Properties of Finite DifferencesI have seen this statement in different variants, but could not find...

April Bush

April Bush

Answered

2022-06-25

Properties of Finite Differences
I have seen this statement in different variants, but could not find a proof:
If the n-th order differences of equally spaced data are non-zero and constant then the data can be represented by a polynomial.
I interpret this statement more formally as follows:
Let f : Z Z be any function such that the finite differences D r , defined by
D r + 1 ( x ) = D r ( x + 1 ) D r ( x ) for r 0 (where D 0 := f by convention)
are such that D r is constant and non-zero for any r r 0 . Then f is a polynomial in x (with Q coefficients) and degree r 0 .
Question: Is this correct as I stated it? How does one prove that such f must be a polynomial?

Answer & Explanation

trajeronls

trajeronls

Expert

2022-06-26Added 21 answers

Every polynomial of degree d can be expressed in the basis given by the Falling Factorials up to falling index d.
That is
p ( x ) = a d x d + a d 1 x d 1 + + a 0 x 0 = = b d x d _ + b d 1 x d 1 _ + + b 0 x 0 _
Then, from the properties of the falling factorial, it is easy to demonstrate that
Δ n p ( 0 ) = k ( 1 ) n k ( n k ) p ( k ) = { 0 d < n d ! b d = d ! a d n = d n ! b n n d
And viceversa, given
Δ d p ( x ) = d ! b d Δ d + 1 p ( x ) = 0
the Newton series assures us that
p ( a + x ) = p ( a ) + x Δ p ( a ) + + x d _ d ! Δ d p ( a )
So we can conclude that
if we are given a sequence of m+1 points equally spaced by h
{ x 0 , x 0 + h , x 0 + 2 h , , x 0 + m h }
and the corresponding sequence of values taken by a function f(x)
{ f ( x 0 + k h ) } k = 0 h
we can put
f ( x 0 + k h ) = f ( h ( x 0 h + k ) ) = g ( k )
denoting the differences as
{ Δ g ( k ) = g ( k + 1 ) g ( k ) Δ n g ( k ) = Δ ( Δ n 1 g ( k ) ) = = ( 0 ) j ( n ) ( 1 ) n j ( n j ) g ( k + j ) | 0 n + k m
we have that the sequence in n of ( 1 ) n Δ n g ( k ) is the Binomial Transform of the sequence of the values g(k+n)
{ ( 1 ) n Δ n g ( k ) } n = 0 m k B i n o m i a l T r a n s f o r m { g ( k + n ) } n = 0 m k
and since the Binomial Transform is self-inverse, we get
g ( k + n ) = ( 0 ) j ( n ) ( n j ) Δ j g ( k ) = ( 0 ) j ( n ) Δ j g ( k ) j ! ( ( n + k ) k ) j _
which is precisely the Newton series.
given n+1 points { k , k + 1 , , k + n } and the respective g(k+j)the Newton series is the unique polynomial of max degree n which interpolates them.
if the differences of order n<m are all equal to each other (at changing k)
Δ n g ( k ) = c 0 | 0 k m n
then all the differences of order higher than n will be null
Δ n + q g ( k ) = Δ q ( Δ n g ( k ) ) = Δ q c = 0 | k
and the relevant Newton series will be the unique polynomial of degree n which interpolates all the m+1 points

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