aa r X i v : . [ m a t h . C V ] M a r On The Indefinite Sum In Fractional Calculus
James Nixon
Abstract.
We present a theorem on taking the repeated indefinite summa-tion of a holomorphic function φ ( z ) in a vertical strip of C satisfying expo-nential bounds as the imaginary part grows. We arrive at this result usingtransforms from fractional calculus. This affords us the ability to indefinitelysum more complicated functions than previously possible; such as holomorphicfunctions of order n ∈ N that have decay at plus or minus imaginary infinity.We then further investigate the indefinite summation operator by restrictingourselves to a space of functions of exponential type. We arrive at a secondrepresentation for the indefinite summation operator, equivalent to the firstpresented, and show we have defined a unique operator on this space. We de-velop a convolution using the indefinite sum that is commutative, associative,and distributive over addition. We arrive at a formula for the complex iteratesof the indefinite sum (the differsum), using this convolution, that resemblesthe Riemann-Liouville differintegral. We close with a generalization of theGamma function.
1. Introduction
Indefinite summation is an operation in mathematics that has a long history.Many mathematicians have investigated the operator, including Ramanujan [ ] andEuler [ ]. The reasoning behind defining indefinite summation is to give mathe-matical sense to the analytic continuation of sums of a holomorphic function. Asin, suppose S ( n ) = P nj =1 f ( j ) for n ∈ N , what happens if we replace n with z ∈ C and require S be holomorphic? There exists techniques for generating the indefi-nite sum of holomorphic functions and each has its advantages and its flaws, wherethe majority of the results are accumulated using Bernoulli polynomials and theLaplace transform. In this paper we will distance ourselves from these methodsand approach the problem with the differintegral. We will try to explain our bestthe advantages of our technique. Mathematics Subject Classification.
Key words and phrases.
Complex Analysis, Gamma Function, Recursion, Complex itera-tions, Hyper-operators.
The method we introduce for finding the indefinite sum of a holomorphic func-tion employs a triple integral transform that comes from fractional calculus. Al-though clunky at first sight, many of the results work out cleanly and with fairease, despite the mess of symbols. The advantage of our method, in comparisonto Ramanujan’s method, is that we achieve a more localized form of the operatoron holomorphic functions; whereas Ramanujan’s method requires the function beanalytic in a half plane of C , we only require it be analytic in a vertical strip of C . Also, unlike Euler-Maclaurin and Ramanujan summation, our method of per-forming indefinite summation only requires knowledge of the function on a verticalline in C (or its values at the positive integers), and none of its derivatives areused in the formula. Our method also allows us to perform an unlimited number ofindefinite summations in a single concise formula, which is much more cumbersomeusing Euler-Maclaurin summation or Ramanujan summation. Further, our methodonly requires an exponential bound as the imaginary part of the argument grows,this implies we can take the indefinite sum of functions of higher order like e z since it has decay at plus or minus imaginary infinity, which is not possible withRamanujan and Euler-Maclaurin summation.A large problem that needs to be addressed when investigating the indefinitesum is the uniqueness of a particular indefinite summation operator. By example,we can always add a one-periodic function to the indefinite sum and still get anotherindefinite sum [ ]. For this reason an additional criterion is added to the indefinitesum so that it satisfies uniqueness. We do not encounter this problem when definingindefinite summation with our transforms because adding a one-periodic functionto our indefinite sum causes our integral expression to blow up at plus or minusimaginary infinity and it no longer converges. This ensures that our indefinite sumis unique, although it satisfies moderately stricter restrictions.We elaborate further on a series of results involving the indefinite summationoperator that we can arrive at. We define a space we can perform indefinite sum-mation in, and from this we develop a second representation of the indefinite sumequivalent to the first. This representation only requires knowledge of the indef-inite sum at the positive integers. From here we are given open ground to talkfreely about interesting formulae which arrive due to the recursive behaviour of theindefinite sum.We define a convolution involving the indefinite sum. It allows us to express the“differsum” in a formula that resembles the Riemann-Liouville differintegral verymuch. We arrive at these results by only observing the behaviour of the indefinitesum at the non-negative integers by using a convenient and powerful lemma. Weclose by first iterating the indefinite sum to an arbitrary complex number in theright half plane, and then the indefinite product of the Gamma function. Thisproduces a generalized function that interpolates z and Γ( z ) while maintaining arecursive pattern.With these formalities we introduce the indefinite summation operator. N THE INDEFINITE SUM IN FRACTIONAL CALCULUS 3
