aa r X i v : . [ m a t h . HO ] F e b Integral Recurrences from A to Z
Robert Dougherty-Bliss [email protected]
February 23, 2021
Abstract
George Boros and Victor Moll’s masterpiece
Irresistible Integrals does well to in-clude a suitably-titled appendix, “The Revolutionary WZ Method,” which gives abrief overview of the celebrated Wilf–Zeilberger method of definite summation.Paradoxically,
Irresistible Integrals does not contain the suitably-titled appendix,“The Revolutionary AZ Method,” which would have been an excellent place togive a brief overview of the Almkvist–Zeilberger method of definite integration !This omission can be forgiven, but once realized it must be rectified. The remark-able AZ machinery deserves to be more widely known to the general public thanit is. We will do our part by presenting a series of case studies that culminate ina—fun but overkill—integral-based proof that e is irrational. Behold, I will stand before thee there upon the rock in Horeb; and thou shalt smite therock and there shall come water out of it, that the people may drink. — Exodus 17:6You have been up all night working out the masterpiece solution to your latest problem.Your answer depends on the integral sequence I ( n ) = Z ∞−∞ x n ( x + 1) n +1 dx, which you desperately need to evaluate. You know that you could break out specialfunctions, contour integrals, or some other method, but you would really just like aquick answer without much fuss.You run to download the file EKHAD from https://sites.math.rutgers.edu/˜zeilberg/tokhniot/EKHAD and read it into Maple with “ read EKHAD; ”. You type the command
AZd(xˆ(2 * n) / (xˆ2 + 1)ˆ(n + 1), x, n, N); and hardly a second has passed when Maple produces the following: -2 n - 1 + (2 n + 2) N, -x
Almkvist–Zeilberger algorithm has told you that your inte-grand satisfies the “recurrence” ( − n − n + 2) N ) x n ( x + 1) n +1 = − ddx x x n ( x + 1) n +1 , where N is the shift operator defined by N f n ( x ) = f n +1 ( x ) . Integrating this equationon ( −∞ , ∞ ) gives the identity ( − n − n + 2) N ) I ( n ) = 0 , which would traditionally be written as I ( n + 1) = 2 n + 12( n + 1) I ( n ) . You are well-aware that the sequence π n (cid:18) nn (cid:19) satisfies the same recurrence and initial condition, so you have just proven that I ( n ) = Z ∞−∞ x n ( x + 1) n +1 dx = π n (cid:18) nn (cid:19) with minimal effort on your part. Such is a normal case study of the AZ algorithm.In general, we often want to understand the sequence of definite integrals I ( n ) = Z F n ( x ) dx. Perhaps we would like to compute the first twenty terms or so to see what I ( n ) lookslike. Sometimes we can ask a computer to churn these out, but other times F n ( x ) isso complicated that even our electronic friends would struggle to keep up for large n . What we need is an efficient algorithm to compute the terms of I ( n ) . We need a recurrence .There are plenty of ad-hoc methods to find a recurrence for I ( n ) . You could inte-grate by parts or differentiate under the integral sign, for example. But these all requireingenuity, insight, and hard work. As Sir Alfred Whitehead once remarked, such inge-nuity is overrated. No one wants to work hard—we want answers!The painless way to discover these recurrences for large classes of integrals isthe Almkvist–Zeilberger algorithm. This is the direct analog of the celebrated Wilf–Zeilberger method of automatic definite summation, but it has received less attentionthan its discrete counterpart. Our goal here is to explore the Almkvist–Zeilberger algo-rithm with a few case studies, leaving the door open for more experimentation.2
A quickstart guide to the AZ algorithm
