INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 7 (2007), #A30

A PROOF OF THE CONTINUED FRACTION EXPANSION OF e^{2/s}

Takao Komatsu^{1}

Graduate School of Science and Technology, Hirosaki University, Hirosaki, 036-8561, Japan

                        komatsu@cc.hirosaki-u.ac.jp

Received: 3/13/07, Accepted: 6/6/07, Published: 6/19/07

Abstract

We give a proof of the continued fraction expansion of e^{2/s}, where s ≥ 3 is an odd integer, by expressing the error between e^{2/s} and its each convergent explicitly in terms of integrals.

1. Introduction

Let α = [a_{0}; a_{1}, a_{2}, …] denote the simple continued fraction expansion of a real α, where

α = a_{0} + 1/α_{1},  a_{0} = ⌊α⌋ ,
α_{n} = a_{n} + 1/α_{n+1},  a_{n} = ⌊α_{n}⌋ (n ≥ 1) .

The n-th convergent of the continued fraction expansion is denoted by p_{n}/q_{n} = [a_{0}; a_{1}, …, a_{n}], and p_{n} and q_{n} satisfy the recurrence relation:

p_{n} = a_{n}p_{n−1} + p_{n−2} (n ≥ 0),  p_{−1} = 1,  p_{−2} = 0,
q_{n} = a_{n}q_{n−1} + q_{n−2} (n ≥ 0),  q_{−1} = 0,  q_{−2} = 1 .

An irrational number α is well approximated by its n-th convergent p_{n}/q_{n}. In fact, for n ≥ 0

\frac{1}{q_{n}(q_{n+1} + q_{n})} < |α − \frac{p_{n}}{q_{n}} | < \frac{1}{q_{n}q_{n+1}}

([4, p. 20]). Precisely speaking, by using the algorithm mentioned above, we can express ([1, Lemma 5.4]) the error as

α − \frac{p_{n}}{q_{n}} = \frac{(−1)^{n}}{q_{n}(α_{n+1}q_{n} + q_{n−1})}.

1This research was supported in part by the Grant-in-Aid for Scientific research (C) (No. 18540006), the Japan Society for the Promotion of Science.

Osler [5] gave a remarkable proof of the simple continued fraction

e^{1/s} = [1; \overline{(2k − 1)s − 1, 1, 1}]_{k=1}^{∞} (s ≥ 2)

by expressing this error explicitly in terms of integrals. Namely, when p_{n}/q_{n} is the n-th convergent of the continued fraction of e^{1/s}, he showed that for n ≥ 0

\frac{p_{3n}}{q_{3n}} − e^{1/s} = −\frac{1}{q_{3n}} ∫_{0}^{1} \frac{x^{n}(x − 1)^{n}}{s^{n+1}n!} e^{x/s}dx ,
\frac{p_{3n+1}}{q_{3n+1}} − e^{1/s} = \frac{1}{q_{3n+1}} ∫_{0}^{1} \frac{x^{n+1}(x − 1)^{n}}{s^{n+1}n!} e^{x/s}dx

and

\frac{p_{3n+2}}{q_{3n+2}} − e^{1/s} = \frac{1}{q_{3n+2}} ∫_{0}^{1} \frac{x^{n}(x − 1)^{n+1}}{s^{n+1}n!} e^{x/s}dx .

This was the direct extension of the result given by Cohn [2] concerning e. A similar expression can be seen in [3] too.

It is known that the continued fraction expansion of e^{2/s} is given by

e^{2/s} = [1; \overline{\frac{(6k − 5)s − 1}{2}, (12k − 6)s, \frac{(6k − 1)s − 1}{2}, 1, 1} ]_{k=1}^{∞} ,

where s > 1 is odd (See [6], §32, (2)). We shall give another proof of the continued fraction expansion of e^{2/s} by showing similar errors explicitly. For convenience, for n ≥ 0 put

A_{n} = (\frac{2}{s})^{3n+1} ∫_{0}^{1} \frac{x^{3n}(x − 1)^{3n}}{(3n)!} e^{2x/s}dx ,
B_{n} = \frac{2^{3n+1}}{s^{3n+2}} ∫_{0}^{1} \frac{x^{3n+1}(x − 1)^{3n+1}}{(3n + 1)!} e^{2x/s}dx ,
C_{n} = (\frac{2}{s})^{3n+3} ∫_{0}^{1} \frac{x^{3n+2}(x − 1)^{3n+2}}{(3n + 2)!} e^{2x/s}dx ,
D_{n} = (\frac{2}{s})^{3n+3} ∫_{0}^{1} \frac{x^{3n+3}(x − 1)^{3n+2}}{(3n + 2)!} e^{2x/s}dx ,

and

E_{n} = (\frac{2}{s})^{3n+3} ∫_{0}^{1} \frac{x^{3n+2}(x − 1)^{3n+3}}{(3n + 2)!} e^{2x/s}dx .

