Cancellations in power series of sine type
CCANCELLATIONS IN POWER SERIES OF SINE TYPE
J. ARIAS DE REYNA
Abstract.
We present a method to study the behavior of a power series oftype f ( x ) := ∞ (cid:88) n =0 ( − n c n x n +1 (2 n + 1)!when x → ∞ .We apply our method to study the function f ( t ) := (cid:90) t dxx (cid:90) x dyy (cid:90) y dzz (cid:8) sin x + sin( x − y ) − sin( x − z ) − sin( x − y + z ) (cid:9) . We will derive various different representations of f ( t ) by means of which itwill be shown that lim t → + ∞ f ( t ) = 0, disproving a conjecture by Z. Silagadze,claiming that this limit equals − π / Introduction.
In Titchmarsh [12, Section 14.32] we find two suggestive equivalents to the Rie-mann Hypothesis: The RH is equivalent to F ( x ) = O ( x + ε ), or alternatively to G ( x ) = O ( x − + ε ) where(1.1) F ( x ) = ∞ (cid:88) n =1 ( − n +1 x n ( n − ζ (2 n ) , G ( x ) = ∞ (cid:88) n =1 ( − n x n n ! ζ (2 n + 1) . The first equivalence is due to Riesz [8], and the second to Hardy and Littlewood[1].Since ζ ( n ) converges to 1 these series can be considered slight modifications ofthe exponential function. As in the case of the exponential the series are convergenteverywhere but the small values they get for x large is the result of an amazingcancellation between large terms of different signs. This phenomenon happens alsoin the case of the sine or cosine series. In these simple cases the many algebraicproperties of the corresponding sums yield the proof of the cancellations. But howcan one treat a case as the series (1.1) above?A problem in MathOverflow leads us to consider the power series(1.2) f ( t ) = ∞ (cid:88) n =1 ( − n (cid:16) n +1 (cid:88) k =1 H k − k (cid:17) t n +1 (2 n + 1)!(2 n + 1)where the H n = (cid:80) nk =1 1 k are the harmonic numbers. The proposer Z. Silagadzeasked for a proof that f ( t ) → − π /
12 (obtained by using some arguments fromphysics).
Date : November 6, 2018.Supported by MINECO grant MTM2012-30748. a r X i v : . [ m a t h . C A ] M a y J. ARIAS DE REYNA
We think that our solution presented here is interesting because it provides anexample of how to treat this type of problems. It must be said that the methodcan be applied to the functions in (1.1), but as we will see in Section 6 it only givesanother path to the connection between these series and the Riemann Hypothesis.Silagadze’s problem in MathOverflow [10] was to compute(1.3) (cid:90) ∞ dxx (cid:90) x dyy (cid:90) y dzz (cid:8) sin x + sin( x − y ) − sin( x − z ) − sin( x − y + z ) (cid:9) for which he conjectured the value − π / t > f ( t ) := (cid:90) t dxx (cid:90) x dyy (cid:90) y dzz (cid:8) sin x + sin( x − y ) − sin( x − z ) − sin( x − y + z ) (cid:9) . and define the integral (1.3) as the lim t →∞ f ( t ). We will prove that this limit = 0.The function f ( t ) defined by the triple integral in (1.4) has many interestingproperties. It extends to an entire function whose power series (1.2) is a slightmodification of the series for sin t . We are interested in its behavior for t → + ∞ .This is a similar problem as the one (equivalent to the Riemann Hypothesis) intro-duced by M. Riesz cited above. We will solve the problem in this relatively simplecase, essentially, by means of a representation of f ( t ) as a Fourier transform (4.5).The transformations we apply to the function f ( t ) conclude with the represen-tation as a Fourier transform in Proposition 4.4. But all our representations areneeded to reach this final one which will solve Silagadze’s problem.Our formulas (2.3) and (4.5) may also be applied to the computation of the tripleintegral. The numerical computation of a triple integral is quite often difficult. Bymeans of the power series or the Fourier representation it can be computed easilyand above all more reliably for relatively small values of t . For large values of t we present explicitly an asymptotic expansion (Section 5) that is very well suitedto compute f ( t ) with high precision for t large. The first term of the asymptoticexpansion f ( t ) = − cos t tt + O (cid:16) log tt (cid:17) shows that f ( t ) has zeros near the zeros of cos t for t large.As an application we will also obtain the x-ray of f ( t ). It shows that the zeroswith small absolute value are real. On average there are two zeros on each intervalof length 2 π . The first zeros are separated by approximately 4, 2, 4, 2, . . . The x-rayalso shows that the general properties of f ( t ) are very similar to those of sin t .To finish the paper we give some details about the Riesz function analogous to f ( t ), also explaining why we may not hope that an analogous analysis will solvethe RH. As expected! 2. The entire function f ( t ) . Proposition 2.1.
