Polynomial Formula for Sums of Powers of Integers

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1 Polynomial Formula for Sums of Powers of Integers K B Athreya and S Kothari left K B Athreya is a retired Professor of mathematics and statistics at Iowa State University, Ames, Iowa. right S Kothari is a Professor in the Electrical Engineering Department at Iowa State University, Ames, Iowa. Keywords Sums of powers, integers, polynomial formula. In this article, it is shown that for any positive integer k, there exist unique real numbers a kr, r=,,...,k +, such that for any integer n k+ n S k,n j k = a kr n r. j= r= The numbers a kr are computed explicitly for r = k +,k,k,...,k 0. This fully determines the polynomials for k =,,...,. The cases k =,, 3 are well known and available in high school algebra books.. Introduction Let k and n be positive integers and S k,n = n j= jk.a well-known story in the history of mathematics is that the great German mathematician Carl Friedrich Gauss, while in elementary school, noted that for k =,S,n n is, when written in reverse order, equal to n+n And when they are added one gets S,n =+n++n+3+n+...+n+ = n +n. This yields S,n = nn+. Shailesh Shirali has told the authors that the great Indian mathematician of ancient times, Aryabhata [], has explicitly mentioned in one of his verses the formulas for the cases k = and k = 3. The three cases, k =,, 3, are now in high school algebra texts. Proofs of these formulas are given using the principle of induction. In this article, we establish that for any integer k there exist unique real numbers a kr,r=,,...,k+, such that for any positive integer n, 7 RESONANCE August 05

2 S k,n GENERAL ARTICLE n k+ j k = a kr n r. j= r= In particular, it is shown that for any k, a kk+ = k +,a kk = ; for k,a kk = k ; for k 3,a kk =0; kk k for k,a kk3 = ; 70 for k 5,a kk =0; kk k k 3k for k,a kk5 = ; 7! for k 7,a kk =0; 3kk...k for k 8,a kk7 = ;!7! for k 9,a kk8 =0; kk...k 8 for k 0,a kk9 =0 ;! and for k,a kk0 =0. It is to be noted that once a formula is proposed, its validity may be verified by the principle of induction. The main problem is to guess what form the formula takes. We guessed that it should be a polynomial based on the known formulas for k =,, 3 and found the explicit polynomial by a recurrence relation. Sury [] has noted that Euler had shown that for any positive integers k and n the sum S k,n can be expressed in terms of Bernoulli polynomials. Sury has also pointed out that Euler s work implies that for each positive integer k the sum S k,n is a polynomial in n. The main contribution of the present article is to determine this polynomial explicitly. We do so from elementary methods, thus making the article accessible to our undergraduate students. RESONANCE August 05 77

3 . An Explicit Formula Fix a positive integer k. For a positive integer n, let n fn S k,n j k j= and set f0 = 0. Then {fn} n 0 satisfies the recurrence relation fn +=fn+n + k,n=0,,, with initial condition f0 = 0. It is clear that {fn} n 0 is uniquely determined by and. Thus, if we find real numbers a kj,j =,...,k + such that g k n k+ r= a kr n r, n 0 3 satisfies and, then g k n hastoequals kn for all n. Since g k 0 = 0 and hence holds, to ensure that g k satisfies it suffices to ensure that the coefficients of powers of n on both sides of the following equation k+ r= are equal. g k n +=g k n+n + k, i.e., a kr n + r = k+ r= This leads to the following conditions: a kr n r +n + k, 5 coeff of n k+ a kk+ = a kk+ k + coeff of n k a kk+ + a kk = a kk RESONANCE August 05

4 k + k coeff of n k a kk+ + a kk k + a kk = a kk + 8 k + k coeff of n k a kk+ + a kk 3 k +a kk + a kk = a kk k + 9 and more generally, k + k coeff of n kr a kk+ + a kk r + r k r a k kr+ k +a kkr = a kkr +.0 r Now provides no information but 7, 8, etc. do. Indeed 7 holds iff k + a kk+ = and 8 holds iff k a kk = k k + a kk+ and 0 holds iff for k r + k k r + k + a kkr+ = a kk+ r r + k a k k... r a kr+ k r +.3 RESONANCE August 05 79

