Euler, Ioachimescu and the trapezium rule. G.J.O. Jameson (Math. Gazette 96 (2012), )

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1 Euler, Iochimescu nd the trpezium rule G.J.O. Jmeson (Mth. Gzette 96 (0), 36 4) The following results were estblished in recent Gzette rticle [, Theorems, 3, 4]. Given > 0 nd 0 < s <, let x n = y n = n n ( + r) [ ( + n) s s], s s ( + r) [ ( + n ) s s] s s u n = (x n + y n ) = y n ( + n ) s Then (x n ) is incresing, (y n ) is decresing, nd both tend to limit, denoted by I s (), s n. Further, when n, n s (I s () x n ), ns (y n I s ()), ns+ (I s () u n ) s. The methods re quite lengthy, nd give the ppernce of being specific to this cse. Here we present different, rgubly simpler, pproch to results of this sort. Without ny extr effort, it delivers more generl version: x n is replced by n f(r) n f(x)dx, 0 where f(x) is function stisfying suitble conditions; in fct, the results gin in both clrity nd simplicity when presented in this wy. The first step, the convergence of (x n ) nd (y n ), is ctully no more thn the fmilir process leding to the existence of Euler s constnt γ. The hrder prt is the derivtion of the other limits. These re obtined in [] using vrious series expnsions. Our method is bsed insted on n estimte for the error in the trpezium rule which is well known to specilists in numericl nlysis, but possibly less so to Gzette reders. The uthor hopes tht it will be of interest to some of them. We describe two proofs, both quite elementry. We consider differentible function f(x) tht is positive nd decresing for x 0 nd stisfies f(x) 0 s x. Let I n = S n = f(0) + f() + + f(n ), n 0 f(x) dx, J r = r+ r f(x) dx, n = S n I n, Γ n = S n+ I n,

2 so tht Γ n = n + f(n). Clerly, I n = J 0 + J + + J n, so n n = [f(r) J r ]. () The bsic results bout these quntities derive from the following very simple inequlity. Since f(x) is decresing, we hve f(r + ) f(x) f(r) for r x r +, hence f(r + ) J r f(r). () By (), it follows tht n 0. Further: Proposition : ( n ) is incresing nd (Γ n ) is decresing. Both converge to the sme limit (sy L) s n, nd n L Γ n for ll n. Proof. We hve n+ n = f(n) J n 0, Γ n+ Γ n = f(n + ) J n 0. So Γ n is bounded below (by 0) nd decresing, hence it tends to limit, L, nd Γ n L for ll n. Since Γ n n = f(n) 0 s n, n lso tends to L. Exmple : The best known exmple of this process is given by f(x) = /(x + ). Then n = n r= log(n + ), nd L is Euler s constnt γ. r of [], Exmple : Let f(x) = /( + x) s, where > 0 nd 0 < s <. Then, in the nottion n = x n, Γ n = y n+, L = I s (). nd hence By (), we hve L = [f(r) J r ], (3) L n = [f(r) J r ], (4) Insted of (), we now consider the trpezium rule estimte for J r, tht is, T r = f(r) + f(r + ). This is the integrl of the liner function greeing with f(x) t r nd r +. In most cses, it is much more ccurte estimte of the integrl thn either f(r) or f(r + ). If f is convex

3 (i.e. curving upwrds), then it is obvious from digrm tht J r T r ; forml proof is contined in Theorem 3 below. A sufficient condition for convexity is f (x) 0 for ll x. It is stisfied by mny functions of the type we re considering, including the specific exmples given bove. Note tht Then hence n T r = f(0) + n f(r) + f(n) = S n f(0) + f(n). r= To tke dvntge of the better pproximtion given by T r, we introduce Λ n = n + Γ n = n + f(n). Λ n = S n + f(n) I n n = (T r J r ) + f(0), L Λ n = (T r J r ). (5) Compre this with (4). For convex f, these formule show t once tht (Λ n ) is incresing nd L Λ n. In the nottion of [], Λ n = u n+. We now estblish n estimte for the error in the trpezium rule, nd use it to derive pir of inequlities for L Λ n which in turn will imply the limits stted in []. The first step is to give n estimte for the error in liner pproximtion to function. This is ctully the cse n = of the more generl result on the polynomil interpolting function t n points (e.g. [, p. 4]). The proof is plesnt ppliction of Rolle s theorem. Proposition : Let f be twice differentible on [, b], nd let p(x) be the liner function greeing with f(x) t nd b. Let q(x) = (x )(b x). Then, given x in (, b), there exists ξ in (, b) such tht p(x) f(x) = q(x)f (ξ). Proof: We prove the sttement for chosen point x 0 in (, b). Let G(x) = p(x) f(x) kq(x), with k chosen so tht G(x 0 ) = 0, hence p(x 0 ) f(x 0 ) = kq(x 0 ). We hve to show tht k = f (ξ) for some ξ. Now G() = G(b) = G(x 0 ) = 0. By Rolle s theorem, pplied twice, there exists ξ in (, b) such tht G (ξ) = 0. Now p (x) = 0 nd q (x) =, so G (x) = f (x) + k for ll x. Hence k = f (ξ), s required. Then Theorem 3: Suppose tht m f (x) M on [, b], nd let T (f) = (b )[f()+f(b)]. m(b )3 T (f) 3 f(x) dx M(b )3.