The function f ( t ) defined for t > by the absolutely convergentintegral (2.2) f ( t ) := (cid:90) t dxx (cid:90) x dyy (cid:90) y dzz (cid:8) sin x + sin( x − y ) − sin( x − z ) − sin( x − y + z ) (cid:9) ANCELLATIONS IN POWER SERIES OF SINE TYPE 3 extends to an entire function with power series expansion (2.3) f ( t ) = ∞ (cid:88) n =1 ( − n (cid:16) n +1 (cid:88) k =1 H k − k (cid:17) t n +1 (2 n + 1)!(2 n + 1) where H n = (cid:80) nk =1 1 k is the n -th harmonic number.Proof. Applying the Mean Value Theorem we get the following bound of the abso-lute value of the integrand in the definition of f ( t )1 x · y (cid:12)(cid:12)(cid:12) sin x − sin( x − z ) z − sin( x − y + z ) − sin( x − y ) z (cid:12)(cid:12)(cid:12) ≤ xy . When integrating over z ∈ (0 , y ) we get something bounded by 2 /x , and integratingthis over y ∈ (0 , x ) we get something bounded by 2. The integral of this over x ∈ (0 , t ) yields something bounded by 2 t . This shows that the integral in (2.2) isabsolutely convergent.Changing variables x = tu , y = tv , z = tw yields f ( t ) = (cid:90) duu (cid:90) u dvv (cid:90) v dww (cid:8) sin( tu )+sin( tu − tv ) − sin( tu − tw ) − sin( tu − tv + tw ) (cid:9) and renaming the variables f ( t ) = (cid:90) dxx (cid:90) x dyy (cid:90) y dzz (cid:8) sin( tx ) + sin( tx − ty ) − sin( tx − tz ) − sin( tx − ty + tz ) (cid:9) . This representation shows that f ( t ) extends to an entire function with power seriesexpansion f ( t ) = ∞ (cid:88) n =0 ( − n A n t n +1 (2 n + 1)!where A n = (cid:90) dxx (cid:90) x dyy (cid:90) y dzz (cid:8) x n +1 + ( x − y ) n +1 − ( x − z ) n +1 − ( x − y + z ) n +1 } . Notice that A = 0. To compute the other A n we get successively (cid:90) y (cid:8) x n +1 − ( x − z ) n +1 } dzz = − n +1 (cid:88) k =1 ( − k (cid:18) n + 1 k (cid:19) y k k x n +1 − k , (cid:90) x dyy (cid:90) y (cid:8) x n +1 − ( x − z ) n +1 } dzz = − n +1 (cid:88) k =1 ( − k (cid:18) n + 1 k (cid:19) x n +1 k , (cid:90) dxx (cid:90) x dyy (cid:90) y (cid:8) x n +1 − ( x − z ) n +1 } dzz = − n + 1 n +1 (cid:88) k =1 ( − k (cid:18) n + 1 k (cid:19) k = 12 n + 1 n +1 (cid:88) k =1 H k k where the last step is proved in Lemma 2.4. J. ARIAS DE REYNA
For the other part of A n we compute (cid:90) y (cid:8) ( x − y ) n +1 − ( x − y + z ) n +1 } dzz = − n +1 (cid:88) k =1 (cid:18) n + 1 k (cid:19) y k k ( x − y ) n +1 − k , (cid:90) x dyy (cid:90) y (cid:8) ( x − y ) n +1 − ( x − y + z ) n +1 } dzz = − n +1 (cid:88) k =1 k (cid:18) n + 1 k (cid:19) (cid:90) x y k ( x − y ) n +1 − k dyy . In the last integral we change variables y = xu yielding= − n +1 (cid:88) k =1 x n +1 k (cid:18) n + 1 k (cid:19) (cid:90) u k (1 − u ) n +1 − k duu == − n +1 (cid:88) k =1 x n +1 k (2 n + 1)! k !(2 n + 1 − k )! ( k − n + 1 − k )!(2 n + 1)! = − n +1 (cid:88) k =1 x n +1 k so that (cid:90) dxx (cid:90) x dyy (cid:90) y (cid:8) ( x − y ) n +1 − ( x − y + z ) n +1 } dzz = − n + 1 n +1 (cid:88) k =1 k . Collecting all our partial results we obtain A n = 12 n + 1 n +1 (cid:88) k =1 H k k − n + 1 n +1 (cid:88) k =1 k = 12 n + 1 n +1 (cid:88) k =1 H k − k . (cid:3) Lemma 2.4.
