Hermite s Rule Surpasses Simpson s: in Mathematics Curricula Simpson s Rule. Should be Replaced by Hermite s

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1 International Matheatical Foru, 4, 9, no. 34, Herite s Rule Surpasses Sipson s: in Matheatics Curricula Sipson s Rule Should be Replaced by Herite s Vito Lapret University of Lublana Faculty of Civil and Geodetic Engineering Departent of Matheatics and Physics Jaova cesta, Lublana, Slovenia 386 Vito.Lapret@fgg.uni-l.si Abstract It is argued that Herite s rule has considerable advantages over Sipson s rule. The forer enables us to estiate not only the integrals but also the sus which is not the case with Sipson s rule. Therefore, Herite s rule should be, in our opinion, included into the first year undergraduate curricula at the university level. It is shown that the coplexities of the derivations are equal for both rules. Moreover, we believe that in atheatics curricula Herite s rule should replace Sipson s rule, because it is a ore powerful tool. Matheatics Subect Classification: 65B5, 65D3, 97-, 97C9 Keywords: Euler s constants, Herite, nuerical integration, nuerical suation, Sipson

2 664 V. Lapret Introduction One of the ost popular ethods of nuerical integration is Sipson s rule, which is applicable to four ties continuously differentiable functions. However, for such functions we have at our disposal also a less known ethod, known as Herite s rule which, however, is ore accurate than Sipson s rule considering the aount of required work. Herite s rule originates fro the Euler-Maclaurin forula of order four [,, 5] and therefore can also be used easily for the approxiation of finite sus and infinite series. 3 It is regrettable that this ethod is not ore widely known and used and we suggest, that in the atheatics curricula Sipson s rule should replace Herite s. The copeting rules We consider four ties continuously differentiable function f: [a, b] R. We set the step h = b a n, n being a positive integer. Sipson s rule. Using positive integer and setting n = and x i = a + ih for i =,,..., n, we forulate the Sipson s rule as [ ] b f(x) dx = h f(a) + f(b) + 4 f(x i ) + f(x i ) 3 a i= i= b a 8 h4 f (4) (ξ) for soe ξ [a, b]. () Herite s rule. The coposite Herite s rule expresses directly the difference between the integral and the Rieann su for a unifor partition of the interval. It says that b a n f(x) dx h f(a + ih) = h [f(b) f(a)] i= h [f (b) f (a)] + r(a, b, h), () Thoas Sipson, 7-76, English atheatician. This is in fact Newton-Cotes quadrature forula of order, which was already known to Kepler in 6 and Cavalieri in 639 and rediscovered by Sipson in 743 [4, p. 9]. 3 Sipson s rule can be used also for nuerical suation [3], however not as directly as the Herite s ethod.

3 Herite s rule surpasses Sipson s 665 where b r(a, b, h) = h 4 W ( ) x a h f (4) (x) dx = b a 7 f (4) (ζ) h 4, (3) a for soe ζ [a, b]. Here, W is -periodic function defined as (i) W (x) := 4 [x( x)] if x < (ii) W (x + ) := W (x) for x R (4) and bounded as follows, W (x) 384 The graph of the function W (x) is shown in Figure. and.3.. for x R. (5) - - Figure : Graph of the function W (x). Using () (5) and the triangle inequality, we obtain the following estiates: b n f(x) dx h f(a + ih) a h f(b) f(a) i= ( h f (b) f (a) + h M ) 4(b a) (6) 6 b a where n f(x) dx h f(a + ih) h f(b) f(a) i= + h ( f (b) f (a) + h M ) 4(b a), (7) 6 M 4 = ax f (4) (x). (8) a x b Forulas (6) and (7) show the rate of convergence of the Rieann su as h.

