The fair soup division and approximating numbers
aa r X i v : . [ m a t h . C A ] J a n THE FAIR SOUP DIVISION AND APPROXIMATING NUMBERS
ALEX RAVSKY
Dedicated to the personnel of canteen at Karpenko Physico-Mechanical Institute, Lviv, Ukraine
Abstract.
We consider a recent The Vee’s fair soup division problem, provide its partial solution,and pose a related open problem.
Food division problems were always interesting for mathematicians, especially before a dinner.For instance, we can divide cakes [2], [7], [3], [5, Prob. 51], [9], pancakes [1, § § SDP ( q ) (where 0 < q < − ”. We consecutively scoop a full ladleof the soup from the bowl and pour it to one of plates. The soup contains two nutritious stuffs.The first of them is evenly dissolved in the volume; the second is floated to the surface and everyscoop takes a part 1 − q of it, so its amount goes down geometrically with a quotient of q .Let ± N be a set of sequences of signs ”+” and ” − ”. We say that ( ∗ ) i ∈ ± N is an asymptoticallyfair division for SDP ( q ) if an order ( ∗ ) i of plates to pour provides asymptotically equal sharesof both stuffs to them. That is, P ni =1 ∗ i o ( n ) and lim n →∞ P ni =1 ∗ i q i = 0 . Moreover, ( ∗ ) i is a boundedly fair division , if P ni =1 ∗ i O (1).There is the following natural Question 1.
For which numbers q ∈ (0 , , SDP ( q ) has an asymptotically (boundedly) fair divi-sion? Answering it, we consider periodic divisions first. It is easy to show the following
Proposition 2.
Let < q < and ( ∗ ) i ∈ ± N be a sequence with a period N . The followingconditions are equivalent: • ( ∗ ) i is an asymptotically fair division for SDP ( q ) ; • ( ∗ ) i is a boundedly fair division for SDP ( q ) ; • P Ni =1 ∗ i P Ni =1 ∗ i q i = 0 . It is easy to check that the shortest period of a periodical asymptotically fair division for
SDP ( q )is 6 (for instance, for a sequence + − − − + + . . . and q = ϕ − , where ϕ is the golden ratio).The last condition of Proposition 2 suggests the following notion. A balanced ± -polynomial is a polynomial P ( x ) of the form P ni =1 ± x i such that P (1) = 0. Then for each q ∈ (0 , SDP ( q ) has a periodic asymptotically fair division iff q is a root of a balanced ± -polynomial.Now we consider general fair divisions. If 0 < q ≤ / q ≥ P ∞ i =2 q i , so SDP ( q ) has noasymptotically fair division.To obtain a partial answer for Question 1, we need the following Mathematics Subject Classification.
Key words and phrases. recreational mathematics, simultaneous approximations, asymptotic approximation,geometric series, series, summation, polynomial.
Lemma 3.
Let ( a n ) be an absolutely convergent series of real numbers such that for any natural k we have | a k − − a k | ≤ ∞ X n = k +1 | a n − − a n | . (1) Then there exists a sequence ( ∗ ) i ∈ ± N of signs such that ∗ k − ∗ k for each natural k and P ∞ i =1 ∗ i a i = 0 .Proof. For each natural k put b k = | a k − − a k | . It suffices to show that we can consecutivelychoose signs for b k , providing P ∞ k =1 ± b k = 0. At the beginning choose a sign ”+” for b and foreach k > b k a sign ”+”, if P k − i =1 ± b i ≥ − ”, otherwise. It is easy to check that(1) provides P ∞ k =1 ± b k = 0. (cid:3) Proposition 4.
For each q ∈ [1 / √ , , SDP ( q ) has a boundedly fair division.Proof. By Lemma 3, it suffices to check that for any any natural k we have q k − − q k ≤ ∞ X i = k +1 q i − − q i = q k +1 q . (cid:3) To improve Proposition 4, we introduce the following notion. A number q ∈ (1 / ,
1) is approx-imating , if there exist non-negative numbers A and N such that for each x ∈ [0 , A ] there exist abalanced ± -polynomial P of degree n ≤ N such that | x − P ( q ) | ≤ Aq n . Proposition 5.
For each approximating number q , SDP ( q ) has a boundedly fair division.Proof. Suppose that the number q is approximating with the constants A and N . Put k = 0. Thenwe can inductively build an increasing sequence ( k n ) of natural numbers such that k n +1 − k n ≤ N for each n , assigning signs ∗ i to numbers q i for k n − < i ≤ k n (a half of the assigned signs are”+” and the other half are ” − ”) assuring P k n i =1 ± q i ≤ Aq k n . The construction assures that ( ∗ ) i isa boundedly fair division for SDP ( q ). (cid:3) This suggests the following
Question 6.
Which numbers q ∈ (1 / , are approximating? The following proposition provides a partial answer to it. Let q ∞ = 0 . . . . be a uniquepositive root of a polynomial Q ∞ ( x ) = x + x + 2 x − Proposition 7.
Each number q ∈ ( q ∞ , is approximating.Proof. For each natural n put P n ( x ) = x − x n + n − X i =2 ( − x ) i = x − x n + x − x n x . Clearly, P n ( x ) is a balanced ± -polynomial. Given a natural number N , put A = P N ( q )1 − q N . A segment[0 , A ] is covered by a family[0 , P ( q )] , [ P ( q ) , P ( q )] , . . . , [ P N − ( q ) , P N ( q )] , [ P N ( q ) , A ]of segments. So in order to show that for each x ∈ [0 , A ] there exist 1 ≤ n ≤ N such that | x − P n ( q ) | ≤ Aq n it suffices to show that that each point of a segment of the family lies sufficientlyclose to a suitable endpoint of the segment. Namely, that Aq ≥ P ( q ), | P n +1 ( q ) − P n ( q ) | ≤ A ( q n + q n +2 ) for each 1 ≤ n ≤ N −
1, and P N ( q ) + Aq N ≥ A . HE FAIR SOUP DIVISION AND APPROXIMATING NUMBERS 3
The last inequality follows from the definition of A . It is easy to check that for each 1 ≤ n ≤ N − | P n +1 ( q ) − P n ( q ) | q n + q n +2 = 2 − q − q q . Put P ∞ ( x ) = x + x x . When N tends to the infinity, P ∞ ( q ) − P N ( q ) / (1 − q N ) tends to zero.Since we can choose N arbitrarily big, it suffices to show that2 − q − q q < P ∞ ( q ) , that is Q ∞ ( q ) >
0, which holds because q > q ∞ . Finally, the inequality Aq ≥ P ( q ) for allsufficiently big N holds because( q + q ) (cid:18) P ∞ ( q ) − P ( q ) q (cid:19) = − q + 2 q > − q ∞ + 2 q ∞ > . (cid:3) Remark 8.
The approach from the proof of Proposition 7 cannot provide that numbers not biggerthan / √ . . . . are approximating. Indeed, suppose that for a given q ∈ (1 / , thereexist a positive number A and balanced ± -polynomials P , . . . , P n of degrees , . . . , n respectivelysuch that for each x ∈ [0 , A ] there exist ≤ i ≤ n such that | x − P i ( q ) | ≤ Aq i . Then A ≤ A ( q + q + · · · + q n ) < A ( q + q + . . . ) = A q − q , which follows q > / √ . References [1] W.G. Chinn, N.E. Steenrod,
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Pidstryhach Institute for Applied Problems of Mechanics and Mathematics National Academyof Sciences of Ukraine
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