Definition . If φ is holomorphic on open Ω ⊂ C then an indefinite sum P of φ is denoted P z φ and is holomorphic on Ω and for z, z + 1 ∈ Ω satisfies, X z φ + φ ( z + 1) = X z +1 φ We note the strongest result on indefinite summation in complex analysis thatwe know of, Ramanujan’s method, which produces an indefinite sum for any holo-morphic function φ of exponential type α < π (as in | φ ( z ) | < Ce α | z | ) [ ]. So farwe have not defined the indefinite sum quite rigorously, as issues of uniqueness needto be addressed, however, we are satisfied with this formal definition until we bringforth our analytic integral expression for the operator.We introduce the backwards difference operator. It appears in combinatoricsand recursion quite frequently. Despite its simple definition, considering it foranalytic functions proves trickier in comparison to the derivative operator (of whichit is a discrete analogue). Definition . If φ ( z ) is holomorphic on open Ω and there is an open subsetΩ ′ ⊂ Ω such that z ∈ Ω ′ ⇒ z, z − ∈ Ω then the backwards difference ▽ z of φ onΩ ′ is given by, ▽ z φ = φ ( z ) − φ ( z − P ▽ = ▽ P = 1 or P = ▽ − . Thus, these operators are inverses of each other.Before progressing we introduce a function that is invaluable to our research,Euler’s Gamma function. Definition . The Gamma function Γ is a meromorphic function on theentire complex plane with poles at the nonpositive integers, for z ∈ C , ℜ ( z ) > z ) = Z ∞ e − t t z − dt The Gamma function satisfies the functional relationship, z Γ( z ) = Γ( z + 1)and converges to the factorial for natural values, Γ( n + 1) = n !. We have imaginaryasymptotics of the Gamma function, which are,(1.2) | Γ( σ + iy ) | ∼ √ πe − σ − π | y | / | y | σ +1 / y → ±∞ There exists a plethora of material on the Gamma function and a quick searchwill give papers unlimited.Before introducing our results, we present our differintegral, which will be thehidden tool we will be applying throughout. For our purposes we only define thedifferintegral centered at zero.
NIXON
Definition . Let f ( x ) be holomorphic in the open sector Ω = { x ∈ C ; π <η < arg( − x ) < η ′ < π ; η, η ′ ∈ R , η ′ − η π } . Assume there exists b ∈ R + dependent on f such that for σ ∈ R + , < σ < b and θ ∈ R , η < θ < η ′ wehave R ∞ | f ( − e iθ x ) | x σ − dt < ∞ . Then, the Weyl differintegral centered at zero d − z dx − z (cid:12)(cid:12)(cid:12) x =0 of f is represented for z , 0 < ℜ ( z ) < b ,(1.3) d − z dx − z (cid:12)(cid:12)(cid:12) x =0 f ( x ) = e iθz Γ( z ) Z ∞ f ( − e iθ x ) x z − dx The differintegral as we’ve defined it is holomorphic in z . We can retrieve f from its differintegral centered at zero by using Mellin’s inversion theorem, f ( − e iθ t ) = 12 πi Z σ + i ∞ σ − i ∞ Γ( ξ ) (cid:16) d − ξ dx − ξ (cid:12)(cid:12)(cid:12) x =0 f ( x ) (cid:17) e − iθξ t − ξ dξ This result shall give us a lot to work with in our arsenal in order to handlethe problem of producing an indefinite sum. The last operator we introduce isthe Twisted Mellin Transform. This operator presents a strong relationship be-tween differentiating a function and performing the backwards difference of thetransformed function. The operator is introduced in [ ], however we found it inde-pendently and write it in a slightly generalized form. Definition . Suppose f is holomorphic in the open sector Ω = { x ∈ C ; − π/ < − η < arg( x ) < η < π/ η ∈ R + } . Assume for σ ∈ R + in the strip b ∈ R + , < σ < b and θ ∈ R , − η < θ < η we have R ∞ e − cos( θ ) x | f ( e iθ x ) | x σ − dx < ∞ .Then, the Twisted Mellin transform Y of f is represented for 0 < ℜ ( z ) < b , Y z f = d − z dx − z (cid:12)(cid:12)(cid:12) x =0 e x f ( − x )The fundamental property of the Twisted Mellin transform is,(1.4) Y z dfdx = ▽ z Y z f which follows by, d − z dx − z (cid:12)(cid:12)(cid:12) x =0 [ e x f ′ ( − x )] = d − z dx − z (cid:12)(cid:12)(cid:12) x =0 h e x f ( − x ) − ddx e x f ( − x ) i = d − z dx − z (cid:12)(cid:12)(cid:12) x =0 e x f ( − x ) − d − ( z − dx − ( z − (cid:12)(cid:12)(cid:12) x =0 e x f ( − x )With these introductions we see we have enough to state our analytic contin-uation of the summation operator. We hope the brevity of our elaboration on theresults outside of this paper we shall use does not affect its comprehension.
