An infinitely differentiable function with compact support: Definition and properties
AAN INFINITELY DIFFERENTIABLE FUNCTION WITHCOMPACT SUPPORT: DEFINITION AND PROPERTIES
J. ARIAS DE REYNA Introduction.
Infinitely differentiable functions of compact support defined on R play an im-portant role in Analysis. Usually, one constructs examples using an idea of Cauchy.For this example the derivatives are cumbersome. This problem makes me searchfor a better example.Looking at a rough plot of such a function and its derivative (see figure 1) I askedif it was possible that the derivative could be formed with two homothetic copiesof the same function translated conveniently. So I posed the following question: t ϕ ( t ) ϕ ′ ( t ) Fig. 1 Does there exist a function ϕ ∈ D ( R ) suchthat:(a) supp( ϕ ) = [ − , ϕ ( t ) > t ∈ ( − , ϕ (0) = 1,(d) and there is a constant k > t ∈ R ϕ (cid:48) ( t ) = k (cid:0) ϕ (2 t + 1) − ϕ (2 t − (cid:1) ?We will prove that there is a unique solution ϕ satisfying the above conditions. For this uniquesolution the value of the constant k is 2. No othervalue of k gives a solution.The function ϕ has many other properties.It can be interpreted as a probability (theorem3), ϕ and some of its translates form a parti-tion of unity (theorem 5), its derivatives can becomputed easily (theorem 4), and the most no-table, it is not a rational function but its valuesat dyadic points are rational numbers that areeffectively computable. Since its derivatives arerelated to the same function, not only the valuesof ϕ but also those of its derivatives ϕ ( k ) ( t ) are rational number at dyadic points.The only reference that we know about this function is a paper [4] by Jessen andWintner (1935) where the function ϕ is defined by means of its Fourier transform,as an example of an infinitely differentiable function, but Jessen and Wintner donot give any other property of this function. Received: November 5, 1980. a r X i v : . [ m a t h . C A ] F e b J. ARIAS DE REYNA Existence and Unicity.
Theorem 1.
There is a unique infinitely differentiable function with compact sup-port ϕ : R → R and such that: (a) supp( ϕ ) = [ − , . (b) ϕ ( t ) > for any t in the open set ( − , . (c) ϕ (0) = 1 . (d) There is a constant k > such that for any t ∈ R ϕ (cid:48) ( t ) = k (cid:0) ϕ (2 t + 1) − ϕ (2 t − (cid:1) and the constant k appearing in (d) is necessarily equal to .Proof. First, assuming that ϕ exists, we will prove the unicity of ϕ and that k = 2.Since ϕ ∈ D ( R ) its Fourier transform is an entire function(1) (cid:98) ϕ ( z ) = (cid:90) R ϕ ( t ) e − πitz dt The Fourier transform of ϕ (cid:48) ( t ), ϕ (2 t + 1) and ϕ (2 t −
1) are2 πiz (cid:98) ϕ ( z ) , e πiz (cid:98) ϕ ( z ) , e − πiz (cid:98) ϕ ( z )respectively. Condition (d) yields(2) (cid:98) ϕ ( z ) = k πzπz (cid:98) ϕ ( z ) . By induction, we obtain from (2) that(3) (cid:98) ϕ ( z ) = (cid:16) k (cid:17) n (cid:104) n (cid:89) h =0 sin πz h πz h (cid:105) (cid:98) ϕ (cid:16) z n +1 (cid:17) . Conditions (a) and (b) imply that (cid:98) ϕ (0) = (cid:82) ϕ ( t ) dt >
