A double inequality for completely monotonic degree of a remainder for an asymptotic expansion of the trigamma function
aa r X i v : . [ m a t h . G M ] M a r COMPLETELY MONOTONIC DEGREE OF REMAINDER OFASYMPTOTIC EXPANSION OF TRIGAMMA FUNCTION
FENG QI
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Abstract.
In the paper, the author compute completely monotonic degreeof a remainder of the asymptotic expansion of the trigamma function. Thisresult verify one of a series of conjectures on completely monotonic degreesof remainders of the asymptotic expansions for the logarithm of the gammafunction and for polygamma functions.
Contents
1. Simple preliminaries 12. Motivations 23. Main result and its proof 44. An open problem 6References 71.
Simple preliminaries
In the literature [1, Section 6.4], the functionsΓ( w ) = Z ∞ s w − e − s d s and ψ ( w ) = [ln Γ( w )] ′ = Γ ′ ( w )Γ( w )for ℜ ( w ) > ψ ′ ( w ), ψ ′′ ( w ), ψ (3) ( w ), and ψ (4) ( w ) are knownas the tri-, tetra-, penta-, and hexa-gamma functions respectively. As a whole, allthe derivatives ψ ( k ) ( w ) for k ≥ φ ( t ) defines on an interval I and satisfies( − k φ ( k ) ( t ) ≥ t ∈ I and k ≥
0, then it is called a completely monotonic function on I .Theorem 12b on [21, p. 161] reads that φ ( t ) is completely monotonic on (0 , ∞ ) if Mathematics Subject Classification.
Primary 33B15; Secondary 26A48, 26A51, 30E15,34E05, 44A10.
Key words and phrases. completely monotonic degree; remainder; asymptotic expansion;trigamma function; completely monotonic function; conjecture; gamma function; polygammafunction.This paper was typeset using
AMS -L A TEX. and only if φ ( t ) = Z ∞ e − ts d σ ( s ) (1.2)converges for all t ∈ (0 , ∞ ), where σ ( s ) is nondecreasing on (0 , ∞ ). The integralrepresentation (1.2) is equivalent to say that φ ( t ) is completely monotonic on (0 , ∞ )if and only if it is a Laplace transform of σ ( s ) on (0 , ∞ ). A result on [5, p. 98] assertsthat, unless φ ( t ) is a trivial completely monotonic function, that is, a nonnegativeconstant on (0 , ∞ ), those inequalities in (1.1) are all strict on (0 , ∞ ). Why do weinvestigate completely monotonic functions? One can find historic answers in twomonographs [20, 21] and closely related references therein.Let φ ( t ) be defined on (0 , ∞ ) and φ ( ∞ ) = lim x →∞ φ ( t ). If t r [ φ ( t ) − φ ( ∞ )] iscompletely monotonic on (0 , ∞ ), but t r + ε [ φ ( t ) − φ ( ∞ )] is not for any number ε > r is called the completely monotonic degree of φ ( t ) with respect to t ∈ (0 , ∞ ); if t r [ φ ( t ) − φ ( ∞ )] is completely monotonic on (0 , ∞ ) for all r ∈ R , then the completelymonotonic degree of φ ( t ) with respect to t ∈ (0 , ∞ ) is said to be ∞ . The notationdeg t cm [ φ ( t )] was designed in [6] to denote the completely monotonic degree r of φ ( t )with respect to t ∈ (0 , ∞ ). Why do we investigate completely monotonic degrees?One can find significant answers in the second paragraph of [17, Remark 1] or inthe papers [15, 16] and closely related references therein.2. Motivations
