rules of calculus for derivative of formal power series

rules of calculus for derivative of formal power series

In this entry, we will show that the rules of calculus
hold for

derivatives

of

formal power series

. While
this could be verified directly in a manner analogous
to what was done for

polynomials

in the parent entry,
we will take a different tack, deriving the results
for power series from the corresponding results for
polynomials. The basis for our approach is the
observation that the ring of formal power series can
be expressed as a limit of

quotients

of the ring of
polynomials:

A

⁢

[

[

x

]

]

=

lim

k

→

∞

⁡

A

⁢

[

x

]

/

⟨

x

k

⟩

Thus, we will proceed in two steps, first extending the
derivative

operation

to the

quotient rings

and showing that
its properties still hold there, then extending it to the
limit and showing that its properties hold there as well.

We begin by noting that the derivative is well-defined
as a map from

A

⁢

[

x

]

/

⟨

x

k

+

1

⟩

to

A

⁢

[

x

]

/

⟨

x

k

⟩

for all

integers

k

≥

0

.

Theorem 1

.

Suppose that

A

is a

commutative ring

,

k

is a non-negative
integer, and that

p

and

q

are elements of

A

⁢

[

x

]

such
that

p

≡

q

modulo

x

k

+

1

. Then

p

′

≡

q

′

modulo

x

k

.

Proof.

By definition of

congruence

,

p

⁢

(

x

)

=

q

⁢

(

x

)

+

x

k

+

1

⁢

r

⁢

(

x

)

for
some polynomial

r

∈

A

⁢

[

x

]

. Taking derivaitves,

p

′

⁢

(

x

)

=

q

′

⁢

(

x

)

+

x

k

⁢

(

k

⁢

r

⁢

(

x

)

+

x

⁢

r

′

⁢

(

x

)

)

, so

p

′

and

q

′

are

equivalent

modulo

x

k

.
∎

It is easy to verify that the sum and product rules hold
in this new

context

:

Theorem 2

.

If

A

is a commutative ring,

k

is a non-negative integer,
and

f

,

g

are elements of

A

⁢

[

x

]

/

⟨

x

k

+

1

⟩

,
then

(

f

+

g

)

′

=

f

′

+

g

′

.

Proof.

Let

p

,

q

be representatives of the

equivalence classes

f

,

g

. Then we have

(

p

+

q

)

′

=

p

′

+

q

′

by the corresponding

theorem

for polynomials. Hence, by definition of quotient,
we have

(

f

+

g

)

′

=

f

′

+

g

′

.
∎

Theorem 3

.

If

A

is a commutative ring,

k

is a non-negative integer,
and

f

,

g

are elements of

A

⁢

[

x

]

/

⟨

x

k

+

1

⟩

,
then

(

f

⋅

g

)

′

=

f

′

⋅

g

+

f

⋅

g

′

.

Proof.

Let

p

,

q

be representatives of the equivalence classes

f

,

g

. Then we have

(

p

⋅

q

)

′

=

p

′

⋅

q

+

p

⋅

q

′

by the corresponding theorem for polynomials. Hence, by
definition of quotient, we have

(

f

⋅

g

)

′

=

f

′

⋅

g

+

f

⋅

g

′

.
∎

When considering the chain rule, we need to note that

composition

does not always pass to the quotient, so we need to restrict the
operands to obtain a well-defined operation. In particular, we
will consider the following two cases:

Theorem 4

.

If

A

is a commutative ring,

p

,

q

,

r

is a n element of

A

⁢

[

x

]

,
and

q

≡

r

modulo

x

k

for some integer

k

>

0

, then

p

∘

q

≡

p

∘

r

modulo

x

k

.

Theorem 5

.

If

A

is a commutative ring,

k

is a non-negative integer, and

p

,

q

,

P

,

Q

are elements of

A

⁢

[

x

]

such that

p

≡

P

modulo

x

k

,

q

≡

Q

modulo

x

k

and

p

⁢

(

0

)

=

0

, then

p

∘

q

≡

P

∘

Q

modulo

x

k

.

[More to come]

Title

rules of calculus for derivative of formal power series

Canonical name

RulesOfCalculusForDerivativeOfFormalPowerSeries

Date of creation

2013-03-22 18:22:35

Last modified on

2013-03-22 18:22:35

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

8

Author

rspuzio (6075)

Entry type

Derivation

Classification

msc 12E05

Classification

msc 11C08

Classification

msc 13P05

Related topic

InvertibleFormalPowerSeries

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