Novel Results on Series of Floor and Ceiling Functions
NNOVEL RESULTS ON SERIES OF FLOOR AND CEILING FUNCTIONS
DHAIRYA SHAH , MANOJ SAHNI , RITU SAHNI Department of Mathematics, Pandit Deendayal Petroleum University, Gandhinagar, 382007, India. Centre for Engineering and Enterprise, Institute of Advanced Research, Gandhinagar, 382007, India. [email protected], [email protected], [email protected] A BSTRACT. We propose and prove a couple of formulas and infinite series involving the floor and the ceiling functions. Formula relating to the difference of floor and ceiling functions is obtained using aforementioned formulas. Partial summations of floor and ceiling of q th roots of natural numbers are equated as simple formulas. Particular cases of the series are taken into consideration and it is proven that both the cases relate to the Riemann-Zeta function. Poles for the both series are mentioned and it is shown that even if both series individually fail to converge at the pole, their difference is convergent at the same. It is shown that our formulas reduce to the Gauss formula and the series reduce to the Riemann-Zeta for a particular value. Further some special cases and scope for future work are discussed. Keywords : Ceiling function; Floor function; Faulhaberβs formula; Riemann-Zeta function; Series I NTRODUCTION One of the immense pleasures in the research of mathematics is finding something new, which was never observed of recorded by humans. Throughout last three centuries, mathematicians such as Gauss, Euler, Bernoulli, Ramanujan, Legendre etc. ventured in the realm of pure mathematics, and especially in the domain of numbers. Some of them discovered a very profound formulas, functions and series of their era. The initial development of partial sum of first π natural numbers was given by Gauss [1] in the 18th century, and it was generalized for all natural powers of natural numbers, known as Faulhaber's formula (or Bernoulli's formula) [2]. In the mid nineteenth century Riemann [3] introduced a working definition for a function of complex variable β²π β² (β π βπ βπ=1 ) (The function with interesting history, already introduced for real variables by Euler in the 18th century). Legendre on the other hand introduced the notion of integer part of π₯ in late 18 th century, and Gauss introduced the [π₯] notation for the same, until in the second half of 20 th century when the terms floor and ceiling functions were coined. The authors, reading the history of number theory and studying the behavior of the Faulhaberβs formula [2], thought on extending the summation to real powers, instead of natural powers. While the work seems interesting, the main problem occurred in exacting the summation