4 666 V. Lapret 3 The applicability of the rules 3. Coputation of definite integrals. Both rules are being used for nuerical integration. This, of course, is the ain purpose of Sipson s rule, as is well known. The presence of the first and the fourth derivative in Herite s rule is only an apparent disadvantage, since both forulas, (6) and (), include the fourth derivatives in their error ters. Moreover, the first derivatives in Herite s rule can be approxiated adequately by the difference quotients [6]. It follows that, with the sae aount of work, an integral can be coputed four ties ore accurately using Herite s rule rather than Sipson s rule. This is evident fro the forulas (6) and () (3). In addition, Herite s rule can be used easily for the approxiation of finite sus and infinite series. 3. Coputation of finite sus. When f 4 (x) becoes and stays sall, relations () (5) becoe very useful for nuerical suation. Naely, for integers and n such that, < n, replacing n by n, and taking a =, b = n, and h = in (), we obtain or f(x) dx n i= f(+i) = f() = = where, considering (3), ρ(, n) = [f(n) f()] [f (n) f ()]+r(, n, ) f(x) dx + [f() + f(n)] + [f (n) f ()] + ρ(, n), (9) W (x ) f (4) (x) dx = W (x) f (4) (x) dx, () and the periodicity of W (x) has been applied. Using (5), we have the error estiate ρ(, n) 384 f (4) (x) dx, () valid for integers < n. Forulas (9) () constitute Herite s suation forula.

5 Herite s rule surpasses Sipson s 667 Exaple. How does the sequence n n = 3 grow? In this exaple we deal with the function f(x) x /3, having the derivatives f (x) 3 x /3, f () (x) 9 x 4/3, f (3) (x) 8 7 x 7/3 >, and f (4) (x) 56 8 x /3 <. Consequently, we get fro (9) the equalities 3 = x /3 dx + ( /3 +n /3) + ( n /3 /3) + ρ(, n) 3 = 3 ( 3 n5 3 ) ( ) n + ( 8 3 ) n 3 + ρ(, n), = where, according to (), we can estiate the reainder ρ(, n) as Therefore 3 = = ρ(, n) () = 384 ( 3 5 n 3 n + < 384 [ f (4) (x)] dx = [ f (3) (x) ] n /3 = n + ) ( n ) 8 3 +ρ(, n), thus ( = ) 8 3 = }{{} = + ( 3 5 n 3 n + :=σ() 3 n + ) 8 3 n + ρ(, n), where 96 3 ρ(, n) , 7 for < n. Hence, the reainder ρ(, n) becoes sall if the paraeter becoes large. For n 9 we have the estiate 8. 7 < ρ(9, n) < In the forula ( 3 3 = σ(9) + 5 n 3 n + = 3 n + ) ρ(9, n) n

6 668 V. Lapret we copute σ(9) = , to conclude that ( 3 3 = n 3 n + = 3 n + ) 8 3 n + δ(n), where the error ter δ(n) is estiated as < δ(n) < 6 for n. Thus, n 3 = can be coputed to at least five decials for n. Here we note that, ore generally, using Herite s suation forula, the generalized haronic nubers H α (n) := n = α could be effectively estiated for every positive integer n and any α [, ). Exaple. How do the sus s(n) := n cos = vary? To answer this question we consider the function f : [, ) R, f(x) cos x x having derivatives f (x) sin x x f () (x) cos x 4x 3/ f (3) (x) sin x 8x f (4) (x) cos x 6x cos x () x 3/ + 3 sin x + 3 cos x 4x 4x 5/ + 3 cos x 5 sin x 5 cos x 4x 5/ 8x 3 8x 7/ 5 sin x 45 cos x 5 sin x 5 cos x + +. (3) 5/ 8x 3 6x 7/ 6x 4 6x 9/ Substituting x = t, we obtain = f(x) dx = n cos t t t dt = ( sin n sin ), (4) for integers n > >. Therefore, using (9) and (), we get f() = ( sin n sin ) + ( cos + cos ) n n + ( sin n cos n n n 3/ + sin + cos ) + ρ(, n), 3/ where, referring to () and (3), we estiate roughly ρ(, n) ( 384 6x + 5 5/ 8x x7/ 6x < ( / / / 6x 9/ ) dx (5) ). (6)

