82 Normat 60:2, 82–90 (2012)
A tale about tails
Tails as tools – tails as toys
Haakon Waadeland
haakonwa@math.ntnu.no
1 Apology.
As you will see rather quickly, this note is not a proper math paper, not even a
survey paper. Almost nothing is new.
It is meant as a final Goodbye to my former colleagues and still friends – with
my warm gratitude.
2 Concepts.
We deal with continued fractions
Œ
K
n=1
(a
n
/b
n
)=
a
1
b
1
+
a
2
b
2
+
a
3
b
3
+
.
.
.
=:
a
1
b
1
+
a
2
b
2
+
a
3
b
3
+ ···
, (1)
where a
n
and b
n
are complex numbers such that all approximants
S
N
(0) =
a
1
b
1
+
a
2
b
2
+ ··· +
a
N
b
N
=
A
n
B
n
(2)
are meaningful. The value Πis accepted. The continued fraction converges iff
lim
næŒ
S
n
(0) exists. The limit is the value of the continued fraction. Again the
value Πis accepted.
The numbers A
N
and B
N
, the way they are normalized, are known to satisfy
certain linear recurrence relations, see e.g.[2] and [3] .
A generalization of (2) is
S
N
(u)=
a
1
b
1
+
a
2
b
2
+ ··· +
a
N
b
N
+ u
=
A
N
+ A
N≠1
u
B
N
+ B
N≠1
u
, (3)
Normat 2/2012 Haakon Waadeland 83
where u is a complex number.
The concept of tail is essential in the theory of continued fractions. For a fixed
N the continued fraction
a
N+1
b
N+1
+
a
N+2
b
N+2
+ ···
(4)
is called the Nth tail of the continued fraction (1), convergent or not. In all newer
books on continued fractions tails and their applications as well as their role in the
theory of continued fractions are presented. The deepest presentation is given in
[4].
In case of convergence, if we in (3) choose u = u
N
= the value (4) of the tail of
(1), then S
N
(u
N
) is the value of the continued fraction (1). This is actually the key
to the use of tails as a tool in continued fractions: Remove the unknown and correct
Nth tail of the continued fraction (1) (like we do to get the Nth approximant) and
replace it by a new tail, incorrect but in some sense "near" the correct tail. We then
get what is called a modified approximant, which in favorable cases is better than
the classical one. We now illustrate this in two s imple cases. First, however, a little
remark on continued fractions: A continued fraction K(a
n
/b
n
) may, if all a
n
,b
n
are
different from 0 , be transformed to a fraction K(c
n
/1) or K(1/d
n
) in a simple way,
see e.g. [2],[3] or [4].
3 Tail as tool.
Example 1.
Take the continued fraction
12 + 0.9
1+
12 + 0.9
2
1+
12 + 0.9
3
1+··· +
12 + 0.9
n
1+···
. (5)
It follows from several theorems, one of them the parabola theorem [2], [3], [4], that
it converge s. For n =5, 10, 20, 30, we compute the values of the approximants.
They are here presented with 10 true digits.
S
5
(0) = 4.799987237 S
10
(0) = 2.816847005
S
20
(0) = 3.112991809 S
30
(0) = 3.130817317
In order to get modified approximants we replace the tails in the given continued
fraction by
12
1+
12
1+···
, (6)
which has the value 3. This gives us the modified approximants
S
5
(3) = 3.151871068 S
10
(3) = 3.129004674
S
20
(3) = 3.131833817 S
30
(3) = 3.131891251
84 Haakon Waadeland Normat 2/2012
Comparison of ordinary and modified approximants in this example illustrates how
a skilled use of tails may accelerate the process of approximating a number by
using continued fractions. (Information: The value of the given continued faction is
3.1318924157 with 11 true digits,) Let X denote the value of the continued fraction
(5): We find in the computation
X ≠ S
30
(0) = 0.001075 ··· ,X≠ S
30
(3) = 0.0000011 ···
In the handbook [1] there is a large number of examples of highly non-trivial ap-
plications of tails to accelerate convergence, not only the one shown here, but
extentions to more general methods, leading to striking results in convergence ac-
celeration. Here we shall, however, restrict ourselves to one very simple example
within the method we have already seen in the previous example. For more ad-
vanced examples we refer to [1].
Example 2
We shall use a story from old days to get a second example of use of tails. We go
back to a paper from 1865 by Julius Worpitzky, published in Jahresbericht from
Friedrichs Gymnasium und Realschule. Worpitzky’s theorem is as follows, slightly
adjusted for our present use: In the continued fraction
p
1+
q
1+
a
3
1+
a
4
1+
a
5
1+···
(7)
let p, q, a
3
,a
4
,a
5
,... all have abs olute value Æ 1/4,with= only in a finite number
of cases. Then the continued fraction, as well as all the tails, will converge to some
value in the disk |w| Æ 1/2. In the following we shall for simplicity assume that p
and q both are real.
We now ask the following question, also raised in [3, p.36]: What can be said about
the Worpitzky set of continued fraction values Ê for fixed values of p and q?We
have
Ê =
p
1+
q
1+w
where w can take any value in |w| Æ 1/2. A simple transformation leads to
w =
(q + 1)Ê ≠ p
p ≠ Ê
.
