A note on a scaling limit of successive approximation for a differential equation with moving singularity.

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1 A note on a scaling limit of successive approximation for a differential equation with moving singularity. Tetsuya Hattori and Hiroyuki Ochiai ABSTRACT We study a method of successive approximation to dφ dx (x) = φ(x)2, a simplest first order nonlinear ordinary differential equation whose solutions have moving singularities. We give a sufficient condition for a series of approximate solutions φ n, n =,, 2,, to have a scaling limit, namely, φ(x) = lim φ n (q x) exists, where q n = φ n (). In other words, the approximation sequence q n n φ n (x), n =,, 2,, approaches the exact solution x in an asymptotically conformal way: φ n (x) q n φ(q n x)as. The crucial condition for the scaling limit to exist is the existence of the limit lim q n+/q n. For a certain choice of φ, the problem is related to a problem of random sequential bisection, through which we find lim q/n n = e /c, where x = c is the unique positive solution to x log 2e x =, x>. Numerical calculations suggest that all the conditions for the scaling limit to exist are satisfied for this choice of φ. Introduction. Consider a simplest first order non-linear ordinary differential equation dφ (.) dx (x) = φ(x)2,x>. A solution φ(x) =(x c) has a singularity, whearas (.) has no singularities. The singularity point x = c is an aribtrary constant, hence it is called a moving singularity. We may, without loss of generality put c =, to fix the arbitrary constant c. Equivalently, we may impose boundary condition at infinity: (.2) φ(x) =x + o(x 2 ),x, to obtain a unique solution φ(x) =x in C ((, )). Applying the method of successive approximation, we obtain a sequence φ n, n =,, 2,,of functions defined recursively, by dφ n+ dx (x) = φ n(x) 2,x>, φ n (x) =x + o(x 2 ),x, n =,, 2,, Graduate School of Mathematics, Nagoya University, Chikusa-ku, Nagoya , Japan. hattori@math.nagoya-u.ac.jp Department of Mathematics, Kyushu University, Hakozaki, Fukuoka , Japan. ochiai@math.kyushu-u.ac.jp

2 or, equivalently, (.3) φ n+ (x) = with an initial approximation x φ n (x ) 2 dx,x>, n =,, 2,, (.4) φ (x) =x + o(x 2 ),x. Successive approximation (.3) gives a sequence of functions converging to a solution φ of the differential equation. We are interested in the rate of convergence of successive approximation to a solution near the moving singularity. We go into this problem by studying the scaling limit, namely, the existence problem of φ(z) = lim q n φ n(qn z), for a sequence of positive numbers q n, n =,, 2,, which diverges to as n. (The scaling factor for x and the scaling factor for φ n should be equal, because φ n (x) x for very large n and x somewhat near.) Possibility of existence of scaling limits for successive approximations to differential equations with moving singularities seems not to have been studied. To be specific, let us denote by C a set of entire functions φ : C C, satisfying φ(x) = x + o(x 2 ), x +, whose coefficients in the Maclaurin series have alternating signs; φ(z) = ( ) k a k z k,a k,,, 2,. In Section 2 we prove the following. Theorem.. Let φ C and let φ n, n =,, 2,, be a sequence defined recursively by (.3) on [, ). Then for each n, φ n are analytically continued to C and φ n C holds. Furthermore, let the coefficients a,k, k =,, 2,,in φ (z) = ( ) k a,k z k, satisfy a,k a k+,, k =, 2, 3,. Then if the sequence q n, n =,, 2,, defined by (.5) q n = φ n (), n =,, 2,. is increasing in n, and the sequence of the ratios of successive terms has a limit greater than ; q n+ (.6) ρ = lim >, q n then the sequence of the entire functions defined by (.7) φ n (z) =q n φ n(q z), n =,, 2,, n converges uniformly on compact sets in C to an entire function φ(z) = ( ) k α k z k defined by (.8) α =, α k = kρ k+ k α k j α j,k =, 2, 3,. j= 2