and keeping the outcome a natural number. As the non-integer powers may omit fractional or irrational values, authors had to break the problem into two, having floor and ceiling functions [4] of real powers. After having the formulas and proving them, authors considered having the same approach with the Riemann-Zeta function [3]. The infinite series with the floor and ceiling function and real powers turned out to be convergent. Considering π β β, π β (0,1], π β β + and π β β, (π π(π ) > ) The following study introduces novel formulas for β βπ π β ππ=1 , β βπ π β ππ=1 , β β(ππ) π β βπ βπ=1 , β β(ππ) π β βπ βπ=1 . Specific values are taken to comprehend the behavior these results. Additionally, we show that with some particular assumptions, the infinite series relate to the Zeta function. . P RELIMINARIES
The following results and definitions are useful for our study: [1]: The partial sums of the series 1 + 2 + 3 + β― + n (first n natural numbers) is given by a simple formula: β π = π(π + 1)2 ππ=1 . ( Bernoulli's) Formula [2]: An expression of the sum of the π powers of the first π positive integers can be equated as β π π = β β ( π+1π ) β π΅ π β π π+1βπππ=0ππ=1 , here π΅ π is Bernoulliβs number of second kind, defined as π΅ π = β 1π + 1 β ππ=0 β (β1) π‘ β (π‘ + 1) π β (ππ‘ ) ππ‘=0 For example, if π = 2 and as π΅ = 1, π΅ = , π΅ = β π = 12 + 1 β β (2 + 1π ) β π΅ π β π which implies β π = 13 β β (3π) β π΅ π β π = 13 π + 12 π + 16 π [3]: The Riemann-Zeta function π(π ) is a function of a complex variable π defined as infinite sum π(π ) = β 1π π β π=1 = 1Ξ(π ) β« π₯ π β1 π π₯ β 1 ππ₯ β0 , where Ξ(π ) = β« π‘ π β1 π βπ‘ ππ‘ β0 The function converges for all complex value of π when π π(π ) > 1 and defines π(π ) = 11 π + 12 π + 13 π + β― [4]: The floor function of any real number π₯ (denoted by βπ₯β ) gives the greatest integer not greater than π₯ . For example, β1.4β = 1, β2β = 2, ββ3.4β = β4 and ββ2β = β2 . The ceiling function (denoted by βπ₯β ), same way gives the smallest integer not smaller than π₯ . For example, β1.4β = 2, β2β = 2, ββ3.4β = β3 and ββ2β = β2 . From above we can see that βπ₯β = βπ₯β = π₯ if and only if π₯ β β€ . R ESULTS AND PROOFS : F
ORMULAS AND S ERIES R ESULTS:
In this section, we propose the results and provide their proofs. Considering π β β, π β (0,1], π β β + and π β β FORMULA 1: ββπ π β ππ=1 = (π + 1)βπ π β β β βπ‘ β βπ π βπ‘=1 FORMULA 2: ββπ π β ππ=1 = πβπ π β + ββπ π β β β β βπ‘ β βπ π βπ‘=1 SERIES 1: β 1β(ππ) π β π βπ=1 = β β(π + 1) π β β βπ π βπ π βπ=1 , π π(π ) > 1π SERIES 2: β 1β(ππ) π β π βπ=1 = β βπ b β β β(π β 1) b βπ π βπ=1 , π π(π ) > 1π Note: Formula 1 and 2 holds for all π β β + , but for π > 1 , value of βπ π β becomes larger than π and running the summation from 1 to βπ π β becomes less economical and hence case when π > 1 is not considered. P ROOFS:
FORMULA 1:
Let
π(π) be the following statement:
π(π): ββπ π β ππ=1 = (π + 1)βπ π β β β βπ‘ β βπ π βπ‘=1 (1) Basic step of induction : π(1) is clearly true.
π(1): 1 = (1 + 1)1 β 1 = 2 Γ 1 β 1 = 1 (β΅ π β (0,1] β β1 π β = β1 β = 1) Inductive step:
Assume
π(π) is true for some π = π . One shall prove
π(π) β π(π + 1)
π(π): ββπ π β ππ=1 = (π + 1)βπ π β β β βπ‘ β βπ π βπ‘=1 (2) It follows that (π + 1): ββπ π β π+1π=1 = (π + 1)βπ π β β β βπ‘ β βπ π βπ‘=1 + β(π + 1) π β (3) Consider π β β such that π = βπ π β . π = βπ π β β π β€ π π < π + 1 β π β€ π < (π + 1) (β΅ π β (0,1]) β βπ β β€ π < β(π + 1) β (β΅ π β β) β βπ β β€ π β€ β(π + 1) β β 1 This splits up in two cases (I): π = β(π + 1) β β 1 & (II): βπ β β€ π < β(π + 1) β β 1 Case (I): π = β(π + 1) β β 1 β π + 1 = β(π + 1) β Here, π β β β π + 1 β β β β(π + 1) β β β β β(π + 1) β = (π + 1) β π + 1 = (π + 1) β (π + 1) π = π + 1 Again π β β β π + 1 β β β (π + 1) π β β β (π + 1) π = β(π + 1) π β (Section 2.4) β β(π + 1) π β = π + 1 = βπ π β + 1 (β΅ π = βπ π β) β β(π + 1) π β = βπ π β + 1 , if π = β(π + 1) β β 1 β΄ From Equation (3)
π(π + 1): ββπ π β = (π + 1)βπ π β β β βπ‘ β βπ π βπ‘=1π+1π=1 + β(π + 1) π β β π(π + 1): ββπ π β = (π + 1)(β(π + 1) π β β 1) β β βπ‘ β β(π+1) π ββ1π‘=1π+1π=1 + β(π + 1) π β β π(π + 1): ββπ π β = (π + 1)β(π + 1) π β + β(π + 1) π β β (π + 1) β β βπ‘ β β(π+1) π ββ1π‘=1π+1π=1 Now, as π + 1 = β(π + 1) β and π + 1 = β(π + 1) π β we get π + 1 = ββ(π + 1) π β β β π(π + 1): ββπ π β = ((π + 1) + 1)β(π + 1) π β β ββ(π + 1) π β β β β βπ‘ β β(π+1) π ββ1π‘=1π+1π=1 β π(π + 1): ββπ π β = ((π + 1) + 1)β(π + 1) π β β β βπ‘ β β(π+1) π βπ‘=1π+1π=1 β΄ π(π) β π(π + 1) for case (I). Case (II): βπ β β€ π < β(π + 1) β β 1 β βπ β + 1 β€ π + 1 < β(π + 1) β βπ β < π + 1 < β(π + 1) β β π < (π + 1) π < π + 1 β β(π + 1) π β = π β β(π + 1) π β = βπ π β = π β΄ From Equation (3)
π(π + 1): ββπ π β = (π + 1)βπ π β β β βπ‘ β βπ π βπ‘=1π+1π=1 + β(π + 1) π β β π(π + 1): ββπ π β = (π + 1)β(π + 1) π β β β βπ‘ β β(π+1) π βπ‘=1π+1π=1 + β(π + 1) π β (β΅ βπ π β = β(π + 1) π β) β π(π + 1): ββπ π β = ((π + 1) + 1)β(π + 1) π β β β βπ‘ β β(π+1) π βπ‘=1π+1π=1 β΄ π(π) β π(π + 1) for case (II). β΄ π(π) β π(π + 1) βπ such that βπ β β€ π β€ β(π + 1) β β 1 Hence
π(π + 1) is true whenever
π(π) is true. Hence by Principle of Mathematical Induction
π(π) is true βπ β β.
FORMULA 2:
Let
π(π) be the following statement:
π(π): ββπ π β ππ=1 = πβπ π β + ββπ π β β β β βπ‘ β βπ π βπ‘=1 (4) Basic step of induction : π(1) is clearly true.
π(1): 1 = 1(1) + 1 β 1 = 1 (β΅ π β (0,1] β β1 β = β1 π β = ββ1 π β β = 1) Inductive step:
Assume
π(π) is true for some π = π . One shall prove
π(π) β π(π + 1)