The Ê-set we are asking for is then
-
-
-
(q + 1)Ê ≠ p
p ≠ Ê
-
-
-
Æ
1
2
.
Standard procedure (including a bit of work) leads to the result, that the Ê-set we
are asking for is the disk
-
-
-
Ê ≠
p(4q + 3)
4(q + 1)
2
≠ 1
-
-
-
Æ
2|pq|
4(q + 1)
2
≠ 1
. (8)
Normat 2/2012 Haakon Waadeland 85
As a numerical example take (as in [3]) p = ≠1/4, q =1/8.Thenthediskis
-
-
-
w +
14
65
-
-
-
Æ
1
65
.
This tells us that if we use ≠
14
65
as an approximation, then the error is less than
1
65
for all continued fractions in question, i. e. all where |a
n
| Æ 1/4 for all n Ø 3.
(A different numerical example is what we get with p =1/5,q = ≠1/6. This leads
to the disk |Ê ≠ 21/80| < 3/80.)
4 Tail as toy
There are many different ways to play with tails. We start with an example of
simplest kind.
Example 3.
Given a continued fraction. The game is to cut it off and replace the removed tail
by a new one. In this example we take the simplest one of all, namely the 1-periodic
continued fraction. Moreover, we assume that the elements are positive and that
the starting continued fraction is a 1-periodic fraction:
u :=
x
1+
x
1+
x
1+···
=
x +1/4 ≠ 1/2
For the purpose of a more "visible" computation we take
x =
9
64
,
in which case we get
u =1/8.
A tail to be used, to replace the removed tail, c omes in this example from a con-
tinued fraction
y = y(t)=
t
1+
t
1+
t
1+···
=
t +1/4 ≠ 1/2,
also 1-periodic with positive elements. The "result of the game" may then be illus-
trated e.g. as folllows
x
1+
x
1+
x
1+
x
1+
x
1+
t
1+
t
1+
t
1+···
.
More generally
c
1
1+
c
2
1+··· +
c
N
1+
d
1
1+
d
2
1+···
. (9)
As we can see, a tail is replaced by a tail of another continued fraction, like we
do to accelerate convergence, see Example 1. But here the purpose as well as the
follow-up is a different one.
86 Haakon Waadeland Normat 2/2012
5 A trip to the garden.
What we are doing here is in a strange way related to what a gardener does, when
he on a rowanberry tree grafts a branch from an apple tree. With some luck the
rowanberry tree with an apple tree branch will give (on that branch) apples. Some
gardeners play with trees and branches in the same way as we play with continued
fractions. (In (9) the set of numbers c
n
corresponds to the rowanberry tree, whereas
the d-set corresponds to the apple tree .) Back to math!
Here we shall place the new tail in turn: after the second, after the third, and after
the fifth term in the given continued fraction. Keeping the numerics we shall call
the (value of) the new continued fraction T (N,t). For t = x =9/64 the t-fraction
y(t) takes the value 1/8, and T (N,9/64) = 1/8 for all vales of N ..
For t =0we have T (N,0) = S
N
(0), the nth approximant of the given contin-
ued fraction. It is a consequence of the theory of continued fraction with postive
elements thst here S
N
(0) < 1/8 for N =2and S
N
(0) > 1/8 for N =3and for
N =5.
We list the three cases as below, with t =0,witht =9/64 and with t = Œ
(implying y = Œ).
N =2
T (2,t)=
x
1+
x
1+y
,
leading to
T (2,t)=
x(1 + y)
1+y + x
=
9(1 + y)
73 + 64y
,
and hence:
t =0: T =9/73 = 0.123287671
t =9/64 : T =1/8=0.125
t = Π: T =9/64 = 0.140625
N =3
T (3,t)=
x
1+
x
1+
x
1+y
leading to
T (3,t)=
x(1 + x)+xy
1+2x +(1+x)y
=
9
64
(73 + 64y)
(82 + 73y)
and hence:
t =0: T =0.125190548
t =9/64 : T =0.125
t = Π: T =0.140625
Normat 2/2012 Haakon Waadeland 87
N =5
x
1+
x
1+
x
1+
x
1+
x
1+y
,
leading to (after some rearrangements) the following expression
x(1 + 3x + x
2
)+y(x +2x
2
)
(1 + 4x +3x
2
)+y(1 + 3x + x
2
)
.
The fifth approximant of the non-teminating fraction, evaluated at the x-value
given above, is then
x(1 + 3x + x
2
)
1+4x +3x
2
=
9.5905
6643.64
=0.125002352
Hence t =0: T =0.125002352.
Moreover, as always: t =9/64: T =1/8.
Finally y =(t)=Πleads to the value
(x +2x
2
)
(1 + 3x + x
2
).
We substitute x =9/64 and get
T (5, Œ)=
9 · 82
59059
=0.124978831.
Remark. We observe that the variation between the values of the continued frac-
tions created by the game is rather small. This goes back to the way the continued
fractions are chosen. A brief presentation of a quite different situation follows.