3 This Theorem gives a sufficient condition for the sequence of successive approximations to exhibit a scaling limit. The conditions on the coefficients in the definition of the class C and in Theorem. may be stronger than necessary. However, as we will see below, we have examples of interest satisfying these conditions, so we impose them to avoid technical complications. The condition (.6) on q n, on the other hand, seems more essential and deserve further study. In fact, existence of the limit in (.6) is still open, but we have some results for the value of ρ, assuming its existence. For a> let O a be a set of non-negative valued functions φ : [, ) [, ) defined on non-negative reals which has an expression (.9) φ(x) = e xt ( F (t)) dt, x, where F : [, ) [, ] is an increasing right continuous function satisfying F (x) Cx a, x>, for some positive constant C. (In particular, we impose φ() <, which implies lim F (x) =.) x In Section 3 we prove the following. Theorem.2. (i) Let a> and φ O a, and let φ n, n =,, 2,, be a sequence defined recursively by (.3) on [, ). If <l<l s (a), then lim l n φ n (l n x)=x, x>, and lim l n φ n () = hold. Here (.) l s (a) = { ((a +)/2) /a, <a<c, e /c, a c, and c is the unique solution to c log 2e =with c>. c (ii) Let <a<c and φ O a, and assume that there exist positive constants C, a, δ, satsfying a a, such that the function F in (.9) correpsonding to φ = φ satisfies F (x) C x a for x δ, then for l>l s (a ), and for each x, lim l n φ n (l n x)=. Numerically, c = and e /c = Theorem. and Theorem.2 imply the following result, which describes the meaning of Theorem.2 in terms of Theorem.. Corollary.3. Assume that φ : C C is in C and its restriction on [, ) is in O a for some a>. Letφ n, n =,, 2,, be a sequence defined recursively by (.3) on C. Then if the assumptions in Theorem. hold, then ρ l s (a). If, in addition, <a<c and there exist positive constants C and δ such that the function F in (.9) correpsonding to φ = φ satisfies F (x) C x a for x δ, then ρ = l s (a). Proof of Corollary.3 assuming Theorem. and Theorem.2. The assumption (.6) implies lim φ n() /n = lim q/n n = ρ. Theorem.2 for x = therefore implies ρ l, if<l<l s (a), hence ρ l s (a). The second part is proved similarly using the second part of Theorem.2. In Section 3 we also give a sufficient condition for the increasing property of q n, n =,, 2, (Proposition 3.3). 3

4 Corollary.3 essentially determines the value of ρ (assuming its existence) in (.6) for the case <a<c, The case a c seems harder. However, the following considerations on a specific choice of φ suggest that ρ = e /c for large a. Put (.)φ (z) = ( exp( z)), z C. z Note that this choice of φ is both in C and in O a for any a>, because z ( exp( z)) = ( ) k (k + )! zk, z C, and x ( exp( x)) = e xt dt, x >. In fact, with this choice, all the assumptions in Theorem., except perhaps (.6) are easily seen to hold. Though we were unable to find a proof for (.6), we have the following result. Theorem.4. Let φ n, n =,, 2,, be a sequence defined recursively by (.3) on [, ), with φ as in (.). Then q n is increasing in n, and lim q/n n e /c, where c is as in Theorem.2. Moreover, if (.6) holds, then lim q/n n = e /c, in particular, ρ = e /c. The specific choice (.) in this Theorem is motivated by studies in random sequential bisection of a rod and binary search trees [3, 7]. We give a proof of Theorem.4 in Section 4. Numerical calculations suggest that q n+ /q n is decreasing in n, hence (.6) does hold. We give details in Appendix. Thus we conclude with the following Conjecture. Conjecture.5. Let φ n, n =,, 2,, be a sequence defined recursively by (.3) on C, with φ as in (.). Then lim φ n() φ n (φ n () z)= φ(z) uniformly on compact sets in C, where φ(z) = ( ) k α k z k is given by (.8). Acknowledgement. The authors would like to thank Prof. K. Uchiyama for comments and discussions, especially for a proof of Lemma 3.5 and Lemma 4.3. T. Hattori would also like to thank Prof. Y. Itoh for describing his results, which partly motivated the present work. The research of T. Hattori is supported by Grant-in-Aid for Scientific Research (C) of the Ministry of Education, Science, Sports and Culture, and the research of H. Ochiai is supported by Grant-in-Aid for Encouragement of Young Scientists of the Ministry of Education, Science, Sports and Culture. 2 Scaling limit. Here we prove Theorem.. First we prove by induction that φ n Cfor all n =,, 2,. Assume that φ n Cfor a non-negative integer n. φ n+ (x) =x + o(x 2 ), x, follows directly from (.3) and φ n (x) =x + o(x 2 ). Also φ n (x) 2 dx < exists, and we can rewrite (.3) as (2.) φ n+ (x) = φ n (x ) 2 dx x φ n (x ) 2 dx, x. 4

