B.1 Derivatives
609
f (x + h) − f (x) =
∞
k=1
h k D k
k!
f (x) ≡ (e
hD
− 1)f (x) ,
with the final step resulting from the definition of the exponential of an operator. We
may solve this equation formally for f (x), to obtain
f (x) = (e
hD
− 1)
−1 [f (x + h) − f (x)] ,
in which (e hD − 1) −1 is an inverse operator. Upon utilizing the series expansion of
the operator function (e D − 1)) −1 in the form
(e
D
− 1)
−1
=
∞
k=0
B k
k!
D
k−1 ,
we obtain f (x) in the form
f (x) =
∞
r=0
B r
r!
(hD)
r−1
[f (x + h) − f (x)] .
If we now separate out the first two terms of the expansion, we find that f (x) is
obtained as
f (x) =
(hD)
−1
−
1
2 +
∞
r=2
B r
r!
(hD)
r−1
[f (x + h) − f (x)]
= (hD)
−1
[f (x + h) − f (x)] −
1
2 [f (x + h) − f (x)]
+
∞
r=2
B r
r!
h
r−1
f
(r−1) (x + h) − f
(r−1) (x)
,
in which f (n) (x) represents the nth derivative of f (x) with respect to x.
By utilizing the antiderivative nature of the inverse differential operator D −1 ,
together with the property B 2k+1 = 0, k ≥ 1, of the Bernoulli numbers, we obtain
the result
f (x) =
1
h
x+h
x
f (t) dt −
1
2 [f (x + h) − f (x)]
+
∞
k=1
B 2k
(2k)!
h
2k−1
f
(2k−1) (x + h) − f
(2k−1) (x)
,