ads Power Series

Section 16.5 Power Series

Subsection 16.5.1 Definition

Earlier in this chapter, we found that a polynomial of degree

n

over a ring

R

is an expression of the form

f (x) = \sum_{i = 0}^{n} a_{i} x^{i} = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{n} x^{n}

where

n \geq 0,

each of the

a_{i}

are elements of

R

and

a_{n} \neq 0 .

In Section 8.5 we defined a generating function of a sequence

s

with terms

s_{0},

s_{1},

s_{2}, \dots

as the infinite sum

G (s, z) = \sum_{i = 0}^{\infty} s_{i} z^{i} = s_{0} + s_{1} z + s_{2} z^{2} + \dots

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The main difference between these two expressions, disregarding notation, is that the latter is an infinite expression and the former is a finite expression. In this section we will extend the algebra of polynomials to the algebra of infinite expressions like

G (s, z),

which are called power series.

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Definition 16.5.1. Power Series.

Let

[R; +, \cdot]

be a ring. A power series over

R

is an expression of the form

f (x) = \sum_{i = 0}^{\infty} a_{i} x^{i} = a_{0} + a_{1} x + a_{2} x^{2} + \dots

where

a_{1}, a_{2}, a_{3}, \dots \in R .

The set of all such expressions is denoted by

R [[x]] .

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Our first observation in our comparison of

R [x]

and

R [[x]]

is that every polynomial is a power series and so

R [x] \subseteq R [[x]] .

This is true because a polynomial

a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{n} x^{n}

of degree

n

R [x],

can be thought of as an infinite expression where

a_{i} = 0

for

i > n .

In addition, we will see that

R [[x]]

is a ring with subring

R [x] .

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R [[x]]

is given a ring structure by defining addition and multiplication on power series as we did in

R [x],

with the modification that, since we are dealing with infinite expressions, the sums and products will remain infinite expressions that we can determine term by term, as was done in with polynomials.

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Definition 16.5.2. Power Series Addition.

Given power series

\begin{matrix} f (x) = \sum_{i = 0}^{\infty} a_{i} x^{i} = a_{0} + a_{1} x + a_{2} x^{2} + \dots and \\ g (x) = \sum_{i = 0}^{\infty} b_{i} x^{i} = b_{0} + b_{1} x + b_{2} x^{2} + \dots \end{matrix}

their sum is

\begin{aligned} f (x) + g (x) & = \sum_{i = 0}^{\infty} (a_{i} + b_{i}) x^{i} \\ = (a_{0} + b_{0}) + (a_{1} + b_{1}) x + (a_{2} + b_{2}) x^{2} + (a_{3} + b_{3}) x^{3} + \dots \end{aligned} .

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Definition 16.5.3. Power Series Multiplication.

Given power series

\begin{matrix} f (x) = \sum_{i = 0}^{\infty} a_{i} x^{i} = a_{0} + a_{1} x + a_{2} x^{2} + \dots and \\ g (x) = \sum_{i = 0}^{\infty} b_{i} x^{i} = b_{0} + b_{1} x + b_{2} x^{2} + \dots \end{matrix}

their product is

\begin{aligned} f (x) \cdot g (x) & = \sum_{i = 0}^{\infty} d_{i} x^{i} where d_{i} = \sum_{j = 0}^{i} a_{j} b_{i - j} \\ = (a_{0} \cdot b_{0}) + (a_{0} \cdot b_{1} + a_{1} \cdot b_{0}) x + (a_{0} \cdot b_{2} + a_{1} \cdot b_{1} + a_{2} \cdot b_{0}) x^{2} + \dots \end{aligned} .

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Example 16.5.4. Some Power Series Calcuations.

Let

\begin{matrix} f (x) = \sum_{i = 0}^{\infty} i x^{i} = 0 + 1 x + 2 x^{2} + 3 x^{3} + \dots and \\ g (x) = \sum_{i = 0}^{\infty} 2^{i} x^{i} = 1 + 2 x + 4 x^{2} + 8 x^{3} + \dots \end{matrix}

be elements in

Z [[x]] .

Let us compute

f (x) + g (x)

and

f (x) \cdot g (x) .