0, so that taking limits for n → ∞ we obtain k = 2 and(4) (cid:98) ϕ ( z ) = (cid:98) ϕ (0) ∞ (cid:89) h =0 sin πz h πz h . If there is a solution to our problem it is unique, because by the inversion formula(5) ϕ ( t ) = (cid:90) R (cid:98) ϕ ( x ) e πitx dx and condition (c) will fix the value of the constant (cid:98) ϕ (0).We will see later that (c) implies (cid:98) ϕ (0) = 1, so that in what follows we will use (cid:98) ϕ ( z ) to denote the function defined in (4) assuming (cid:98) ϕ (0) = 1.Now we will show that the solution ϕ exists. We start from the function (cid:98) ϕ ( z )defined in (4). Since the infinite product converges uniformly in compact sets, thefunction (cid:98) ϕ ( z ) is entire. Equation (2) may be used to expand it in power series(6) (cid:98) ϕ ( z ) = ∞ (cid:88) k =0 ( − k c k (2 k )! (2 πz ) k , where the c k are rational numbers defined by the recurrence(7) (2 k + 1)2 k c k = k (cid:88) h =0 (cid:18) k + 12 h (cid:19) c h . NFINITELY DIFFERENTIABLE FUNCTION 3
From equation (7) we obtain that the numbers c k are positive. Also we have(8) c k = F k (2 k + 1)(2 k − · · · k (cid:89) n =1 (2 n − − , where F k are natural numbers, F = 1, F = 1, F = 19, F = 2915, F = 2 788 989.Using the known formulassin zz = ∞ (cid:89) n =1 cos z n , and sin πzπz = ∞ (cid:89) n =1 (cid:16) − z n (cid:17) , we obtain(9) (cid:98) ϕ ( z ) = ∞ (cid:89) m =1 (cid:16) cos πz m (cid:17) m = ∞ (cid:89) m =1 (cid:16) − z m (cid:17) v ( m ) , where v ( m ) is the greatest exponent such that 2 v ( m ) divides m .It is clear that (cid:98) ϕ restricted to R is infinitely differentiable. We will show alsothat it is a rapidly decreasing function.Let f ( x ) = (sin x ) /x . For x ∈ R ∗ , we have | f ( x ) | ≤ | sin x | ≤
1. For all n | x n (cid:98) ϕ ( x ) | = (cid:12)(cid:12)(cid:12) x n ∞ (cid:89) h =0 f ( πx/ h ) (cid:12)(cid:12)(cid:12) ≤ (cid:12)(cid:12)(cid:12) x n n − (cid:89) h =0 f ( πx/ h ) (cid:12)(cid:12)(cid:12) ≤ n ) π − n . It is easy to see that there is a constant M r ≥ r ∈ N such that | ∂ r f ( πx/ h ) | ≤ π r − hr M r . Applying the rule to differentiate an infinite product and the same idea used aboveto bound | x n (cid:98) ϕ ( x ) | we obtain | x n ∂ r (cid:98) ϕ ( x ) | ≤≤ (cid:88) S r ! s ! · · · s t ! (cid:88) H (cid:12)(cid:12)(cid:12) t (cid:89) i =1 ∂ s i f ( πx/ h i ) (cid:12)(cid:12)(cid:12) (cid:12)(cid:12)(cid:12) x n (cid:89) h (cid:54) = h i f ( πx/ h ) (cid:12)(cid:12)(cid:12) ≤ (cid:88) S r ! s ! · · · s t ! M s · · · M s t (cid:16)(cid:88) H π r − s h −···− s t h t (cid:17) n + t ) π − n < ∞ where the sum extended to S refers to all sets { s , . . . , s t } of natural numbers suchthat s + · · · + s t = r and s i ≥ H to all sets { h , . . . , h t } of t distinct natural numbers.Once we have proved that (cid:98) ϕ is a test function in Schwartz space we define ϕ by means of equation (5). It follows that ϕ is infinitely differentiable and rapidlydecreasing. Since (cid:98) ϕ satisfies (2) with k = 2, we obtain that ϕ satisfies condition (d)with k = 2. We will show that ϕ also satisfies conditions (a), (b) and (c). Insteadof using Paley-Wiener’s Theorem we prefer to use another method, which gives ussome additional information.Let µ n be the Radon measure in R whose Fourier transform is(10) F ( µ m ) = m (cid:89) k =1 (cid:16) cos πx k (cid:17) k . Since(11) F (cid:0) δ − k − + δ − − k − (cid:1) = cos πx k , J. ARIAS DE REYNA µ m is the convolution product(12) µ m = ∞ ∗ k =1 (cid:0) δ − k − + δ − − k − (cid:1) k where the powers have also the meaning of convolution products.It is clear that the total variation (cid:107) µ m (cid:107) = 1, µ m ≥ µ m ) ⊂ [ − , ∞ (cid:88) k =1 k k +1 = 1 . Lemma 1.
