A note on the asymptotics for incomplete Betafunctions
aa r X i v : . [ m a t h . C A ] J a n A NOTE ON THE ASYMPTOTICS FOR INCOMPLETEBETAFUNCTIONS
JAN-CHRISTOPH SCHLAGE-PUCHTA
Abstract.
We determine the asymptotic behaviour of certain incompleteBetafunctions. Introdution and results
For integers k, ℓ ≥ P k,ℓ = ( k + ℓ )!( k − ℓ ! Z k/ ( k + ℓ ) t k − (1 − t ) ℓ dt. By repeated partial integration we obtain the representation P k,ℓ = (cid:18) ℓℓ + k (cid:19) k + ℓ k − X ν =0 (cid:18) k + ℓν (cid:19) (cid:18) kℓ (cid:19) ν . Vietoris[5] showed that P k,ℓ ≤ . Alzer and Kwong [1] proved that P k,ℓ ≥ holdsfor all k, ℓ . Interest in bounds for P k,ℓ stems from application to statistics, see [4].In this note we consider the asymptotic behaviour of P k,ℓ . We prove the follow-ing. Theorem 1.
We have (1) P k,l = e − k k − X ν =0 k ν ν ! + O (cid:18) k ℓ (cid:19) , (2) P k,ℓ = 1 − e − ℓ ℓ X ν =0 ℓ ν ν ! + O (cid:18) ℓ k (cid:19) and (3) P k,ℓ = 12 + O p min( k, ℓ ) ! . Note that the first two estimates are good if one of the parameters k, ℓ is rathersmall, whereas the third one gives information in the general case.Comparing (1) and (2) with Vietoris’ result that P k,ℓ ≤ we see that ℓ − X ν =0 ℓ ν ν ! < e ℓ < ℓ X ν =0 ℓ ν ν ! . Note that equality is impossible as e is transcendental. A more precise version ofthis inequality has been asked by Ramanujan (Question 294) and was answered by Karamata [2]. For a detailed discussion of this result, Uhlmann’s inequalities [4]and Vietoris bound we refer the reader to the historical notes by Vietoris [6].From Theorem 1 we deduce the following.
Corollary 1.
The only accumulation points of the set { P k,ℓ : k, ℓ ∈ N } are andthe real numbers e − k P k − ν =1 k ν ν ! and − e − ℓ P ℓν =1 ℓ ν ν ! .Proof. Suppose that ( k n ) , ( ℓ n ) are integer sequences such that the pairs ( k n , ℓ n ) areall distinct and that the sequence ( P k n ,ℓ n ) converges. If both k n , ℓ n tend to infinity,then by (3) we have that P k n ,ℓ n tends to . If one of these sequences does not tendto infinity, then we can pass to a subsequence and assume that k n or ℓ n is constant.Then our claim follows from (1) or (2). (cid:3) Together with Vietoris bound we obtain the following.
Corollary 2.
We have lim ℓ →∞ P k,ℓ ≥ − √ πk . Proof.
We have lim ℓ →∞ ( P k,ℓ + P ℓ,k ) = 1 − k k e k k ! , using Stirling’s formula in the form k ! = (cid:0) ke (cid:1) k √ πke θ/ k with 0 ≤ θ ≤
1, and P ℓ,k ≤ , we conclude lim ℓ →∞ P k,ℓ ≥ − k k e k k ! ≥ − √ πk . (cid:3) Preliminary estimates
For the proof of (3) we use another representation of P k,ℓ , which is due to Raab[3]. Theorem 2.
We have P k,ℓ = U k,ℓ V k,ℓ , where U k,ℓ = exp Z ∞ t (cid:18) e t − − t + 12 (cid:19) (cid:16) e − ( k + ℓ ) t − e − kt − e − ℓt (cid:17) dt,V k,ℓ = √ ℓ π ∞ X ν =1 c ν ( k/ℓ ) √ ν ( ν + ℓ ) , and c ν ( x ) = exp Z ∞ t (cid:18) e t − − t + 12 (cid:19) (cid:16) e − ν (1+ x ) t − e − νxt − e − νt (cid:17) dt. We first compute the occurring integrals.
Lemma 1.
We have for x ≥ Z ∞ t (cid:18) e t − − t + 12 (cid:19) e − xt dt = 112 x + O (cid:18) x (cid:19) and Z ∞ t (cid:18) e t − − t + 12 (cid:19) e − xt dt = 12 log x + C + O ( x ) for < x ≤ , where C is some constant. NOTE ON THE ASYMPTOTICS FOR INCOMPLETE BETAFUNCTIONS 3
Proof.