The Wilf–Zeilberger method of definite summation is a breakthrough in automaticsummation techniques. Roughly, the Wilf–Zeilberger method can automatically prove(and semi-automatically discover) most commonly occurring summation identites ofthe form S ( n ) = X k f ( n, k ) = RHS ( n ) . One piece of the puzzle is that, whenever f ( n, k ) is a “suitable” function, it satifiesa specific type of inhomogenous linear recurrence with polynomial coefficients in n .Exactly, there exists a nonnegative integer d and polynomials p j ( n ) such that d X j =0 p j ( n ) f ( n + j, k ) = G ( n, k + 1) − G ( n, k ) , where G ( n, k ) is some function with G ( n, ±∞ ) = 0 . Summing over k yields therecurrence d X j =0 p j ( n ) S ( n + j ) = G ( n, ∞ ) − G ( n, −∞ ) = 0 . This method has been (rightly) advertised from here to the Moon and back. See thearticle, [8], the book [7], the lecture notes [11], and the lively
Monthly article [6].The Almkvist–Zeilberger algorithm is to definite integrals what the Wilf–Zeilbergermethod is to definite sums. The input to the algorithm is a “suitable” function F n ( x ) with a discrete parameter n . The output is a linear recurrence operator L ( N, n ) withpolynomial coefficients in N and n , and function R ( n, x ) , rational in n and x , suchthat L ( N, n ) F n ( x ) = ddx R ( n, x ) F n ( x ) . Explicitly, there is a nonnegative integer d and polynomials p k ( n ) such that d X k =0 p k ( n ) F n + k ( x ) = ddx R ( n, x ) F n ( x ) . The left-hand side is independent of x except for the F n ( x ) , so integrating this equationon [0 , , say, gives L ( N, n ) Z F n ( x ) = R ( n, F n (1) − R ( n, F n (0) . If F n (0) = F n (1) = 0 and R ( n, x ) is well-behaved, then I ( n ) = R F n ( x ) satisfies L ( N, n ) I ( n ) = 0 , meaning that we have discovered a recurrence for the sequence of integrals I ( n ) . Theonly thing to verify is that F n ( x ) is “suitable,” and that R ( n, x ) is well-behaved on theregion of integration. 3hat functions are “suitable”? The requirement is that F n ( x ) is hypergeometric in n and x , meaning that there exist fixed rational functions R ( n, x ) and R ( n, x ) suchthat F n +1 ( x ) /F n ( x ) = R ( n, x ) F ′ n ( x ) /F n ( x ) = R ( n, x ) . This is all that the algorithm needs to produce its identity.The version of the Almkvist–Zeilberger algorithm that we will use is implementedin the procedure
AZd(f, x, n, N) in the Maple package
EKHAD referenced inthe introduction. It takes an expression f in the continuous variable x and discreteparameter n . The symbol N stands for the “shift” operator N on the set of sequencesby N a ( n ) = a ( n + 1) . For example, the Fibonacci numbers F ( n ) satisfy ( N − N − F ( n ) = 0 . Now, let us get on to the case studies.
Let us begin humbly, by evaluating an integral that we already know.
Proposition.
For each integer n ≥ , I ( n ) = Z ∞ e − x x n dx = n ! . Proof.
Type the following into Maple:
AZd(exp(-x) * xˆn, x, n, N);
This produces:
N - n - 1, -x
That is, the Almkvist–Zeilberger algorithm has told us that ( N − ( n + 1)) f n ( x ) = − ddx e − x x n +1 . (1)Since the antiderivative of the right-hand side vanishes for x = 0 and x = ∞ , integrat-ing on [0 , ∞ ) gives ( N − ( n + 1)) I ( n ) = 0 , and since I (0) = 1 , we have I ( n ) = n ! . 4 “A Complicated Integral” This is from Section 3.8 of [4].
Proposition. I ( n ) = Z ∞ x n ( x + 1) n + r +1 dx = (cid:20) r (cid:18) r + nn (cid:19)(cid:21) − . Proof.
Type the following into Maple:
AZd(xˆn / (x + 1)ˆ(n + r + 1), x, n, N);
This produces: (n + 1) + (-n - r - 1) N, x
And for r > , integrating the implied identity (( n + 1) − ( n + r + 1) N ) x n ( x + 1) n + r +1 = ddx x x n ( x + 1) n + r +1 (2)yields (( n + 1) − ( n + r + 1) N ) I ( n ) = 0 . The sequence ( r (cid:0) r + nn (cid:1) ) − satisfies the same recurrence and initial condition (check!). Proposition.
The integral sequence I ( n ) = Z ( x (1 − x )) n dx satisfies ( N − n + 12(2 n + 3) ) I ( n ) = 0 . Proof.
Type the following into Maple:
AZd((x * (1 - x))ˆn, x, n, N);
This produces: n + 1 + (-4 n - 6) N, (-1 + 2 x) (-1 + x) x ( N − n + 12(2 n + 3) )( x (1 − x )) n = ddx (2 x − x − x ( x (1 − x )) n , (3)on [0 , yields the result, since the antiderivative of the right-hand side vanishes at x = 0 and x = 1 .The recurrence implies that I ( n ) begins as follows: / , / , / , / , / , / , / , . . . Corollary. I ( n ) = 1(2 n + 1) (cid:0) nn (cid:1) Proof.