π(π): ββπ π β ππ=1 = πβπ π β + ββπ π β β β β βπ‘ β βπ π βπ‘=1 (5) It follows that
π(π + 1): ββπ π β π+1π=1 = πβπ π β + ββπ π β β β β βπ‘ β βπ π βπ‘=1 + β(π + 1) π β (6) Consider π β β such that π = βπ π β . π = βπ π β β π β 1 < π π β€ π β (π β 1) < π β€ π (β΅ π β (0,1]) β β(π β 1) β < π β€ βπ β (β΅ π β β) β β(π β 1) β + 1 β€ π β€ βπ β This splits up in two cases I): π = βπ β & (II): β (π β 1) β + 1 β€ π < βπ β Case (I): π = βπ β β π = π (β΅ π β β β βπ β β β β βπ β = π ) β π + 1 = π + 1 β π + 1 > π β (π + 1) π > π β β(π + 1) π β > βπβ β β(π + 1) π β > π (β΅ π β β) β β(π + 1) π β = π + 1 β β(π + 1) π β = βπ π β + 1 (β΅ π = βπ π β) β β(π + 1) π β = βπ π β + 1 , if π = βπ β. β΄ From Equation (6)
π(π + 1): ββπ π β π+1π=1 = πβπ π β + ββπ π β β β β βπ‘ β βπ π βπ‘=1 + β(π + 1) π β β π(π + 1): ββπ π β π+1π=1 = π(β(π + 1) π β β 1) + β(π + 1) π β + ββπ π β β β β βπ‘ β β(π+1) π ββ1π‘=1 β π(π + 1): ββπ π β π+1π=1 = (π + 1)β(π + 1) π β β π + ββπ π β β β β βπ‘ β β(π+1) π ββ1π‘=1 (Adding and subtracting ββ(π + 1) π β β) Also. for case (I) we have βπ π β = π and βπ β = π β ββπ π β β = π β π(π + 1): ββπ π β π+1π=1 = (π + 1)β(π + 1) π β β π + π + ββ(π + 1) π β β β ββ(π + 1) π β β β β βπ‘ β β(π+1) π ββ1π‘=1 β π(π + 1): ββπ π β π+1π=1 = (π + 1)β(π + 1) π β + ββ(π + 1) π β β β ββ(π + 1) π β β β β βπ‘ β β(π+1) π ββ1π‘=1 β π(π + 1): ββπ π β π+1π=1 = (π + 1)β(π + 1) π β + ββ(π + 1) π β β β β βπ‘ β β(π+1) π βπ‘=1 β΄ π(π) β π(π + 1) for case (I). Case (II): β (π β 1) β + 1 β€ π < βπ β β β(π β 1) β + 2 β€ π + 1 < βπ β + 1 β β(π β 1) β < π + 1 β€ βπ β β (π β 1) < (π + 1) π β€ π β β(π + 1) π β = π β β(π + 1) π β = βπ π β = π β΄ From Equation (6)
π(π + 1): ββπ π β π+1π=1 = πβπ π β + ββπ π β β β β βπ‘ β βπ π βπ‘=1 + β(π + 1) π β β π(π + 1): ββπ π β π+1π=1 = πβ(π + 1) π β + ββ(π + 1) π β β β β βπ‘ β β(π+1) π βπ‘=1 + β(π + 1) π β π(π + 1): ββπ π β π+1π=1 = (π + 1)β(π + 1) π β + ββ(π + 1) π β β β β βπ‘ β β(π+1) π βπ‘=1 β΄ π(π) β π(π + 1) for case (II). β΄ π(π) β π(π + 1) βπ such that β(π β 1) β + 1 β€ π β€ βπ β Hence
π(π + 1) is true whenever
π(π) is true. Hence by Principle of Mathematical Induction
π(π) is true βπ β β.
SERIES 1: β 1β(ππ) π β π = βπ=1 β { β(π + 1) π β β βπ π βπ π } βπ=1 π β (0,1], π β β + , π π(π ) > 1π Let π = β(ππ) π β for some π β β and let π(π) denote function that gives number of consecutive integers, π , for which π is the particular natural number (i.e. no. of repetition of π ). Then β 1β(ππ) π β π = βπ=1 β π(π)π π βπ=1 Now, π = β(ππ) π β β π β€ (ππ) π < π + 1 β π β€ (ππ) < (π + 1) (β΅ π β (0,1]) β π π β€ π < (π + 1) π β βπ π β β€ π < β(π + 1) π β (β΅ π β β) It follows that π is at least β π π β and at most strictly less than β (π+1) π β β΄ No of consecutive integer π such that π = β(ππ) π β is β (π+1) π β β β π π β β΄ π(π) = β(π + 1) π β β βπ π β β΄ β 1β(ππ) π β π = βπ=1 β { β(π + 1) π β β βπ π βπ π } βπ=1 ERIES 2: β 1β(ππ) π β π = βπ=1 β { βπ b β β β(π β 1) b βπ π } βπ=1 π β (0,1], π β β + , π π(π ) > 1π Let π = β(ππ) π β for some π β β and let π(π) denote function that gives number of consecutive integers, π , for which π is the particular natural number (i.e. no. of repetition of π ). Then β 1β(ππ) π β π = βπ=1 β π(π)π π βπ=1 Now, π = β(ππ) π β β π β 1 < (ππ) π β€ π β (π β 1) < (ππ) β€ π (β΅ π β (0,1]) β (π β 1) π < π β€ π π β β(π β 1) b β < π β€ βπ b β (β΅ π β β) It follows that π is at least strictly greater than β (πβ1) b β and at most β π b β. β΄ No of consecutive integer π such that π = β(ππ) π β is β π b β β β (πβ1) b β β΄ π(π) = βπ b β β β(π β 1) b β β΄ β 1β(ππ) π β π = βπ=1 β { βπ b β β β(π β 1) b βπ π } βπ=1 . C OROLLARIES AND PROOFS C OROLLARIES: COROLLARY 1:
Subtracting formula 1 from formula 2 with some basic modification we get: β(1 β (βπ π β β βπ π β)) ππ=1 = β (1 β (βπ‘ β β βπ‘ β)) βπ π βπ‘=1 , π β (0,1] Also, taking π = , π β β , corollary 1 reduces to: β (βπ β β βπ β) ππ=1 = π β βπ β We are considering π = , π β β for the following: COROLLARY 2:
Formula 1 reduces to: β βπ β ππ=1 = (π + 1) βπ β β
1π + 1 β β ( π + 1π‘ ) β π΅ π‘ β βπ β π+1βπ‘ππ‘=0 COROLLARY 3:
Formula 2 reduces to: β βπ β ππ=1 = (π) βπ β + βπ β π β
1π + 1 β β ( π + 1π‘ ) β π΅ π‘ β βπ β π+1βπ‘ππ‘=0 COROLLARY 4:
Taking b = 1, Series 1 becomes: β 1βπ β π βπ=1 = β (ππ‘ ) π(π β π‘) πβ1π‘=0 COROLLARY 5:
Taking b = 1, Series 2 becomes: β 1βπ β π βπ=1 = β(β1) πβπ‘+1 (ππ‘ ) π(π β π‘) πβ1π‘=0 , π π(π ) > π Both series in corollaries 4 and 5 have poles at
π π(π ) = π , but following can be simply observed: β ( 1βπ β π β 1βπ β π ) βπ=1 = β (ππ‘ ) β π(π β π‘) β [1 + (β1) πβπ‘ ] πβ2π‘=0 , π β 1 (This shows that even if both series may individually have poles at π = π but their difference is convergent.) P ROOFS OF C OR OLLARIES:
Corollaries 1,2 and 3 needs no proof, as Corollary 1 is simply subtraction of Formula 1 and 2. And Corollaries 2 and 3 are simply the case when π is of form . OROLLARY 4:
Consider the following improper integral (Known as the Gamma Function)
Ξ(π ) = β« π‘ π β1 π βπ‘ ππ‘ β0 Take π‘ = π₯ βπ β β ππ‘ = βπ β ππ₯ , π₯ β 0 as π‘ β 0 and π₯ β β as π‘ β β β΄ Ξ(π ) = β« (π₯ βπ β) π β1 π βπ₯βπ β βπ β ππ₯ β0 β΄ Ξ(π ) = βπ β π β« π₯ π β1 π βπ₯βπ β ππ₯ β0 β΄ 1βπ β π = 1Ξ(π ) β« π₯ π β1 π βπ₯βπ β ππ₯ β0 β΄ β 1βπ β π βπ=1 = β 1Ξ(π ) β« π₯ π β1 π βπ₯βπ β ππ₯ β0βπ=1 (7) Right Hand Side (RHS) of the equation (7) β 1Ξ(π ) β« π₯ π β1 π βπ₯βπ β ππ₯ β0βπ=1 The number of repetitions of βπ β is given by the function π(π) with π = and π = 1 (Section 4, Series 1). β΄ β 1Ξ(π ) β« π₯ π β1 π βπ₯βπ β ππ₯ β0βπ=1 = β 1Ξ(π ) β« π₯ π β1 {β(π + 1) π β β βπ π β}π βππ₯ ππ₯ β0βπ=1 = β 1Ξ(π ) β« π₯ π β1 {(π + 1) π β π π }π βππ₯ ππ₯ β0βπ=1 = β 1Ξ(π ) β« π₯ π β1 {β (ππ‘ ) π π‘πβ1π‘=0 } π βππ₯ ππ₯ β0βπ=1 = β β { 1Ξ(π ) β« π₯ π β1 (ππ‘ ) π π‘ π βππ₯ ππ₯ β0 } βπ=1πβ1π‘=0 = β { 1Ξ(π ) (ππ‘ ) β β« π₯ π β1 π π‘ π βππ₯ ππ₯ β0βπ=1 } πβ1π‘=0 β { 1Ξ(π ) (ππ‘ ) β β« π₯ π β1 π π‘βπ π π π βππ₯ ππ₯ β0βπ=1 } πβ1π‘=0 = β { 1Ξ(π ) (ππ‘ ) β π π‘βπ β« π₯ π β1 π π π βππ₯ ππ₯ β0βπ=1 } πβ1π‘=0 = β { 1Ξ(π ) (ππ‘ ) β π π‘βπ Ξ(π ) βπ=1 } πβ1π‘=0 (β΅ β« π₯ π β1 π π π βππ₯ ππ₯ β0 = β« (ππ₯) π β1 π βππ₯ πππ₯ β0 = Ξ(π )) = β {Ξ(π )Ξ(π ) (ππ‘ ) β π π‘βπ βπ=1 } πβ1π‘=0 = β {(ππ‘ ) β 1π π βπ‘βπ=1 } πβ1π‘=0 = β (ππ‘ ) π(π β π‘) πβ1π‘=0 β΄ β 1Ξ(π ) β« π₯ π β1 π βπ₯βπ β ππ₯ β0βπ=1 = β (ππ‘ ) π(π β π‘) πβ1π‘=0 Therefore, from equation (7) β΄ β 1βπ β π βπ=1 = β (ππ‘ ) π(π β π‘) πβ1π‘=0 Using the definition of the Zeta function, this can also be re-written in the integral form as: β΄ β 1βπ β π βπ=1 = β« π(π₯, π , π)π π₯ β 1 ππ₯ β0 . where π(π₯, π , π) = β(Ξ(π β π‘)) β1 (ππ‘ ) π₯ π βπ‘β1πβ1π‘=0 COROLLARY 5:
Again, take the Gamma Function
Ξ(π ) = β« π‘ π β1 π βπ‘ ππ‘ β0 Take π‘ = π₯ βπ β β ππ‘ = βπ β ππ₯ , π₯ β 0 as π‘ β 0 and π₯ β β as π‘ β β β΄ Ξ(π ) = β« (π₯ βπ β) π β1 π βπ₯βπ β βπ β ππ₯ β0 β΄ Ξ(π ) = βπ β π β« π₯ π β1 π βπ₯βπ β ππ₯ β0 β π = 1Ξ(π ) β« π₯ π β1 π βπ₯βπ β ππ₯ β0 β΄ β 1βπ β π βπ=1 = β 1Ξ(π ) β« π₯ π β1 π βπ₯βπ β ππ₯ β0βπ=1 (8) Right Hand Side (RHS) of the equation (8) β 1Ξ(π ) β« π₯ π β1 π βπ₯βπ β ππ₯ β0βπ=1 The number of repetitions of βπ β is given by the function π(π) with π = and π = 1 (Section 4, Series 2). β΄ β 1Ξ(π ) β« π₯ π β1 π βπ₯βπ β ππ₯ β0βπ=1 = β 1Ξ(π ) β« π₯ π β1 {βπ π β β β(π β 1) π β}π βππ₯ ππ₯ β0βπ=1 = β 1Ξ(π ) β« π₯ π β1 {π π β (π β 1) π }π βππ₯ ππ₯ β0βπ=1 = β 1Ξ(π ) β« π₯ π β1 {β(β1) πβπ‘+1 (ππ‘ ) π π‘πβ1π‘=0 } π βππ₯ ππ₯ β0βπ=1 = β β { 1Ξ(π ) β« π₯ π β1 (β1) πβπ‘+1 (ππ‘ ) π π‘ π βππ₯ ππ₯ β0 } βπ=1πβ1π‘=0 = β {(β1) πβπ‘+1 Ξ(π ) (ππ‘ ) β β« π₯ π β1 π π‘ π βππ₯ ππ₯ β0βπ=1 } πβ1π‘=0 = β {(β1) πβπ‘+1 Ξ(π ) (ππ‘ ) β β« π₯ π β1 π π‘βπ π π π βππ₯ ππ₯ β0βπ=1 } πβ1π‘=0 = β {(β1) πβπ‘+1 Ξ(π ) (ππ‘ ) β π π‘βπ β« π₯ π β1 π π π βππ₯ ππ₯ β0βπ=1 } πβ1π‘=0 = β {(β1) πβπ‘+1 Ξ(π ) (ππ‘ ) β π π‘βπ
Ξ(π ) βπ=1 } πβ1π‘=0 (β΅ β« π₯ π β1 π π π βππ₯ ππ₯ β0 = β« (ππ₯) π β1 π βππ₯ πππ₯ β0 = Ξ(π )) = β {(β1) πβπ‘+1 β Ξ(π )Ξ(π ) (ππ‘ ) β π π‘βπ βπ=1 } πβ1π‘=0 = β {(β1) πβπ‘+1 (ππ‘ ) β 1π π βπ‘βπ=1 } πβ1π‘=0 = β(β1) πβπ‘+1 (ππ‘ ) π(π β π‘) πβ1π‘=0 β΄ β 1Ξ(π ) β« π₯ π β1 π βπ₯βπ β ππ₯ β0βπ=1 = β(β1) πβπ‘+1 (ππ‘ ) π(π β π‘) πβ1π‘=0 herefore, from equation (8) β΄ β 1βπ β π βπ=1 = β(β1) πβπ‘+1 (ππ‘ ) π(π β π‘) πβ1π‘=0 Using the definition of the Zeta function, this can also be re-written in the integral form as: β΄ β 1βπ β π βπ=1 = β« π(π₯, π , π)π π₯ β 1 ππ₯ β0 . where π(π₯, π , π) = β(β1) πβπ‘+1 (Ξ(π β π‘)) β1 (ππ‘ ) π₯ π βπ‘β1πβ1π‘=0 R ESULTS AND S PECIAL C ASES