5 The integrand in the last term is entire, hence we can analytically continue φ n+ to the whole complex plain C as an entire function, using this expression. Put φ n (z) = ( ) k a n,k z k, z C, n=,, 2,. Inserting this in (.3) we find (2.2) and (2.3) a n+, = φ n+ () = q n+ = φ n (x) 2 dx (> ), n =,, 2,, a n+,k = k a n,k j a n,j ( ), k=, 2, 3,,n=,, 2,. k j= This proves that φ n+ C. By induction, φ n Cfor all n. Let us introduce a notation which we use throughout this section. For r>, let M r be a map on a space of infinite sequences M r : a = {a k k =,, 2, } M r (a) ={M r (a) k k =,, 2, } defined by M r (a) = and (2.4) M r (a) k = kr k+ k a k j a j, k =, 2, 3,. j= For sequences a = {a k k =,, 2, } and b = {b k k =,, 2, } we write a b if a k b k, k =,, 2,. Obviously we have, for a non-negative sequence a, (2.5) M r (a) M r (a), if <r r. Define α(r) ={α(r) k k =,, 2, } by (2.6) M r (α(r)) = α(r). Then α = {α k k =,, 2, } in (.8) is α = α(ρ). Next we assume all the assumptions in Theorem.. In particular, we assume a n, = q n q n+ = a n+,. Then, by induction in n using (2.2) we see that (2.7) a n,k a k+ n,,,, 2,, n =,, 2,. In fact, we assumed this for n = in Theorem.. Assume that (2.7) holds for some n. Then using (2.3) a n+,k a k+ n, ak+ n+,, k =, 2, 3,. Hence (2.7) holds also for n +. Put φ n (z) = ( ) k ᾱ n,k z k, z C, n=,, 2,. Then (.7) implies (2.8) ᾱ n,k = q k n a n,k, k =,, 2,, 5