For any natural number n we have (2.5) n (cid:88) k =1 ( − k (cid:18) nk (cid:19) k = − n (cid:88) k =1 H k k . Proof.
To prove a n = b n it is sufficient to prove a = b and a n − a n − = b n − b n − .In our case the equality for n = 1 is checked easily. So, we have to show n (cid:88) k =1 ( − k +1 (cid:18) nk (cid:19) k − n − (cid:88) k =1 ( − k +1 (cid:18) n − k (cid:19) k = H n n . Since (cid:0) nk (cid:1) = (cid:0) n − k − (cid:1) + (cid:0) n − k (cid:1) , the equality is equivalent to( − n +1 n + n − (cid:88) k =1 ( − k +1 (cid:18) n − k − (cid:19) k = H n n or, multiplying by n (2.6) n (cid:88) k =1 ( − k +1 (cid:18) nk (cid:19) k = H n . This is formula 0.155.4 in Gradshteyn and Ryzhik [2, p. 4]. (cid:3)
ANCELLATIONS IN POWER SERIES OF SINE TYPE 5 Plot of the function f ( t ) . The power series expansion of f ( t ) is analogous to the one for sin t . Thesepower series are not well suited for computation, because of the violent cancellationbetween large terms. Nevertheless, one may use it using high precision in thecomputation of the terms to get approximate values. Assuming t >
0, each term ofthe power series (2.3) is in absolute value less than t n +1 / (2 n + 1)!. Therefore theerror committed by summing only the first N > t terms of the series (2.3) is lessthan 2 − N . It follows that to compute f ( t ) with error less than 2 ε we need onlycompute the sum of the first N terms of the series with error less than ε , taking N large enough so that N ≥ log(1 /ε )2 log 2 , N > t. All the terms of the series are less than t n +1 / (2 n + 1)! < ( et n +1 ) n +1 ≤ e t . So, wemust compute each term with an error less than ε/N , for which it will suffice tocompute each term working with a precision P := t + log( N/e )log 10 decimal digits . Of course this will be difficult for t very large, but today we may easily computewith thousands of digits of precision. �� �� �� �� ��� ��� ��� - ��� - ��� - ��������� We may say that the computation of the power series is much easier than thecomputation of the multiple integral (2.2) or any other multiple integral giving f ( t )considered in this paper. This is true even when t is small.4. Computation of the limit of f ( t ) . First we prove the following integral representation of f ( t ): Proposition 4.1.
For t > we have (4.2) f ( t ) = (cid:90) t du (cid:90) (cid:90) (cid:16) x sin uu (1 − x )(1 − xy ) − sin( ux ) u (1 − x )(1 − y ) + sin( uxy ) u (1 − y )(1 − xy ) (cid:17) dx dy. Proof.
Starting from the power series expansion it is easy to get f ( t ) = ∞ (cid:88) n =1 ( − n n +1 (cid:88) k =1 k k − (cid:88) j =1 j t n +1 (2 n + 1)!(2 n + 1)= (cid:90) t du (cid:90) dx (cid:90) dy ∞ (cid:88) n =1 ( − n n +1 (cid:88) k =1 x k − k − (cid:88) j =1 y j − u n (2 n + 1)! . J. ARIAS DE REYNA If C is a circle with radius r > t , we may express the factorial by means of theResidues Theorem(4.3) f ( t ) = (cid:90) t du (cid:90) dx (cid:90) dy πi (cid:90) C e z z ∞ (cid:88) n =1 ( − n n +1 (cid:88) k =1 x k − k − (cid:88) j =1 y j − ( u/z ) n dz. (Since | u/z | < f ( t ) = (cid:90) t du (cid:90) dx (cid:90) dy πi (cid:90) C e z z u x yz − u xz − u x z − u x yz ( u + z )( u x + z )( u x y + z ) dz. By the Residue Theorem we obtain12 πi (cid:90) C e z z u x yz − u xz − u x z − u x yz ( u + z )( u x + z )( u x y + z ) dz = x sin uu (1 − x )(1 − xy ) − sin( ux ) u (1 − x )(1 − y ) + sin( uxy ) u (1 − y )(1 − xy )establishing our claim (4.2). Observe that since our integral representation (4.2)appears after computing one of the integrals in the absolutely convergent integral(4.3), the integrals in our new representation are also absolutely convergent. (cid:3) Notice that in the representation (4.2) we may interchange the integral in u withthe integral in ( x, y ), but then it is not easy to justify the interchange of the limitin t with the integrals, because the integrand is not dominated by an integrablefunction. If we would proceed formally in this way we would easily get that thelimit is 0, but, as said before, this is not allowed. Therefore we follow another path. Proposition 4.4.