Let ( µ m ) be the sequence of measures defined in (12) . This sequence ofmeasures converges in the weak-* topology σ ( M b ( R ) , C ∗ ( R )) towards the measure ϕλ with density ϕ with respect to Lebesgue measure λ .Proof. Denote by C ∗ ( R ) the Banach space of complex valued bounded functionsdefined on R . Since the measures µ m are on the unit ball of the dual space, whichis weakly compact, there is a measure µ that is a weak cluster point to the sequence µ m . Since F ( µ m ) → F ( ϕλ ) pointwise, we have F ( µ ) = F ( ϕλ ). Since F is injectivein the space of bounded Radon measures, we obtain µ = ϕλ . Therefore ϕλ is theonly weak cluster point, so that it is the weak limit of the sequence µ m . (cid:3) Since µ m → ϕλ with weak convergence, it follows that ϕ satisfies condition (a)and, since ϕ is continuous it follows that ϕ ( x ) ≥ x ∈ R .Now we know that (cid:82) ϕ ( t ) dt = (cid:98) ϕ (0) = 1. This fact, together with the fact thatsupp( ϕ ) = [ − ,
1] yields ϕ (0) = (cid:90) − ϕ (cid:48) ( t ) dt = (cid:90) − (cid:0) ϕ (2 t + 1) − ϕ (2 t − (cid:1) dt = 2 (cid:90) ϕ (2 t + 1) dt = (cid:90) ϕ ( u ) du = 1 . and ϕ satisfies condition (c).It remains to show that ϕ satisfies (b). By the same reasoning as above we havefor every x ∈ ( − , ϕ ( x ) = 2 (cid:90) x − ϕ (2 t + 1) dt. Therefore ϕ ( x ) is not decreasing in ( − ,
0) ( since ϕ (cid:48) ( x ) ≥ ϕ is an evenfunction, ϕ ( x ) > ϕ ( t ) > t ∈ ( − x, x ). If ϕ ( x ) > ϕ (( x − / >
0, therefore ϕ ( t ) > t ∈ ( − , (cid:3) Other expressions for ϕ . We have seen two possible definitions of ϕ : the expression (5) and that givenin Lemma 1. We will give another two. One as the limit of a sequence of stepfunctions and another by means of an integral. We need some previous notationsand definitions.Let p n be the sequence of polynomials defined by the recurrence(14) p = 1; p n ( x ) = p n − ( x )(1 + x ) n . NFINITELY DIFFERENTIABLE FUNCTION 5
It is easy to see that(15) p n ( x ) = n (cid:89) k =1 (cid:16) − x k − x (cid:17) The degree g n of p n is given by the equations(16) g = 0 , g n = 2 g n − + n. Therefore(17) g n n = 12 + 22 + · · · + n n . Equations (12) and (14) show that µ n is the measure obtained when we substituteeach power x m by δ m − gn n +1 in the polynomial2 − ( n +12 ) p n ( x ) . For each n ∈ N , let ϕ n be the step function obtained from the polynomial 2 − ( n +12 ) p n ( x )substituting each power x m by the characteristic function of the interval (cid:104) m − − g n n +1 , m + 1 − g n n +1 (cid:105) multiplied by 2 n . We have then: Theorem 2. ϕ is the limit of the sequence of step functions ϕ m .Proof. It suffices to observe that for a characteristic function f of an interval withdyadic extremes, we havelim m →∞ µ m ( f ) = lim m →∞ (cid:90) ϕ m f = (cid:90) ϕf, and the fact, easily proved, that ϕ m is monotonous non decreasing in ( − ,
0) andmonotonous not increasing in (0 , ϕ m (0) = 1. (cid:3) It is easy to see that(18) p m +1 ( x ) = p m ( x )(1 + x + x + · · · + x m +1 − )This gives us an easy algorithm to obtain the ϕ m , and also shows that(19) p m ( x ) = (1 + x )(1 + x + x + x ) · · · (1 + x + · · · + x m − ) . Therefore we have a combinatorial interpretation of the coefficient of x r in p m ( x ): The coefficient of x r in p m ( x ) is the number of partitions of r , in m parts r = s + s + · · · + s m such that ≤ s i ≤ i − Theorem 3.
Let σ = (cid:78) ∞ k =1 λ k be the measure defined on [0 , N , λ k being theLebesgue measure on [0 , . For − ≤ x ≤ we have ϕ ( x ) = σ (cid:110) ( x k ) : 0 ≤ ∞ (cid:88) k =1 x k k ≤ x + 1 (cid:111) Proof.