RESULTS: (I):
Taking π = 1 in formula 1 and 2, both of them reduces to the Gauss formula. (Take formula 1 for example) β βπ β ππ=1 = (π + 1) βπ β β β π βπ=1 β β π ππ=1 = (π + 1)π β β π ππ=1 (β΅ βπ₯ β = π₯ , βπ₯ β β) β 2 β π ππ=1 = (π + 1)π β β π ππ=1 = (π + 1)π2 (II): Taking π = 1 and π = 1 in series 1 and 2, both reduce to Riemann-Zeta function. (Take series 1 for example) β 1β(1 β π) β π = βπ=1 β { β(π + 1) π } βπ=1 = β 1π π βπ=1 = π(π ) S PECIAL C ASES:
Set (1)
For π = (I): Formula 1 becomes: βββπβ ππ=1 = 6πββπβ β 2ββπβ β 3ββπβ + 5ββπβ6 (II): Formula 2 becomes: βββπβ ππ=1 = 6πββπβ β 2ββπβ + 3ββπβ β ββπβ6 One can go further with π = , , , β¦ in both formulas. Following sets are special cases of series 1 and 2 which are observed solely by intuition. et (2): For π = , π π(π ) > 2 series 1 and 2 are observed to be: π = 2 (I): Series 1: β 1ββ(2π)β π βπ=1 = β 2 βπ2β + 1π π βπ=1 (II):
Series 2: β 1ββ(2π)β π βπ=1 = β 2 βπ2βπ π βπ=1 π = 3 (I):
Series 1: β 1ββ(3π)β π βπ=1 = β 2 βπ3β + 1π π βπ=1 (II):
Series 2: β 1ββ(3π)β π βπ=1 = β β2π3 βπ π βπ=1 π = 4 (I):
Series 1: β 1ββ(4π)β π βπ=1 = β 1β2βπβ π βπ=1 = β βπ2β + (β1) π π π βπ=1 (II): Series 2: β 1ββ(4π)β π βπ=1 = β 1β2βπβ π βπ=1 = β βπ2βπ π βπ=1
Set (3):
For π = , π π(π ) > 3 series 1 and 2 are observed to be: π = 2 (I): Series 1: β 1β β(2π) β π βπ=1 = β 3π(π + 1)2 + βπ2β β βπ β 12 βπ π βπ=1 (II): Series 2: β 1β β(2π) β π βπ=1 = β 3π(π β 1)2 + βπ2β β βπ β 12 βπ π βπ=1 π = 3 (I): Series 1: β 1β β(3π) β π βπ=1 = β π(π + 1) + βπ3β β βπ β 13 βπ π βπ=1 (II): Series 2: β 1β β(3π) β π βπ=1 = β π(π β 1) + βπ3β β βπ β 13 βπ π βπ=1 π = 4 (I): Series 1: β 1β β(4π) β π βπ=1 = β β3π(π + 1)4 β + βπ2β β βπ β 12 βπ π βπ=1 (II): Series 2: β 1β β(4π) β π βπ=1 = β β3π(π β 1)4 β + βπ2β β βπ β 12 βπ π βπ=1 Set (4):
For π = , π π(π ) > 4 series 1 and 2 are observed to be: π = 2 (I): Series 1: β 1β β(2π) β π βπ=1 = β π(π + 1)(2π + 1) + 2 βπ2β + 1π π βπ=1 (II): Series 2: β 1β β(2π) β π βπ=1 = β (π β 1)π(2π β 1) + 2 βπ2βπ π βπ=1 Set (5):
For π = , π π(π ) > 5 series 1 and 2 are observed to be: π = 2 (I): Series 1: β 1β β(2π) β π βπ=1 = β 52 π(π + 1)(π + π + 1) + βπ2β β βπ β 12 βπ π βπ=1 (II): Series 2: β 1β β(2π) β π βπ=1 = β 52 (π β 1)π(π β π + 1) + βπ2β β βπ β 12 βπ π βπ=1 One can go further with π = , , , β¦ with different values of π in both formulas. C ONCLUSION AND O PEN P ROBLEMS