For any complex t we have (4.5) f ( t ) = (cid:90) (cid:16)
12 log (1 − x ) + ∞ (cid:88) n =1 (1 − x ) n − n (cid:17) sin( tx ) x dx. Proof.
We take t > f ( t ) = (cid:90) (cid:90) dx dy (cid:90) t (cid:16) x sin uu (1 − x )(1 − xy ) − sin( ux ) u (1 − x )(1 − y ) + sin( uxy ) u (1 − y )(1 − xy ) (cid:17) du. Now subdivide the inner integral in three and change variables appropriately f ( t ) = (cid:90) (cid:90) dx dy (cid:16)(cid:90) t x sin uu (1 − x )(1 − xy ) du − (cid:90) tx sin uu (1 − x )(1 − y ) du + (cid:90) txy sin uu (1 − y )(1 − xy ) du (cid:17) . We use Iverson’s notation [5], so that, for any proposition P , the symbol [ P ] is 1 if P is true and 0 if it is false. In this way we may write f ( t ) = (cid:90) (cid:90) dx dy (cid:90) t (cid:16) x [ u < t ] sin uu (1 − x )(1 − xy ) du − [ u < tx ] sin uu (1 − x )(1 − y ) + [ u < txy ] sin uu (1 − y )(1 − xy ) (cid:17) du. ANCELLATIONS IN POWER SERIES OF SINE TYPE 7
To simplify the notation we put a := u/t , so that always 0 < a <
1. Interchangingthe integrals we obtain f ( t ) = (cid:90) t sin uu du (cid:90) (cid:90) (cid:16) x (1 − x )(1 − xy ) du − [ a < x ](1 − x )(1 − y ) + [ a < xy ](1 − y )(1 − xy ) (cid:17) dx dy and will compute the inner double integral in x and y . This is J ( a ) := (cid:90) (cid:90) (cid:16) x (1 − x )(1 − xy ) du − [ a < x ](1 − x )(1 − y ) + [ a < xy ](1 − y )(1 − xy ) (cid:17) dx dy. We subdivide the square (0 , into three disjoint sets S := { ( x, y ) : 0 < x ≤ a, < y < } , S := { ( x, y ) : a < x < , < xy ≤ a } ,S := { ( x, y ) : a < xy } . In S the integrand is = x (1 − x )(1 − xy ) , and in S it is= x (1 − x )(1 − xy ) − − x )(1 − y ) = − − y )(1 − xy ) . In S the integrand is equal to= x (1 − x )(1 − xy ) − − x )(1 − y ) + 1(1 − y )(1 − xy ) = 0so that J ( a ) = (cid:90) dy (cid:90) a x (1 − x )(1 − xy ) dx − (cid:90) a dx (cid:90) a/x − y )(1 − xy ) dy = 12 log (1 − a ) − π ∞ (cid:88) n =1 (1 − a ) n n . This yields f ( t ) = (cid:90) t (cid:16)
12 log (1 − u/t ) + ∞ (cid:88) n =1 (1 − u/t ) n − n (cid:17) sin uu du. A change of variables u = tx yields (4.5) for t >
0, and it is clear that the two sidesare entire functions. (cid:3)
Proposition 4.6.
We have lim t →∞ f ( t ) = lim t → + ∞ (cid:90) (cid:16)
12 log (1 − x ) + ∞ (cid:88) n =1 (1 − x ) n − n (cid:17) sin( tx ) x dx = 0 . Proof.