7 Herite s rule surpasses Sipson s 669 Thus, we have = where cos ( = σ() + sin n + cos n n sin n 4n σ() = cos ( + sin + cos cos ) n + ρ(, n), 4n 3/ + sin 4 + cos 4 3/ (7) ). (8) Using (6), we estiate ρ(7, n) < 7 5. Moreover, using (8), we calculate σ(7) = Hence, according to (7), we obtain the expression where = cos = s(n) + δ(n) s(n) sin n + cos n n sin n 4n cos n 4n 3/ and < δ(n) <.5 4, for n 8. Figure shows the graph of the sequence s(n) together with the graph of the function s(x) of continuous variable x Figure : The graphs of the sequence s(n) and the function s(x). 3.3 Coputation of infinite sus. It is not difficult to deduce fro relations (9) () a theore relating the convergence of a series = f() and the integral f(x) dx. The result is as follows:

8 67 V. Lapret Theore [integral test 4 ]. If f C 4 [, ), f (4) (x) dx converges, and f (n) exist, then: the finite liits λ := li n N n f(n) and λ := li n N n (a) The series = f() converges. The sequence n f(x) dx converges. (b) If the series = f() converges, then λ = and f() = f() + li n N n = = f(x) dx + f() where ρ () = ρ(, ), consequently ρ () = W (x) f (4) (x) dx and ρ () λ f () + ρ (), (9) f (4) (x) dx. () Exaple 3. Does the series cos = converge? Referring to integral test theore and returning to Exaple we see, that for the function f(x) cos x x, we have f (4) (x) dx < and λ = λ =. However, according to (4), li f ( x) dx does not exist since every point of n the interval [, ] is the liit point of the sequence n sin n, see e.g. [7] or [], and sin ( n ) sin n. Exaple 4. Show that the series cos = converges and copute its su! Indeed, the series results fro the function f : [, ) R, f(x) cos x x having the derivatives f (x) sin x x 3/ f () (x) cos x 4x f (3) (x) sin x 8x 5/ f (4) (x) cos x 6x 3 Thus, we can roughly estiate cos x () x + 5 sin x + cos x 4x 5/ x cos x 33 sin x 6 cos x 8x 3 8x 7/ x 4 7 sin x 87 cos x 79 sin x 4 cos x + +. () 8x 7/ 6x 4 6x 9/ x 5 f (4) (x) Φ(x) := 6x x7/ 6x x9/ x. (3) 5 4 The onotonicity of the function f is not assued!

9 Herite s rule surpasses Sipson s 67 Therefore f (4) (x) dx converges and, considering (), we estiate ρ () 384 f (4) (x) dx 384 Φ(x) dx / / 64. (4) 4 To exaine the iproper integral f(x) dx we use the substitution x = t and integrate by parts five ties n cos t f(x) dx = dt t [ sin t = cos t t t + 4 n for integers n > >. Obviously, convergent, and the estiate sin t t 6 4 sin t t 3 + cos t t 4 + ] 48 sin t n t 5 sin t t 6 dt, (5) sin t t 6 dt sin t dt t 6 dt is absolutely convergent, hence dt t 6 = 5 5/ (6) holds. Consequently, considering the expression (5), f(x) dx is also convergent and we have where I() sin + cos f(x) dx = I() + δ (), (7) + 4 sin 3/ cos 48 sin. (8) 5/ Moreover, referring to (5) and (6), the error δ() is roughly estiated as δ () = 4 sin t dt t 6 4 dt t = 48, (9) 6 5/ for. Now, we have all the needed eleents to use the forula (9). Indeed, fro (9), (7), (8) and () we conclude that S := = cos = σ() + (), (3)