2. The Analytic Continuation of the Summation Operator
We introduce a few lemmas on the absolute convergence of certain modifiedMellin transforms. These lemmas shall follow quickly and are applied one afterthe other to the triple integral transform that will be our method of analytically
N THE INDEFINITE SUM IN FRACTIONAL CALCULUS 5 continuing the summation operator. The first lemma is inspired by the Paley-Wiener theorem on the Fourier transform, where they show holomorphic functionsbounded on a horizontal strip in C by x have a Fourier transform bounded by anexponential in the same horizontal strip [ ]. Our result appears much more fittingto our differintegral however. We prove the result using the differintegral, but all ittakes is some clever re-maneuvering to make it apply on the Mellin transform–whichwe leave to the interested reader. Lemma . Let φ be holomorphic in the strip < ℜ ( z ) = σ < b for b, σ, ∈ R + and let | lim ℜ ( z ) → φ ( z ) | < ∞ . Suppose that | φ ( z ) | < Ce α |ℑ ( z ) | for α ∈ R , α < π/ and C ∈ R + . Then f ( x ) = πi R σ + i ∞ σ − i ∞ Γ( ξ ) φ ( ξ ) x − ξ dξ is holomorphicin the sector α − π/ < arg( x ) < π/ − α and R ∞ | f ( e iθ x ) | x σ − dx < ∞ for θ ∈ R , α − π/ < θ < π/ − α .Conversely, let f be holomorphic in the sector α ∈ R , α − π/ < arg( x ) <π/ − α . Suppose for σ ∈ R + in the strip b ∈ R + , < σ < b and θ ∈ R , α − π/ <θ < π/ − α we have R ∞ | f ( e iθ x ) | x σ − < ∞ . Then φ ( z ) = z ) R ∞ f ( x ) x z − dx isholomorphic in the strip < ℜ ( z ) < b and | φ ( z ) | < M e α |ℑ ( z ) | for M ∈ R + . Proof.
We prove the first direction. Since | φ ( z ) | < Ce α |ℑ ( z ) | , for α − π/ <θ < π/ − α and x ∈ R + , f ( e iθ x ) = πi R σ + i ∞ σ − i ∞ Γ( ξ ) φ ( ξ ) e − iθξ x − ξ dξ . f is holomor-phic because πi R σ + inσ − in Γ( ξ ) φ ( ξ ) x − ξ dξ → πi R σ + i ∞ σ − i ∞ Γ( ξ ) φ ( ξ ) x − ξ dξ uniformly in x in the open sector. We also know that | f ( e iθ x ) | < x − σ π R σ + i ∞ σ − i ∞ | Γ( ξ ) φ ( ξ ) e − iθξ | dz = C σ x − σ where the integral is absolutely convergent again because of the boundson φ and the asymptotics of the Gamma function (1.2). Now we can shift σ inthe open strip and this does not affect convergence so that | f ( e iθ x ) | < Cx − ǫ as x → ǫ > | f ( e iθ x ) | < Cx − b as x → ∞ for C ∈ R + and C =sup σ ∈ [ ǫ,b ] { π R σ + i ∞ σ − i ∞ | Γ( ξ ) φ ( ξ ) e − iθξ | dξ } . This ensures that R ∞ | f ( e iθ x ) | x σ − dx < ∞ for 0 < σ < b and α − π/ < θ < π/ − α .For the other direction we haveΓ( z ) φ θ ( z ) = Z ∞ f ( e iθ x ) x z − dx = e − iθz Z ∞ f ( x ) x z − dx = e − iθz Γ( z ) φ ( z )which follows by contour integration. Now take the absolute value to see that | φ ( z ) | < C σ e θ ℑ ( z ) | Γ( z ) | where α − π/ < θ < π/ − α and C σ = R ∞ | f ( e iθ x ) | x σ − dx .By the asymptotics of the Gamma function (1.2), | φ ( z ) | < M e ( π/ − θ ) |ℑ ( z ) | whichby taking | θ | = π/ − α we get | φ ( z ) | < M e α |ℑ ( z ) | , where φ = φ and M ∈ R + .We note we have dropped the term | y | − σ − / in the inverse Gamma function’sasymptotics because it decays to zero as y → ±∞ . (cid:3) What we have just shown is subtle but very useful. We have not seen itproved before and so we decided to show a proof–though we have our suspicionsa proof is already out there. It shows we have an isomorphism between spaces ofholomorphic functions (the space f is in to the space φ is in) which is something wecan manipulate freely. Our next lemma comes from asymptotics for the incompleteGamma function and is the core of the convergence of our triple integral. NIXON
Lemma . Let f ( x ) be holomorphic in the sector Ω = { x ∈ C ; α − π/ < arg( x ) < π/ − α } . Suppose for σ ∈ R + in the strip b ∈ R + , < σ < b and θ ∈ R , α − π/ < θ < π/ − α we have R ∞ | f ( e iθ x ) | x σ − dx < ∞ . Then for λ ∈ R + and < σ < b , Z ∞ e − λt t σ − (cid:16) Z t e λx | f ( e iθ x ) | dx (cid:17) dt < ∞ Proof.