Given the continued fraction in Example 1, and let the "interrupting" continued
fraction be as the y(t) from Example 3. With the same notations as before we get
T (2,t)=
12.9
1+
12.81
1+y
,
leading to (since y(0) = 0,y(12) = 3)
T (2, 0) = 0.9341..., T (2, 12) = 3.0606..., T (2, Œ) = 12.9.
T (3,t)=
12.9
1+
12.81
1+
12.729
1+y
T (3.0) = 6.67335‘ T (3.12) = 3.1744440 T (3, Œ)=0.9341
88 Haakon Waadeland Normat 2/2012
6 A contribution from the real world
In the previous example the functions or continued fractions were picked merely to
serve the purpose of being toys in a game with continued fractions. In the example
to come here the function and its continued fraction expansion are picked right out
of the middle of the mathematics everyday toolbox. We have here used a rather
well known (but not very well known) continued fraction expansion of the confluent
hypergeometric funcion being equal to the exponential function
exp(z)=
1
F
1
(1; 1; z)=
+
z
2+
z
≠3+
z
2+
z
≠5+
z
2+
z
≠7+
...
This function (of z) is, as well known, analytic in the whole plane. Our game
with this continued fraction is, however, such that we keep z =1fixed. As long as
nothing else is changed it has the value e. Our game is to replace all numbers 2 by
a variable u, and study the function
g(u):=
1
1+
1
≠1+
1
u +
1
≠3+
1
u +
1
≠5+
1
u +
1
≠7+
...
Here we use an old theorem by Sleszinsky–Pringsheim: The c ontinued fraction
K(a
n
/b
n
) converges if for all n
|b
n
| Ø |a
n
| +1.
In our case all a
n
=1. Hence, if |u| Ø 1 the continued fraction converges. Deeper
results can be proved, but we will not go into that here.
We will be aiming at asymptotic properties , and replace g(u) by w ú g(1/w). For
that purpose we study the function w ú g(1/w) in a neighborhood of the origin
w =0.
A proper discussion of the power se ries shows that approximation number 2
gives
wg(1/w)=1+
2
3
w +
1
9
w
2
+ O(w
3
)
Going back to u again,we find the relation
g(u)=u +
2
3
+
1
9u
+ O(u
≠2
)
Geometric interpretation of the asymptotic investigation:
Theorem 1 The graph of the function g(u) approaches the straight line
g = u +
2
3
when u æŒ.
Approximation number 5 leads to
g = u +
2
3
+
1
9u
≠
2
135u
2
≠
1
405u
3
+
2
1701u
4
+ O
1
1
u
5
2
.
Normat 2/2012 Haakon Waadeland 89
7 Crossing the border. A tail used as a passport
Continued fractions of the form
F
1
z
1+G
1
z +
F
2
z
1+G
2
z +
F
3
z
1+G
3
z + ···
(10)
are called Thron fractions, or simply T-fractions. Here F
k
and G
k
are complex
constants and z a complex variable. The Thron fractions have several interesting
properties, in particular on applications, see [1].
In the present section we shall restrict ourselves to a very special e xample:
z
1 ≠ z +
z
1 ≠ z +
z
1 ≠ z + ···
(11)
Simple recursive computation shows that the nth approximant is
z
!
1 ≠ (≠z)
n
"
1 ≠ (≠z)
n+1
(12)
This shows that in |z| < 1 the continued fraction by convergence defines the
function
f(z):=z, (13)
whereas in |z| > 1 it defines the function
g(z):=≠1. (14)
Take any approximation and modify it with the relevant tail value. Keep in mind
that the two formulas for the approximation, (11) reduced to the first n terms and
12 as it stands, are equal formulas. This is true for g as well as for f.
Modification of f.
In the "(11)-version" of the approximation, add in the last denominator the
proper value of the tail, which is z. Then the whole expression "telescopes" down
to the term z, valid in the whole plane, except at z = Œ. We thus have analytic
continuation of f to the whole plan minus Π.
z =
z
1 ≠ z +
z
1 ≠ z + ··· +
z
1 ≠ z + z
Modification of g. As above, except that the proper tail value now is ≠1.The
telescoping will here lead to the value ≠1, which is now the analytic continuation
of g to the whole Riemann sphere minus the origin z =0.
≠1=
z
1 ≠ z +
z
1 ≠ z + ··· +
z
1 ≠ z ≠ 1
90 Haakon Waadeland Normat 2/2012
References
1. A. Cuyt, W.B.Jones, V.B.Petersen, B.Verdonk and H.Waadeland,Handbook of
Continued Fractions for Special Functions, Springer Science 2008.
2. W.B.Jones and W.J.Thron,Continued Fractions.Analytic Theory and Applica-
tions, Addison-Wesley 1980.
3. L.Lorentzen and H.Waadeland,Continued Fractions with Applications, North-
Holland 1992.
4. L.Lorentzen and H. Waadeland,Continued Fractions Vol. 1: Convergence Theory,
Atlantis Press 2008.
See further O.Perron, Die Lehre von den Kettenbruechen, Teubner 1957.