6 which, with (2.3) and (2.4), satisfies (2.9) ᾱ n+ = M qn+ /q n (ᾱ n ),n=,, 2,, where we put ᾱ n = {ᾱ n,k k =,, 2, }. We now prove the following (2.) and (2.): { φ n n =,, 2, } is a set of uniformly bounded and equicontinuous functions (2.) on {z C z /2}. (2.) lim ᾱn,k = α k,,, 2,, Applying a standard argument using Ascoli Arzelà Theorem, we see that (2.) and (2.) imply that φ n (z) converges uniformly to φ(z) on z /2. Note that (.3) implies (2.2) φ n+ (z) = q n q n+ q nz/q n+ φn (z ) 2 dz, z C, n =,, 2,. This with (.6) then imply that φ n (z) actually converges uniformly to φ(z) on any compact sets as n, hence φ is entire, and our proof will be complete. We are left with proving (2.) and (2.). To prove (2.), note that (2.7) and (2.8) (with a n, = q n ) imply ( ) ᾱ n,k, k, n =,, 2,. Hence, if z /2 then φ n (z) ᾱ n,k z k 2, n=,, 2,, implying uniform boundedness. Let ɛ>. If z i /2, i =, 2, and z z 2 <ɛ/4, then φ n (z ) φ n (z 2 ) ᾱ n,k z k z2 k z z 2 2 k+ k<ɛ, which implies equicontinuity. Thus (2.) is proved. Finally, we prove (2.). Let γ = {γ k k =,, 2, }, be a non-negative sequence satisfying γ =, and for r> define ã(r) n = {ã(r) n,k k =,, 2, }, n =,, 2,,by (2.3)ã(r) n = M n r (γ), n=,, 2,. By (2.5), we see that ã(r) n is decreasing in r. Lemma 2.. in (2.6). For r> and for any γ in (2.3), ã(r) n,k = α(r) k if n k, where α(r) is as Proof. By definition, ã(r) n, = α(r) =,n =,, 2,, hence in particular the claim holds for n =. Assume that the claim holds for some n and for all k satisfying k n. Then for k n + we have ã(r) n+,k = kr k+ k j= ã(r) n,k j ã(r) n,j = kr k+ k α(r) k j α(r) j = α(r) k, j= hence the claim holds for n +. 6

7 Let us proceed with the proof of (2.), and let <ɛ<. The assumption (.6) implies that there exists n such that ( ɛ)ρ q n+ ( + ɛ)ρ, n n, q n which further implies, by induction, with (2.9) and (2.5), (2.4)M n (+ɛ)ρ (ᾱ n ) ᾱ n +n = M n ρ (ᾱ n ) M n ( ɛ)ρ (ᾱ n ), n =,, 2,. Put γ =ᾱ n in (2.3). Comparing (2.4) with (2.3) we have, ã(ρ( + ɛ)) n = Mρ(+ɛ) n (ᾱ n ) ᾱ n+n Mρ( ɛ) n (ᾱ n )=ã(ρ( ɛ)) n, n =,, 2,. With Lemma 2. we further have, α(ρ( + ɛ)) k ᾱ n+n,k α(ρ( ɛ)) k, n k, Hence α(ρ( + ɛ)) k lim inf ᾱn,k lim sup ᾱ n,k α(ρ( ɛ)) k, k =,, 2,. Noting that <ɛ< is arbitrary and α(r) k is a polynomial in r, we have (2.). 3 Monotonicity arguments. Here we prove Theorem.2. Put Ω={f : [, ) [, ) decreasing, right continuous, f() =, lim f(t) =}. t If φ O a, then the corresponding F appearing in the expression (.9) satisfies F Ω. We shall first rewrite the recursion equation (.3) in terms of corresponding F s. For l> and f Ω define R l (f) : [, ) [, ) byr l (f)() = and (3.) R l (f)(t) = lt lt f(s)f(lt s) ds, t >. Lemma 3.. (i) R l (Ω) Ω. (ii) Let l> and f Ω and g Ω. Iff(t) g(t), t>, then R l (f)(t) R l (g)(t), t>. Proof. Note that (3.2) R l (f)(t) = f(lt s 2 )f(lt+s 2 ) ds, t >. Continuity and non-negativity of R l (f) follows from (3.2) and f Ω. Then the decreasing property and R l (f)(t), t>, follows. lim R l (f)(t) = then follows from (3.2) and the t dominated convergence theorem. Hence (i) is proved. Using (3.2) and non-negativity of f and g, (ii) also follows easily. 7

