IT IS A POLYNOMIAL
Alexander Abian
Can you give as clear and as simple as possible proof of the following
in the Real Analysis?
Let f be an infinitely differentiable function on [0, 1] such
that at every x in [0, 1] an n-th order derivative of f is zero,
where n depends on x. Prove that f is a polynomial.
--
--------------------------------------------------------------------------
ABIAN TIME-MASS EQUIVALENCE FORMULA T = A m^2 in Abian units
ALTER EARTH'S ORBIT AND TILT - STOP GLOBAL DISASTERS AND EPIDEMICS
ALTER THE SOLAR SYSTEM. REORBIT VENUS INTO A NEAR EARTH-LIKE ORBIT
TO CREATE A BORN AGAIN EARTH (1990)
Adrian Cable
Alexander Abian wrote:
>
> Can you give as clear and as simple as possible proof of the following
> in the Real Analysis?
>
> Let f be an infinitely differentiable function on [0, 1] such
> that at every x in [0, 1] an n-th order derivative of f is zero,
> where n depends on x. Prove that f is a polynomial.
Huh? Aren't you a Professor Emeritus of mathematics at Iowa State
university?
Thanks, cheers,
Adrian Cable.
JamesKmpd4
Well, so is Escultura at the University of the Phillipines!
Larry Taylor
JamesKmpd4 wrote:
>
> >
> >Alexander Abian wrote:
> >> <snip>
> >Huh? Aren't you a Professor Emeritus of mathematics at Iowa State
> >university?
>
> Well, so is Escultura at the University of the Phillipines!
>
You are kidding, I sincerely hope.
Larry Taylor
ull...@math.okstate.edu
In article <199805021638...@ladder01.news.aol.com>#1/1,
james...@aol.com (JamesKmpd4) wrote:
>
> >
> >Alexander Abian wrote:
> >>
> >> Can you give as clear and as simple as possible proof of the following
> >> in the Real Analysis?
> >>
> >> Let f be an infinitely differentiable function on [0, 1] such
> >> that at every x in [0, 1] an n-th order derivative of f is zero,
> >> where n depends on x. Prove that f is a polynomial.
> >
> >university?
>
> Well, so is Escultura at the University of the Phillipines!
>
> http://www.geocities.com/Vienna/6404/sinprez.html
>
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ull...@math.okstate.edu
In article <354B02...@csubak.edu>#1/1,
> > >university?
> >
> > Well, so is Escultura at the University of the Phillipines!
> >
>
Hard to tell who's kidding about what these days. In any
case those who're tweaking Abian about this should give a proof
of the result first - seems likely to me that he knows how to
prove it and the post was a challenge.
(Or it could be that his proof is wrong - I can _almost_
construct a counterexample...<g>. In any case it seems unlikely
that his proof is going to involve the sort of wild nonsense as
in EEE's gibberish.)
David C. Ullrich
JamesKmpd4
Larry Taylor wrote:
>JamesKmpd4 wrote:
>>
>> >
>> >Alexander Abian wrote:
>> >> <snip>
>> >Huh? Aren't you a Professor Emeritus of mathematics at Iowa State
>> >university?
>>
>> Well, so is Escultura at the University of the Phillipines!
>>
>
>You are kidding, I sincerely hope.
>
>Larry Taylor
Unfortunately I am not. Take a look at his home page.
Michael Hovdan
Alexander Abian wrote:
> Can you give as clear and as simple as possible proof of the following
> in the Real Analysis?
>
> Let f be an infinitely differentiable function on [0, 1] such
> that at every x in [0, 1] an n-th order derivative of f is zero,
> where n depends on x. Prove that f is a polynomial.
>
> --
>
> --------------------------------------------------------------------------
>
> ABIAN TIME-MASS EQUIVALENCE FORMULA T = A m^2 in Abian units
> ALTER EARTH'S ORBIT AND TILT - STOP GLOBAL DISASTERS AND EPIDEMICS
> ALTER THE SOLAR SYSTEM. REORBIT VENUS INTO A NEAR EARTH-LIKE ORBIT
> TO CREATE A BORN AGAIN EARTH (1990)
A Taylor expansion around f(0.5) should result in the polynomial you are
seeking.
jons...@telcel.net.ve
In article <6iec3p$rd9$1...@news.iastate.edu>#1/1,
ab...@iastate.edu (Alexander Abian) wrote:
>
>
> Can you give as clear and as simple as possible proof of the following
> in the Real Analysis?
>
> Let f be an infinitely differentiable function on [0, 1] such
> that at every x in [0, 1] an n-th order derivative of f is zero,
> where n depends on x. Prove that f is a polynomial.
If I remember well there is a proof in "A Primer of Real Functions", by
Ralph P. Boas. The main idea is the following:
Let 0 <= a < b <= 1 and define A_k={x \in [a,b] : f^(k)(x) = 0}.