This paper introduces a set of a new type of partial summation formulas and convergent infinite series. Very fundamental yet powerful principle of mathematical induction is used to prove the formulas. Two functions π(π) and π(π) are derived to get the number of repetitions of β(ππ) π β and β(ππ) π β for consecutive natural numbers π and are used in getting the equivalent series for β β(ππ) π β βπ βπ=1 and β β(ππ) π β βπ βπ=1 . These functions are also used in proving the relation of corollaries 4 and 5 with the zeta function. A set of results and special cases helps to understand the behaviour of the formulas and series at particular values. The extensions done in the paper may be helpful in improving the measurement quality as well expanding the domain of research. Problem 7.1:
Finding the explicit formulas for (π) β β π π β π ππ=1 & ( ππ ) β β π π β π ππ=1 , π β β, π β β + . Putting π = 1 for varying π in (π) & (ππ) both reduces to the Faulhaberβs formula ( Section 2.2 ). And if π =1 is taken, for varying π , then both reduces to the Formula 1 and Formula 2 accordingly ( Section 3 ). Problem 7.2(A):
Are the following assumptions true? If
π π(π ) > π then (I) β 1β(2π) β π βπ=1 = β π(π β 1)2 {β (β (π π‘ (1 + (β1) πβπ‘ )2 ) ππ=1 ) πβ2π‘=βπ2βββπ2β } + π¦ π π π βπ=1 π¦ π = { 2 βπ2 β + 1 , π ππ£ππβπ2 β β βπ β 12 β , π πππ II) β 1β(2π) β π βπ=1 = β π(π β 1)2 {β (β (π π‘ (1 + (β1) πβπ‘ )2 ) πβ1π=1 ) πβ2π‘=βπ2βββπ2β } + π§ π π π βπ=1 π§ π = { 2 βπ2 β , π ππ£ππβπ2 β β βπ β 12 β , π πππ (B): If these assumption holds true, one can go on to find the equivalent for β 1β(ππ) β π βπ=1 & β 1β(ππ) β π βπ=1 , π π ( π ) > π , π β β\ { } . Problem 7.3:
Does the following application hold true? Consider the following double series π π = β β π π,ππ π π π` π=1ππ=1 If π π,π = 1 then the upper bound of time complexity (Big-O) for the given series is Ξ (β π π+π+1π+1 β) . Problem 7.4:
The Riemann hypothesis is a very well-known unsolved problem, involves the Zeta function. Series 1 and 2 reduce to the Riemann-Zeta function for π = 1 and corollaries 4 & 5 relate to the same for all values except π = π = 1 . Can these series or corollaries be useful in solving the Hypothesis? A CKNOWLEDGEMENT The authors would like to acknowledge Dr. Nishant Doshi of Pandit Deendayal Petroleum University, Gandhinagar, Gujarat, India, for bringing us a very interesting exercise, from which we were able to work on all the results. Authors would like to thank Kenneth Beitler for contacting us and for giving a very helpful suggestion for proving the results. We are especially grateful to have guidance of Dr. Anand Sengupta and Dr. Atul Dixit of Indian Institute of Technology, Gandhinagar by the means of video conference and discussions. Also, we are thankful to Mrs. Meghna Parikh and Mr. Bhashin Thakore for valuable discussion throughout the preparation. R EFERENCES [1]
T. A. Apostol, Introduction to Analytic Number Theory, Springer, 1976. [2]
J. H. Conway, R. Guy, The Book of Numbers. Springer, 1996. [3]