Normat 60:2, 91–95 (2012) 91
Reflections on a CF-expansion of 2+2
1/3
Haakon Waadeland
haakonwa@math.ntnu.no
1 Introduction
Continued fractions are essential in the paper. An informal presentation of how
they are written is as follows:
a
1
b
1
+
a
2
b
2
+
a
3
b
3
+
.
.
.
=:
a
1
b
1
+
a
2
b
2
+
a
3
b
3
+ ···
If the a
n
’s , b
n
’s or both are again continued fractions, we get a branched continued
fraction.
The background for the present paper is a branched continued fraction from our
handbook [1], representing the number 2+2
1/3
. In the work with the handbook
the formula (1) below was found on Internet. But later, in the concluding work
on the handbook, the formula was no longer there. Nevertheless, we included it
(although reluctantly). Some weeks after publication of the book, we got a letter
from Domingo Gomez in Venezuela, asking us where we had found the formula.
We told him about Internet, and expressed our apologies for the lack of references.
He then sent a new letter, containing among other things, a reference to [2], which
we should have seen, since we already had that book.
The following result, presented on page 184 in [1] is found in [2], [3]:
The number 2+2
1/3
can be represented by a branched continued fraction: The
equation
C =3+
1
3+
C
3+
C
3+
C
3+···
, (1)
has C =2+2
1/3
as a solution. In (1) we replace on the right hand side repeatedly
C by the same continued fraction, then
2+2
1/3
= C =3+
1
3+
3+
1
3+
C
3+
C
3+···
3+
3+
1
3+
C
3+
C
3+···
3+···
.
92 Haakon Waadeland Normat 2/2012
The first approximants are
C
0
=3,C
1
=3+
1
3
,C
2
:= 3 +
1
3+
C
0
3
,
C
3
:= 3 +
1
3+
C
1
3+
C
0
3
=3+
1
3+C
1
(C
2
≠ 3)
,
and generally
C
n
:= 3 +
1
3+
C
n≠2
3+··· +
C
0
3
=3+
1
3+C
n≠2
(C
n≠1
≠ 3)
.
Numerical values of the first approximants:
C
0
=3,C
1
=3+
1
3
=3.3333 ..., C
2
=3+
1
4
=3.25,C
3
=3+
6
23
=3.2608 ...
C
4
=3.2598870 ...,C
5
=3.25991189 ...,C
6
=3.25992366 ...
Since C =3.25992104989 ... we see that already the approximants of order 4, 5, 6
are very good.
2 Why a =3?
A natural (and vague) question here is: Let a be a positive number. In the formula
C = a +
1
a +
C
a +
C
a +
C
a + ···
(2)
is a =3the only value for which we get such a nice result? In case of YES, then
WHY? An attempt to come up with a possible answer starts with the observation
that the continued fraction (2) converges for all C outside the ray (≠Œ, ≠a
2
/4).In
the following we assume that C is located outside this ray. Actually, we are looking
for positive solutions C. Let S be the value of the continued fraction
S :=
C
a +
C
a +
C
a + ···
. (3)
Then we have
S :=
C
a + S
,
leading to
S := ≠
a
2
+
Ú
C +
a
2
4
,
and hence, from (2)
C := a +
1
a
2
+
Ò
C +
a
2
4
. (4)
Normat 2/2012 Haakon Waadeland 93
Observe that any solution C has to be Ø a. A rearrangement followed by a squaring
gives
!
1
C ≠a
≠
a
2
"
2
= C +
a
2
4
,
and finally
1
(C ≠a)
2
≠
a
C ≠a
≠ C =0.