Since [a,b] is the union of all the A_k's, by Baire's Theorem some A_n
is not nowhere dense. Hence the closure of A_n contains a (non empty)
open interval B where (by continuity) f^(n) is identically 0. Hence there
exists a family F of open subintervals of [0,1] whose union is dense in
[0,1], and such that f is a polynomial in each interval of the family.
By a second application of Baire's theorem it is possible to prove that
UF = (a,b), etc.
Best Regards,
Jose H. Nieto
ull...@math.okstate.edu
In article <6iipnq$nih$1...@nnrp1.dejanews.com>#1/1,
jons...@telcel.net.ve wrote:
>
> In article <6iec3p$rd9$1...@news.iastate.edu>#1/1,
> ab...@iastate.edu (Alexander Abian) wrote:
> >
> >
> > Can you give as clear and as simple as possible proof of the following
> > in the Real Analysis?
> >
> > Let f be an infinitely differentiable function on [0, 1] such
> > that at every x in [0, 1] an n-th order derivative of f is zero,
> > where n depends on x. Prove that f is a polynomial.
>
> If I remember well there is a proof in "A Primer of Real Functions", by
> Ralph P. Boas. The main idea is the following:
>
> Let 0 <= a < b <= 1 and define A_k={x \in [a,b] : f^(k)(x) = 0}.
>
> Since [a,b] is the union of all the A_k's, by Baire's Theorem some A_n
> is not nowhere dense. Hence the closure of A_n contains a (non empty)
> open interval B where (by continuity) f^(n) is identically 0.
This far I got (assuming an implicit "for some n" above).
> Hence there
> exists a family F of open subintervals of [0,1] whose union is dense in
> [0,1], and such that f is a polynomial in each interval of the family.
Yes,
> By a second application of Baire's theorem it is possible to prove that
> UF = (a,b), etc.
How does that go?
David C. Ullrich
ull...@math.okstate.edu
In article <354C52AD...@hypercon.com>#1/1,
mho...@writeme.com wrote:
>
>
> Alexander Abian wrote:
>
> > Can you give as clear and as simple as possible proof of the following
> > in the Real Analysis?
> >
> > Let f be an infinitely differentiable function on [0, 1] such
> > that at every x in [0, 1] an n-th order derivative of f is zero,
> > where n depends on x. Prove that f is a polynomial.
> A Taylor expansion around f(0.5) should result in the polynomial you are
> seeking.
No, the problem with that is that an infinitely differentiable
function need not equal its Taylor series.
Actually there are other problems with that, but never mind,
that's the big one. IF we were assuming that f was real-analytic,
ie that f was equal to its Taylor series, then I could prove the
thing. But we're not assuming it's real-analytic.
Robert Israel
Alexander Abian wrote:
> Can you give as clear and as simple as possible proof of the following
> in the Real Analysis?
>
> Let f be an infinitely differentiable function on [0, 1] such
> that at every x in [0, 1] an n-th order derivative of f is zero,
> where n depends on x. Prove that f is a polynomial.
It brings back memories: this problem (the "polynomial problem") was going
around U. of Chicago ca. 1970.
Suppose f is not a polynomial. Let
C = {x: there is no neighbourhood of x on which f is a polynomial}.
A_n = {x: f^(n)(x) = 0}.
Clearly C is closed and nonempty, and A_n closed with union_n A_n = [0,1].
Applying the Baire Category Theorem to C, A_n intersect C has nonempty interior
in C for some n, i.e. there exist x0 in C and delta > 0 such that
C intersect (x0 - delta, x0 + delta) is contained in A_n.
Suppose f^(k)(x0) <> 0 for some k > n. By Taylor's Theorem, we get f^(n)(x) <> 0
for 0 < |x - x0| < eta (for some eta > 0). Taking eta < delta, this implies
C intersect (x0 - eta, x0 + eta) = { x0 }, but then f is a polynomial on
(x0 - eta, x0) and on (x0, x0 + eta) and hence on (x0 - eta, x0 + eta),
contradicting x0 in C. So we must conclude that f^(k)(x0) = 0 for all k >= n.
The same is true for all points in C intersect (x0 - delta, x0 + delta).
Now suppose c in (x0 - delta, x0 + delta) \ C. Then f is a polynomial on some
neighbourhood of c. Let [a,b] be the maximal interval containing c on which
f is a polynomial. Then a or b (say b) is in
(x0 - delta, x0 + delta) intersect C, so that f^(k)(b) = 0 for all k >= n.
Now if f has degree p on [a,b], f^(p) <> 0 on [a,b] so p < n, and then
f^(k)(c) = 0 for all k >= n.
We conclude that f^(k)(x) = 0 for all k >= n and all x in (x0 - delta, x0 + delta). But then f is a polynomial on (x0 - delta, x0 + delta), contradiction.
Robert Israel isr...@math.ubc.ca
Department of Mathematics (604) 822-3629
University of British Columbia fax 822-6074
Vancouver, BC, Canada V6T 1Z2