approximating algebraic numbers with linear recurrences

approximating algebraic numbers with linear recurrences

Linear recurrences

can be used to obtain

rational approximations

for real

algebraic numbers

. Suppose that

ρ

is the root of a

polynomial

p

⁢

(

x

)

=

∑

k

=

0

n

c

k

⁢

x

k

with

rational

coefficients

c

k

and
further assume that the roots of

p

are distinct and that all the
other roots of

p

are strictly smaller in

absolute value

than

ρ

.

Consider the recursion

∑

k

=

0

n

c

k

⁢

a

m

+

k

=

0

.

As discussed in the parent entry, the solution of this recurrence
is

a

m

=

∑

k

=

1

n

b

k

⁢

r

k

m

,

where

r

1

,

…

,

r

n

are the roots of

p

— let us agree that

r

1

=

ρ

— and the

b

k

’s are determined by the initial conditions
of the recurrence. If

b

1

≠

0

, we have

a

m

+

1

a

m

=

b

1

⁢

ρ

m

+

1

+

∑

k

=

2

n

b

k

⁢

r

k

m

+

1

b

1

⁢

ρ

m

+

∑

k

=

1

n

b

k

⁢

r

k

m

=

ρ

⁢

1

+

∑

k

=

2

n

(

b

k

b

1

)

⁢

(

r

k

ρ

)

m

+

1

1

+

∑

k

=

2

n

(

b

k

b

1

)

⁢

(

r

k

ρ

)

m

.

Because

|

r

k

ρ

|

<

1

when

k

>

1

, we have

lim

m

→

∞

⁡

(

r

k

/

ρ

)

m

=

0

, and hence

lim

m

→

∞

⁡

a

m

+

1

a

m

=

ρ

.

To illustrate this method, we will compute the

square root

of two. Now,
we cannot use the equation

x

2

-

2

=

0

because it has two roots,

+

2

and

-

2

, which are equal in absolute value. So what
we shall do instead is to use the equation

(

x

-

1

)

2

=

2

, or

x

2

-

2

⁢

x

-

3

=

0

, whose roots are

1

-

2

and

1

+

2

. Since

|

1

+

2

|

>

|

1

-

2

|

, we can use our method to approximate
the larger of these roots, namely

1

+

2

; to approximate

2

,
we subtract

1

from our answer. The recursion we should use is

a

m

+

2

-

2

⁢

a

m

+

1

-

a

m

=

0

or, moving terms around,

a

m

+

2

=

2

⁢

a

m

+

1

+

a

m

.

If we choose

b

1

=

-

b

2

=

1

/

(

2

⁢

2

)

, then we have

a

0

=

b

1

⁢

(

1

+

2

)

0

+

b

2

⁢

(

1

-

2

)

0

=

b

1

+

b

2

=

0

a

1

=

b

1

⁢

(

1

+

2

)

1

+

b

2

⁢

(

1

-

2

)

1

=

b

1

+

b

2

+

(

b

1

-

b

2

)

⁢

2

=

1

.

Starting with these values, we obtain the following

sequence

:

a

0

=

0

a

1

=

1

a

2

=

2

a

3

=

5

a

4

=

12

a

5

=

29

a

6

=

70

a

7

=

169

a

8

=

408

a

9

=

985

a

10

=

2738

a

11

=

5741

a

12

=

13860

Therefore, as our approximations to

2

, we have

1

,

3

2

,

7

5

,

17

12

,

41

29

,

99

70

,

239

169

,

577

408

,

1753

985

,

3003

2738

,

8119

5741

.

By making suitable transformations, one can compute all the roots
of a polynomial using this technique. A way to do this is to start
with a rational number

h

which is closer to the desired root than
to the other roots, then make a change of variable

x

=

1

/

(

y

+

h

)

.

As an example, we shall examine the roots of

x

3

+

9

⁢

x

+

1

.
Approximating the polynomial by leaving off the

constant term

, we
guess that the roots are close to

0

,

+

3

⁢

i

, and

-

3

⁢

i

. Since the
two

complex roots

are

conjugates

, it suffices to find one of them.

To look for the root near

0

, we make the change of variable

x

=

1

/

y

to obtain

y

3

+

9

⁢

y

2

+

1

=

0

, which gives the recursion

a

k

+

3

+

9

⁢

a

k

+

2

+

a

k

=

0

. Picking some initial values, this
recursion gives us the sequence

0

,

0

,

1

,

-

9

,

81

,

-

738

,

6723

,

-

61236

,

557766

,

-

5090401

,

…

,

whence we obtain the approximations

-

1

9

,

-

1

9

,

-

81

738

,

-

738

6723

,

-

6723

61236

,

-

61236

557766

,

-

557766

5090401

,

…

.

To look for the root near

3

⁢

i

, we make the change of variable

x

=

1

/

(

y

+

3

⁢

i

)

to obtain

y

3

+

(

9

+

9

⁢

i

)

⁢

y

2

+

(

-

27

+

54

⁢

i

)

⁢

y

+

82

-

27

⁢

i

=

0

, which gives the recursion

a

k

+

3

+

(

9

+

9

⁢

i

)

⁢

a

k

+

2

+

(

-

27

+

54

⁢

i

)

⁢

a

k

+

1

+

(

82

-

27

⁢

i

)

⁢

a

k

=

0

.

Picking some initial values, this
recursion gives us the sequence

0

,

0

,

1

,

-

9

-

9

⁢

i

,

27

+

108

⁢

i

,

-

82

-

945

⁢

i

,

-

225

+

11196

⁢

i

,

44415

-

127953

⁢

i

,

…

,

Title

approximating algebraic numbers with linear recurrences

Canonical name

ApproximatingAlgebraicNumbersWithLinearRecurrences

Date of creation

2013-03-22 16:53:03

Last modified on

2013-03-22 16:53:03

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

16

Author

rspuzio (6075)

Entry type

Definition

Classification

msc 11B37

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