Then, our main theorem is stated as follows.

INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 7 (2007), #A30 3

Theorem 1.1. Let p_{n}/q_{n} be the n-th convergent of the continued fraction of e^{2/s}. Then, for n ≥ 0

\frac{p_{5n}}{q_{5n}} − e^{2/s} = −\frac{1}{q_{5n}} A_{n} ,
\frac{p_{5n+1}}{q_{5n+1}} − e^{2/s} = −\frac{1}{q_{5n+1}} B_{n} ,
\frac{p_{5n+2}}{q_{5n+2}} − e^{2/s} = −\frac{1}{q_{5n+2}} C_{n} ,
\frac{p_{5n+3}}{q_{5n+3}} − e^{2/s} = \frac{1}{q_{5n+3}} D_{n} ,

and

\frac{p_{5n+4}}{q_{5n+4}} − e^{2/s} = \frac{1}{q_{5n+4}} E_{n} .

2. The Continued Fraction of e^{2/s} The proof of the main theorem is based upon the following identities. Lemma 2.1.

A_{n} = −E_{n−1} − D_{n−1} ,  (2.1)
B_{n} = \frac{(6n + 1)s − 1}{2}A_{n} − E_{n−1} ,  (2.2)
C_{n} = (12n + 6)sB_{n} + A_{n} ,  (2.3)
D_{n} = −\frac{(6n + 5)s − 1}{2}C_{n} − B_{n} ,  (2.4)

and

E_{n} = D_{n} − C_{n} .  (2.5)

These identities correspond with the desired relations:

p_{5n} = p_{5n−1} + p_{5n−2},  q_{5n} = q_{5n−1} + q_{5n−2},
p_{5n+1} = \frac{(6n + 1)s − 1}{2}p_{5n} + p_{5n−1},  q_{5n+1} = \frac{(6n + 1)s − 1}{2}q_{5n} + q_{5n−1},
p_{5n+2} = (12n + 6)sp_{5n+1} + p_{5n},  q_{5n+2} = (12n + 6)sq_{5n+1} + q_{5n},
p_{5n+3} = \frac{(6n + 5)s − 1}{2}p_{5n+2} + p_{5n+1},  q_{5n+3} = \frac{(6n + 5)s − 1}{2}q_{5n+2} + q_{5n+1},
p_{5n+4} = p_{5n+3} + p_{5n+2},  q_{5n+4} = q_{5n+3} + q_{5n+2}.

INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 7 (2007), #A30 4

We shall prove Theorem 1.1 and Lemma 2.1 simultaneously. Proof. When n = 0, the relations in Theorem 1.1 are true. Indeed,

A_{0} = \frac{2}{s} ∫_{0}^{1} e^{2x/s}dx = e^{2/s} − 1 = q_{0}e^{2/s} − p_{0} .

B_{0} = \frac{2}{s^{2}} ∫_{0}^{1} x(x − 1)e^{2x/s}dx = \frac{1}{2s}[(2x^{2} − 2(s + 1)x + s^{2} + s)e^{2x/s}]_{0}^{1}

= \frac{s − 1}{2}e^{2/s} − \frac{s + 1}{2} = q_{1}e^{2/s} − p_{1} .

C_{0} = (\frac{2}{s})^{3} \frac{1}{2} ∫_{0}^{1} x^{2}(x − 1)^{2}e^{2x/s}dx

= \frac{1}{s^{2}} [(2x^{4} − 4(s + 1)x^{3} + 2(3s^{2} + 3s + 1)x^{2}

− 2s(3s^{2} + 3s + 1)x + 3s^{4} + 3s^{3} + s^{2})e^{2x/s}]_{0}^{1} = (3s^{2} − 3s + 1)e^{2/s} − (3s^{2} + 3s + 1) = q_{2}e^{2/s} − p_{2} .