The series expansion of e x yields for t → t (cid:18) e t − − t + 12 (cid:19) = 1 t (cid:18)
11 + t/ t / O ( t ) − t (cid:19) = 1 /
12 + O ( t ) . For t → ∞ this expression tends to 0, in particular, it is bounded for all positive t .Hence for x → ∞ the integral in question becomes112 Z ∞ e − xt dt + O (cid:18)Z ∞ te − xt dt (cid:19) = 112 x + O (cid:18) x (cid:19) . For x → e − xt = 1 − xt + O ( x t ) and obtain Z t (cid:18) e t − − t + 12 (cid:19) e − xt dt = Z t (cid:18) e t − − t + 12 (cid:19) dt + O ( x ) = C + O ( x )and Z ∞ t ( e t − e − xt dt = Z ∞ t ( e t − dt − x Z ∞ e t − dt + O ( x ) . As a function of x , the integral R ∞ e − xt t dt defines a function that is differentiablefrom the right in 0 and has bounded second derivative in (0 , ∞ ), hence, for x ≥ O ( x ).Finally Z ∞ e − xt t dt = Z ∞ x e − s s ds = Z x dt t + Z e − t − t dt + Z ∞ e − t t dt + Z x − e − t t dt = −
12 log x + C + O ( x ) . Combining these estimates our claim follows. (cid:3)
From this we obtain
Lemma 2.
We have U k,ℓ = 1 − (cid:18) k + 1 ℓ − k + ℓ (cid:19) + O (cid:18) k, ℓ ) (cid:19) . Proof.
From Lemma 1 we obtain U k,ℓ = exp (cid:18) (cid:18) k + ℓ − k − ℓ (cid:19) + O (cid:18) k, ℓ ) (cid:19)(cid:19) , inserting the Taylor series for exp and using the fact that k, ℓ ≥ (cid:3) Next we compute c ν ( x ). Lemma 3. If νx ≥ , then c ν ( x ) = 1 + 112 ν (1 + x ) − νx + 112 ν + O (cid:18) ν min(1 , x ) (cid:19) . If νx ≤ , then c ν ( x ) = K √ νx + O (cid:18)r xν + ( νx ) / (cid:19) for some constant K . JAN-CHRISTOPH SCHLAGE-PUCHTA
Proof. If νx >
1, then we apply Lemma 1 to obtain c ν ( x ) = exp (cid:18) ν ( x + 1) − νx + 112 ν + O (cid:18) ν min(1 , x ) (cid:19)(cid:19) = 1 + 112 ν ( x + 1) − νx + 112 ν + O (cid:18) ν min(1 , x ) (cid:19) . If νx <
1, then ν ( x + 1) ≥
1, and we obtain c ν ( x ) = exp (cid:18) ν ( x + 1) + 12 log( νx ) + C + 112 ν + O (cid:18) ν + νx (cid:19)(cid:19) = K √ νx + O (cid:18)r xν + ( νx ) / (cid:19) . (cid:3) We now compute V k,ℓ . Lemma 4.
We have V k,ℓ = 12 + O (cid:18) √ k + 1 √ ℓ (cid:19) Proof.
We have X ν ≤ ℓ/k c ν ( k/ℓ ) √ ν ( ν + ℓ ) ≪ X ν ≤ ℓ/k p νk/ℓ √ ν ( ν + ℓ ) ≪ √ ℓk thus, V k,ℓ = √ ℓ π X ν>ℓ/k √ ν ( ν + ℓ ) ν k + ℓℓ − ν kℓ + 112 ν + O (cid:18) ν min(1 , k /ℓ ) (cid:19)! + O (cid:18) √ k (cid:19) = √ ℓ π X ν>ℓ/k √ ν ( ν + ℓ ) − ν kℓ + O (cid:18) ν min(1 , k /ℓ ) (cid:19)! + O √ k + X ν>ℓ/k ν / √ ℓ = √ ℓ π X ν>ℓ/k √ ν ( ν + ℓ ) − ν kℓ + O (cid:18) ν min(1 , k /ℓ ) (cid:19)! + O (cid:18) √ k + 1 √ ℓ (cid:19) If k > ℓ we obtain V k,ℓ = √ ℓ π ∞ X ν =1 √ ν ( ν + ℓ ) + O √ k + 1 √ ℓ + √ ℓ ∞ X ν =1 ν / ( ν + ℓ ) ! = 12 π Z ∞ dt √ t ( t + 1) + O (cid:18) √ k + 1 √ ℓ (cid:19) = 