Let us first note that f (0) < C so that R t e x f ( x ) dx < ∞ . Performintegration by parts: Z ∞ e − λt t σ − Z t e λx | f ( e iθ x ) | dx dt = Z ∞ e λx | f ( e iθ x ) | Z ∞ x e − λt t σ − dt dx From looking at [ ] we have an asymptotic expansion of the incomplete Gammafunction in the integral, R ∞ x e − λt t σ − dt ∼ (cid:16) xλ (cid:17) σ e − λx P ∞ k =0 b k ( λx − σ ) k +1 which implieswe can find a C ∈ R + such that R ∞ x e − λt t σ − dt < Ce − λx x σ − . Where this followsbecause we can factor 1 /x from the sum and the rest of the series is O (1) as x → ∞ .This shows, Z ∞ e λx | f ( e iθ x ) | Z ∞ x e − λt t σ − dt dx < C Z ∞ | f ( e iθ x ) | x σ − dx This gives the result. (cid:3)
We can now give our theorem on the analytic continuation of the summationoperator. We may not have given enough motivation as to why the integral trans-forms we present are the ones we are looking for, but the recursive behaviour issatisfied and from a familiarity with fractional calculus it seems less questionableand more apparent. The expression may seem to have been plucked from the airsince we have tried to separate the direct attachment to the differintegral, howeversomeone who is familiar with the transforms will clearly see that we are conjugating(in terms of abstract algebra) the integral operator by the Twisted Mellin trans-form, where the inverse Twisted Mellin transform is calculated using the inverseMellin transform.
Theorem . Let φ ( z ) be holomorphic in the strip b ∈ R + , < ℜ ( z ) < b , b > and let | lim ℜ ( z ) → φ ( z ) | < ∞ . Let | φ ( z ) | < Ce α |ℑ ( z ) | for α ∈ R , α < π/ and C ∈ R + . Then the indefinite sum of φ , P z φ is represented for < σ < b and < ℜ ( z ) < b by, (2.1) X z φ = 1Γ( z ) Z ∞ e − t t z − (cid:16) Z t e x πi (cid:16) Z σ + i ∞ σ − i ∞ Γ( ξ ) φ ( ξ ) x − ξ dξ (cid:17) dx (cid:17) dt or written more compactly, (2.2) X z φ = Y z Z x Y − x φ N THE INDEFINITE SUM IN FRACTIONAL CALCULUS 7
Proof.