5 for r =0,,,...k. It follows from 7, 8, 9 and 0 that a kr,r=,,, k + are uniquely determined as 0 provides a recursive determination of a kkr+ from the knowledge of a kj for j =k +,...,k r +. Thus, 7 9 yield for any k, a kk+ = k +, a kk = k ; and for k,a kk = k ; and 0 determines a kj for j k. This proves the theorem given in the abstract. Let us call it Theorem A: Theorem A. Let k be a positive integer. Then, there exist unique numbers a kr, r =,,...,k + such that for any integer n>, S k,n n j k = j= k+ r= a kr n r. 3. Another Proof of Theorem A We provide another proof of Theorem A at the end of this equation by first establishing the following: Theorem B. For positive integers k and n let g k n n j= jk.then i g n= nn+ ii For k, n, k +g k n = k k + n r g r n r r= +n k+ +k +n k n. 730 RESONANCE August 05

6 Proof. i This result follows from Gauss s argument: g n= n = n +n implying g n=+n++n n +=n +n. ii For k, n, k + r k k + = r k r= = r= k r= n r g r n k + r n r n j r j= n j r j= where, if n =,weset n j= jr =0forr, n k+ = k + j r k +j k j k+ r j= r=0 n n = j + k+ +n + k + j= n + j= j k+ j= = g k+ n + n + k +g k n n k +g k+ n n k+ = n +k +g k n k +n k n k+ yielding ii. ProofofTheoremA.ByTheoremBi,g n isa polynomial of degree two. Then Theorem B ii implies j k RESONANCE August 05 73

7 that g n is a polynomial of degree three and by induction g k n is a polynomial of degree k +. Further, Theorem B ii also shows that the leading coefficient in g k n isk +.Onecanusethesameformulato show that the coefficient of n k in g k nis. Remark. We now derive explicit expressions for g k n fork =, 3,, 5. From Theorem B ii we deduce that 3 3g n = n g n + n 3 +3n n = 3 n yielding nn + = n 3 + n 3 + n, + n 3 +3n n g n n j= j = n3 3 + n + n. Next, g 3 n = n g n + n g n +n +n 3 n nn + = n n + n n + n +n +n 3 n = n + n 3 + n 3 +n = n +n 3 + n, 73 RESONANCE August 05

8 yielding g 3 n= n + n3 + n. Next, 5 5 5g n = n g n + n g n 5 + n 3 g 3 n + n 5 +5n n 3 nn + = 5 n n +0 n n + n n +0 n 3 + n3 + n + n 5 +5n n = n 5 + n n n n 0 = = n 5 + n 5 + n30 + n 0+n, yielding g n= n5 5 + n + n3 3 n 30. Next, g 5 n = n g n + n g n + n 3 g 3 n + n g n 3 +n +n 5 n n = n + n RESONANCE August

9 n +5 n n + n n +0 n 3 + n3 + n n +5 n n + n3 3 n 30 +n +n 5 n 5 = n + n n 0 0 +n n n = n +3n 5 + n 5 + n3.0 n + n.0, yielding g 5 n= n + n5 + n 5 n. This process can be continued and g k n canbecomputed recursively for all integers k. In Section, we compute g k n fork =,, 3,...,.. A Matrix Method Now that we know that a kr,r=,,...k +are determined uniquely, here is a matrix inversion method to find them. Let Then, since k+ r= ψ k l= l r k,l=,,...,k +. r= a kr l r = S k,l = ψ k l, l=,,...,k +, 5 73 RESONANCE August 05