8 Lemma 3.2. Let a>and φ O a, and let φ n, n =,, 2,, be a sequence defined recursively by (.3) on [, ). Then φ n O a, n =,, 2,, and each φ n has an expression (3.3) φ n (x) = e xt ( F n (t)) dt, x, with F n Ω, and F n, n =,, 2,, satisfies (3.4) F n+ = R ( F n ), n =,, 2,. Proof. By assumption there exists F Ω such that (3.5) φ (x) = e xt ( F (t)) dt, x. Define F n : [, ) [, ), n =,, 2,, recursively by (3.4), and put (3.6) φ n (x) = Lemma 3.(i) implies e xt ( F n (t)) dt, x, n=,, 2,. (3.7) F n Ω, n =, 2, 3,. By definition, φ O a implies F (t) Ct a, t>. On the other hand, if, for some n, there exists C n > (independent of t) such that F n (t) C n t a, t>, then (3.2) implies that there exists C n+ > such that F n+ (t) C n+ t 2a C n+ t a, <t. We have already seen that F n+ Ω, hence F n+ is bounded. Therefore there exists C > such that F n+ (t) C t a, t>. By induction, similar estimates hold for all n =,, 2,. This with (3.7) and (3.6) implies that φ n O a, n =,, 2,. Therefore, if we can prove that φ n = φ n for all n, the proof of Lemma 3.2 is complete. Since this holds by definition for n =, it suffices to prove that φ n satisfies the same recursion relation (.3) as φ n. Using (3.) in (3.4), we have, from (3.6), φ n+ (x) = = = = x x du dy t dt e xt ( F n (s))( F n (t s)) ds t dv u + v e x(u+v) ( F n (u))( F n (v)) du dv e y(u+v) ( F n (u))( F n (v)) ( φ n (y)) 2 dy, x, which proves that φ n satisfies (.3). Using Lemma 3.(ii), we can now state a sufficient condition for q n, n =,, 2,,tobe increasing. Proposition 3.3. Let a>, φ O a, and F be as in (3.5), and let φ n, n =,, 2,,be a sequence defined recursively by (.3) on [, ). If ( F )(t) R ( F )(t), t>, then q n = φ n (), n =,, 2,, is increasing. 8

9 Proof. Lemma 3.2 implies that there exists F n, n =,, 2,, such that F n = R n ( F ) Ω, φ n (x) = e xt ( F n (t)) dt, x, n =,, 2,. By assumption, F (t) F (t), t>. Applying R on both sides and using Lemma 3.(ii), we inductively obtain ( ) F (t) F (t) F 2 (t),t>. Integrating by t>, we see that q n = φ n () is increasing. We next turn to a proof of Theorem.2. Lemma 3.4. Assume that F : [, ) [, ) satisfies F Ω and F (t) Ct a, t>, for some positive constants C and a. Then F n, n =,, 2,, defined by (3.4) satisfies lim F n(l n t)= ( ) a + /a, t>, if <l<l s (a), where l s (a) = sup l(a ) and l(a) =. <a <a 2 Remark. By elementary calculus, we see that l s (a) in this Lemma is equal to l s (a) of (.) in Theorem.2. Proof of Lemma 3.4. For a> define f a Ωby (3.8) f a (t) = max{( t a ), }, t >. Let a>, <t, and <l l(a), and put We have A = {s (, ] f a (lt( + s)/2) }. (, ] \ A = {s (, ] l a t a ( + s) a 2 a }. Since f a is decreasing, we have, Therefore {s (, ] f a (lt( s)/2) = } {s (, ] f a (lt( + s)/2) = }. R l ( f a )(t) = f a (lt s 2 ) f a (lt +s 2 ) ds = ( l a t a ( s) a 2 a )( l a t a ( + s) a 2 a ) ds = A + + [ l a t a ( s) a 2 a l a t a ( + s) a 2 a ] ds (,]\A A [l a t a ( s) a 2 a +l a t a ( + s) a 2 a ] ds l 2a t 2a ( s 2 ) a 2 2a ds. 9

