convergents to a continued fraction

convergents to a continued fraction

Definition.

The

n

th

convergent

to a

continued fraction

[

a

0

,

a

1

,

a

2

,

a

3

,

…

]

is the value of the

fraction

obtained by cutting off the fraction after

a

n

, i.e. the value of

[

a

0

,

a

1

,

…

,

a

n

]

.

Write

p

0

=

a

0

p

1

=

a

0

⁢

a

1

+

1

q

0

=

1

q

1

=

a

1

so that

[

a

0

]

=

a

0

=

p

0

q

0

=

a

0

+

1

a

1

=

a

0

⁢

a

1

+

1

a

1

=

p

1

q

1

For

n

>

1

, define

p

n

=

a

n

⁢

p

n

-

1

+

p

n

-

2

(1)

q

n

=

a

n

⁢

q

n

-

1

+

q

n

-

2

Theorem 1

.

The

n

th

convergent to

[

a

0

,

a

1

,

a

2

,

a

3

,

…

]

is given by

[

a

0

,

…

,

a

n

-

1

,

a

n

]

=

p

n

q

n

where

p

n

,

q

n

are defined as above.

Proof.

Induction. For

n

>

1

,

[

a

0

,

…

,

a

n

-

1

,

a

n

]

=

[

a

0

,

…

,

a

n

-

1

+

1

a

n

]

=

(

a

n

-

1

+

1

a

n

)

⁢

p

n

-

2

+

p

n

-

3

(

a

n

-

1

+

1

a

n

)

⁢

q

n

-

2

+

q

n

-

3

=

a

n

⁢

(

a

n

-

1

⁢

p

n

-

2

+

p

n

-

3

)

+

p

n

-

2

a

n

⁢

(

a

n

-

1

⁢

q

n

-

2

+

q

n

-

3

)

+

q

n

-

2

=

a

n

⁢

p

n

-

1

+

p

n

-

2

a

n

⁢

q

n

-

1

+

q

n

-

2

=

p

n

q

n

∎

Theorem 2

.

For

n

≥

1

, the numbers

p

n

-

1

q

n

-

1

and

p

n

q

n

are a

Farey pair

; in fact,

p

n

⁢

q

n

-

1

-

p

n

-

1

⁢

q

n

=

(

-

1

)

n

-

1

(2)

and thus

p

n

q

n

-

p

n

-

1

q

n

-

1

=

(

-

1

)

n

-

1

q

n

-

1

⁢

q

n

(3)

Proof.

This is again a simple induction. The statement is true for

n

=

1

. For

n

>

1

, we have

p

n

⁢

q

n

-

1

-

p

n

-

1

⁢

q

n

=

(

a

n

⁢

p

n

-

1

+

p

n

-

2

)

⁢

q

n

-

1

-

p

n

-

1

⁢

(

a

n

⁢

q

n

-

1

+

q

n

-

2

)

=

p

n

-

2

⁢

q

n

-

1

-

p

n

-

1

⁢

q

n

-

2

=

-

(

-

1

)

n

-

2

=

(

-

1

)

n

-

1

∎

Note that if

[

a

0

,

a

1

,

…

]

is a

simple continued fraction

, then the above theorem implies that

gcd

⁡

(

p

n

,

q

n

)

=

1

, since any common factor of

p

n

and

q

n

must divide

(

-

1

)

n

.

Theorem 3

.

For

n

≥

2

,

p

n

⁢

q

n

-

2

-

p

n

-

2

⁢

q

n

=

(

-

1

)

n

⁢

a

n

(4)

p

n

q

n

-

p

n

-

2

q

n

-

2

=

(

-

1

)

n

⁢

a

n

q

n

-

2

⁢

q

n

(5)

Proof.

Similar to the proof of the above theorem.
∎

Theorem 4

.

If

[

a

0

,

a

1

,

…

]

is a simple continued fraction, then

q

n

≥

n

and, for

n

>

3

,

q

n

>

n

.

Proof.

This follows directly from the iterative definition for the

q

i

and the fact that the

a

i

are

positive integers

.
∎

These results easily imply the following important

convergence

theorem:

Theorem 5

.

For any continued fraction, the even convergents

p

2

⁢

n

/

q

2

⁢

n

are strictly

monotonically increasing

, and the odd convergents

p

2

⁢

n

+

1

/

q

2

⁢

n

+

1

are strictly

monotonically decreasing

. In

addition

, every odd convergent is

greater

than each even convergent. If the continued fraction is simple, then the limit of the odd convergents is equal to the limit of the even convergents, and thus the continued fraction has a well-defined value equal to their common limit.

Proof.