10 it follows that A k a k a k a kk+ = ψ k ψ k ψ k k +, where A k is the k + k + matrix with entries in the lth row given by l, l,...,l k+, l=,,...,k +. It can be checked that A k x x k+ = 0 0 implies that x =0,x =0,...,x n+ =0. This implies that A k is invertible and hence, for any integer k, a k a k a kk+ 5. An Example = A k ψ k ψ k ψ kk+. 7 We now give an example where the function g k narises in a counting situation. Let S n be the set {0,,,...,n} of integers. How many closed intervals [i,j] are there where i,j S n and i< j? Thereareexactlyn intervals of length one each. These are [i,i +]fori =0,,, n. There are exactly n intervals of length two each. These are [i,i+], i=0,,...,n. And in general there are exactly n k + intervals of length k each. These are RESONANCE August

11 [i,i+ k], i=0,,...,nk. Thus the total number of closed intervals [i,j] withi,j in S n and i<jis n+n +...+nk This is precisely g n. Next, consider the set S n, {i,j i,j S n }. How many squares are there with all vertices in S n, and sides parallel to the axes? There are exactly n such squares whose side length is one. These are with vertices {i,j, i +,j, i,j +, i +,j +} with 0 i n, 0 j n. Next, for any integer l, l n there are exactly n l + squares whose side length is l with vertices all in S n,. These are the ones with vertices {i,j, i+l, j, i,j+l, i+l, j+l} with 0 i n l, 0 j n l. Thus, the total number of squares with all vertices in S n, is n l= n l + = n j= j which is precisely g n. Next, consider the set S n,3 {i,j,k,i,j,k S n }.How many cubes are there with all vertices in S n,3 and faces parallel to the coordinate planes? There are exactly n 3 such cubes whose side length is one. These are the ones with vertices {i,j,k, i +,j,k, i,j +,k, i,j,k +, i+,j+,k,i,j+,k+,i+,j,k+,i+,j+,k+} with 0 i n, 0 j n, 0 k n. Similarly, there are n l + 3 cubes of side length l with vertices in S n,3. Thus, the total number of cubes with vertices in S n,3 is n l= nl+3 = n j= j3 which is precisely g 3 n. Next, for any integer k>3consider the set S n,k {i,i,...,i k i j S n,j =,,...,k}. How many k-dimensional cubes are there with all vertices in S n,k and faces parallel to the coordinate hyperplanes? Arguing as before, the number of such cubes with side length l and vertices in S n,k is n l + k. Thus, the total number of k-dimensional cubes with all vertices in S n,k is n l= n l +k = n j= jk which is precisely g k n. 73 RESONANCE August 05

12 . Determination of a kj for j =k+,k,...,k0 Returning to 7 9 we note that a kk+ = Next, for k 3 k +, a kk = k, a kk = k. a kk = k 3 akk+ k+ = kk 3! k akk 3 k akk k! =0. For k, a kk3 = For k 5, a kk = k ak k+ k+ 5 = kk k = k 5 kk k. 70 akk+ k+ k ark k3 ak k k 3! 5! 3! k k akk 5 akk k ak k3 k = kk k k 3 5!! k3! + 0 =0. RESONANCE August

13 For k, k k + k a kk5 = a kk+ a kk 7 k a kk 5 / k 3 k 5 a k k3 3 = kk k k 3k! 7! 5! ! kk k k 3k =! 7. For k 7, k k + k a kk = a kk+ a kk k k 3 a kk a kk3 / k 5 k a kk5 = kk...k 5 7! 8!! + 70 = kk...k 5 0 = 0.! 7! For k 8, k k + a kk7 = a kk+ 8 9 k k a kk a kk RESONANCE August 05

14 k 3 a kk3 5 / k 5 k 7 a kk5 3 = kk...k 8! 9! 7! ! 3kk...k =. 7!! For k 9, k k + k a kk8 = a kk+ a kk k k 3 a kk a kk3 8 / k 5 k 7 k 5 a kk5 a kk7 = kk...k 7 9! 0! 8! = +!!!! + 3 7!! 5! kk...k ! =0. For k 0, a k k9 = k k + a kk+ 0 RESONANCE August