12 + O (cid:18) √ ℓ (cid:19) . We have √ ℓ π X ν>ℓ √ ν ( ν + ℓ ) − ν kℓ + O (cid:18) ν min(1 , k /ℓ ) (cid:19)! = √ ℓ π X ν>ℓ √ ν ( ν + ℓ ) + O (cid:18) √ k (cid:19) , NOTE ON THE ASYMPTOTICS FOR INCOMPLETE BETAFUNCTIONS 5 and for ℓ/k ≤ N ≤ ℓ X N ≤ ν ≤ N ν / ( ν + ℓ ) ≤ √ N ℓ as well as X N ≤ ν ≤ N ν / ( ν + ℓ ) ≤ N / ℓ and therefore √ ℓ X ν ≥ ℓ/k √ ν ( ν + ℓ ) · ν kℓ = ℓ / k X ν ≥ ℓ/k ν / ( ν + ℓ ) = O (cid:18) √ k (cid:19) and √ ℓ X ν ≥ ℓ/k √ ν ( ν + ℓ ) · ν min( k /ℓ ) = ℓ / k X ν ≥ ℓ/k ν / ( ν + ℓ ) = O (cid:18) √ k (cid:19) . Using these estimates we obtain √ ℓ π X ℓ/k ≤ ν ≤ ℓ √ ν ( ν + ℓ ) − ν kℓ + O (cid:18) ν min(1 , k /ℓ ) (cid:19)! = √ ℓ π X ℓ/k ≤ ν ≤ ℓ √ ν ( ν + ℓ ) + O (cid:18) √ k (cid:19) . For k ≤ ℓ we obtain V k,ℓ = √ ℓ π X ν ≥ ℓ/k √ ν ( ν + ℓ ) + O (cid:18) √ k (cid:19) = 12 π Z ∞ /k dt √ t ( t + 1) + O (cid:18) √ k (cid:19) = 12 π Z ∞ dt √ t ( t + 1) + O (cid:18) √ k (cid:19) = 12 + O (cid:18) √ k (cid:19) , and the proof is complete. (cid:3) JAN-CHRISTOPH SCHLAGE-PUCHTA Proof of Theorem 1
We first prove (1). Note that this inequality is only interesting if the error termis o (1), in particular we may assume that k < ℓ . Under this assumption we have P k,ℓ = (cid:18) ℓℓ + k (cid:19) k + ℓ k − X ν =0 (cid:18) k + ℓν (cid:19) (cid:18) kℓ (cid:19) ν = k − X ν =0 (cid:18) ℓℓ + k (cid:19) k + ℓ − ν k ν ν ! exp ν X µ =1 log k + ℓ − µk + ℓ ! = k − X ν =0 (cid:18) ℓℓ + k (cid:19) k + ℓ − ν k ν ν ! exp − ν X µ =1 µk + ℓ + O (cid:18) µ ( k + ℓ ) (cid:19)! = k − X ν =0 (cid:18) ℓℓ + k (cid:19) k + ℓ − ν k ν ν ! exp (cid:18) O (cid:18) ν ℓ (cid:19)(cid:19) = k − X ν =0 (cid:18) ℓℓ + k (cid:19) k + ℓ k ν ν ! exp (cid:18) O (cid:18) k ℓ (cid:19)(cid:19) = exp (cid:18) ( k + ℓ ) log ℓℓ + k (cid:19) k − X ν =0 k ν ν ! exp (cid:18) O (cid:18) k ℓ (cid:19)(cid:19) = exp (cid:18) − k + O (cid:18) k ℓ (cid:19)(cid:19) k − X ν =0 k ν ν != e − k k − X ν =0 k ν ν ! + O (cid:18) k ℓ (cid:19) . The proof of (2) is quite similar. Analogously to the previous case we may assumethat ℓ < k . P k,ℓ = (cid:18) ℓℓ + k (cid:19) k + ℓ (cid:18) ℓ + kℓ (cid:19) k + ℓ − k + ℓ X ν = k (cid:18) k + ℓν (cid:19) (cid:18) kℓ (cid:19) ν ! = 1 − (cid:18) ℓℓ + k (cid:19) k + ℓ ℓ X µ =0 (cid:18) k + ℓµ (cid:19) (cid:18) kℓ (cid:19) k + ℓ − µ = 1 − (cid:18) kℓ + k (cid:19) k + ℓ ℓ X µ =0 (cid:18) k + ℓµ (cid:19) (cid:18) ℓk (cid:19) µ = 1 − exp (cid:18) − ℓ + O (cid:18) ℓ k (cid:19)(cid:19) ℓ X µ =0 ℓ µ µ ! (cid:18) k + ℓk (cid:19) µ = 1 − e − ℓ ℓ X µ =0 ℓ µ µ ! . Finally (3) follows from Theorem 2, Lemma 2 and Lemma 4.
NOTE ON THE ASYMPTOTICS FOR INCOMPLETE BETAFUNCTIONS 7
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