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Re: Request Algebra Proof Of Ruiz Identity
Posted:
Feb 7, 2006 4:35 PM
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Hi, I think that idea by Rob Johnson it's Ok !
Let us try a proof by means of Algebra.
Suppose in the following that x_0,x_1,...,x_n
are distinct points on the real axis.
We use following facts:
1) If P_n and Q_n are polynomials of degree =< n such that
P_n(x_j)=Q_n(x_j) , in {0,1,...,n} ,
then P_n=Q_n , that is P_n(x)=Q_n(x) for all x.
2) Consider that {x_0,x_1,...,x_n} are in D , D being
a subset in R:=(-infty,infty).
If f:D--->R then there exists a unique polynomial L_nf of degree =< n
,
such that (L_nf)(x_j) =f(x_j) for all j in {0,1,...,n}.
This is a well-know polynomial, namely Lagrange interpolation
polynomial.
In order to prove this fact, see 1) .
3) There are known [Representations of L_nf ] :
Denote w(x)=(x-x_0)(x-x_1)...(x-x_n) and observe that
w'(x_k)=(x_k-x_0)...(x_k-x_{k-1})(x_k-x_{k+1})...(x_k-x_n) .
Then, with notation
l_k(x)=w(x)/((x-x_k)(w'(x_k)) = x^n/w'(x_k) + c_1x^{n-1} +....
we have
3.1) (L_nf)(x)=SUM_{i=0 to i=n} l_i(x)*f(x_i)= D_n[f]x^n+ ....
where
(3.1.1) D_n[f]= SUM_{i=0to i=n} f(x_i)/w'(x_i) .
Another representation is following: denote by V_n the Vandermonde
determinant corresponding to the points {x_0,x_1,...,x_n} , that is
V_n :=
| 1 x_0 ... x_0 ^n|
| 1 x_1 ... x_1 ^n|
... ... ... ...
| 1 x_n ... x_n ^n|
= Prod_{0=<i<j=<n}(x_j-x_i) .
Likewise consider the functional determinant
(W_nf)(x) :=
| 1 x_0 ....x_0^n f(x_0)|
| 1 x_1 ... x_1^n f(x_1)|
... .... ... ....
|1 x_n ... x_n^n f(x_n)|
| 1 x ... x^n 0 |
It's clear that W_nf is a polynomial of degree =<n .
On the other hand
(3.2.1) (P_nf)(x) := - (W_nf)(x) / V_n =
= E_n[f]x^n+....
where
(3.2.2) V_n* E_n[f]:=
| 1 x_0 x_0^2 ...x_0^{n-1} f(x_0) |
| 1 x_1 x_1^2 ... x_1^{n-1} f(x_1)|
... ... ... ... ... ...
| 1 x_n x_n^2 ...x_n^{n-1} f(x_n)| .
But (P_nf)(x_j) =f(x_j) , j in {0,1,...,n). Using 1) we conclude that
(3.2.3) (L_nf)(x) = (P_nf)(x) , and as consequence
==============================
(*) E_n(f) = D_n[f] .
==============================
see (3.1.1) and (3.2.2) . Select in (*)
f(x)= H(x):=A_0x^n+A_1x^{n-1}+ ....+A_n .
Because for f(x)=e_j(x)=x^j , j in {0,1,...,n-1}, we have
see (3.2.2) ,
E_n[e_j] =0 , j=0,1,...,n-1 (two columns are equal),
and E_n[e_n]=1 . Using the liniarity of E_n[f] we give
E_n[H] = A_0 . Using (*) we conclude that
================================
(**) D_n[f]:= SUM_{i=0 to i=n} H(x_i)/w'(x_i) = A_0
================================
for any polynomial H (x)=A_0x^n+ .... .
Now the solution of your problem: choose the points
====================
x_k=k , k in {0,1,...,n} and the function
H(x):=H_1(x)= (x-n)^n = 1*x^n+ ...
>From (**) you find your identity.
-------------------------------------------------------------
REMARKS:
1) Such identities are well-known in literature (perhaps must
be attributed to Gauss).In the book by John Riordan
"Combinatorial Identities" there is said that for case H_2(x)=x^n
is attributed to ...?? .[I do'nt remember to whom , but try to find in
the book.]
------------------------------------------------------------
2) I have observe that in case when
H(x):= H_3(x)= B_0x^{n+1}+A_0x^n+...
then (**) furnishes us
===============================
D_n[H_3]= B_0(x_0+x_1+...+x_n) +A_0 .
==============================
3) When H(x)=H_4(x)=C_0x^{n+2}+B_0x^{n+1}+A_0x^{n}+ ...
then (**) is
=====================================
D_n[H_4] = (C_0/2)*( S_2 +S_1^2 )+B_0*S_1 +A_0 , where
====================================
S_2=x_0^2+...x_n^2 , S_1=x_0+x_1+...+x__n
------------------------------------------------------------------------
4) (**) is a generalization of your identity;
----------------------------------------------------------------------
5) If you want to find a more general identity tahn (**) , try to use
Lagrange-Hermite interpolation . Perhaps,help/Alex
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