First the sum:

\begin{aligned} f (x) + g (x) & = \sum_{i = 0}^{\infty} i x^{i} + \sum_{i = 0}^{\infty} 2^{i} x^{i} \\ = \sum_{i = 0}^{\infty} (i + 2^{i}) x^{i} \\ = 1 + 3 x + 6 x^{2} + 11 x^{3} + \dots \end{aligned}

The product is a bit more involved:

\begin{aligned} f (x) \cdot g (x) & = (\sum_{i = 0}^{\infty} i x^{i}) \cdot (\sum_{i = 0}^{\infty} 2^{i} x^{i}) \\ = (0 + 1 x + 2 x^{2} + 3 x^{3} + \dots) \cdot (1 + 2 x + 4 x^{2} + 8 x^{3} + \dots) \\ = 0 \cdot 1 + (0 \cdot 2 + 1 \cdot 1) x + (0 \cdot 4 + 1 \cdot 2 + 2 \cdot 1) x^{2} + \dots \\ = x + 4 x^{2} + 11 x^{3} + \dots \\ = \sum_{i = 0}^{\infty} d_{i} x^{i} where d_{i} = \sum_{j = 0}^{i} j 2^{i - j} \end{aligned}

We can compute any value of

d_{i},

with the amount of time/work required increasing as

i

increases.

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def d(i):
    s=0
    for j in range(1,i+1):
        s+=j*2^(i-j)
    return s
d(20)

A closed-form expression for

d_{i}

would be desirable. Using techniques from Chapter 8, the formula is

d_{i} = 2^{i + 1} - i - 2,

which we leave it to the reader to derive. Hence,

f (x) \cdot g (x) = \sum_{i = 0}^{\infty} (2^{i + 1} - i - 2) x^{i}

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Subsection 16.5.2 Properties, Units

We have seen that addition and multiplication in

R [[x]]

is virtually identical to that in

R [x] .

The following theorem parallels Theorem 16.3.8, establishing the ring properties of

R [[x]] .

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Theorem 16.5.5. Properties of Power Series.

Let

[R; +, \cdot]

be a ring. Then:

$R [[x]]$ is a ring under the operations of power series addition and multiplication, which depend on the operations in R.
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If R is a commutative ring, then $R [[x]]$ is a commutative ring.
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If R is a ring with unity, $1,$ then $R [[x]]$ is a ring with unity (the unity in R[x] is $1 + 0 x + 0 x^{2} + \dots$ ).
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If R is an integral domain, then $R [[x]]$ is an integral domain.
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If F is a field, then $F [[x]]$ is not a field. However, $F [[x]]$ is an integral domain.
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We are most interested in the situation when the set of coefficients is a field. The theorem above indicates that when

F

is a field,

F [[x]]

is an integral domain. A reason that

F [[x]]

is not a field is the same as one that we can cite for

F [x],

namely that

x

does not have multiplicative inverse in

F [[x]] .

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With all of these similarities, one might wonder if the rings of polynomials and power series over a field are isomorphic. It turns out that they are not. The difference between

F [x]

and

F [[x]]

becomes apparent when one studies which elements are units in each. First we prove that the only units in

F [x]

are the nonzero constants; that is, the nonzero elements of

F .

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Theorem 16.5.6. Polynomial Units.

Let

[F; +, \cdot]

be a field. Polynomial

f (x)

is a unit in

F [x]

if and only if it is a nonzero constant polynomial.

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Proof.

(⇒)

Let

f (x)

be a unit in

F [x] .

Then

f (x)

has a multiplicative inverse, call it

g (x),

such that

f (x) \cdot g (x) = 1 .

Hence, the

\deg (f (x) \cdot g (x)) = \deg (1) = 0 .

But

\deg (f (x) \cdot g (x)) = \deg f (x) + \deg g (x) .

\deg f (x) + \deg g (x) = 0,

and since the degree of a polynomial is always nonnegative, this can only happen when the

\deg f (x) = \deg g (x) = 0 .

Hence,

f (x)

is a constant, an element of

F,

which is a unit if and only if it is nonzero.

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(⇐)

f (x)

is a nonzero element of

F,

then it is a unit since

F

is a field. Thus it has an inverse, which is also in

F [x]

and so

f (x)

is a unit of

F [x] .

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Before we proceed to categorize the units in

F [[x]],

we remind the reader that two power series

a_{0} + a_{1} x + a_{2} x^{2} + \dots

and

b_{0} + b_{1} x + b_{2} x^{2} + \dots

are equal if and only if corresponding coefficients are equal,

a_{i} = b_{i}

for all

i \geq 0 .

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Theorem 16.5.7. Power Series Units.

Let

[F; +, \cdot]

be a field. Then

f (x) = \sum_{i = 0}^{\infty} a_{i} x^{i}

is a unit of

F [[x]]

if and only if

a_{0} \neq 0 .

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Proof.

