Computation and Properties of Maps
129
5.2.1 The Structure 1 D 1
Consider the vector space R
2 of ordered pairs (a 0 , a 1 ), a 0 , a 1 ∈ R in which
an addition and a scalar multiplication are defined in the usual way:
(a 0 , a 1 ) + (b 0 , b 1 ) = (a 0 + b 0 , a 1 + b 1 ),
(5.12)
t · (a 0 , a 1 ) = (t · a 0 , t · a 1 ),
for a 0 , a 1 , b 0 , b 1 ∈ R. Besides the above addition and scalar multiplication, a
multiplication between vectors is introduced in the following way:
(a 0 , a 1 ) · (b 0 , b 1 ) = (a 0 · b 0 , a 0 · b 1 + a 1 · b 0 ),
(5.13)
for a 0 , a 1 , b 0 , b 1 ∈ R. With this definition of a vector multiplication the set of
ordered pairs becomes an algebra, denoted by 1 D 1 .
In the same way as in the case of complex numbers, one can identify (a 0 , 0)
as the real number a 0 . Where in the complex numbers, (0, 1) was a root of
−1, here it has another interesting property:
(0, 1) · (0, 1) = (0, 0),
which follows directly from eq. (5.13). So (0, 1) is a root of 0. Such a property
suggests thinking of d = (0, 1) as something infinitely small, small enough that
its square vanishes. Because of this we call d = (0, 1) the differential unit.
The first component of the pair (a 0 , a 1 ) is called the real part, and the second
component is called the differential part. Using this notation it becomes clear
that elements of 1 D 1 can be written as a 0 + a 1 · d, and multiplication amounts
to multiplying the polynomials (a 0 + a 1 · d) and (b 0 + b 1 · d) and keeping only
terms linear in d.
It is easy to verify that (1, 0) is a neutral element of multiplication, because
according to eq. (5.13)
(1, 0) · (a 0 , a 1 ) = (a 0 , a 1 ) · (1, 0) = (a 0 , a 1 ).
It turns out that (a 0 , a 1 ) has a multiplicative inverse if and only if a 0 is
nonzero. In case a 0 = 0 the inverse is
(a 0 , a 1 )
−1 =
1
a 0
, −
a 1
a 2
0
.
(5.14)
Using equations it is easy to check that in fact (a 0 , a 1 )
−1
· (a 0 , a 1 ) = (1, 0).
An outstanding result of the methods of differential algebras is that differentiation becomes an algebraic problem, and the differential part of the
difference
f (x + d) − f (x)
equals the conventional derivative. Thus, given any differentiable function f,
we can compute its derivatives by just evaluating the formula and thus obtain
f
(x) = D [f (x + d) − f (x)] = D [f (x + d)] ,
(5.15)
129
5.2.1 The Structure 1 D 1
Consider the vector space R
2 of ordered pairs (a 0 , a 1 ), a 0 , a 1 ∈ R in which
an addition and a scalar multiplication are defined in the usual way:
(a 0 , a 1 ) + (b 0 , b 1 ) = (a 0 + b 0 , a 1 + b 1 ),
(5.12)
t · (a 0 , a 1 ) = (t · a 0 , t · a 1 ),
for a 0 , a 1 , b 0 , b 1 ∈ R. Besides the above addition and scalar multiplication, a
multiplication between vectors is introduced in the following way:
(a 0 , a 1 ) · (b 0 , b 1 ) = (a 0 · b 0 , a 0 · b 1 + a 1 · b 0 ),
(5.13)
for a 0 , a 1 , b 0 , b 1 ∈ R. With this definition of a vector multiplication the set of
ordered pairs becomes an algebra, denoted by 1 D 1 .
In the same way as in the case of complex numbers, one can identify (a 0 , 0)
as the real number a 0 . Where in the complex numbers, (0, 1) was a root of
−1, here it has another interesting property:
(0, 1) · (0, 1) = (0, 0),
which follows directly from eq. (5.13). So (0, 1) is a root of 0. Such a property
suggests thinking of d = (0, 1) as something infinitely small, small enough that
its square vanishes. Because of this we call d = (0, 1) the differential unit.
The first component of the pair (a 0 , a 1 ) is called the real part, and the second
component is called the differential part. Using this notation it becomes clear
that elements of 1 D 1 can be written as a 0 + a 1 · d, and multiplication amounts
to multiplying the polynomials (a 0 + a 1 · d) and (b 0 + b 1 · d) and keeping only
terms linear in d.
It is easy to verify that (1, 0) is a neutral element of multiplication, because
according to eq. (5.13)
(1, 0) · (a 0 , a 1 ) = (a 0 , a 1 ) · (1, 0) = (a 0 , a 1 ).
It turns out that (a 0 , a 1 ) has a multiplicative inverse if and only if a 0 is
nonzero. In case a 0 = 0 the inverse is
(a 0 , a 1 )
−1 =
1
a 0
, −
a 1
a 2
0
.
(5.14)
Using equations it is easy to check that in fact (a 0 , a 1 )
−1
· (a 0 , a 1 ) = (1, 0).
An outstanding result of the methods of differential algebras is that differentiation becomes an algebraic problem, and the differential part of the
difference
f (x + d) − f (x)
equals the conventional derivative. Thus, given any differentiable function f,
we can compute its derivatives by just evaluating the formula and thus obtain
f
(x) = D [f (x + d) − f (x)] = D [f (x + d)] ,
(5.15)