10 67 V. Lapret where σ() = cos 43 cos and, according to (9) and (4), sin + 5 cos 97 sin + 4 3/ 48 sin (3) 5/ () = δ () + ρ () δ () + ρ () () := / / 64, 4 (3) for. Since, for exaple, () < 6.5 5, we copute directly, using (3), the su σ() = Hence, using (3), we estiate.338 < S <.336. (33) Thus, S =.33...; the su of the series in question is calculated to three decial places. Figure 3 shows the graph of the sequence n s(n) := n converging slowly towards the su S = cos = = = cos s(n) S Figure 3: The graph of the sequence n s(n) = n cos = converging to S. We note that the ain part of the error () results fro the error δ() of the coputation of the iproper integral f(x) dx. However, if we coputed this integral ore accurately by other eans or by increasing the nuber of repetitions of the integration by parts, then in the forula = cos = cos x x dx + cos + sin cos + 43/ + ρ()

11 Herite s rule surpasses Sipson s 673 we could take even lower values of and still achieve a rather good accuracy. For exaple, even = 7 is adequate since ρ(7) < 5 5. This exaple exhibits Herite s rule as an iportant convergence accelerating tool. Naely, the sequence of partial sus s(n) := n cos = converges very slowly. Indeed, using the forulas (9) (), (5) and (), we find with σ() defined in (3), with the suand s(n) sin n n s(n) = σ() + s(n) + (, n) (34) 3 cos n n 95 sin n 4n 3/ and with the error (, n) estiated as + 43 cos n + n (, n) ρ(, n) + δ(, n) f (4) (x) n sin t dx t 6 dt 48 sin n, (35) n 5/ 384 f (4) (x) dx + 4 dt t 6 (3) (), for n >. Thus, using (3), we estiate (5, n) < 6, for n 5, and copute, using (3), σ(5) = Therefore, according to (34) and (35), we have s(n) = sin n 3 cos n 95 sin n n n 4n 3/ + 43 cos n 48 sin n + + δ(n), (36) n n 5/ where, for n 5, the estiate < δ(n) < 5 holds. This way we obtain s(95) = >.387. Hence, referring to (33) we estiate s(95) S > =.9. Thus, the relative error of the approxiation s(n) S can be larger than.5%, even when n is larger than one illion. Figure 4 shows the graph of the sequence n s(n) obtained using (36).

12 674 V. Lapret Figure 4: Slow convergence of the sequence of partial sus s(n). 3.4 Introduction of Euler s constant for a function. For an integrable function f :[, ) R we define the sequence γ n := f() = f(x) dx. (37) If f satisfies certain conditions the sequence γ n converges. In particular, for f C 4 [, ), using (9), we obtain γ n = [f() + f(n)] + [f (n) f ()] If, additionally, finite liits f( ) := li n N n W (x) f (4) (x) dx. (38) f(n) and f ( ) := li f (n) exist, n N n and the integral f (4) (x) dx is convergent, we deduce fro (38), that finite liit γ := li n γ n, exists. It is known as the Euler s constant for f. Assuing all the conditions discussed above and using (38), we obtain γ = [f() + f( )] + [f ( ) f ()] Now, fro (38) and (39) we get i.e. γ γ n = [f( ) f(n)] + [f ( ) f (n)] W (x) f (4) (x) dx. (39) n W (x) f (4) (x) dx, γ = γ n + [f( ) f(n)] + [f ( ) f (n)] + ρ (n). (4)

13 Herite s rule surpasses Sipson s 675 Referring to (), the error ter ρ (n) = is estiated, considering (), as n ρ (n) 384 W (x) f (4) (x) dx, (4) n f (4) (x) dx. (4) The relations (4) (4) constitute a tool for nuerical approxiation of Euler s constant for a given function f. The knowledge of Euler s constant for f relieves the coputation of partial sus resulting fro f. Naely, using (37) and (4), we have f() = γ + = f(x) dx + [f(n) f( )] + [f (n) f ( )] ρ (n). (43) In addition, assuing that the previously entioned conditions concerning the function f are fulfilled, we have the following equivalence 5 : f() converges. The sequence n = Additionally, if I := li n f(x) dx exists, then f(k) = γ + I. k= f(x) dx converges. Exaple 5. How does the sequence n H α (n) := n = α, of the so-called generalized haronic nubers with paraeter α R, vary? This question has an easy answer for α >. Indeed, H α (n) arising fro the function f α (x) x α having the derivatives f α(x) αx (α+), f α () (x) α(α + )x (α+), f α (3) (x) α(α + )(α + )x (α+3) and f α (4) (x) α(α + )(α + )(α + 3)x (α+4) >. Hence, for α >, we get f α ( ) = f α( ) = and (4) f α (x) dx <. Therefore, the sequence γ α (n) := n = f α() f α(x) dx is convergent, i.e. Euler s constant γ α ( ) := li γ α (n) for the n 5 whether f has constant sign or not