It is well-known in [1, p. 257, 6.1.40] and [1, p. 260, 6.4.11] thatln Γ( w ) ∼ (cid:18) w − (cid:19) ln w − w + 12 ln(2 π ) + ∞ X k =1 B k k (2 k − w k − (2.1)and ψ ( n ) ( w ) ∼ ( − n − (cid:20) ( n − w n + n !2 w n +1 + ∞ X k =1 B k (2 k + n − k )! w k + n (cid:21) (2.2)as w → ∞ in | arg w | < π for n ≥
0, where B k for k ≥ we w − − w ∞ X k =1 B k w k (2 k )! , | w | < π. Stimulated by the asymptotic expansions (2.1) and (2.2), many mathematiciansconsidered and investigated the remainders R n ( t ) = ( − n " ln Γ( t ) − (cid:18) t − (cid:19) ln t + t −
12 ln(2 π ) − n X k =1 B k k (2 k −
1) 1 t k − and its derivatives, where n ≥
0. For detailed information, please refer to thepapers [2, 11], [3, Theorem 8], [8, Section 1.4], [9, Theorem 2], [10, Theorem 2.1],[13, Theorem 3.1], and closely related references therein.In [17, Section 4] and [18, 19], the author posed, modified, and finalized thefollowing conjectures:(1) the completely monotonic degreesdeg t cm (cid:20) ln t − t − ψ ( t ) (cid:21) = 1 , (2.3) OMPLETELY MONOTONIC DEGREE OF A REMAINDER 3 deg t cm (cid:20) t + 12 t + 16 t − ψ ′ ( t ) (cid:21) = 3 , (2.4)deg t cm (cid:20) ψ ′ ( t ) − t − t − t + 130 t (cid:21) = 4 (2.5)with respect to t ∈ (0 , ∞ ) are valid;(2) when m = 0, the completely monotonic degrees of R n ( t ) with respect to t ∈ (0 , ∞ ) satisfydeg t cm [ R ( t )] = 0 , deg t cm [ R ( t )] = 1 , (2.6)and deg t cm [ R n ( t )] = 2( n − , n ≥
2; (2.7)(3) when m = 1, the completely monotonic degrees of − R ′ n ( t ) with respect to t ∈ (0 , ∞ ) satisfydeg t cm [ − R ′ ( t )] = 1 , deg t cm [ − R ′ ( t )] = 2 , (2.8)and deg t cm [ − R ′ n ( t )] = 2 n − , n ≥
2; (2.9)(4) when m ≥
2, the completely monotonic degrees of ( − m R ( m ) n ( t ) withrespect to t ∈ (0 , ∞ ) satisfydeg t cm (cid:2) ( − m R ( m )0 ( t ) (cid:3) = m, deg t cm (cid:2) ( − m R ( m )1 ( t ) (cid:3) = m + 1 , (2.10)and deg t cm (cid:2) ( − m R ( m ) n ( t ) (cid:3) = m + 2( n − , n ≥ . (2.11)In [3, Theorem 1], [7, Theorem 1], [18, Theorem 2.1], [19, Theorem 2.1], and [22,Theorem 3], the first conjecture in (2.8), which is equivalent to (2.3), was uncon-sciously or consciously verified again and again.In [10, Theorem 2.1], it was proved thatdeg t cm (cid:2) R n ( t ) (cid:3) ≥ n, n ≥ . This result is weaker than those conjectures in (2.7).In [4, Theorem 1], [17, Theorem 2], and [22, Theorem 4], the second conjecturein (2.8) was proved once again.In [18, Theorem 2.1] and [19, Theorem 2.1], those five conjectures in (2.6), (2.8),and (2.9) were confirmed.In [22, Theorems 1 and 2], it was acquired thatdeg t cm (cid:2) ( − R ′′ ( t ) (cid:3) = 2 and deg t cm (cid:2) ( − R ′′ ( t ) (cid:3) = 3 . These two results confirm conjectures in (2.10) just for the case m = 2. The latteris equivalent to the conjecture (2.4).The main aim of this paper is to prove the conjecture (2.5) which is equivalentto deg t cm (cid:2) ( − R ′′ ( t ) (cid:3) = 4 , (2.12)a special case m = n = 2 of the conjecture in (2.11). F. QI Main result and its proof
The conjecture (2.5), or say, the identity (2.12), a special case m = n = 2 of theconjecture in (2.11), can be confirmed by the following theorem. Theorem 3.1.