Let ν k be the measure in [ − , N ν k = ∞ (cid:79) m =1 (cid:0) δ − m − k + 12 δ − − m − k (cid:1) ( k = 1 2, . . . , ) and let ( t k, , t k, , . . . ) denote the variables in the space [ − , N . J. ARIAS DE REYNA
Let µ be the measure defined on { , } N as the product of the measure assigning0 and 1 measure 1 / ν k = f k ( µ ) the image measure, with f k { , } N → [ − , N given by f k ( ε , ε , . . . ) =( t k, , t k, , . . . ) where t k,m = (cid:40) − m − k when ε m = 1 , − − m − k when ε m = 0 .µ is also the image measure of Lebesgue measure on [0 ,
1] by the application g : [0 , → { , } N defined by g ( x ) = ( ε , ε , . . . ) if x = (cid:80) ∞ m =1 ( ε m / m ) with ε m ∈ { , } . The function g is well defined only almost everywhere but this isno difficulty.The measure ϕ ( t ) dt is the limit of the µ m , therefore for all integrable f , (cid:90) f ( t ) ϕ ( t ) dt = (cid:90) f (cid:16)(cid:88) t k,m (cid:17) d ∞ (cid:79) k =1 ν k . Since each ν k is an image measure the last integral can be transformed in an integralon [0 , N with respect to the measure σ = ⊗ ∞ k =1 λ .The relation f k ◦ g ( x k ) = ( t k, , t k, , . . . ) implies x k = (cid:80) ∞ m =1 ( ε m / m ) with ε m ∈{ , } , t k,m = 2 − m − k if ε m = 1 and t k,m = − − m − k when ε m = 0. Therefore (cid:88) m t k,m = ∞ (cid:88) m =1 ε m − m − k − (cid:16) ∞ (cid:88) m =1 − m − k − ∞ (cid:88) m =1 ε m − m − k (cid:17) = x k − k +1 − − k From this we get (cid:90) f ( t ) ϕ ( t ) dt = (cid:90) f (cid:16) ∞ (cid:88) k =1 x k − k +1 − (cid:17) dσ. Taking f ( t ) = χ [ − , x +1] ( t ) with − ≤ x ≤ ϕ ( x ) = (cid:90) − ≤ (cid:80) ∞ k =1 x k − k +1 − ≤ x +1 dσ = (cid:90) ≤ (cid:80) ∞ k =1 x k − k ≤ x +1 dσ = σ (cid:110) ( x k ) : 0 ≤ ∞ (cid:88) k =1 x k − k ≤ x + 1 (cid:111) In other words we have proved the Proposition:
Let x k be independent randomvariables uniformly distributed in [0 , , ϕ ( x ) (with − ≤ x ≤ ) is equal to theprobability that the sum (cid:80) x k − k be ≤ x + 1. (cid:3) Properties.
Theorem 4.
Let θ ( t ) = ∞ (cid:88) k =0 ( − s ( k ) ϕ ( t − k − where s ( k ) denotes the sum of the digits of k when written in base 2. Then (a) θ is an infinitely differentiable function. (b) θ (cid:48) ( t ) = 2 θ (2 t ) . (c) For t ∈ [ − , , ϕ ( k ) ( t ) = 2( k +12 ) θ (2 k t + 2 k ) . NFINITELY DIFFERENTIABLE FUNCTION 7
Proof.
The sum in the definition of θ ( t ) is locally finite, therefore θ is infinitelydifferentiable and its derivative is θ (cid:48) ( t ) = ∞ (cid:88) k =0 ( − s ( k ) (cid:0) ϕ (2 t − k − − ϕ (2 t − k − − (cid:1) = 2 ∞ (cid:88) k =0 (cid:0) ( − s ( k ) ϕ (2 t − k ) − − ( − s ( k ) ϕ (2 t − k + 1) − (cid:1) t θ ( t ) using the definition of s ( k ) this yields(21) θ (cid:48) ( t ) = 2 θ (2 t ) . By repeated differentiation of (21) we obtain(22) θ ( k ) ( t ) = 2( k +12 ) θ (2 k t ) . For t ∈ [ − ,
1] we have ϕ ( t ) = θ ( t + 1) so that(23) ϕ ( k ) ( t ) = 2( k +12 ) θ (2 k t + 2 k ) , if t ∈ [ − , . (cid:3) This proves that on any dyadic point t = q/ n the Taylor expansion is a poly-nomial(24) T ( t, x ) = n (cid:88) k =0 ϕ ( k ) ( t ) k ! x k and for q odd the degree of T ( t, x ) is n . Corollary.
The function ϕ is not analytic on any point of the interval [ − , . Theorem 5.
For u > and t ∈ R we have (25) (cid:88) k ∈ Z ϕ ( t + uk ) = (cid:88) k ∈ Z u (cid:98) ϕ (cid:16) ku (cid:17) e πik u . Proof.