15 k k a kk a kk 0 9 k 3 k 5 a kk3 a kk 7 5 / k 7 k 9 a kk7 3 = kk...k 8 0!! 70 7!! 5! + 3 7!!3! 9! + kk...k 8 0 =.! For k, k k + k a kk0 = a kk+ a kk k k 3 a kk a kk3 0 8 k 5 k 7 a kk5 a kk7 / k 9 k 0 a kk9 = + kk...k 9! = 0 9 8! ! kk...k 9! 0= RESONANCE August 05

16 Summarizing the above we have: for k a kk+ = a k+ kk = for k a kk = k for k 3 a kk =0 for k a kk3 = kkk 70 for k 5 a kk =0 for k a kk5 = kkkk3k! for k 7 a kk =0 for k 8 a kk7 = 3 kk...k 7!! for k 9 a kk8 =0 for k 0 a kk9 = kk...k80! for k a kk0 =0. 7 This, in turn, yields for integers n, S,n S,n S 3,n S,n S 5,n S,n S 7,n S 8,n S 9,n S 0,n = n + n = n3 3 + n + n = n + n3 + n = n5 5 + n + n3 3 n 30 = n + n5 + n 5 n = n7 7 + n + n5 n3 + n = n8 8 + n7 + n 7 n 7 + n = n9 9 + n8 + n7 3 n n3 9 n 30 = n0 0 + n9 + n83 n n 3 0 n = n + n0 + n95 n7 + n 5 n n RESONANCE August 05 7

17 S,n S,n = n + n + n0 n8 8 + n n 33 + n 5 = n3 3 + n + n n 9 + n7 7 n n n. A natural conjecture from our calculations is that for any k r +,k,rpositive integers, a kkr =0.It may be noted that we have verified it for r =,,...,5 and k r Concluding Remarks 7. Asymptotic Behavior of S k,n For Large n It follows from that lim n S k,n = n k+ n j= n k+ j k a kj n j j= j= n k j k j= = a kk+ + a kjn j j= = j k n k+ = k k + + k +. j= n k+ a kj n j n k+ This also follows from the Riemann sum approximation n j k n = n f k+ k j n n, where f kx=x k, 0 x j= j= which converges to 0 f k xdx = k +. 7 RESONANCE August 05

18 Similarly, the second order behavior is given by n j k = n k+ k + n + k j= j= n n j k n k+ k +. And so on. j= 7. No Induction Involved Suggested Reading a kj n j n k+ For the cases k =,, 3, the proof in elementary texts involves merely the verification of the formula using the principle of induction. Our proofs of the polynomial formula in Theorem A involves no induction but are from first principles. 7.3 Other Treatments There are many papers on this subject. Some of them are listed in the Suggested Reading. [] Walter Eugene Clark, Translation of Aryabhatiya of Aryabhata, The University of Chicago Press, verse, pp.37, 930. [] B Sury, Bernoulli numbers and the Riemann Zeta function, Resonance, No.7, pp.5, 003. [3] A F Beardon, Sums of powers of integers, Am. Math. Monthly, Vol. 03, pp.0 3, 99. [] A W F Edwards, Sums of powers of integers: A Little of the History, The Mathematical Gazette, Vol., No.35, 98. [5] H K Krishnapriyan, Eulerian polynomials and Faulhaber's result on sums of powers of integers, The College Mathematics Journal, Vol., pp.8 3, 995. [] Robert Owens, Sums of powers of integers, Mathematics Magazine, Vol.5, pp.38 0, 99. [7] S Shirali, On sums of powers of integers, Resonance, Indian Academy of Sciences, July 007. Address for Correspondence K B Athreya Department of Mathematics & Statistics Iowa State University Ames, Iowa 500 USA kbathreya@gmail.com RESONANCE August 05 73