D_{0} = (\frac{2}{s})^{3} \frac{1}{2} ∫_{0}^{1} x^{3}(x − 1)^{2}e^{2x/s}dx

= −\frac{1}{2s^{2}} [(4x^{5} − (10s + 8)x^{4} + (20s^{2} + 16s + 4)x^{3} − (30s^{3} + 24s^{2} + 6s)x^{2}

+ (30s^{4} + 24s^{3} + 6s^{2})x − (15s^{5} + 12s^{4} + 3s^{3}))e^{2x/s}]_{0}^{1}

= \frac{15s^{3}}{2} + 6s^{2} + \frac{3s}{2} − (\frac{15s^{3}}{2} − 9s^{2} + \frac{9s}{2} − 1) e^{2/s} = p_{3} − q_{3}e^{2/s} .

E_{0} = (\frac{2}{s})^{3} \frac{1}{2} ∫_{0}^{1} x^{2}(x − 1)^{3}e^{2x/s}dx

= −\frac{1}{2s^{2}} [(4x^{5} − (10s + 12)x^{4} + (20s^{2} + 24s + 12)x^{3} − (30s^{3} + 36s^{2} + 18s + 4)x^{2}

+ (30s^{4} + 36s^{3} + 18s^{2} + 4s)x − (15s^{5} + 18s^{4} + 9s^{3} + 2s^{2}))e^{2x/s}]_{0}^{1}

= \frac{15s^{3}}{2} + 9s^{2} + \frac{9s}{2} + 1 − (\frac{15s^{3}}{2} − 6s^{2} + \frac{3s}{2})e^{2/s} = p_{4} − q_{4}e^{2/s} .

Suppose that Theorem 1.1 is true up to some integer n − 1(≥ 0). Since

\frac{d}{dx}(x^{3n}(x − 1)^{3n}e^{2x/s}) = 3nx^{3n−1}(x − 1)^{3n}e^{2x/s} + 3nx^{3n}(x − 1)^{3n−1} + \frac{2}{s}x^{3n}(x − 1)^{3n}e^{2x/s} ,

by integrating from 0 to 1 we get (2.1). Hence,

p_{5n} − q_{5n}e^{2/s} = (p_{5n−1} + p_{5n−2}) − (q_{5n−1} + q_{5n−2})e^{2/s}
= E_{n−1} + D_{n−1} = −A_{n} .

INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 7 (2007), #A30 5

Since

\frac{d}{dx}(x^{3n}(x − 1)^{3n+1}e^{2x/s})

= 3nx^{3n−1}(x − 1)^{3n+1}e^{2x/s} + (3n + 1)x^{3n}(x − 1)^{3n}e^{2x/s} + \frac{2}{s}x^{3n}(x − 1)^{3n+1}e^{2x/s}

= −3nx^{3n−1}(x − 1)^{3n}e^{2x/s} + (6n + 1 − \frac{2}{s})x^{3n}(x − 1)^{3n}e^{2x/s} + \frac{2}{s}x^{3n+1}(x − 1)^{3n}e^{2x/s}

and

\frac{d}{dx}(x^{3n+1}(x − 1)^{3n+1}e^{2x/s})

= (3n + 1)x^{3n}(x − 1)^{3n+1}e^{2x/s} + (3n + 1)x^{3n+1}(x − 1)^{3n}e^{2x/s} + \frac{2}{s}x^{3n+1}(x − 1)^{3n+1}e^{2x/s}

= 2(3n + 1)x^{3n+1}(x − 1)^{3n}e^{2x/s} − (3n + 1)x^{3n}(x − 1)^{3n}e^{2x/s} + \frac{2}{s}x^{3n+1}(x − 1)^{3n+1}e^{2x/s} ,

by integrating from 0 to 1 and canceling the term of x^{3n+1}(x − 1)^{3n}e^{2x/s} we get

\frac{s}{2}(−3n) ∫_{0}^{1} x^{3n−1}(x − 1)^{3n}e^{2x/s}dx + \frac{(6n + 1)s − 1}{2} ∫_{0}^{1} x^{3n}(x − 1)^{3n}e^{2x/s}dx

− \frac{1}{s(3n + 1)} ∫_{0}^{1} x^{3n+1}(x − 1)^{3n+1}e^{2x/s}dx = 0 .

Thus, we have (2.2). Hence,

p_{5n+1} − q_{5n+1}e^{2/s} = (\frac{(6n + 1)s − 1}{2}p_{5n} + p_{5n−1}) − (\frac{(6n + 1)s − 1}{2}q_{5n} + q_{5n−1}) e^{2/s}

= −\frac{(6n + 1)s − 1}{2}A_{n} + E_{n−1} = −B_{n} .

