Download Representation of One as the Sum of Unit Fractions 1 Introduction

International Mathematical Forum, Vol. 6, 2011, no. 1, 25 - 30
Representation of One
as the Sum of Unit Fractions
Yuya Dan
Faculty of Business Administration
Matsuyama University
4-2 Bunkyo Matsuyama, Japan 790-8578
[email protected]
Abstract
One is expressed as the sum of the reciprocals of a certain set of
integers. We give an elegant proof to the fact applying the polynomial
theorem and basic calculus.
Mathematics Subject Classification: 11B75, 97I40
Keywords: unit fraction, Diophantine equation, mathematical analysis
1
Introduction
Let us consider the representation of one as the sum of unit fractions. For
examples, we can take 2, 3 and 6 for
1 1 1
+ + = 1,
2 3 6
(1)
1
1 1 1 1
+ + + +
= 1.
3 4 4 8 24
(2)
and 3, 4, 4, 8 and 24 for
It is well known that any positive rational number can be written as the sum
of unit fractions. In the paper, we give a part of solutions to the Diophantine
equation
1
1
1
+
+ ···+
= 1,
x1 x2
xn
where the xj are not necessarily distinct integer for j = 1, 2, . . . , n.
(3)
Yuya Dan
26
Table 1: Possible combinations of α for n = 6
α1
6
4
3
2
2
1
1
0
0
0
0
α2
0
1
0
2
0
1
0
3
1
0
0
α3
0
0
1
0
0
1
0
0
0
2
0
α4
0
0
0
0
1
0
0
0
1
0
0
α5
0
0
0
0
0
0
1
0
0
0
0
α6
0
0
0
0
0
0
0
0
0
0
1
denominator
6! = 720
4! · 2 = 48
3! · 3 = 18
2! · 2! · 22 = 16
2! · 4 = 8
2·3=6
5
3! · 23 = 48
2·4=8
2! · 32 = 18
6
To explain our result for n = 6, we find all possible combinations of
α = (α1 , α2 , α3 , α4 , α5 , α6 ) ∈ N6
(4)
α1 + 2α2 + 3α3 + 4α4 + 5α5 + 6α6 = 6,
(5)
such that
where αj ∈ N = {0, 1, 2, . . . } for j = 1 to 6. Next, we take the quantities
6
αj !j αj = α1 !1α1 · α2 !2α2 · · · · · α6 !6α6
(6)
j=1
for each possible α. Then, we can calculate the sum of reciprocals of the above
quantities
1
1
1
1 1 1
1
1
1
1
1
+
+
+
+ + + +
+ +
+
(7)
720 48 18 16 8 6 5 48 8 18 6
which is equal to 1.
The example is generalized to our main result:
Theorem 1.1. For any positive integer n,
n
1
= 1,
α !j αj
α∈S j=1 j
(8)
n
where the summation over Sn runs through all possible α = (α1 , α2 , . . . , αn ) in
Nn such that
n
jαj = n.
(9)
j=1
Representation of one as the sum of unit fractions
2
27
Preliminaries
Lemma 2.1 (Polynomial theorem). Let n and m be positive integers.
For any x = (x1 , x2 , . . . , xn ) in Rn ,
(x1 + x2 + · · · + xn )m =
m!
xα
α!
(10)
|α|=m
where α! = α1 !α2 ! · · · αn ! for α = (α1 , α2 , . . . , αn ) in Nn and the summation
runs through all possible α in Nn such that |α| = α1 + α2 + · · · + αn = m.
Proof. Each coefficient of
xα =
n
α
xj j
(11)
j=1
in the right-hand side for some α in Nn with |α| = m is equal to the number
of combinations of the products among x1 , x2 , . . . , xn .
Lemma 2.2. Given a polynomial
f (x) =
n
aj xj .
(12)
j=0
Then, the j th coefficient of f (x) can be expressed by
aj =
1 (j)
f (0),
j!
(13)
where f (j) stands for the j th derivative.
Proof. f is infinitely differentiable, since the j-times differential function of xk
is
j
k!
d
xk−j
xk =
(14)
dx
(k − j)!
if j ≤ k, and
d
dx
j
xk = 0
(15)
if j > k. We have
f
(j)
(x) =
n
k=0
ak
d
dx
j
xk =
n
k=j
k!
ak xk−j
(k − j)!
(16)
Yuya Dan
28
for any j between 0 and n, then
f (j) (0) = j!aj ,
(17)
which implies the conclusion of the lemma
aj =
1 (j)
f (0).
j!
(18)
Lemma 2.3. Let n be a positive integer. We put
n
1 n
1 2
,
g(x) = x + x + · · · + x
2
n
(19)
then g (n) (0) = n!.
Proof. Put
n
g(x) = x
1
1
1 + x + · · · + xn−1
2
n
n
= xn h(x),
(20)
where
h(x) =
1
1
1 + x + · · · + xn−1
2
n
n
.
(21)
Leibniz rule implies
g
(n)
(x) =
n
j=0
=
n
j=0
n!
(n − j)!j!
d
dx
n!n!
xj ·
(n − j)!j!j!
n−j
n
x ·
d
dx
j
d
dx
j
h(x)
(22)
h(x).
Therefore, we obtain
g (n) (0) = n!h(0) = n!.
(23)
Representation of one as the sum of unit fractions
3
29
Proof of the main result
We begin with the relation
m
1 2
1 n
x1 + x2 + · · · + xn
=
2
n
α
α
m!
1 2 2
1 n n
α1
x ·
x
x
·····
α! 1
2 2
n n
|α|=m
n
1
jα
= m!
xj j .
α
j
α !j
j=1 j
|α|=m
(24)
by Lemma ??. Putting m = n and xj = t for j = 1, 2, . . . , n implies
n
n
1 jαj
1 n
1 2
= n!
t .
t + t +···+ t
2
n
α !j αj
j=1 j
(25)
|α|=n
Compare to the coefficients of tn in both side of the identity. We obtain n!
from the left-hand side of the identity by Lemma ?? and Lemma ??. On the
other hand, the coefficient of tn in the right-hand side is the sum of all terms
with tn , which is written by
n
α∈Sn
1
,
αj
α
!j
j
j=1
(26)
where the summation runs through all possible α in Sn defined by
Sn = {α ∈ Nn ; α1 + 2α2 + · · · + nαn = n}.
(27)
Hence, we obtain
n
α∈Sn
1
=1
αj
α
!j
j
j=1
(28)
which completes the proof of our main result.
4
Concluding Remarks
In this paper, a part of reciprocal bases of one is investigated from analytic
point of view. In particular, the polynomial theorem and the multi-index
analysis play an important role in the proof. Although these are not all of
solutions to the Diophantine equation, one is presented as the sum of the
reciprocal numbers of a certain set of integers.
Yuya Dan
30
References
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Received: August, 2010