For C ”= a we get the cubic equation
H := C
3
≠ 2aC
2
+(a + a
2
)C ≠1 ≠ a
2
=0. (5)
The solution is well known already from the time of the renaissance (Cardano and
others). We let MAPLE do the work for us and describe the result as follows: With
P :=
1
36a
2
≠ 8a
3
+ 108 + 12
≠3a
4
+ 54a
2
+ 81
2
1/3
,Q:=
1
1
3
a ≠
1
9
a
2
2
. (6)
the roots are:
x
1
=
P
6
≠
6Q
P
+
2a
3
,x
2
=
≠P
12
+
3Q
P
+
2a
3
+
Ô
3
2
I
1
P
6
+
6Q
P
2
;
x
3
=
≠P
12
+
3Q
P
+
2a
3
≠
Ô
3
2
I
1
P
6
+
6Q
P
2
;
Following Maple we use I instead of i. For a =3we have Q =0, and the roots are
very simple. W e have
P := 6 · 2
(1/3)
,
and hence the roots are
2
(1/3)
+2,e
(2Ifi/3)
· 2
(1/3)
+2,e
(4Ifi/3)
· 2
(1/3)
+2.
Only the first one is of interest to us. This gives an answer to the two questions
raised. The two additional solutions are irrelevant for us. By going to other types
of continued fractions similar questions may be studied.
3 Another example.
For a c ubic equation with real coefficients like the equation (5) there are different
cases to study: one real root and two complex conjugate roots, three real roots,
where in special cases two or three may coincide.
The positive solutions >aof (5) are possible C-values in (2), but these are by
far not as nice as in the case by Domingo Gomez. We are not going into a discussion
of the different cases. We choose as an example the c ase where the factor of I is 0,
i.e. when
P
6
+
6Q
P
=0,
94 Haakon Waadeland Normat 2/2012
or
P
2
+ 36Q =0, (7)
or, in terms of a:
Z :=
1
36a
2
≠ 8a
3
+ 108 + 12
≠3a
4
+ 54a
2
+ 81
2
(2/3)
+ 12a ≠ 4a
2
.
The solutions R of the equation Z =0give the a-values for which x
2
= x
3
.We
find
a = R :=
Ò
9+6
Ô
3, ≠
Ò
9+6
Ô
3,
and stick to the positive value
R =
Ò
9+6
Ô
3=4.403669476 (8)
in the following. We find
x
1
=
P
6
≠
6Q
P
+
2a
3
=1+
Ô
3+
1
1 ≠
Ô
3
3
2
Ò
9+6
Ô
3=4.593260526 (9)
and
x
2
= x
3
=
≠P
12
+
3Q
P
+
2a
3
= ≠
1
2
1
1+
Ô
3
2
+
1
2
1
1+
Ô
3
3
2
Ò
9+6
Ô
3=2.107039213
(10)
The e quation (5) is thus satisfied with a = R from (8) and C = x
1
from (9) or
C = x
2
= x
3
from (10). Only the first one is a solution of (2):
C = a +
1
a +
C
a +
C
a +
C
a + ···
with a =
9+6
Ô
3 and C =1+
Ô
3+
!
1 ≠
Ô
3
3
"
9+6
Ô
3 .
4 A short remark on a different approach
For a fixed positive a we write the equation (5) in the following way:
(C ≠a)
2
+ a =
1+a
2
C
To make notation more familiar we replace C by x and the two expressions by y.
We can then solve the equation graphically in the following way:
y =(x ≠ a)
2
+ a (Parabola)
y =
1+a
2
x
(Hyperbola)
The x-values of the intersection of the two curves are the C-values we are looking
for.
Normat 2/2012 Haakon Waadeland 95
References
1. A.Cuyt, V.Brevik Petersen, B.Verdonk, H.Waadeland, W.B.Jones, Handbook of
Continued Fractions for Special Functions, Springer, 2008.
2. Steven R. Finch, Mathematical Constants, p. 3-4, vol 94 of Encyclopedia of
Mathematics and its Applications, Cambridge University Press, 2003.
3. Domingo Gomez Morin, La Quinta Operacion Aritmetica, ISBN: 980-12-1671-9