The completely monotonic degree of the function Q ( t ) = ψ ′ ( t ) − t − t − t + 130 t with respect to t ∈ (0 , ∞ ) is . In other words, the identity (2.12) , or say, theidentity deg t cm [ Q ( t )] = 4 (3.1) is valid.Proof. Making use of the integral representations1 w τ = 1Γ( τ ) Z ∞ s τ − e − ws d s, ℜ ( w ) , ℜ ( τ ) > ψ ( k ) ( w ) = ( − k +1 Z ∞ s k − e − s e − ws d s, ℜ ( w ) > , k ≥ Q ( t ) = Z ∞ (cid:18) s − e − s − − s − s
12 + s (cid:19) e − ts d s , Z ∞ h ( s ) e − ts d s,h ′ ( s ) = s + (cid:0) s − s + 90 (cid:1) e s − s (cid:0) s + 60 (cid:1) e s − s − e s − ,h ′′ ( s ) = e s (cid:0) s − (cid:1) + 3 (cid:0) s + 20 s + 30 (cid:1) e s − (cid:0) s − s + 30 (cid:1) e s − s + 1060( e s − ,h (3) ( s ) = se s + (90 − s ) e s − se s − s + 45) e s + s e s − ,h (4) ( s ) = " e s + 5(6 s − e s + 10(33 s − e s +10(33 s + 35) e s + 5(6 s + 25) e s − e s − = 130( e s − ∞ X k =7 " k + (110 k − k + 5(3 k − k − +(165 k + 350)2 k + 30 k + 125 k ! s k = 160( e s − (cid:18) s s
14 + 193 s
84 + 85 s
42 + 5065 s · · · (cid:19) > , ∞ ). Consecutively utilizing the L’Hˆospital rule giveslim s → + h ( s ) = 0 , lim s → + h ′ ( s ) = 0 , lim s → + h ′′ ( s ) = 0 , lim s → + h (3) ( s ) = 0 . OMPLETELY MONOTONIC DEGREE OF A REMAINDER 5
Accordingly, consecutively integrating by parts results in t Q ( t ) = − t Z ∞ h ( s ) d (cid:0) e − ts (cid:1) d s d s = − t (cid:20) h ( s ) e − ts (cid:12)(cid:12) s = ∞ s =0 + − Z ∞ h ′ ( s ) e − ts d s (cid:21) = t Z ∞ h ′ ( s ) e − ts d s = − t Z ∞ h ′ ( s ) d (cid:0) e − ts (cid:1) d s d s = − t (cid:20) h ′ ( s ) e − ts (cid:12)(cid:12) s = ∞ s =0 + − Z ∞ h ′′ ( s ) e − ts d s (cid:21) = t Z ∞ h ′′ ( s ) e − ts d s = − t Z ∞ h ′′ ( s ) d (cid:0) e − ts (cid:1) d s d s = − t (cid:20) h ′′ ( s ) e − ts (cid:12)(cid:12) s = ∞ s =0 + − Z ∞ h (3) ( s ) e − ts d s (cid:21) = t Z ∞ h (3) ( s ) e − ts d s = − Z ∞ h (3) ( s ) d (cid:0) e − ts (cid:1) d s d s = − (cid:20) h (3) ( s ) e − ts (cid:12)(cid:12) s = ∞ s =0 + − Z ∞ h (4) ( s ) e − ts d s (cid:21) = Z ∞ h (4) ( s ) e − ts d s. Consequently, the function t Q ( t ) is completely monotonic on (0 , ∞ ). This meansthat deg t cm [ Q ( t )] ≥ . For ℜ ( w ) > k ≥
1, we have the recurrent relation ψ ( k − ( w + 1) = ψ ( k − ( w ) + ( − k − ( k − w k . See [1, p. 260, 6.4.6]. From this, it follows thatlim t → + (cid:2) t k ψ ( k − ( t ) (cid:3) = lim t → + (cid:18) t k (cid:20) ψ ( k − ( t + 1) − ( − k − ( k − t k (cid:21)(cid:19) = ( − k ( k − k ≥