We first show convergence of the triple integral. This will not betoo difficult as the bulk of the work was placed in the lemmas before. If f ( x ) = πi R σ + i ∞ σ − i ∞ Γ( ξ ) φ ( ξ ) x − ξ dξ it is holomorphic in the sector α − π/ < arg( x ) < π/ − α and satisfies the conditions of Lemma 2.1, so we know, R ∞ | f ( e iθ x ) | x σ − dx < ∞ for σ ∈ R + in the strip 0 < σ < b . Therefore f satisfies the conditions of Lemma 2.2,which implies R ∞ e − t t σ − R t e x | f ( e iθ x ) | dx dt < ∞ . Therefore the triple integraltransform is convergent. Also, R ∞ e − cos( θ ) t t σ − | R e iθ t e e iθ x f ( x ) dx | dt < ∞ becauseall of these transforms converge in sectors, so in the last step we are justified inapplying the Twisted Mellin transform.Now we show that this is in fact an indefinite sum of φ . Let us assume that1 < ℜ ( z ) < b so that ▽ z P z φ exists and we get, by applying Equation (1.4), ▽ z X z φ = ▽ z h z ) Z ∞ e − t t z − Z t e x f ( x ) dx dt i = 1Γ( z ) Z ∞ e − t t z − ddt (cid:16) Z t e x f ( x ) dx (cid:17) dt = 1Γ( z ) Z ∞ e − t t z − e t f ( t ) dt = 1Γ( z ) Z ∞ f ( t ) t z − dt = φ ( z )Therefore we know that this operator satisfies ▽ P = 1. Let us now showit happens on the right as well. We first note that πi R σ + i ∞ σ − i ∞ Γ( ξ )( φ ( ξ ) − φ ( ξ − x − ξ dξ = f ( x ) + f ′ ( x ) and that R t e x ( f ( x ) + f ′ ( x )) dx = e t f ( t ) − f (0), X z ( ▽ z φ ( z )) = 1Γ( z ) Z ∞ e − t t z − Z t e x ( f ( x ) + f ′ ( x )) dx dt = 1Γ( z ) Z ∞ ( f ( t ) − f (0) e − t ) t z − dt = φ ( z ) − f (0)= φ ( z ) + A Therefore we have shown ▽ P φ = φ and P ▽ φ = φ + A (where A ∈ C , andthis constant is much like a constant of integration, as in R tc g ′ ( t ) dt = g ( t ) + A ) sothat it is natural to say ▽ − = P , which is a defining property of P and assuresthat P z φ + φ ( z + 1) = P z +1 φ . This gives the result. (cid:3) More advantageously we can show that P z is exponentially bounded as itsimaginary argument grows so that we may consider P z P z φ . Going on in such a NIXON manner we can produce all the natural iterates of the indefinite summation opera-tor.
Lemma . Let φ ( z ) be holomorphic in the strip b ∈ R + , < ℜ ( z ) < b for b > and let | lim ℜ ( z ) → φ ( z ) | < ∞ . Assume | φ ( z ) | < Ce α |ℑ ( z ) | for α ∈ R , α <π/ and C ∈ R + . Then, | P z φ | < M e α |ℑ ( z ) | for M ∈ R + . Proof.
Observe the representation of the indefinite sum in (2.1) and observethat the final integral transform applied is the Mellin transform. Observe further, if f ( x ) = πi R σ + i ∞ σ − i ∞ Γ( ξ ) φ ( ξ ) x − ξ dξ we know f is holomorphic in the sector α − π/ < arg( x ) < π/ − α . If α − π/ < θ < π/ − α then, R ∞ t z − e − t R t e x f ( x ) dx dt = e iθz R ∞ t z − e − e iθ t R t e e iθ x f ( e iθ x ) dx dt . We have done a substitution through bothintegrals which is justified since all these transforms converge in sectors. We knowthat θ is arbitrary and the final function does not depend on θ by consideringcontour integrals. We leave the details to be filled by the reader. These integralsare absolutely convergent, again by Lemma 2.1 and Lemma 2.2. By Lemma 2.2 wealso see that | Z ∞ t z − e − e iθ t Z t e e iθ x f ( e iθ x ) dx dt | Z ∞ t σ − e − t cos( θ ) Z t e x cos( θ ) | f ( e iθ x ) | dx dt< K Z ∞ | f ( e iθ x ) | x σ − dx = C σ for some K ∈ R + . This shows that, X z φ = e iθz Γ( z ) Z ∞ e − e iθ t t z − Z t e e iθ x f ( e iθ x ) dx dt so we have | P z φ | < C σ e ( π/ − θ ) |ℑ ( z ) | which gives us | P z φ | < M e α |ℑ ( z ) | for an M ∈ R + . This shows the result. (cid:3) With this we can state our corollary on the fact we can take any natural iterateof the indefinite sum and it will remain holomorphic in the same strip.
Corollary . If φ ( z ) is holomorphic in the strip b ∈ R + , < ℜ ( z ) < b for b > and | φ ( z ) | < Ce α |ℑ ( z ) | for α ∈ R , α < π/ and C ∈ R + and | lim ℜ ( z ) → φ ( z ) | < ∞ then φ can be indefinitely summed an arbitrary amount oftimes (2.3) X z ... ( n times ) ... X z φ = Y z Z x ... ( n times ) ... Z x Y − x φ Proof.
