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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 [1] L. Brenton and R. Bruner, ”On Recursive Solutions of a Unit Fraction Equation,” J. Austral. Math. Soc. Ser. A 57, No. 3, pp.341–356 (1994). [2] L. Brenton and R. Hill, ”On the Diophantine Equation 1 = Σ1/ni + 1/Πni and a Class of Homologically Trivial Complex Surface Singularities,” Paciﬁc J. Math. Vol. 133, No. 1, pp.41–67 (1988). [3] N. Burshtein, ”On Distinct Unit Fractions Whose Sum Equals 1,” Discrete Mathematics, Vol. 300, No. 1–3, pp.213–217 (2005). [4] P. Erdös and R. Graham, ”Unit Fractions,” Old and New Problems and Results in Combinatorial Number Theory, Monographie No. 28, L’Enseign. Math. Univ. de Genevé, pp.30–44 (1980). [5] P. Erdös and S. Stein, ”Sums of Distinct Unit Fractions,” Proc. Amer. Math. Soc. 14, pp.126–131 (1963). [6] R. K. Guy, Unsolved Problems in Number Theory, Springer-Verlag, New York (1981). [7] G. H. Hardy and E. M. Wright, An Inrtoduction to the Theory of Numbers, Oxford University Press (2008). [8] T. Salat and J. Tomanova, ”On the Class of all Reciprocal Bases for Integers,” Acta Mathematica Universitatis Comenianae, Vol. LXXVI, No. 2, pp.257–261 (2007). Received: August, 2010