1. If t ε Q ( t ) for ε > , ∞ ), then its secondderivative should be positive on (0 , ∞ ), that is, (cid:2) t ε Q ( t ) (cid:3) ′′ = t ε +2 (cid:2) Q ( t ) ε + (cid:0) tQ ′ ( t ) + 7 Q ( t ) (cid:1) ε + t Q ′′ ( t ) + 8 tQ ′ ( t ) + 12 Q ( t ) (cid:3) = t ε +2 (cid:20)(cid:18) ψ ′ ( t ) − t − t − t + 130 t (cid:19) ε + 60 t ψ ′′ ( t ) + 210 t ψ ′ ( t ) − t − t − t − t ε + 30 t ψ (3) ( t ) + 240 t ψ ′′ ( t ) + 360 t ψ ′ ( t ) − t − t + 230 t (cid:21) = t ε +2 Q ( t ) (cid:20) ε + 60 t ψ ′′ ( t ) + 210 t ψ ′ ( t ) − t − t − t − t Q ( t ) ε + 30 t ψ (3) ( t ) + 240 t ψ ′′ ( t ) + 360 t ψ ′ ( t ) − t − t + 230 t Q ( t ) (cid:21) should be positive on (0 , ∞ ). For this, the discriminant∆( t ) = (cid:20) t ψ ′′ ( t ) + 210 t ψ ′ ( t ) − t − t − t − t Q ( t ) (cid:21) − t ψ (3) ( t ) + 240 t ψ ′′ ( t ) + 360 t ψ ′ ( t ) − t − t + 230 t Q ( t )should be negative on (0 , ∞ ). But, by virtue of the limit (3.2),∆( t ) = − t ψ ′ ( t ) (cid:2) t ψ (3) ( t ) + 60 t ψ ′′ ( t ) − t − t − t + 49 (cid:3) − t (cid:2) ψ ′′ ( t ) (cid:3) − t (cid:0) t + 15 t + 5 t − (cid:1) ψ (3) ( t ) − t + 900 t + 1875 t + 150 t − t − t − t − t (cid:2) ψ ′ ( t ) (cid:3) − t (cid:0) t + 75 t + 35 t − (cid:1) ψ ′′ ( t ) − [30 t ψ ′ ( t ) − t − t − t + 1] → , t → + . This leads to a contradiction. Consequently, we arrive atdeg t cm [ Q ( t )] ≤ . The identity (3.1), or say, the identity (2.12), is thus proved. The proof of Theo-rem 3.1 is complete. (cid:3) An open problem
For a given point ( a, b ) on the first quadrant { ( x, y ) : x, y ≥ } , let y = f ( x ) bea differentiable function on [0 , a ] with a > f (0) = 0, and f ( a ) = b >
0. Denotetan θ ( x ) = f ′ ( x ) and S a,b ( f ) = Z a cos θ ( x )d x = Z a cos arctan f ′ ( x )d x = Z a p f ′ ( x )] d x. What is the minimuminf (cid:8) S a,b ( f ) : f (0) = 0 , f ( a ) = b, f ( x ) ∈ C ([0 , a ]) (cid:9) ?What is the differentiable function f m ( x ) on [0 , a ] such that S a,b ( f m ) attains theabove infimum? OMPLETELY MONOTONIC DEGREE OF A REMAINDER 7
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Institute of Mathematics, Henan Polytechnic University, Jiaozuo 454010, Henan,China, College of Mathematics, Inner Mongolia University for Nationalities, Tongliao028043, Inner Mongolia, China, School of Mathematical Sciences, Tianjin PolytechnicUniversity, Tianjin 300387, China
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