The Calculus of Finite Differences
John Stalker
March 15, 2016
Theorem 1 Suppose x1 < x2 < · · · < xm . The following three conditions
are equivalent:
1. As formal power series in D,
m n−1
X
X
cj,k D k exp(xj D) = O(D n ),
j=1 k=0
2. For any polynomial p of degree less than n,
m n−1
X
X
cj,k p(k) (xj ) = 0.
j=1 k=0
3. There is a function r such that if f ∈ C n ([x1 , xm ]) then
m n−1
X
X
cj,k f (k) (xj ) =
j=1 k=0
Z
xm
x1
r(x)f (n) (x) dx.
Condition 3 clearly implies Condition 2, since the integrand vanishes if f is
a polynomial p of degree less than n.
We now show that Conditions 1 and 2 are equivalent. Condition 1 means
that
m n−1
X
X
cj,k D k exp(xj D) = 0
j=1 k=0
in the ring R1 of polynomials in D, modulo the ideal generated by D n . Consider the vector space V of polynomials in x of degree less than n. d/dx is
an endomorphism of this vector space. Its n’th power is zero, but all lower
1
powers are linearly independent. Let R2 be the ring of endomorphisms generated by d/dx. Then R1 and R2 are isomorphic, the isomorphism taking D to
d/dx. Henceforth we identify R1 and R2 via this isomorphism. We therefore
drop the subscripts, and write D for d/dx. Having made this identification,


m n−1
X
X
j=1 k=0

cj,k D k exp(xj D)p (x) =
m n−1
X
X
cj,k p(k) (x + xj ).
j=1 k=0
Suppose Condition 1 holds. Then
m n−1
X
X
cj,k D k exp(xj D)p = 0
j=1 k=0
for all p, and hence


and thus
m n−1
X
X
j=1 k=0

cj,k D k exp(xj D)p (0) = 0
m n−1
X
X
cj,k p(k) (xj ) = 0.
j=1 k=0
Suppose Condition 2 holds. If p is a polynomial of degree less than n
then so is q, defined by q(x) = p(x + y) for all y. So
m n−1
X
X
cj,k q (k) (xj ) = 0,
j=1 k=0
or
m n−1
X
X
cj,k p(k) (xj + y) = 0.
j=1 k=0
In other words,


m n−1
X
X
j=1 k=0

cj,k D k exp(xj D)p (y) = 0
for all y. Thus
m n−1
X
X
cj,k D k exp(xj D)p
j=1 k=0
2
is the zero polynomial. But p was an arbitrary element of V , so
m n−1
X
X
cj,k D k exp(xj D) = 0
j=1 k=0
as an endomorphism of V . In other words, Condition 1 holds.
It remains to prove that Condition 2 implies Condition 3. This we will
prove with the explicit r
r(x) =
m n−1
X
X
cj,k bn−k−1 (xj − x)
j=1 k=0
where
(−z)i /i! if z ≥ 0
0
if z < 0.
If i = 0 and z = 0 then we interpret 00 as 1.
By Condition 2,
bi (z) =
m n−1
X
X
cj,k p(k) (xj ) = 0.
j=1 k=0
for all polynomials p of degree less than n. This holds in particular for
p(x) =
n−1
X
l=0
f (l) (x1 )
(x1 − x)l .
l!
In other words,
m n−1
X
X n−1
X
cj,k
j=1 k=0 l=k
f (k) (x1 )
(xj − x1 )l−k = 0
(l − k)!
or, shifting indices,
m n−1
X
X n−k−1
X
l=1 k=0
i=0
cl,k
f (i+k) (x1 )
(xl − x1 )i = 0
i!
Now
Z
xm
x1
r(x)f
(n)
(x) dx =
m−1
X Z xj+1
j=1
3
xj
r(x)f (n) (x) dx
so
xm
Z
x1
r(x)f (n) (x) dx =
m−1
m
X n−1
XX
Z
cl,k
xj+1
xj
j=1 k=0 l=1
bn−k−1 (xl − x)f (n) (x) dx
If j ≥ l then the integrand on the right is zero. Otherwise, choose xj < α <
β < xj+1 and integrate by parts n − k times,
Z
β
α
bn−k−1 (xl − x)f (n) (x) dx =
n−k−1
X h
bi (xl − α)f (i+k) (α)
i=0
i
− bi (xl − β)f (i+k) (β) .
Taking limits as α tends to xj and β tends to xj+1 ,
Z
xj+1
xj
bn−k−1 (xl − x)f (n) (x) dx =
n−k−1
X
i=0
1h
(xj+1 − xl )i f (i+k) (xj+1 )
i!
i
− (xj − xl )i f (i+k) (xj ) .
We then sum over j,
m−1
X Z xj+1
xj
j=1
bn−k−1 (xl − x)f
(n)
(x) dx = f
(k)
(xl ) −
n−k−1
X
f
(i+k)
i=0
(x1 − xl )i
(x1 )
i!
Multiplying by ck,l and summing over k and l,
Z
xm
x1
r(x)f (n) (x) dx =
m n−1
X
X
cj,k f (k) (xj )
j=1 k=0
m n−1
X
X n−k−1
X
+
l=1 k=0
i=0
cl,k
(i + k)! (i+k)
f
(x1 )(xl − x1 )i .
i!
The second sum is, as we have seen, zero, so
Z
xm
x1
r(x)f (n) (x) dx =
m n−1
X
X
cj,k f (k) (xj )
j=1 k=0
and Condition 3 holds, with
r(x) =
m n−1
X
X
cj,k bn−k−1 (xj − x).
j=1 k=0
4