10 Performing the first integration in the right hand side we have R l ( f a )(t) l a 2 a + ta l(a) a 2 a + ta = t a = f a (t), if <t and <l l(a). For t> we have R l ( f a )(t) = f a (t). Hence (3.9) R l ( f a )(t) f a (t), t>, <l l(a). For a> define T a : Ω ΩbyT a (f)(t) =f(at), t>. Then (3.)R l = R T l, and (3.)R l T a = T a R l. The assumptions on F implies F (t) f a (C /a t), t>. Since F is bounded, this implies that for any <a a there exists a positive constant C such that F (t) f a (C t)= T C ( f a )(t), t>. Lemma 3. and (3.9) with (3.) and (3.) then imply F n (t) =R l n ( F )( l n t) T C R l n ( f a )( l n t) T C ( f a )( l n t)= f a (C l n t), t>, < l l(a ), <a a. Therefore if <l<l s (a), choose l >land a a such that l <l(a ) to find lim F n(l n t) lim f a (C l n l n t)= f a (+) =, t>. Lemma 3.5. Assume that a, b, l satisfy <a<c, a<b<2a, l s (a) l<l s (b), and define f a,+ : [, ) [, ] by f a,+ (t) = min{( t a + Ct b ), }, t, where C is a constant satisfying ( ) πγ( + a)(l/2) 2a (b a)/a C max{,c,c 2 }, C = 2Γ(3a/2) ( (l/l s (b)) b, C 2 = a ( ) b a (b a)/a. ) b b Then R l (f a,+ (t)) f a,+ (t), t. Remark. If <a<c, l satisfying the assumptions exists, but if a c such l does not exist, because then l s (b) =l s (a) =e /c.

11 Proof. Note that l s (b) < 2 for all b>, as is obvious from (.). Note also that C C 2 implies t a + Ct b, t, hence f a,+ is non-negative on [, ). Define t > byct b a =. C and b>aimply t. Also it is easy to see that f a,+ (t) =,t t. By definition, f a,+ (t), t, hence R l (f a,+ )(t), t. Therefore the statement holds for t t. In the following we assume <t<t. Using f a,+ (t) t a + Ct b and (3.2), we see that ( ) l a t a + I 2 + I 3, R l (f a,+ )(t) ( l s (a) ( l 2a ( ) ) l b I 2 = Ct b ( u C 2) 2 ) a du t 2a b +, l s (b) ( ) lt a+b I 3 = C ( u 2 ) ( ( a u) b a ( + u) b a + C 2 ( ) ) lt b a ( u 2 ) b a du. 2 Using t<t = C /(b a) and C C (noting a<b<2a), we find I 2 Ct b. Using Ct b a = ( ) lt b a and l<l s (b) < 2, we have C, which further implies, for u, 2 ( ) lt b a ( u) b a ( + u) b a + C ( u 2 ) b a 2 ( u) b a ( + u) b a +( u 2 ) b a 2 ( u 2 ) b a +( u 2 ) b a ( u 2 ) b a ( 2+), hence I 3. We therefore have R l (f a,+ )(t) ( ) l a t a + Ct b t a + Ct b f a,+(t). l s (a) Proof of Theorem.2. Let <l<l s (a). By assumptions, φ n has an expression φ n (x) = e xt ( F n (t)) dt, x, where F n is as in (3.3). Then Lemma 3.4 and dominated convergence theorem imply lim l n φ n (l n x) = lim e xs ( F n (l n s)) ds = For x =, Lemma 3.4 and Fatou s Lemma imply lim inf which implies, ( F n (l n s)) ds lim inf ( F n(l n s)) ds =, e xs ds = x, x >. lim l n φ n () =. The second part of the Theorem is proved similarly using Lemma 3.5 in place of Lemma 3.4.