Observe Theorem 2.1 and Lemma 2.3. Theorem 2.1 ensures we canindefinitely sum φ and then Lemma 2.3 ensures that that indefinite sum satisfies theoriginal conditions of Theorem 2.1. The representation (2.3) follows by inductionand by applying Lemma 2.2 repeatedly. (cid:3) Now before ending this section and conversing solely of functions of exponentialtype, we see that functions like e z as well as e z are indefinitely summable. We N THE INDEFINITE SUM IN FRACTIONAL CALCULUS 9 can find more complicated functions of order greater than one that are indefinitelysummable if they behave nicely as the imaginary argument grows.
3. The exponential space E and the indefinite sum As we’ve defined the indefinite sum at this point, we’ve only used functionsbounded by an exponential as the imaginary part grows. We now consider boundingthe function by an exponential as the real part of z grows as well. For this reason, wedefine an exponential space in which the indefinite sum operates. The exponentialbounds will ensure we have a unique operator on the space and will show our secondrepresentation as well as our first representation are equivalent where they intersectin definition. Definition . Suppose f ( z ) is a holomorphic function on the right halfplane z ∈ C , ℜ ( z ) > | lim ℜ ( z ) → f ( z ) | < ∞ . Assume there exists α, ρ, C ∈ R + , α < π/ , such that | f ( z ) | < Ce α |ℑ ( z ) | + ρ |ℜ ( z ) | . The space E contains allsuch functions f that satisfy these conditions. E contains polynomials, exponentials with a base that have a real part greaterthan zero, rationals with poles in the left half plane, and other special functions.Aside, on notation, we will write α f for the bound of f as the imaginary part growsand we will write ρ f for the bound of f as the real part grows (when distinctionbetween functions is required). For our purposes we will be using the indefinitesum we defined in Theorem 3.1. Let f belong to E and let σ ∈ R + , σ > X z f = 1Γ( z ) Z ∞ t z − e − t Z t e x πi Z σ + i ∞ σ − i ∞ Γ( ξ ) f ( ξ ) x − ξ dξ dx dt We know that if | f ( z ) | < C ℜ ( z ) e α |ℑ ( z ) | that this representation converges andthat | P z f | < M ℜ ( z ) e α |ℑ ( z ) | . Therefore P is defined for any element of E . We alsohave the convenient formula, X n f = n X j =1 f ( j )which follows by the defining recursion of P and because P f = 0 since lim z → d − z dx − z (cid:12)(cid:12)(cid:12) x =0 e x R x e t g ( t ) dt = e x R x e t g ( t ) dt (cid:12)(cid:12)(cid:12) x =0 = 0. This limit is allowed to be taken because | e − x R x e t g ( t ) dt |
Let us say | f | < Ce α |ℑ ( z ) | + ρ |ℜ ( z ) | . We know that P z f is bounded byan exponential and of type less than π/ X z f = X z −⌊ℜ ( z ) ⌋ f + ⌊ℜ ( z ) ⌋ X j =1 f ( z − ⌊ℜ ( z ) ⌋ + j ) which comes from the defining property of P and shows that, | X z f | < ⌊ℜ ( z ) ⌋ X j =0 M e α |ℑ ( z ) | + ρ |ℜ ( z ) −⌊ℜ ( z ) ⌋ + j | < M |⌊ℜ ( z ) ⌋ + 1 | e α |ℑ ( z ) | + ρ |ℜ ( z ) | And therefore since |⌊ℜ ( z ) ⌋ + 1 | < Ce ǫ |ℜ ( z ) | for some ǫ > (cid:3) With this we are now able to talk freely of the indefinite sum acting on aspace. We can also give a more compact form for the operator which follows byRamanujan’s master theorem.
Corollary . Let f ∈ E and define the entire function ϑ ( x ) = P ∞ n =0 (cid:16) P n +1 j =1 f ( j ) (cid:17) x n n ! .Then for z ∈ C , ℜ ( z ) > , X z f = 1Γ(1 − z ) (cid:16) ∞ X n =0 (cid:0) n +1 X j =1 f ( j ) (cid:1) ( − n n !( n + 1 − z ) + Z ∞ ϑ ( − x ) x − z dx (cid:17) With these results we are prepared to analyze the indefinite sum in this spacemore intricately.
4. On Indefinite Sum Convolution
Beginning this section we will prove a lemma that appears elsewhere but weshow it using techniques from fractional calculus. This result will be applied re-peatedly on various transforms involving the indefinite sum which will give us apowerful way of proving properties of the indefinite sum over the complex plane bylooking at its values on the positive integers.
Lemma . Let f ∈ E and for n ∈ N assume f ( n ) = 0 , then f = 0 . Proof.
