From: "David C. Ullrich" <ullrich@math.okstate.edu>
Subject: Re: extremum points of real valued functions
Date: 25 Jun 2000 00:07:44 GMT
Newsgroups: sci.math
Summary: [missing]

artur.steiner@usa.net wrote in article <8j36i4$b76$1@nnrp1.deja.com>...
> In a very good book on Real Analysis, by Robert Bartle, there's a proof
> of that famous theorem which says that, if f is a function from R to R
> and a is an element of it's domain such that the first n-1 derivatives
> are zero and the derivative of order n is different from zero, then:
> 
> If n is even and the nth derivative is >0, then f has a relative
> minimum at a;
> If n is even and the nth derivative is <0, then f has a relative
> maximum at a;
> If n is odd, then f has neither a minimum nor a maximum at a.
> 
> The proof is based on Taylor's Theorem and the author assumes the nth
> derivative of f exists and is continuous on a neighborhood of a. I'm
> not quite sure, but I think such assumptions are too strong. I think
> it's not necessary the existence of such neighborhood. Isn't the simple
> existence of the nth derivative of f at a a sufficient condition for
> the theorem to be valid?

	Yes. Actually you can give a version of Taylor's theorem that's true
under weaker hypotheses than often stated. (I read Chung say this and
say he was giving Hardy's(?) version - I never bothered to look at what
standard references include it. I really should, could put up a webpage
on <red>BAD</red> calculus books...)

	If f is just differentiable at 0 then 

1.   f(x) = a + bx + o(x),

where a = f(0), b = f'(0) and "o(phi(x))" denotes
"some function g(x) such that

g(x) / phi(x) -> 0 as x -> wherever."

Equation (1) is an obvious consequence of Taylor's
theorem as often stated for continuously differentiable
functions, but in fact it's valid supposing just that
f is differentiable at the origin (it's just the definition
of differentiability rearranged). Same thing works for
n > 1, by induction:

	Say f(0) = f'(0) = f''(0) = 0. Then f' and its
derivative both vanish at 0, so applying the
above with f' in place of f shows that

f'(x) = o(x).

But this immediately implies that 

2.   f(x) = o(x^2)

by the Mean Value Theorem. QED.

(The same argument shows by induction that if
f(0) = ... = f^(n)(0) = 0 then f(x) = o(x^n). This
shows that in general if f^(n)(0) exists then
f has a Taylor expansion with error term o(x^n).)

	In fact we didn't even need to assume that
f' was continuous to get (2) (which I thought I
was going to be assuming when I started typing
this, actually). We assume literally nothing but
the existence of f''(0) (which of course implies
that f' be defined at every point in some 
neighborhood of 0, which is exactly enough
to be able to apply MVT.)

	Now about the extrema: If f^(n)(0) exists
and is non-zero but all the lower-order derivatives
vanish then

f(x) = ax^n + o(x^n)

for some non-zero a, hence if n is odd [etc].

> Thank you
> Artur
> 
> 
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