Notice that

\frac{d}{dx}(x^{3n+2}(x − 1)^{3n+2}e^{2x/s})

= (3n + 2)x^{3n+1}(x − 1)^{3n+2}e^{2x/s} + (3n + 2)x^{3n+2}(x − 1)^{3n+1}e^{2x/s} + \frac{2}{s}x^{3n+2}(x − 1)^{3n+2}e^{2x/s}

= (6n + 4)x^{3n+2}(x − 1)^{3n+1}e^{2x/s} − (3n + 2)x^{3n+1}(x − 1)^{3n+1}e^{2x/s} + \frac{2}{s}x^{3n+2}(x − 1)^{3n+2}e^{2x/s} ,

\frac{d}{dx}(x^{3n+2}(x − 1)^{3n+1}e^{2x/s})

= (3n + 2)x^{3n+1}(x − 1)^{3n+1}e^{2x/s} + (3n + 1)x^{3n+2}(x − 1)^{3n}e^{2x/s} + \frac{2}{s}x^{3n+2}(x − 1)^{3n+1}e^{2x/s}

= (6n + 3)x^{3n+1}(x − 1)^{3n+1}e^{2x/s} + (3n + 1)x^{3n+1}(x − 1)^{3n}e^{2x/s} + \frac{2}{s}x^{3n+2}(x − 1)^{3n+1}e^{2x/s}

INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 7 (2007), #A30 6

and

\frac{d}{dx}(x^{3n+1}(x − 1)^{3n+1}e^{2x/s})

= 2(3n + 1)x^{3n+1}(x − 1)^{3n}e^{2x/s} − (3n + 1)x^{3n}(x − 1)^{3n}e^{2x/s} + \frac{2}{s}x^{3n+1}(x − 1)^{3n+1}e^{2x/s} .

Integrating these three equations from 0 to 1 and canceling the terms of x^{3n+2}(x − 1)^{3n+1}e^{2x/s} and x^{3n+1}(x − 1)^{3n}e^{2x/s}, we get

(\frac{2}{s})^{2} ∫_{0}^{1} x^{3n+2}(x − 1)^{3n+2}e^{2x/s}dx

= (12n + 6)(3n + 2) ∫_{0}^{1} x^{3n+1}(x − 1)^{3n+1}e^{2x/s}dx + (3n + 2)(3n + 1) ∫_{0}^{1} x^{3n}(x − 1)^{3n}e^{2x/s}dx . Thus, we have (2.3). Hence,

p_{5n+2} − q_{5n+2}e^{2/s} = ((12n + 6)sp_{5n+1} + p_{5n}) − ((12n + 6)sq_{5n+1} + q_{5n})e^{2/s}
= −(12n + 6)sB_{n} − A_{n} = −C_{n} .

Notice that

\frac{d}{dx}(x^{3n+3}(x − 1)^{3n+2}e^{2x/s})

= (3n + 3)x^{3n+2}(x − 1)^{3n+2}e^{2x/s} + (3n + 2)x^{3n+3}(x − 1)^{3n+1}e^{2x/s} + \frac{2}{s}x^{3n+3}(x − 1)^{3n+2}e^{2x/s} ,

\frac{d}{dx}(x^{3n+3}(x − 1)^{3n+1}e^{2x/s})

= (3n + 3)x^{3n+2}(x − 1)^{3n+1}e^{2x/s} + (3n + 1)x^{3n+3}(x − 1)^{3n}e^{2x/s} + \frac{2}{s}x^{3n+3}(x − 1)^{3n+1}e^{2x/s}

= (6n + 4 + \frac{2}{s}) x^{3n+2}(x − 1)^{3n+1}e^{2x/s} + (3n + 1)x^{3n+2}(x − 1)^{3n}e^{2x/s} + \frac{2}{s}x^{3n+2}(x − 1)^{3n+2}e^{2x/s} ,

\frac{d}{dx}(x^{3n+2}(x − 1)^{3n+2}e^{2x/s})

= (6n + 4 − \frac{2}{s}) x^{3n+2}(x − 1)^{3n+1}e^{2x/s} − (3n + 2)x^{3n+1}(x − 1)^{3n+1}e^{2x/s} + \frac{2}{s}x^{3n+3}(x − 1)^{3n+1}e^{2x/s}

and

\frac{d}{dx}(x^{3n+2}(x − 1)^{3n+1}e^{2x/s})

= (3n + 2)x^{3n+1}(x − 1)^{3n+1}e^{2x/s} + (3n + 1)x^{3n+2}(x − 1)^{3n}e^{2x/s} + \frac{2}{s}x^{3n+2}(x − 1)^{3n+1}e^{2x/s} .