(⇒)

f (x)

is a unit of

F [[x]],

then there exists

g (x) = \sum_{i = 0}^{\infty} b_{i} x^{i}

F [[x]]

such that

\begin{aligned} f (x) \cdot g (x) & = (a_{0} + a_{1} x + a_{2} x^{2} + \dots) \cdot (b_{0} + b_{1} x + b_{2} x^{2} + \dots) \\ = 1 \\ = 1 + 0 x + 0 x^{2} + \dots \end{aligned} .

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Since corresponding coefficients in the equation above must be equal,

a_{0} \cdot b_{0} = 1,

which implies that

a_{0} \neq 0 .

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(⇐)

Assume that

a_{0} \neq 0 .

To prove that

f (x)

is a unit of

F [[x]]

we need to find

g (x) = \sum_{i = 0}^{\infty} b_{i} x^{i}

F [[x]]

such that

f (x) \cdot g (x) = \sum_{i = 0}^{\infty} d_{i} x^{i} = 1 .

If we use the formula for the coefficients

f (x) \cdot g (x)

and equate coefficients, we get

\begin{array}{lll} d_{0} = a_{0} \cdot b_{0} = 1 & \Rightarrow & b_{0} = a_{0}^{- 1} \\ d_{1} = a_{0} \cdot b_{1} + a_{1} \cdot b_{0} = 0 & \Rightarrow & b_{1} = - a_{0}^{- 1} \cdot (a_{1} \cdot b_{0}) \\ d_{2} = a_{0} b_{2} + a_{1} b_{1} + a_{2} b_{0} = 0 & \Rightarrow & b_{2} = - a_{0}^{- 1} \cdot (a_{1} \cdot b_{1} + a_{2} \cdot b_{0}) \\ ⋮ & ⋮ & ⋮ \\ d_{s} = a_{0} \cdot b_{s} + a_{1} \cdot b_{s - 1} + \dots + a_{s} \cdot b_{0} = 0 & \Rightarrow & b_{s} = - a_{0}^{- 1} \cdot (a_{1} \cdot b_{s - 1} + a_{2} \cdot b_{s - 2} + \dots + a_{s} \cdot b_{0}) \end{array} .

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Therefore the powers series

\sum_{i = 0}^{\infty} b_{i} x^{i}

is an expression whose coefficients lie in

F

and that satisfies the statement

f (x) \cdot g (x) = 1 .

Hence,

g (x)

is the multiplicative inverse of

f (x)

and

f (x)

is a unit.

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Example 16.5.8.

Let

f (x) = 1 + 2 x + 3 x^{2} + 4 x^{3} + \dots = \sum_{i = 0}^{\infty} (i + 1) x^{i}

be an element of

Q [[x]] .

Then, by Theorem 16.5.7, since

a_{0} = 1 \neq 0,

f (x)

is a unit and has an inverse, call it

g (x) .

To compute

g (x),

we follow the procedure outlined in the above theorem. Using the formulas for the

b_{i}^{'}

s, we obtain

\begin{matrix} b_{0} = 1 \\ b_{1} = - 1 (2 \cdot 1) = - 2 \\ b_{2} = - 1 (2 \cdot (- 2) + 3 \cdot 1) = 1 \\ b_{3} = - 1 (2 \cdot 1 + 3 \cdot (- 2) + 4 \cdot 1) = 0 \\ b_{4} = - 1 (2 \cdot 0 + 3 \cdot 1 + 4 \cdot (- 2) + 5 \cdot 1) = 0 \\ b_{5} = - 1 (2 \cdot 0 + 3 \cdot 0 + 4 \cdot (1) + 5 \cdot (- 2) + 6 \cdot 1) = 0 \\ ⋮ \end{matrix}

For

s \geq 3,

we have

b_{s} = - 1 (2 \cdot 0 + 3 \cdot 0 + \dots (s - 2) \cdot 0 + (s - 1) \cdot 1 + s \cdot (- 2) + (s + 1) \cdot 1) = 0

Hence,

g (x) = 1 - 2 x + x^{2}

is the multiplicative inverse of

f (x) .

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Certainly

F [[x]]

contains a wider variety of units than

F [x] .

Yet

F [[x]]

is not a field, since

x \in F [[x]]

is not a unit. So concerning the algebraic structure of

F [[x]],

we know that it is an integral domain that contains

F [x] .

If we allow our power series to take on negative exponents; that is, consider expressions of the form

f (x) = \sum_{i = - \infty}^{\infty} a_{i} x^{i}

where all but a finite number of terms with a negative index equal zero. These expressions are called extended power series. The set of all such expressions is a field, call it

E .

This set does contain, for example, the inverse of

x,

which is

x^{- 1} .

It can be shown that each nonzero element of

E

is a unit.

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Exercises 16.5.3 Exercises

1.