Both sequences satisfy the same recurrence and initial condition (check!).This fact gives us an interesting identity. One way to try and evaluate I ( n ) is byapplying the binomial theorem to the integrand: I ( n ) = Z x n (1 − x ) n dx = Z n X k =0 (cid:18) nk (cid:19) ( − k x n + k dx = n X k =0 (cid:18) nk (cid:19) ( − k n + k + 1 . This remaining sum is complicated, but we can pair it with our previous corollary toget another.
Corollary. n X k =0 (cid:18) nk (cid:19) ( − k n + k + 1 = 1(2 n + 1) (cid:0) nn (cid:1) Our final case study is a slightly more complicated sequence of integrals. We will notbe able to derive a closed form, but we will derive a wealth of other information.
Proposition.
The integral sequence I ( n ) = Z ( x (1 − x )) n e − x dx satisfies ( N + 2(2 n + 3)( n + 2) N − ( n + 1)( n + 2)) I ( n ) = 0 . roof. Let f n ( x ) be the integrand. The Almkvist–Zeilberger algorithm produces the“calculus exercise” ( N + 2(2 n + 3)( n + 2) N − ( n + 1)( n + 2) f n ( x )= ddx ( − nx − x + 3 nx − x − nx + 5 x − x ) f n ( x ) , (4)and integrating this proves the proposition.The recurrence is hopelessly complicated; we probably won’t be able to solve it.But it does produce the following initial terms: − e , − e , −
426 + 1158 e , − e , . . . This data is very suggestive! It leads us to conjecture that I ( n ) = a n + b n e − for some integers a n and b n . This is true by virtue of the recurrence: if I ( n ) = a n + b n e − and I ( n + 1) = a n +1 + b n +1 e − , then I ( n + 2) = − n + 3)( n + 2) I ( n + 1) + ( n + 1)( n + 2) I ( n )= − n + 3)( n + 2)( a n +1 + b n +1 e − ) + ( n + 1)( n + 2)( a n + b n e − )= a n +2 + b n +2 e − , where we take a n +2 = − n + 3)( n + 2) a n +1 + ( n + 1)( n + 2) a n b n +2 = − n + 3)( n + 2) b n +1 + ( n + 1)( n + 2) b n . That is, a n and b n are sequences of integers which satisfy the same recurrence that I ( n ) satisfies, only the initial conditions are different: a = − a = 14 b = 3 b = − . Better yet, note that − a b = 2402465304= 0 . . . . ≈ e − . That is, − a n /b n seems to be a good approximation to e − !To see why this is, we must go back to the initial integral. For ≤ x ≤ , we have x (1 − x ) ≤ / , therefore ≤ I ( n ) = Z e − x ( x (1 − x )) n dx ≤ n Z e − x dx, I ( n ) goes to zero exponentially quickly. Therefore | a n + b n e − | → exponentially quickly, meaning that | a n b n + e − | = | ( − a n b n ) − e − | → . In words, − a n /b n gives an exponentially-good rational approximation of e − . Todouble check, we can use our recurrence to compute a /b : − a b = 4932941648663833516994295346011418332399206400001340912564441170249019237618446466016434749440000= 0 . . . . ≈ . . . . = e − . Better still , this remarkable approximation − a n /b n ≈ e − is too good to be true in thefollowing sense. Proposition.
Let α be a real number. If there exist sequences of integers a n and b n such that | b n | → ∞ and | α − a n b n | ≤ C | b n | δ for some positive constants C and δ , then α is irrational.Proof. If α = a/b is rational, then | α − a n b n | = | ( b n a − ba n ) /b || b n | ≥ C ′ | b n | for some positive constant C ′ . But the inequality C ′ | b n | ≤ C | b n | δ is impossible if | b n | → ∞ .This fact together with our approximation − a n /b n ≈ e − gives us an unnecessarilycomplicated proof that e is irrational. Proposition. e is irrational with δ = 1 .Proof. Let a n and b n be the approximating sequences induced by I ( n ) = Z e − x ( x (1 − x )) n dx.
