Re: On a basic proof in interpolation: usage of the Fundamental Algebra
- From: Robert Israel <israel@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx>
- Date: Sun, 06 Jan 2008 18:05:34 -0600
rg_linux1@xxxxxxxx writes:
This message concerns a proof given for an identity in Interpolation
(Approximation) theory given in
B.I. Kvasov, "Method of Shape-Preserving Spline Approximation", World
Scientific, Singapore, 2000, p. 9, Lemma 1.1.
My main concern and interrogation is on the use of the Fundamental
Theorem of Algebra in the proof of that Lemma 1.1 on p. 9 of the book
by Kvasov. My goal is not to criticize Dr. Kvasov book and proofs
there in but it is rather to understand better the use of the
Fundamental Theorem of Algebra he made in his proof.
The Lemma states that for any polynomial P_k(x) of degree k <= N we
have that
-- N
\
P_k (x) = / P_k(x_j) l_j(x)
-- j=0
where l_j(x) are the Lagrange coefficient polynomials defined as
follow:
l_j(x) = w_N(x)/((x-x_j) w'_N(x_j)) with w_N(x) = (x-x_o)(x-x_1) ...
(x-x_N) as it can be found in any standard Num Analysis textbook.
....
One can say as given in the book of Kvasov that F_k,N(x) has N+1
roots. In his book, Kvasov adds to complete the proof that by the
Fundamental Theorem of Algebra and because F_k,N(x) has N+1 roots,
F_k,N(x) must be identically equal to zero. This where I have an
interrogation on the use of the Fundamental Theorem of Algebra in
manner Kvasov completes the proof as it is difficult for me to put
side-by-side what the Fundamental Theorem of Algebra states that the
number of roots for a polynomial of degree N, P_N(x) = 0 which is at
most N. The converse is also true that if a polynomial of degree N,
P_N(x), has N roots then P_N(x) = 0. How can we justify using the
Fundamental Theorem of Algebra for a polynomial of degree N that has N
+1 roots to state that it is identically equal to zero?
Actually I don't think it requires the FTA at all. It's true over any field,
not just an algebraically closed one: if a polynomial p(x) has n roots
a_1, ..., a_n, then p(x) is divisible by (x-a_1)...(x-a_n); therefore
if it is not the 0 polynomial, it must have degree at least n. Conversely,
if a polynomial has more roots than its degree, it must be 0.
--
Robert Israel israel@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
.
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