Integrating these four equations from 0 to 1 and eliminating the terms of x^{3n+3}(x − 1)^{3n+1}e^{2x/s}, x^{3n+2}(x − 1)^{3n+1}e^{2x/s} and x^{3n+2}(x − 1)^{3n}e^{2x/s}, we get

\frac{4}{s} ∫_{0}^{1} x^{3n+3}(x − 1)^{3n+2}e^{2x/s}dx

= − (12n + 10 − \frac{2}{s}) ∫_{0}^{1} x^{3n+2}(x − 1)^{3n+2}e^{2x/s}dx − (3n + 2) ∫_{0}^{1} x^{3n+1}(x − 1)^{3n+1}e^{2x/s}dx .

Thus, we have (2.4). Hence,

p_{5n+3} − q_{5n+3}e^{2/s} = (\frac{(6n + 5)s − 1}{2}p_{5n+2} + p_{5n+1}) − (\frac{(6n + 5)s − 1}{2}q_{5n+2} + q_{5n+1}) e^{2/s}

= −\frac{(6n + 5)s − 1}{2}C_{n} − B_{n} = D_{n} .

Since x^{3n+3}(x − 1)^{3n+2} − x^{3n+2}(x − 1)^{3n+2} = x^{3n+2}(x − 1)^{3n+3}, we get D_{n} − C_{n} = E_{n}, which is (2.5). Hence,

p_{5n+4} − q_{5n+4}e^{2/s} = (p_{5n+3} + p_{5n+2}) − (q_{5n+3} + q_{5n+2})e^{2/s} = D_{n} − C_{n} = E_{n} .

□

3. The Continued Fraction of e^{2}

Let \frac{p_{n}^{∗}}{q_{n}^{∗}} be the n^{th} convergent of the continued fraction of e^{2} = [7; \overline{3k − 1, 1, 1, 3k, 12k + 6} ]_{k=1}^{∞}. Then, for n ≥ 0 we have

\frac{p_{n}^{∗}}{q_{n}^{∗}} = \frac{p_{n+2}}{q_{n+2}} ,

where p_{n}/q_{n} is the n-th convergent of the continued fraction of e^{2/s} mentioned above. Thus, by replacing p_{n}/q_{n} by p_{n−2}^{∗}/q_{n−2}^{∗} (n ≥ 2), Theorem 1.1 with Lemma 2.1 holds for s = 1.

4. Additional Comments

Some results in our theorem can be derived directly from Osler’s results. By the relation for i = 1, 2, …

[ …, 1, a_{i} − \frac{1}{2}, 1, …] = [ …, 1, a_{3i−2}, 4a_{3i−1} + 2, a_{3i}, 1, …] ,

if we replace s by s/2 in the continued fraction [1; \overline{(2k − 1)s − 1, 1, 1}]_{k=1}^{∞}, then we have

[1; \overline{(2k − 1)s/2 − 1, 1, 1}]_{k=1}^{∞} = [1; \overline{ks − \frac{s + 1}{2} − \frac{1}{2}, 1, 1}]_{k=1}^{∞}

= [1; \overline{\frac{(6k − 5)s − 1}{2}, (12k − 6)s, \frac{(6k − 1)s − 1}{2}, 1, 1} ]_{k=1}^{∞} .

Therefore, if we replace s by s/2 and n by 3n in the Osler’s integral for p_{3n} − q_{3n}e^{1/s}, we get our integral for p_{5n} −q_{5n}e^{2/s}. If we replace s by s/2 and n by 3n+2 in the Osler’s integral for p_{3n+1} − q_{3n+1}e^{1/s}, we get our integral for p_{5n+3} − q_{5n+3}e^{2/s}. If we replace s by s/2 and n by 3n + 2 in the Osler’s integral for p_{3n+2} − q_{3n+2}e^{1/s}, we get our integral for p_{5n+4} − q_{5n+4}e^{2/s}.

5. Acknowledgement

This work was initiated when the author stayed in the Department of Mathematics and Statistics, University of Tennessee at Martin in 2006. In special, he thanks Sarah Holliday, Bill Austin, Louis Kolitsch, and Chris Caldwell. The author also thanks the anonymous referee for some helpful comments.

References

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[3] C. S. Davis, On some simple continued fractions connected with e, J. London Math. Soc, 20 (1945), 194–198.

[4] A. Ya. Khinchin, Continued fractions, Dover, New York, 1997.

[5] T. Osler, A proof of the continued fraction expansion of e^{1/M}, Amer. Math. Monthly 113 (2006), 62–66.

[6] O. Perron, Die Lehre von den Kettenbrüchen, Chelsea Publishing Company, New York, 1950.