The left hand side of (25) is locally finite, therefore the sum is infinitelydifferentiable. It is a periodic function of t with period u . Therefore it has aFourier series expansion (cid:88) k ∈ Z ϕ ( t + uk ) = (cid:88) k ∈ Z a k e πik u J. ARIAS DE REYNA where a n = 1 u (cid:90) u (cid:88) k ∈ Z ϕ ( t + uk ) e − πin tu dt = (cid:88) k ∈ Z u (cid:90) u ϕ ( t + uk ) e − πin tu dt = (cid:88) k ∈ Z u (cid:90) u ( k +1) uk ϕ ( v ) e − πin v − uku dv = 1 u (cid:90) ϕ ( v ) e − πiv nu dv = 1 u (cid:98) ϕ (cid:16) nu (cid:17) . (cid:3) Some particular cases of (25) are interesting:(26) (cid:88) k ∈ Z ϕ (cid:16) t + kn (cid:17) = n for n ∈ N . Furthermore(27) (cid:88) k ∈ Z ϕ ( t + k ) = 1 . which is equivalent to(28) ϕ ( t ) + ϕ ( t −
1) = 1 , for t ∈ [0 , . Also, from (25) it follows that(29) (cid:88) k ∈ Z ϕ ( t + 2 k ) = 12 (cid:88) k ∈ Z (cid:98) ϕ (cid:16) k (cid:17) e πikt , which is no more than the Fourier expansion(30) ϕ ( t ) = 12 + ∞ (cid:88) k =0 (cid:98) ϕ (cid:16) k + 12 (cid:17) cos(2 k + 1) πt, valid for t ∈ [ − ,
1] and which has good convergence properties.The product (9) implies that the sign of the coefficient (cid:98) ϕ ((2 k +1) /
2) is the parityof 1 + v (1) + 1 + v (2) + · · · + 1 + v ( k ) = k + v ( k !) = s ( k ), therefore also equalto the sign of θ ( k ).Equation (25) is not only a Fourier expansion, it is also Poisson’s formula appliedto ϕ ( t + x ). For t = 0 it yields(31) (cid:88) m ∈ Z ϕ ( ma ) = (cid:88) m ∈ Z a (cid:98) ϕ (cid:16) ma (cid:17) , and using the knowledge about the support of ϕ , this implies(32) a + 2 aϕ ( a ) = (cid:88) m ∈ Z a (cid:98) ϕ (cid:16) ma (cid:17) , for 12 ≤ a ≤ . Values at dyadic points.
First we determine the values of ϕ (1 − − n ). Theorem 6.
For each natural number n we have (33) (cid:90) t n − ϕ ( t ) dt = ( n − n ) ϕ (1 − − n ) . (34) (cid:90) t n ϕ ( t ) dt = c n . NFINITELY DIFFERENTIABLE FUNCTION 9 where c n are the rational numbers that appear in the expansion (6) of ϕ .Proof. We can check, by differentiation, that in the sequence of functions f ( t ) = ϕ ( t ) , f ( t ) = ϕ (cid:16) t − (cid:17) , f ( t ) = 2 ϕ (cid:16) t − − (cid:17) ,f k ( t ) = 2( k ) ϕ (cid:16) t k − k − k − − · · · − (cid:17) each function is a primitive in [ − ,
1] of the previous one and all vanish at the point t = −
1. So integrating by parts (cid:90) t n ϕ ( t ) dt = ( − n (cid:90) − t n ϕ ( t ) dt = ( − n (cid:90) − t n f ( t ) dt = − ( − n n (cid:90) − t n − f ( t ) dt = ( − n ( − n n ! (cid:90) − f n ( t ) dt = n ! f n +1 (0) = n ! 2( n +12 ) ϕ (1 − − n − ) . Moreover ∞ (cid:88) n =0 x n n ! (cid:90) +1 − t n ϕ ( t ) dt = (cid:90) +1 − e xt ϕ ( t ) dt = (cid:98) ϕ (cid:16) ix π (cid:17) = ∞ (cid:88) k =0 c k (2 k )! x k , and this proves (34). (cid:3) From the two formulas we obtain(35) ϕ (1 − − n − ) = 2 − ( n +12 )2(2 n )! F n (2 n + 1)(2 n − · · · n (cid:89) k =1 (2 k − − , where F k are the integers defined in (8).We may compute in a similar way all the numbers ϕ (1 − − n ). With this objectivenotice that (cid:90) ϕ ( t ) e − πixt