This is an example of the Riemann-Lebesgue Lemma, once it has been shownthat the function g ( x ) := 1 x (cid:16)
12 log (1 − x ) + ∞ (cid:88) n =1 (1 − x ) n − n (cid:17) is in L [0 , (cid:3) J. ARIAS DE REYNA Asymptotic Expansion.
The representation (4.4) as a Fourier integral yields by known methods an as-ymptotic expansion for f ( t ). First notice that the dilogarithm function Li ( z ) canbe defined in the cut plane C (cid:114) [1 , ∞ ) by(5.1) Li ( z ) = − (cid:90) z log(1 − u ) duu where the path of integration is the segment joining 0 and z . For z in the planewith two cuts along the real axis ( −∞ ,
0] and [1 , + ∞ ) the dilogarithm satisfies theEuler functional equation (cf. Lewin [6, (1.12), p. 5])(5.2) Li ( z ) + Li (1 − z ) = π − log z log(1 − z ) . Here and in the sequel we will denote by log w the principal branch of the logarithm. Definition 5.3.
Let Ω be the complex plane with the two cuts ( −∞ ,
0] and [1 , + ∞ ).We define the function g ( z ) holomorphic in Ω by(5.4) g ( z ) := (cid:16)
12 log (1 − z ) − π (1 − z ) (cid:17) z = (cid:16)
12 log (1 − z ) − Li ( z ) − log z log(1 − z ) (cid:17) z . It is easier to give explicitly the asymptotic expansion of the Fourier transform(5.5) J ( t ) := (cid:90) g ( x ) e ixt dx. Since f ( t ) = Im J ( t ) its asymptotic expansion can be obtained easily from the oneof J ( t ). Proposition 5.6.
For any t > we have f ( t ) = Im J ( t ) where (5.7) J ( t ) = i (cid:90) ∞ g ( iy ) e − ty dy − ie it (cid:90) ∞ g (1 + iy ) e − ty dy. Proof.
By the definition of g ( z ) and (4.4) we have f ( t ) = Im (cid:90) g ( x ) e ixt dx. We apply Cauchy’s Theorem to a rectangle with vertices at 0, 1, 1 + iR , iR with R > R → + ∞ . In this way the integral in (0 ,
1) can be converted intothe two infinite integrals in the Proposition. The bounds are easy. (cid:3)
Since the two integrals in Proposition 5.6 are Laplace integrals we may applyWatson’s Lemma [3, 17.03, p. 501] to get their asymptotic expansions. In thiscase we have the additional difficulty of logarithmic singularities at the extremes ofthe integrals at 0 and 1. We follow the path in Lyness [7]. To get the asymptoticexpansion we need the behavior of g ( z ) near z = 0 and z = 1. ANCELLATIONS IN POWER SERIES OF SINE TYPE 9
Lemma 5.8.
For | z | < in Ω we have (5.9) g ( z ) = ∞ (cid:88) n =0 (cid:16) H n +1 n + 1 − n + 1) (cid:17) z n + (cid:16) ∞ (cid:88) n =0 z n n + 1 (cid:17) log z := ∞ (cid:88) n =0 A n z n + (cid:16) ∞ (cid:88) n =0 B n z n (cid:17) log z. For z ∈ Ω , with | − z | < we have (5.10) g ( z ) = − ∞ (cid:88) n =0 Ψ (cid:48) ( n + 1)(1 − z ) n + 12 (cid:16) ∞ (cid:88) n =0 (1 − z ) n (cid:17) log (1 − z ):= ∞ (cid:88) n =0 C n (1 − z ) n + (cid:16) ∞ (cid:88) n =0 D n (1 − z ) n (cid:17) log (1 − z ) where Ψ( z ) = Γ (cid:48) ( z ) / Γ( z ) is the digamma function.Proof. Near z = 0 we have g ( z ) = g ( z ) + g ( z ) log z where by (5.4) we have g ( z ) = (cid:16)
12 log (1 − z ) − Li ( z ) (cid:17) z , g ( z ) = − log(1 − z ) z .g ( z ) and g ( z ) are holomorphic at z = 0. Expanding in power series at z = 0 weget (5.9).Near z = 1 we have g ( z ) = g ( z ) + g ( z ) log (1 − z ) with g ( z ) and g ( z ) holo-morphic at z = 1. By (5.4) g ( z ) = − π z + Li (1 − z ) z , g ( z ) = 12 z . The functions g ( z ) and g ( z ) are holomorphic at z = 1 and expanding them inpower series we obtain (5.10). We notice the equalities.(5.11) Ψ( n + 1) = H n − γ, Ψ (cid:48) ( n + 1) = π − n (cid:88) k =1 k . (cid:3) Proposition 5.12.