14 676 V. Lapret function f α exists for any α >. Furtherore, considering (4), (5) and (4), and the generalized integral ean value theore, we have ρ α(n) = for soe ξ, ϑ [, ]. n = ϑ W (x) f (4) α (x) dx = W (ξ) [ f (3) α (n) ] α(α + )(α + ) 384 n α+3 <, (44) Let K 3 be given constant and n an integer such that n e K. Then, using the Taylor series expansion of the exponential function, we have n α = e α ln n e αk = (αk) = > αk + (αk)! + (αk)3 6 = αk ) ( + Kα + K 3 α αk ( + 3α + α ) = K α(α + )(α + ), α >. Thus, referring to (44), we obtain the unifor estiate ρ α(n) < true for α >, K 3 and n e K. This way, for exaple, for n and any α >. Now, according to (4), we have and therefore also 9Kn 3, (45) ρ α(n) < 9, (46) γ α ( ) = γ α (n) n + α α n + α+ ρ α(n) (47) H α (n) = γ α ( ) + We distinguish two cases: α = and α. dx x + α n α α n α+ ρ α(n). (48) A) α = : The nuber C := γ ( ) is called Euler-Mascheroni constant. Referring to (47) and (44), we have C = H (n) ln n n + n ϑ 64 n, ϑ. (49) 4

15 Herite s rule surpasses Sipson s 677 Setting n = in (49), we estiate < C < Hence, C is coputed to nine decial places, C = and, according to (49), we estiate C + ln n + n n < H (n) < C + ln n + n n + 64 n. 4 B) α R + \ {}: In this case, referring to (48), we have γ α ( ) = γ α (n) + ρ α (n), where γ α (n) H α (n) ( n α ) α n + α α n. α+ Thus, for exaple, considering (44), γ / ( ) = H / (n) + ( n ) n + 4n n 5 ϑ 4 n 3 n. Putting n = in this equation, we obtain the inequality < γ / < Thus, we coputed γ / ( ) to eight decial places as γ / ( ) = and we have H / (n) = ( γ / ( ) ) + n + n 4n n + 5 ϑ 4 n 3 n, for every n N and for soe ϑ = ϑ(n) [, ]. The function α γ α ( ) varies for larger n, thanks to (38) or (46), approxiately like any of the functions α γ α (n). Additionally, for α >, the sequence n γ α (n) is strictly decreasing [, Th. ]. Figure 5 shows, on the left, the graphs of the functions α γ α (n) and, on the right, the graph of the function α γ α ( ) Γ Α n n n Α Γ Α Figure 5: The graphs of the functions α γ α (n) (left) and the graph of the function α γ α ( ) (right). Exaple 6. The functions f (x) cos x x and f (x) cos x, not being x onotonous, satisfy all the required conditions for the existence of their Euler s constants, thanks to () (3) and () (). copute the. We leave it to the reader to Α

16 678 V. Lapret 4 The derivations of the rules 4. Sipson s rule. To obtain the basic Sipson s rule we first interpolate a given four ties continuously differentiable function ϕ(t), defined on the interval [ h, h] with h >, by quadratic function, the Lagrange interpolating polynoial, L(t) a t + b t + c, (5) (a, b and c being constants) with prescribed values L( h) = ϕ( h), L() = ϕ() and L(h) = ϕ(h). (5) Figure 6 below illustrates this interpolation. Due to the interpolating condiy y L t y φ t φ φ h φ h t h h Figure 6: Interpolation ϕ(t) L(t). tions (5), coefficients a, b and c in (5) are deterined by the syste of linear equations having partial solution a h b h + c = ϕ( h) c = ϕ( ) a h + b h + c = ϕ( h ) a = [ϕ( h) ϕ() + ϕ(h)] and c = ϕ(). h Consequently, since ϕ(t) L(t), we obtain, using (5), h h ϕ(t) dt h h = h 3 L(t) dt (5) = 3 a h3 + ch [ϕ( h) + 4ϕ() + ϕ(h)].