4 Proof: T (f) = p(x) dx, where p(x) is s bove. Proposition, Write b = h nd substitute x = y: q(x) dx = mq(x) p(x) f(x) Mq(x). h So [p(x) f(x)] dx lies between mh3 nd Mh3. 0 y(h y) dy = [ hy 3 y3] h 0 = 6 h3. Now q(x) 0 on [, b], so by Note: We puse to sketch second, eqully ttrctive, proof of Theorem 3. c = ( + b) nd f(x) dx = I(f). Integrtion by prts gives (x c)f (x) dx = = [ ] b (x c)f(x) f(x) dx (b )[f(b) + f()] I(f) = T (f) I(f). Write Now integrte by prts the other wy round! With q(x) s bove, we hve q (x) = +b x = (c x), so we cn use q(x) s the ntiderivtive of x c. We obtin since q() = q(b) = 0. T (f) I(f) = sttement follows s before. [ = ] b q(x)f (x) + q(x)f (x) dx. q(x)f (x) dx Agin the integrnd lies between mq(x) nd Mq(x), nd the Clerly, if f (x) is decresing, then Theorem 3 pplies with M = f () nd m = f (b). Hence for T r nd J r defined s bove, if f (x) is decresing, then We cn now stte our bsic result on L Λ n. Theorem 4: Suppose tht f (r + ) T r J r f (r). () f(x) is positive nd decresing for x 0, (b) f(x) nd f (x) tend to 0 s x, (c) f (x) is positive nd decresing for x 0. Then f (n + ) L Λ n f (n ). 4

5 (Note tht f (x) is negtive.) Proof: By (5) nd Theorem 3, provided tht r= f (r) converges, we hve f (r) L Λ n f (r). + Since f (x) is decresing, () (pplied to successive intervls) gives f (r) f (x) dx = f (n ) nd + f (r) n n+ f (x) dx = f (n + ). The list of conditions on f(x) might seem rther long, but it is very esily seen tht they re ll stisfied by /( + x) s (where s > 0). Actully, with bit of effort, one cn show tht the condition f (x) 0 follows from the others. Exmple 3: To pply this to γ, tke f(x) = /(x + ), nd replce n by n in Theorem 4. Then Λ n = n r= r + n log n, nd we deduce tht γ = Λ n + R n, where (n + ) R n (n ). Even with n quite smll, this gives good pproximtion to γ. For exmple, when n = 4, the resulting lower nd upper bounds for γ re nd The sme work pplies, rther more directly, to the estimtion of the til of convergent series. Suppose tht f stisfies (), (b), (c) nd tht f(x) dx is convergent. Then, by (), n= f(n) is convergent (this is the integrl test for convergence ). Write Sn = f(n), In = f(x) dx. It is fmilir fct tht I n pproximtes S n in some sense. We cn now describe this pproximtion rther ccurtely. Clerly, T r = S n f(n) nd J r = I n. So the proof of Theorem 4 gives: Theorem 5: Under these conditions, Sn = In + f(n)+r n, where f (n+) R n f (n ). Exmple 4: Let f(x) = /x s, where s >. The sum of the series n= /ns is the Riemnn zet function ζ(s). In the nottion of Theorem 5, S n = (s )n s + n s + R n, 5 n

6 where s (n + ) s+ R n (n ) s+. Note: If f (x) is convex, then refinement of the second proof of Theorem 3 (which we will not describe here) leds to T r J r [f (r + ) f (r)]. This mens tht the f (n ) in Theorem 4 cn be replced by f (n), so tht the n cn be replced by n in Exmples 3 nd 4. Finlly, we return to the limits stted t the beginning. We ssume two further conditions which re clerly stisfied by /( + x) s. Agin, we will not tke ny trouble investigting the extent to which some of the conditions re implied by the others. Theorem 6: Let f(x) be s in Theorem 4, nd ssume further: (d) f (x + ) f (x) s x, (e) f (x) f(x) 0 s x. Then, when n, L Λ n f(n) 0, L Λ n f (n) L n f(n),, Γ n L f(n). Proof. The first sttement follows t once from Theorem 4 nd (d), since (d) lso implies tht f (x )/f (x) s x. Condition (e) now gives the second sttement. The lst two sttements follow t once, since n = Λ n f(n) nd Γ n = Λ n + f(n). This illustrtes nicely the fct tht Λ n is better pproximtion to L thn n or Γ n. These limits reproduce the ones from []. Consider, for exmple, the first one. With f(x) = /( + x) s nd the nottion of [], we hve so tht L Λ n f (n ) = (I s() u n ) ( + n )s+, s n s+ sn s+ L Λ n (I s () u n ) = ( + n ) s+ f (n ) s s n. A further limit stted in [] is lim n n s+ (z n I s ()) = s/6, where z n = (x n + y n ) (Theorem 3). Recll tht y n = Γ n. We leve it s n exercise for the reder to show tht under our conditions, lim n (Γ n Γ n )/f (n) = nd to derive the limit just stted. 6

7 References. Alin Sintmrin, Regrding generlistion of Iochimescu s constnt, Mth. Gzette 94 (00), Lee W. Johnson nd R. Den Riess, Numericl Anlysis, Addison-Wesley (98). Dept. of Mthemtics nd Sttistics, Lncster University, Lncster LA 4YF, UK e-mil: 7

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