Take 0 < σ < g ( x ) = πi R σ + i ∞ σ − i ∞ Γ( ξ ) f (1 − ξ ) x − ξ dξ . Bya simple exercise in contour integration we have g ( x ) = P ∞ n =0 f ( n + 1) x n n ! = 0 thisimplies that f = 0. (cid:3) We see this result follows quickly, however its power is abundant. Firstly, itimplies if f ( n ) = f ( n ) for n ∈ N then f ( z ) = f ( z ) for all z ∈ C ℜ ( z ) > . We willnot waste time in putting it to use. With this we define a new convolution that isassociative, commutative, and spreads across addition. To be clear, we will makethe following statement on notation, P z f = P z f ( s ) ▽ s , where the additionalterms declare the variable we are performing the sum across. This notation allowsus to nest more variables which is required for what comes next. Definition . Suppose f ( z − , g ( z − ∈ E and α f + 2 α g < π/
2, then f × g = X z +1 f ( s − g ( z + 1 − s ) ▽ s . N THE INDEFINITE SUM IN FRACTIONAL CALCULUS 11
We note this is definitively a convergent expression if we fix z and take theindefinite sum from Theorem 2.1. The motivation for introducing this convolutionis because ( f × g )( n ) = P nj =0 f ( j ) g ( n − j ) which is quite a familiar expression.Furthermore we know that the space E is almost completely determined by how itsfunctions behave on the naturals so this is bound to be valuable in some sense. Weshow some properties of this convolution acting in the space E . Theorem . Suppose f, g, h ∈ E then: (1) If α f + 2 α g < π/ , then if f × g = h we have h ∈ E . (2) If α f + 2 α g < π/ and α f + α g < π/ , then f × g = g × f . (3) If α f + 2( α g + 2 α h ) < π/ and then f × ( g × h ) = ( f × g ) × h . (4) If α f + 2 α g < π/ and α f + 2 α h < π/ then f × ( g + h ) = ( f × g )+ ( f × h ) . (5) If α f + 2 α g < π/ then f × g = 0 implies that f = 0 or g = 0 . (6) Let ( z + 1) s − = Γ( z + s )Γ( z +1) for s ∈ C ℜ ( z ) > and s , s ∈ C ℜ ( z ) > we have s ) ( z + 1) s − × s ) ( z + 1) s − = s + s ) ( z + 1) s + s − . Proof.
We prove each result one by one.(1) First take f × g = P z −⌊ℜ ( z ) ⌋ f ( s − g ( z + 1 − s ) ▽ s + P ⌊ℜ ( z ) ⌋ j =1 f ( z −⌊ℜ ( z ) ⌋ − j ) g ( ⌊ℜ ( z ) ⌋ + 1 − j ). Therefore since | f ( s − g ( z + 1 − s ) | 1. By associativity the result followsfor integers so that n − ( z + 1) n − × m − ( z + 1) m − = n + m − ( z +1) n + m − . Now apply Lemma 4.1 on s and s since Γ( z + s )Γ( z +1)Γ( s ) ∈ E in s , s when ℜ ( z ) , ℜ ( s ) , ℜ ( s ) > 0. The result follows from this.This shows all the cases. (cid:3) With this we are prepared to show a new representation of complex iterationsof the indefinite sum. To this end we define a new operator called the differsum.This term is novel to us and is used to express the operators similarity to thedifferintegral. Definition . Let φ ( z ) be holomorphic in the open set Ω. A differsum ▽ sz of φ is holomorphic in s on open Ω ′ and holomorphic in z for z ∈ Ω, satisfying: (1) For n ∈ N we have ▽ nz φ = ▽ z ▽ z . . . ( n times ) . . . ▽ z φ and ▽ − nz φ = P z P z . . . ( n times ) · · · P z φ .(2) For s , s , s + s ∈ Ω ′ , ▽ s z ▽ s z φ = ▽ s + s z φ .The definitive properties of the differsum is what we would expect of an iterateof the backwards difference operator/indefinite sum. It appears as an object thatwould be very difficult to analyze. Thankfully though, the deep connection betweenrecursion and our differintegral allows us to speak freely of the differsum. Theorem . Suppose f ( z ) ∈ E and q ∈ C , ℜ ( q ) > , then ▽ − qz f ( z ) = q ) P z f ( s )( z + 1 − s ) q − ▽ s . Proof. We make use of the fact ▽ − qz +1 f ( z − 1) = q ) ( f ( z ) × ( z + 1) q − )( z ),where this is well defined because f ∈ E and because ( z + 1) q − ∈ E in z and as theimaginary part of z grows it is bounded by M ǫ e ǫ |ℑ ( z ) | for any ǫ > 0. The proof of thisis by induction for q an integer. It is clearly true for q = n = 1. Assume for n . Since n − ( z + 1) n − × n ! ( z + 1) n and since ▽ − ( n +1) z +1 f ( z − 1) = ▽ − nz +1 f ( z − × f ( z ) × n − ( z +1) n − ) × f ( z ) × ( n − (( z +1) n − × n ! f ( z ) × ( z +1) n . Nowsince, for q , q ∈ C , we have q ) ( z +1) q − × q ) ( z +1) q − = q + q ) ( z ) q + q − we know that ▽ − q z +1 ▽ − q z f ( z − 1) = ▽ − ( q + q ) z +1 f ( z − z we have for ℜ ( q ) > ▽ − qz f ( s ) = 1Γ( q ) X z f ( s )( z + 1 − s ) q − ▽ s (cid:3) To close we make use of our expression for the differsum. We take the transformon log(1+ s ). This will produce a generalization of the Gamma function. This comesabout in the following sense, P z log(1 + s ) ▽ s = log(Γ(1 + z )), which follows byLemma 4.1. Exponentiate and we have a well defined expression for Γ. Furthermorethe iterates of the indefinite sum once exponentiated will produce functions Γ ( z ) = z, Γ ( z ) = Γ( z ) , Γ ( z ) , Γ ( z ) , ... such that Γ n ( z )Γ n +1 ( z ) = Γ n +1 ( z + 1). Thesefunctions are normalized at one so that Γ n (1) = 1.We wish to be more brazen than this however and to develop the complexiterates of this recursion. Namely for z, q ∈ C we have if ℜ ( q ) > ℜ ( z ) > q ( z ) = e q ) P z − log(1+ s )( z − s ) q − ▽ s satisfies the recursion Γ q ( z )Γ q +1 ( z ) = Γ q +1 ( z + 1). This function expands theΓ function considerably and allows for a broad generalization of the Barnes G-function. In spirit, Γ satisfies the same recursion as the Barnes G-function–we areunsure if the two agree however. 5. Final Remarks We close hoping the reader has seen the connection between fractional calculusand recursion. The indefinite sum behaves quite neatly and we have only scratched N THE INDEFINITE SUM IN FRACTIONAL CALCULUS 13 the surface of its structure and uses. We state that this paper is sister to [ ]–on thedifferintegral and complex iterations. The two problems appear as an intrigue in it-eration and show the power of fractional calculus and its familiarity and connectionwith iteration in general. For a more clear application of this, observe [ ]. We aresatisfied with the brevity of this paper as it is meant to show a pure mathematicalcuriousity which follows through fractional calculus.Although only briefly mentioned, our generalization of the Gamma functionoffers many questions. Can this functions be analytically continued to a mero-morphic function for q ∈ C and z ∈ C ? What is its Weierstrass factorization inaccordance to its multiplicative inverses zeroes? Explicitly, what is the Weierstrassfactorization of q ( z ) in z dependent on q ?We are aware of more advanced techniques that apply to more complicatedrecursions. Many linear algebraic recursions in T f ( z ) = f ( z + 1) can be solvedby these methods. Notably we solved for complex values of the recursion − T − .We can do this for more complicated recursions by creating a similar isomorphismsbetween spaces that the Twisted Mellin transform induces. We can also performfractional iterations of the recursion in a manner similar to how we produced thediffersum. These results are very exotic and further capitulate the importance ofthe differintegral. 6. Acknowledgements The author is indebted to the University of Toronto where he is a student whomakes constant use of their mathematics library, of which these results would’venever come to fruition without. References [1] Eric Delabaere, Ramanujan’s Summation , Algorithms Seminar, (2001-2002).[2] Chelo Ferreira, Jos´ e L. L´ o pez, Ester Prez Sinusa, Incomplete gamma functionfor large values of their variables , Advances in applied mathematics, (2004).[3] Elias M. Stein and Rami Shakarchi, Complex Analysis , Princeton UniversityPress, (2003).[4] Masaaki Sugihara, Justification of a formal derivation of the Euler-Maclaurinsummation formula , Analytic Extension Formulas and their Applications, Vol-ume 9, (2001), pp 251-261.[5] Zuoquin Wang, The Twisted Mellin Transform , arXiv, (2007).[6] James Nixon, Complex Iterations and Bounded Analytic Hyper-Operators ,arXiv, (2015). Toronto, Canada E-mail address ::