17 Herite s rule surpasses Sipson s 679 Hence, we study the error ter or the so called reainder h R(h) := ϕ(t) dt h [ϕ( h) + 4ϕ() + ϕ(h)]. (5) 3 h Following [], we introduce the auxiliary function S(h) := R(h) R() h 5, h [, ]. (53) Considering 5, we have h S(h) ϕ(x) dx + h ϕ(x) dx (54) h 3 [ϕ( h) + 4ϕ() + ϕ(h)] R() h5. Using the Newton-Leibniz theore we find the derivatives S (h) 3 [ϕ( h) + ϕ(h)] 4 3 ϕ() + h 3 [ϕ ( h) ϕ (h)] 5 R() h 4 S () (h) 3 [ ϕ ( h) + ϕ (h)] h [ ϕ () ( h) + ϕ () (h) ] R() h 3 3 S (3) (h) h [ ϕ (3) (h) ϕ (3) ( h) ] 6 R() h. (55) 3 Due to (53) and (5) we have S() = S() =. Hence, using the Rolle s theore, we see that there exists ξ (, ) such that S (ξ ) =. However, due to (55) we have S () = as well, and we can apply the Rolle s theore again to assure that S () (ξ ) = for soe ξ (, ξ ). Siilarly, since S () () =, due to (55), we deduce that S (3) (η) = for soe η (, ξ ). considering (55), we obtain the equation = η 3 [ ϕ (3) (η) ϕ (3) ( η) ] 6R() η, η. Therefore, Thus, applying the Lagrange s theore on finite increent for function ϕ (3), we have = η 3 ϕ(4) (ξ) 6R() η (56) for soe ξ ( η, η). Fro this equation, since η, we conclude R() = ϕ(4) (ξ) 9.

18 68 V. Lapret Referring to (5), we have proved (6) when a =, b =, h = and n = : ϕ(t) dt = 3 [ϕ( ) + 4ϕ() + ϕ()] ϕ(4) (ξ). (57) 9 For four ties continuously differentiable function ψ :[a, b] R we substitute x = x(t) := b a t + a+b and ϕ(t) := ψ(x(t)) to obtain, considering the expression (57), b for soe ξ (a, b). a ψ(x) dx = b a = b a 6 ϕ(t) dt [ ψ(a) + 4ψ( a+b ) + ψ(b)] ψ(4) (ξ) 9 ( ) b a 5 (58) Now we construct the coposite Sipson s rule by dividing interval [a, b] into subintervals [x i, x i+ ], where x i = a + i h and h = b a, for i =,,..., and N. For each subinterval [x i, x i+ ] we use the forula (58) according to which there exists, for every i {,,..., }, soe ξ i (x i, x i+ ), such that xi+ f(x) dx = h x i 3 [f(x i) + 4f(x i+ ) + f(x i+ )] f (4) (ξ i ) 9 Since [a, b] = [x i, x i+ ], we obtain the equalities i= b a f(x) dx = i= = i= xi+ x i f(x) dx which we siplify to the identity b af(x) dx = h 3 [ h 5. { h 3 [f(x i) + 4f(x i+ ) + f(x i+ )] f } (4) (ξ i ) h 5, 9 f(a) + f(b) + 4 i= ] f(a + (i )h) + f(a + ih) h5 f (4) (ξ i ). (59) 9 i= i=