dt = 12 πix + (cid:90) ϕ (cid:48) ( t ) e − πixt πix dt = 12 πix − (cid:90) ϕ (2 t − e − πixt πix dt = 12 πix (cid:16) − e − πix (cid:98) ϕ (cid:16) x (cid:17)(cid:17) . Therefore(36) (cid:90) e xt ϕ ( t ) dt = ∞ (cid:88) n =0 x n n ! (cid:90) t n ϕ ( t ) dt = − x (cid:16) − e x (cid:98) ϕ (cid:16) ix π (cid:17)(cid:17) from which we obtain ϕ (1 − − n ). Another way to compute these numbers is to use(37) f ( x ) = 1 + x (cid:90) e xt ϕ ( t ) dt = e x (cid:98) ϕ (cid:16) ix π (cid:17) , together with the fact that(38) f (2 x ) = e x − x f ( x ) . Therefore(39) f ( x ) = ∞ (cid:88) n =0 d n n ! x n , where d = 1 and we have the recurrence(40) ( n + 1)(2 n − d n = n − (cid:88) k =0 (cid:18) n + 1 k (cid:19) d k . It follows that there are integers G n such that(41) d n = G n ( n + 1)! n (cid:89) k =1 (2 k − − . The numbers d n , equation (33) and(42) d n = n (cid:90) t n − ϕ ( t ) dt determine the values of ϕ (1 − − n ).We may prove now the following theorem: Theorem 7.
The function ϕ takes rational values at each dyadic point.Proof. Let t = q/ n with | q | < n . We compute ϕ ( q − n ). Since ϕ and all itsderivatives vanish at the point −
1, Taylor’s theorem with the rest in integral formgives us ϕ ( q − n ) = (cid:90) t − ( t − x ) n n ! ϕ ( n +1) ( x ) dx. Applying our formula for the derivatives of ϕ we obtain ϕ ( t ) = 1 n ! 2( n +22 ) (cid:90) t − ( t − x ) n θ (2 n +1 (1 + x )) dx. Since for 2 h ≤ n +1 (1 + x ) ≤ h + 1) we have θ (2 n +1 (1 + x )) = ( − s ( h ) ϕ (2 n +1 (1 + x ) − h − n +1 (1 + x ) − h − u we obtain ϕ ( t ) = 1 n ! 2( n +22 )2 − n − q +2 n − (cid:88) h =0 ( − s ( h ) (cid:90) − (cid:16) t − u n +1 − h + 12 n +1 + 1 (cid:17) n ϕ ( u ) du = 1 n ! 2 − ( n +12 ) q +2 n − (cid:88) h =0 ( − s ( h ) (cid:90) − (cid:0) q − h ) + 2 n +1 − − u (cid:1) n ϕ ( u ) du = 1 n ! 2 − ( n +12 ) q +2 n − (cid:88) h =0 ( − s ( h ) n (cid:88) k =0 (cid:18) nk (cid:19)(cid:0) q − h )+2 n +1 − (cid:1) n − k ( − k (cid:90) R u k ϕ ( u ) du. This formula, together with equality (cid:90) − u n ϕ ( u ) du = (cid:0) − n (cid:1) (cid:90) u n ϕ ( u ) du and (34) proves our theorem, and we obtain ϕ ( q − n ) = 2 q +2 n − (cid:88) h =0 (cid:98) n/ (cid:99) (cid:88) k =0 ( − s ( h ) k +12 ) − ( n +12 )( n − k )! (cid:0) q − h )+2 n +1 − (cid:1) n − k ϕ (1 − − k − ) (cid:3) NFINITELY DIFFERENTIABLE FUNCTION 11
For the computation we may first obtain the common denominator of ϕ ( q − n ) fora fixed n , and using (30) it is possible then to compute the exact value of ϕ ( q − n ).For n = 5 the common denominator is 33 177 600 = 2 and we obtain q
33 177 600 ϕ ( q/ q
33 177 600 ϕ ( q/ q
33 177 600 ϕ ( q/ References [1] N. Bourbaki,
Fonctions d’une variable r´eelle , Hermann, Paris, 1958.[2] E. Hewitt and K. Stromberg,
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Distribution functions and the Riemann zeta function , Trans. Amer.Math. Soc. (1935), 48–88. Univ. de Sevilla, Facultad de Matem´aticas, c/ Tarfia, sn, 41012-Sevilla, Spain
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