The following asymptotic expansion is valid for t → + ∞ (5.13) J ( t ) ∼ ∞ (cid:88) n =0 i n +1 A n n ! t n +1 + ∞ (cid:88) n =0 i n +1 B n (cid:16) Ψ( n + 1) − log t + πi (cid:17) n ! t n +1 − ie it ∞ (cid:88) n =0 ( − i ) n C n n ! t n +1 − ie it ∞ (cid:88) n =0 ( − i ) n D n (cid:110) Ψ (cid:48) ( n +1)+ (cid:16) γ − H n +log t + πi (cid:17) (cid:111) n ! t n +1 where the coefficients A n , B n , C n and D n are defined in Lemma 5.8.Proof. By the results in Lyness [7] the asymptotic expansion of J ( t ) is obtained byintegrating term by term the expression (5.7) after substituting the power series at the extremes at 0 and 1, respectively. Explicitly J ( t ) ∼ ∞ (cid:88) n =0 i n +1 A n (cid:90) ∞ y n e − ty dy + ∞ (cid:88) n =0 i n +1 B n (cid:90) ∞ y n (cid:16) log y + πi (cid:17) e − ty dy − ie it ∞ (cid:88) n =0 ( − i ) n C n (cid:90) ∞ y n e − ty dy − ie it ∞ (cid:88) n =0 ( − i ) n D n (cid:90) ∞ y n (cid:16) log y − πi (cid:17) e − ty dy Computing the integrals we obtain (5.13). (cid:3)
Corollary 5.14.
The first order terms of the asymptotic expansion of f ( t ) are (5.15) − cos t tt + (cid:16) π sin t − γ cos t − (cid:17) log tt + (cid:16) π cos t − γ cos t γπ sin t − − γ (cid:17) t where γ is Euler’s constant. Riesz type functions
We have proved that the function f ( t ) given by the power series (1.2) tends to0 by writing it as a Fourier transform of an L function. As an example of whathappens in the case of a Riesz-type function we just consider Riesz’s F ( x ). Rieszproves [8](6.1) F ( x ) = ∞ (cid:88) n =1 ( − n +1 x n ( n − ζ (2 n ) = x (cid:88) µ ( n ) n e − xn . The Riemann Hypothesis is equivalent to F ( x ) = O ( x + ε ) for any ε >
0. Rieszshows that F ( x ) is not O ( x α ) for any α < /
4. Therefore, it is not true that F ( x )converges to 0 when x → ∞ . Riesz proves that F ( x ) x − / converges to 0. Withsome work we may reprove this. Noticing that(6.2) √ t e − t = 1 √ π (cid:90) + ∞ cos( arctan x )(1 + x ) / cos( xt ) dx, t > F ( x ) √ x = 1 √ π (cid:90) + ∞ (cid:16) ∞ (cid:88) n =1 µ ( n ) n cos( arctan( n t ))(1 + n t ) / (cid:17) cos( xt ) dt. We apply summation by parts and use M ( x ) := (cid:80) n ≤ x µ ( n ) = O ( xe − c √ log x ) toshow that(6.4) g ( t ) := ∞ (cid:88) n =1 µ ( n ) n cos( arctan( n t ))(1 + n t ) / is in L (0 , ∞ ). This implies that F ( x ) x − / = o (1).We do not give the details of the proofs, but they are clearly more complicatedthan those given by Riesz.We can also obtain in this way that M ( x ) = O ( x a ) with ≤ a < ζ ( s ) does not vanish on Re s > a . Again quite a difficult way to get this simpleresult. ANCELLATIONS IN POWER SERIES OF SINE TYPE 11 The x-ray of the function f ( t ) . This represents f ( z ) on the square ( − , × ( − , f ( z )takes real values, in blue lines where f ( z ) is purely imaginary. At t = 0 we observea triple zero. Table 1 contains some more zeros of f ( t ).8. Acknowledgement.
I wish to express my thanks to Jan van de Lune ( Hallum, The Netherlands )for his encouragements and linguistic assistance.This work has been supported by MINECO grant MTM2012-30748.
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Facultad de Matem´aticas, Univ. de Sevilla, Apdo. 1160, 41080-Sevilla, Spain
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