19 Herite s rule surpasses Sipson s 68 By supposition, the function f (4) is continuous on the interval [a, b]. Consequently there exist [9, p. 7, Theore 3.] Therefore µ := in a x b f (4) (x) and µ := ax a x b f (4) (x). (6) for every i {,,..., }. Thus, or µ µ µ f(ξ i ) µ i= i= f (4) (ξ i ) µ f (4) (ξ i ) µ. This estiate, together with (6) and [9, p. 6, Theore 3.3], eans that there exists soe ξ [a, b] such that f (4) (ξ i ) = f (4) (ξ). i= Fro this equality and (59) follows the coposite Sipson s rule (6). 4. Herite s rule. For the four ties continuously differentiable function ϕ : [, ] R we have, integrating by parts, ϕ(x) dx = ϕ(x) (x ) dx = [ϕ() + ϕ()] ( x ) ϕ (x) dx, (6) ( x ) ϕ (x) dx = ϕ (x) ( x x + ) dx = [ϕ () ϕ ()] (6) ( x x + ) ϕ (x) dx,

20 68 V. Lapret and ( x x + ) ϕ (x) dx = = ( 6 x3 4 x + x) ϕ (3) (x) dx = = ϕ (x) ( 6 x3 4 x + x) dx ( 6 x3 4 x + x) ϕ (3) (x) dx (63) ϕ (3) (x) ( 4 x4 x3 + 4 x) dx ( 4 x4 x3 + 4 x) ϕ (4) (x) dx. (64) Relation (6) is the basic Euler-Maclaurin forula of order [5, pp., 4]. Fro (6) and (6) we obtain, using Bernoulli polynoial B (x) x x +, the basic Euler-Maclaurin forula of order 6 ϕ(x) dx = [ϕ() + ϕ()] [ϕ () ϕ ()] + B (x) ϕ (x) dx. (65) We reark that this equation can be transfored, of course, to the basic trapeze rule ϕ(x) dx = [ϕ() + ϕ()] + (x x) ϕ (x) dx. Moreover, fro (65) and (63) we get the basic Euler-Maclaurin forula of order 3 ϕ(x) dx = [ϕ() + ϕ()] [ϕ () ϕ ()] 6 B 3 (x) ϕ (3) (x) dx, (66) where B 3 (x) x 3 3 x + x is the third Bernoulli polynoial. Finally, fro (66) and (64) we find the basic Herite s rule 6 ϕ(x) dx = [ϕ() + ϕ()] [ϕ () ϕ ()] + [x( x)] ϕ (4) (x) dx. (67) 4 6 This rule is closely related with the basic Euler-Maclaurin forula of order 4 [5, Th., p. 4].

21 Herite s rule surpasses Sipson s 683 For four-ties continuously differentiable function ψ :[, n] R, and n being integers, < n, we use auxiliary, four ties continuously differentiable functions u i :[, ] R, u i (t) := ψ(i +t), which help us to activate the basic Herite s rule (67), i i ψ(x) dx = u i (t) dt = [u i() + u i ()] [u i() u i()] + [t( 4 t)] u (4) i (t) dt = [ψ(i ) + ψ(i)] [ψ (i) ψ (i )] + [t( 4 t)] ψ (4) (i + t) dt, true for integer i { +,..., n}. Suing, with respect to i { +,..., n}, the left and the right hand sides of the forulas above, we get, ψ(x) dx = i=+ i i = i=+ + i=+ ψ(x) dx [ψ(i ) + ψ(i)] i=+ [t( 4 t)] ψ (4) (i + t) dt. Consequently, due to telescoping of suands, we obtain, ψ(i) = ψ(x) dx + [ψ() + ψ(n)] i= where the reainder ρ(, n) = [ψ (i) ψ (i )] + [ψ (n) ψ ()] + ρ(, n), (68) i=+ [t( 4 t)] ψ (4) (i + t) dt, (69) can be siplified by introducing the -periodic function W (x) defined as (i) W (x) := 4 [x( x)] for x [, ] (ii) W (x + ) := W (x) for x R. (7)