Let

f (x) = \sum_{i = 0}^{\infty} a_{i} x^{i}

and

g (x) = \sum_{i = 0}^{\infty} b_{i} x^{i}

be elements of

R [[x]] .

Let

f (x) \cdot g (x) = \sum_{i = 0}^{\infty} d_{i} x^{i} = 1 .

Apply basic algebra to

(a_{0} + a_{1} x + a_{2} x^{2} + \dots) \cdot (b_{0} + b_{1} x + b_{2} x^{2} + \dots)

to derive the formula

d_{s} = \sum_{i = 0}^{s} a_{i} b_{s - i}

for the coefficients of

f (x) \cdot g (x) .

Hence, to show that

f (x) \cdot g (x) = \sum_{s = 0}^{\infty} (\sum_{i = 0}^{s} a_{i} b_{s - i}) x^{s}

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2.

Prove that for any integral domain $D,$ the following can be proven: $f (x) = \sum_{i = 0}^{\infty} a_{i} x^{i}$ is a unit of $D [[x]]$ if and only if $a_{0}$ is a unit in $D .$
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Compare the statement in part a to that in Theorem 16.5.7.
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Give an example of the statement in part a in $Z [[x]] .$
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3.

Use the formula for the product to verify that the expression

g (x)

of Example 16.5.8 is indeed the inverse of

f (x) .

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4.

Determine the inverse of $f (x) = 1 + x + x^{2} + \dots = \sum_{i = 0}^{\infty} x^{i}$ in $Q [[x]] .$
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Repeat part a with $f (x)$ taken in $Z_{2} [[x]] .$
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Use the method outlined in Chapter 8 to show that the power series $f (x) = \sum_{i = 0}^{\infty} x^{i}$ is the rational generating function $\frac{1}{1 - x} .$ What is the inverse of this function? Compare your results with those in part a.
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5.

Determine the inverse of $h (x) = \sum_{i = 0}^{\infty} 2^{i} x^{i}$ in $Q [[x]] .$
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Use the procedures in Chapter 8 to find a rational generating function for $h (x)$ in part a. Find the multiplicative inverse of this function.
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Answer.

$\begin{matrix} b_{0} = 1 \\ b_{1} = (- 1) (2 \cdot 1) = - 2 \\ b_{2} = (- 1) (2 \cdot (- 2) + 4 \cdot 1) = 0 \\ b_{3} = (- 1) (2 \cdot 0 + 4 \cdot (- 2) + 8 \cdot 1) = 0 \end{matrix}$

All other terms are zero. Hence, $f (x)^{- 1} = 1 - 2 x$

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$\begin{aligned} f (x) & = 1 + 2 x + 2^{2} x^{2} + 2^{3} x^{3} + \dots \\ = (2 x)^{0} + (2 x)^{1} + (2 x)^{2} + (2 x)^{3} + \dots \\ = \frac{1}{1 - 2 x} \end{aligned}$

The last step follows from the formula for the sum of a geometric series.

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6.

Let

a (x) = 1 + 3 x + 9 x^{2} + 27 x^{3} + \dots = \sum_{i = 0}^{\infty} 3^{i} x^{i}

and

b (x) = 1 + x + x^{2} + x^{3} + \dots = \sum_{i = 0}^{\infty} x^{i}

both in

R [[x]] .

What are the first four terms (counting the constant term as the $0^{th}$ term) of $a (x) + b (x) ?$
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Find a closed form expression for $a (x) .$
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What are the first four terms of $a (x) b (x) ?$
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7.

Write as an extended power series:

${(x^{4} - x^{5})}^{- 1}$
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${(x^{2} - 2 x^{3} + x^{4})}^{- 1}$
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Answer.

$\begin{aligned} {(x^{4} - x^{5})}^{- 1} & = (x^{4} (1 - x))^{- 1} \\ = x^{- 4} \frac{1}{1 - x} \\ = x^{- 4} (\sum_{k = 0}^{\infty} x^{k}) \\ = \sum_{k = - 4}^{\infty} x^{k} \end{aligned} .$

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$\begin{aligned} {(x^{4} - 2 x^{3} + x^{2})}^{- 1} & = {(x^{2} (x^{2} - 2 x + 1))}^{- 1} \\ = x^{- 2} {(1 - 2 x + x^{2})}^{- 1} \\ = x^{- 2} (\sum_{k = 0}^{\infty} (k + 1) x^{k}) \\ = \sum_{k = - 2}^{\infty} (k + 2) x^{k} \end{aligned} .$

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8.

Derive the closed form expression

d_{i} = 2^{i + 1} - i - 2

for the coefficients of the product

f (x) \cdot g (x)

in Example 16.5.4.

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