", ISBN 0977117022, 1981
 Michael Grossman. "Bigeometric Calculus: A System with a ScaleFree 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/(1x^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 Riemannintegrable 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^{(N1)}(u) = (c_N/1!)u + c_{N1}/0!,
f^{(N2)}(u) = (c_N/2!)u^2 + c_{N1}u/1! + c_{N2}/0!,
.................................................
f^{(Nk)}(u) = (c_N/k!)u^k + c_{N1}u^{k1}/(k1)1! + ... + c_{Nk}/0!.
So, for k = N
f^{(0)}(u) := f(u) = (c_N/N!)u^N + c_{N1}u^{N1}/(N1)1! + ... + c_0/0!.
We set c_k/k! = a_k, so that
f(u) = a_N.u^N + a_{N1}.u^{N1} + ... + 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_{N1}.x^{N1} + ... + 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 nontrivial 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 nontrivial 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.