22 684 V. Lapret Indeed, considering (69) and (7), and substituting x = i + t, we have ρ(, n) = = = = i i i i W (t) ψ (4) (i + t) dt W (x + i) ψ (4) (x) dx W (x) ψ (4) (x) dx W (x) ψ (4) (x) dx. (7) [ (x( ) ] () The second derivative x) (6x 6x + ) has equal values for x = and x =. Hence, W (x) is twice continuously differentiable over the entire R. For a four ties continuously differentiable function f:[a, b] R we define the four ties continuously differentiable auxiliary function ψ : [, n] R, ψ(t) := f(a + t h), where h = b a and the integer n satisfies n. We have n the derivatives ψ () (t) h f () (a+th) for t [, n] and =,, 3, 4. Referring to (68) and (7), we obtain ψ(t) dt = ψ(i) i= + [ψ() + ψ(n)] [ψ (n) ψ ()] W (t) ψ (4) (t) dt. (7) Perforing the substitution x = a + th and considering the definition of ψ, we find the relation b a f(x) dx h = f(a + ih) [f(a) + f(b)] h [f (b) f (a)] i= b which proves Herite s rule () (5). + a W ( ) x a h h 4 f (4) (x) dx h,

23 Herite s rule surpasses Sipson s Concluding rearks We favor Herite s rule in coparison with Sipson s for the following reasons:. Herite s rule can be used easily in two copleentary ways: for estiating integrals using sus and for estiating sus using integrals. This is not possible with Sipson s rule.. On the basis of Herite s rule it is straightforward to deduce an integral test coparing the convergence of the integral f(x) dx and the series f(). 3. Herite s rule introduces in a natural way the notion of Euler s constant for a function and enables its coputation. 4. Concerning nuerical integration of periodic function, Herite s rule is both ore useful and sipler than Sipson s rule, if the length of integration interval is equal to the period of the integrand. 5. The reainder in Herite s rule can be estiated, a priori, four ties ore accurately than the reainder in Sipson s rule for approxiately equal nubers of function evaluations. 6. Coposite Herite s rule () deonstrates very clearly the difference between an integral and its integral su together with its estiates (6) (7). 7. Herite s rule can be derived within the introductory course of calculus in the context of the fundaental theore of calculus (Newton-Leibniz theore). This way, Herite s rule becoes iediately applicable in cases when the priitive is not an eleentary function and in addition, illustrates that analytical and coputational ethods copleent each other. 8. The derivation of Herite s rule is certainly not ore coplex than the corresponding derivation of Sipson s rule.

24 686 V. Lapret References [] P. J. Davis and P. Rabinowitz, Methods of Nuerical Integration, Acadeic Press, Chestnut Hill, MA, 984. [] R. L. Graha, D. E. Knuth, and O. Patashnik, Concrete Matheatics, Addison-Wesley, Reading, MA, 994. [3] Kreinski, Using Sipson s rule to approxiate sus of infinite series, College Math. J., 8 (997), pp [4] R. Kress, Nuerical Analysis, Springer-Verlag, New York (998). [5] V. Lapret, The Euler-Maclaurin and Taylor forulas: twin eleentary derivations, Matheatics Magazine, 74 (), pp. 9. [6] V. Lapret, An Invitation to Herite s Integration and Suation: A Coparison between Herite s and Sipson s Rules, SIAM Rev. 46 (4), No., pp [7] Ogilvy, Stanley C. (969). The Sequence {sin n}, Math. Mag. 4 (969), p. 94. [8] A. Poorten, Life on the Edge, Aer. Math. Monthly 9 (), pp [9] M.H. Protter and C.B. Morrey A First Course in Real Analysis, Springer- Verlag, New York Inc, 99. [] J. Sándor, On generalized Euler constants and Schlöilch Leonnier type inequalities, J. Math. Anal. Appl. 38 (7), pp [] Staib, John H.; Deos, Miltiades S. (967). On the Liit Points of the Sequence {sin n}, Math. Mag. 4 (967), pp. 3. [] E. Süli and D.F. Mayers, An Introduction to Nuerical Analysis, Cabridge University Press, Cabridge (3). [3] S. Wolfra, Matheatica, version 6., Wolfra Research, Inc., Received: February, 9