12 4 Rod bisection. Here we prove Theorem.4. By explicit calculation, we see that x ( e x )= e xt ( F (t)) dt with {, x<, (4.) F (x) =, x. Define F n : [, ) [, ], n =,, 2,, recursively by (3.4). Then Lemma 3.2 implies that φ n in Theorem.4 is given by the formula (3.3). Lemma 4.. q n = φ n () is increasing in n. Proof. By explicit calculation, we have, <x<, F (x) = 2 2 x, x<2,, x 2. In particular, we have F (x) F (x), x>, hence Proposition 3.3 implies that q n is increasing. Note that (4.) and (3.4) inductively imply (4.2) F n (x) =,x>2 n, n =,, 2,. Lemma 4.2. lim q/n n exists and satisfies lim q/n n e /c. Proof. It is noted in eq. (5.) of [7] that F n (x) (for the choice (4.)) is the probability that at the nth stage of random sequential bisection of a rod of length x, all the pieces have length shorter than. Namely, one starts with a rod of length x and breaks it into two pieces randomly with uniform distribution. Then one breaks each of the resulting two pieces randomly (with no correlation between the pieces), and so on, and see, after n steps, whether all the 2 n pieces are shorter than unit length. By performing partial integration on (4.3) q n = φ n () = ( F n (t)) dt, and noting (4.2) to deal with the boundary terms, we have, for n>, q n = tf n(t) (4.4) dt = t σ n(t) dt, where, in the second equality, we made a change of variable x /x and put σ n (t) = df n(/t) = dt t 2 F n ( t ). 2

13 F n (/t) is the probability that all the pieces have length shorter than at the nth stage of random sequential bisection of a rod of length /t. In other words, it is the probability that starting from a rod of unit length, X n, the longest piece at nth stage, is shorter than t. Therefore σ n is the probability density of X n. Then (4.4) implies that q n is the expectation value of /X n. If we denote the expectations by E[ ], then q n =E[/X n ]. Now consider the longest piece X n,m at n + mth stage among descendants from the longest piece X n at nth stage. Clearly X n+m X n+m. Note also that X n,m /X n and X n are independent and the former is equal in distribution to X m. Therefore Hence q n+m =E[ X n+m ] E[ X n+m ]=E[ X n X n,m ]E[ X n ]=E[ X m ]E[ X n ]=q m q n. (4.5) q n+m q n q m, n,m =, 2, 3,. Put p n = log q n, n =, 2, 3,. Then p n+m p n + p m, n =, 2, 3,. Using standard arguments on subadditivity, we deduce that lim n p n = inf n n p n. Therefore, the limit lim exp(p n/n) = lim q n /n = inf q n /n n exists. Note that Theorem.2 is applicable to the present choice (4.) of F with any a>. Choose a c. Suppose that e /c = l s (a) > lim q/n n and choose l so that e /c >l> lim q/n n. Then Theorem.2 for x = implies lim (l qn /n )n = lim l n φ n () =, which further implies lim q/n n l, which contradicts the choice of l. Therefore e /c lim q/n n. Lemma 4.3. If q n satisfies (.6), then lim q/n n e /c. Proof. In [3] it is essentially proved that (4.6) lim F n(l n x)=, if l>e /c, x >. (In the reference, (4.6) may not be explicit, but this property of F n is essentially used there to prove that H k / log k c, in probability, as k. Here H k is the height of a binary search tree with k nodes constructed by standard insertions from a random permutation of k positive integers. In fact, it is easy to see that (4.6) and Lemma 2. in [3] imply H k / log k c in probability. One should note the correspondence F n (t) = Prob[ Z n t ] between our notation and the notation in the reference.) Hence dominated convergence implies, for each x>, (4.7) lim l n φ n (l n x)=,l>e /c. 3