8e have | a n + b n e − | ≤ n Z e − x = C n . The sequence b n satisfies the recurrence ( N + 2(2 n + 3)( n + 2) N − ( n + 1)( n + 2)) I ( n ) = 0 . It turns out—see [10]—that this reveals considerable asymptotic information about b n .In particular, if we rewrite the recurrence as a polynomial in n , the leading coefficientis N − . The only solution to N − is N = 1 / , and this implies that / n ≤ C ′ | b n | for some constant C ′ . Thus | a n + b n e − | ≤ C ′ | b n | , or | a n b n + e − | ≤ C ′ | b n | δ , where δ = 1 . The claim follows from the previous proposition.Proving that e is irrational is an easy exercise, but our proof gives a quantitativemeasure on the irrationality of e . Given a real α , the irrationality measure of α , denoted µ ( α ) is defined to be the smallest real µ such that (cid:12)(cid:12)(cid:12)(cid:12) α − pq (cid:12)(cid:12)(cid:12)(cid:12) > q µ + ǫ holds for any ǫ > and all integers p and q with q sufficiently large. A number isirrational iff it has irrationality measure > . In fact, irrationality measures are forrational numbers, for algebraic numbers, and ≥ for transcendental numbers. It iswell-known that almost every number has irrationality measure , but it is notoriouslydifficult to prove this for naturally ocurring, specific constants.Our proof implies the obvious lower bound µ ( e ) ≥ δ = 2 . With a little morework—see [9]—we can show that µ ( e ) ≤ /δ = 2 , so µ ( e ) = 2 .It has become a game to provide better and better upper bounds for the irrationalitymeasure of famous constants. For example, µ ( π ) ≥ since π is transcendental, butwe do not know the exact value. The current “world record” upper bound is held byZeilberger and Zudilin, who showed in [12] that µ ( π ) ≤ . . . . Ignoring the many technical details, their proof is very similar to ours. The basic ideais to find a rapidly-decaying sequence I ( n ) such that I ( n ) = a n + πb n for integers a n and b n , then show that b n has nice asymptotic properties. This styleof proof was notably used by Fritz Beukers in [3] where he elegantly proved that theconstants ζ (2) = X k ≥ k ζ (3) = X k ≥ k are irrational by considering integrals of the form Z Z x n y n (1 − x ) n (1 − y ) n − xy dx dy and Z Z Z x n y n z n (1 − x ) n (1 − y ) n (1 − z ) n (1 − (1 − xy ) z ) n +1 dx dy dz, respectively. (The irrationality of ζ (3) was first shown in stunning fashion by RogerAp´ery; see [9] and [2].)To find their approximating sequences, Zeilberger and Zudilin tweaked integralssimilar to the ones above, adding parameters to the integrands and performing an ex-haustive computer search to find those parameters which gave the empirically bestupper bound. This method continues to provide possible avenues for constructive irra-tionality proofs; see [5] and [13].It is too late for us to become famous proving that ζ (3) is irrational. In fact, nothingwe have done here is “new” or “groundbreaking.” We should be content to have somenew tools to play with. But you never know: One day you might just plug the rightintegrand into the Almkvist–Zeilberger algorithm to prove that(FAMOUS CONSTANT)is irrational.Until then, have fun! References [1] Almkvist, G. and Zeilberger, D., 1990. The method of differentiating under theintegral sign. J. Symb. Comput., 10(6), pp.571-592.[2] Ap´ery, R., 1979. Irrationalit´e de ζ (2) et ζ (3) . Ast´erisque, 61(11-13), p. 1.[3] Beukers, F., 1979. A note on the irrationality of ζ (2) and ζ (3) . Bulletin of theLondon Mathematical Society, 11(3), pp.268-272.[4] Boros, G. and Moll, V., 2004. Irresistible integrals: symbolics, analysis and exper-iments in the evaluation of integrals. Cambridge University Press.[5] Dougherty-Bliss, R., Koutschan, C. and Zeilberger, D., 2021. Tweaking the Beuk-ers Integrals In Search of More Miraculous Irrationality Proofs A La Apery. arXivpreprint arXiv:2101.08308.[6] Nemes, I., Petkov˘sek, M., Wilf, H.S. and Zeilberger, D., 1997. How to do Monthlyproblems with your computer. The American Mathematical Monthly, 104(6),pp.505-519. 107] Petkov˘sek, M., Wilf, H., and Zeilberger, D., 1997. A = B . Wellesley, Mass, Mas-sachusetts: AK Peters.[8] Tefera, A., 2004. What is... a Wilf-Zeilberger pair. AMS Notices, 57.[9] Van der Poorten, A., 1979. A proof that Euler missed... The Mathematical Intelli-gencer, 1(4), pp.195-203.[10] Wimp, J. and Zeilberger, D., 1985. Resurrecting the asymptotics of linear recur-rences. Journal of mathematical analysis and applications, 111(1), pp.162-176.[11] Zeilberger, D., 1995. Three recitations on holonomic systems and hypergeometricseries. J. Symb. Comput., 20(5/6), pp.699-724.[12] Zeilberger, D. and Zudilin, W., 2020. The irrationality measure of ππ