This is basically obvious from the previous observations. Write

c

n

for the

n

th

convergent, i.e.

c

n

=

p

n

q

n

Each

q

i

is

positive

, so

c

n

-

c

n

-

2

=

(

-

1

)

n

⁢

a

n

q

n

⁢

q

n

-

2

is positive for

n

even and negative for

n

odd. This proves the observations about monotonicity. Also,

c

n

-

c

n

-

1

=

(

-

1

)

n

-

1

q

n

⁢

q

n

-

1

is positive for

n

odd, so that

c

2

⁢

n

+

1

>

c

2

⁢

n

(6)

Now, if for some

j

,

k

we had

c

2

⁢

j

+

1

≤

c

2

⁢

k

, then

j

≠

k

by (

6

). If

k

<

j

, then since the even convergents increase,

c

2

⁢

j

+

1

≤

c

2

⁢

k

<

c

2

⁢

j

, while if

j

<

k

, then since the odd convergents decrease,

c

2

⁢

k

+

1

<

c

2

⁢

j

+

1

<

c

2

⁢

k

. In either case, this contradicts (

6

).

As to the statement about simple continued fractions, it is clear that the even (odd) convergents

converge

since they form a monotonically increasing (

decreasing

)

sequence

that is bounded below (above). But

|

p

2

⁢

n

q

2

⁢

n

-

p

2

⁢

n

-

1

q

2

⁢

n

-

1

|

=

1

q

2

⁢

n

⁢

q

2

⁢

n

-

1

≤

1

2

⁢

n

⁢

(

2

⁢

n

-

1

)

→

0

and thus the limits are identical.
∎

Next we prove the following theorem regarding the connection between the “tail” of a continued fraction, its convergents, and its value:

Theorem 6

.

If

x

=

[

a

0

,

a

1

,

…

]

is a simple continued fraction, write

t

n

=

[

a

n

,

a

n

+

1

,

…

]

for

n

≥

0

(the

n

th

complete convergent

). Then

x

=

p

n

-

2

+

t

n

⁢

p

n

-

1

q

n

-

2

+

t

n

⁢

q

n

-

1

Proof.

This is another simple proof by induction. Note that

t

n

=

[

a

n

,

a

n

+

1

,

…

]

=

a

n

+

1

t

n

+

1

so that

t

n

+

1

=

1

t

n

-

a

n

Then

t

n

+

1

⁢

p

n

+

p

n

-

1

t

n

+

1

⁢

q

n

+

q

n

-

1

=

1

t

n

-

a

n

⁢

p

n

+

p

n

-

1

1

t

n

-

a

n

⁢

q

n

+

q

n

-

1

=

1

t

n

-

a

n

⁢

(

a

n

⁢

p

n

-

1

+

p

n

-

2

)

+

p

n

-

1

1

t

n

-

a

n

⁢

(

a

n

⁢

q

n

-

1

+

q

n

-

2

)

+

q

n

-

1

=

t

n

⁢

p

n

-

1

+

p

n

-

2

t

n

⁢

q

n

-

1

+

q

n

-

2

=

x

∎

Finally, we derive a bound on how well the convergents approximate the value of the continued fraction:

Theorem 7

.

If

x

=

[

a

0

,

a

1

,

…

]

is a simple continued fraction, then

|

x

-

p

n

q

n

|

<

1

q

n

2

Proof.

x

-

p

n

q

n

=

p

n

-

1

+

t

n

+

1

⁢

p

n

q

n

-

1

+

t

n

+

1

⁢

q

n

-

p

n

q

n

=

q

n

⁢

p

n

-

1

+

t

n

+

1

⁢

p

n

⁢

q

n

-

p

n

⁢

q

n

-

1

-

t

n

+

1

⁢

p

n

⁢

q

n

q

n

⁢

(

q

n

-

1

+

t

n

+

1

⁢

q

n

)

=

q

n

⁢

p

n

-

1

-

p

n

⁢

q

n

-

1

q

n

⁢

(

q

n

-

1

+

t

n

+

1

⁢

q

n

)

=

(

-

1

)

n

q

n

⁢

(

q

n

-

1

+

t

n

+

1

⁢

q

n

)

But

t

n

+

1

>

a

n

+

1

, so that

q

n

-

1

+

t

n

+

1

⁢

q

n

>

q

n

-

1

+

a

n

+

1

⁢

q

n

=

q

n

+

1

and thus

|

x

-

p

n

q

n

|

=

1

q

n

⁢

(

q

n

-

1

+

t

n

+

1

⁢

q

n

)

<

1

q

n

⁢

q

n

+

1

<

1

q

n

2

since the

q

i

are strictly increasing.
∎

References

1

G.H. Hardy & E.M. Wright,

An Introduction to the Theory of Numbers

, Fifth Edition, Oxford Science Publications, 1979.

Title

convergents to a continued fraction

Canonical name

ConvergentsToAContinuedFraction

Date of creation

2013-03-22 18:04:20

Last modified on

2013-03-22 18:04:20

Owner

rm50 (10146)

Last modified by

rm50 (10146)

Numerical id

8

Author

rm50 (10146)

Entry type

Theorem

Classification

msc 11Y65

Classification

msc 11J70

Classification

msc 11A55

Defines

convergent

Defines

complete convergent

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