14 As noted in the Introduction, φ in Theorem.4 satisfies all the assumptions in Theorem. except perhaps (.6). Hence if (.6) also holds, the consequences of Theorem. hold. In particular, the uniform convergence of φ n (z) =qn φ n(qn z) (in a neighborhood of z = ) and φ n () =, n Z +, imply, with (3.3), ( ɛ >) δ >, n N; ( n n )( x δ) e xt ( F n (q n t))dt <ɛ. If q = lim q/n n >l, then l n q n for sufficiently large n, Since F n (t) is increasing in t, this implies, for sufficiently large n, l n φ n (l n x) ɛ, x δ. Comparing with (4.7), we see that l e /c. This holds for all l<q, hence q e /c. Lemma 4., Lemma 4.2, and Lemma 4.3 prove Theorem.4. Remark. The arguments in [3] use, in particular, the Biggins-Kingman-Hammersley theorem, which in turn is based on arguments of Chernov type inequalities together with the law of large numbers for superconvolutive sequences [, 2, 5, 6]. The arguments are strong enough to control q n /n, but unfortunately cannot control q n to prove the existence of lim q n+/q n. A Appendix. We have no explicit example for which all the assumptions in Theorem. are proven to hold. Even for the most promising case (.), we lack a proof of (.6). We give some of our numerical results for the ratio q n+ /q n in Table. Table : Numerical results for q n /q n. The number of sample points is N = 32. The digits shown are stable between N = 6 and N = 32 results. n q n+ /q n Numerical values are obtained by discretizing F n (i.e., represent the function by its values at a finite number, say N, of points), and performing numerical integration (i.e., approximating by a discrete sum of N terms) of (3.4), starting from (4.). The results suggest that q n+ /q n is decreasing in n, hence (.6) does hold. Our data is also consitent with lim q n+/q n = e /c =.26. In spite of the promising numerical results, a proof of (.6) seems not very easy. Numerical results suggest that q n+ /q n is decreasing in n. In fact q 2 /q q /q, or equivalently, q 2 q 2 does hold, by substituting n = m = in (4.5). However, q 3 /q 2 q 2 /q,orq 3 q q2 2, already seems rather hard. For example, recall the proof of Lemma 4.2 and for n =, 2, 3, and k =,, 2,, 2 n, let Z n,k be the length ratio of left piece of rod at n-th bisection stage, to the k-th piece at n -st stage. Start with two rods and perform two (independent) bisection stages Z n,k and Z n,k. Denote by X n and X n the length of the longest pieces, respectively, at 4

15 n-th stage. For example, X = max{z, Z }. Then q n q m =E[X n X m ]. Now consider conditional expectations and Q 3 (Z,Z 2,Z 2,Z ) =E[X 3 X Z,Z 2,Z 2,Z ] Q 22 (Z,Z 2,Z 2,Z )=E[X 2X 2 Z,Z 2,Z 2,Z ]. Our expectation is E[ Q 3 ]=q 3 q q 2 2 =E[Q 22 ]. A sufficient condition for this to hold is (A.) Q 3 Q 22, a.s. (?) However, it turns out that (taking an obvious contiuous versions of conditional expectations) while xq 22 (/2,x,/2, /2) = 6( log 2) = 4.9, lim xq 3 (/2,x,/2, /2) = 8 log 2 = 5.5, x 2 x, and 2 Q 3(/2, /2, /2, /2) = 8 32 log 2 = 4.4, 3 so that (A.) does not hold. This shows some difficulties encountered in an attempt to prove monotonicity of q n+ /q n. References [] J. D. Biggins, The first and last-birth problems for a multitype age-dependent branching process, Adv. Appl. Probab. 8 (976) [2] J. D. Biggins, Chernoff s theorem in the branching random walk, J. Appl. Probab. 4 (977) [3] L. Devroye, A note on the height of binary search trees, Journ. Assoc. Computing Machinery 33 (986) [4] W. Feller, An introduction to probability theory and its applications, John Wiley and Sons, 966. [5] J. M. Hammersley, Postulates for subadditive processes, Ann. Probab. 2 (974) [6] J. F. C. Kingman, The first-birth problem for an age-dependent branching process, Ann. Probab. 3 (975) [7] M. Sibuya, Y. Itoh, Random sequential bisection and its associated binary tree, Ann. Inst. Stat. Math. 39A (987)