Successive Derivatives and Integer Sequences

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1 Journal of Integer Sequences, Vol 4 (20, Article 73 Successive Derivatives and Integer Sequences Rafael Jaimczu División Matemática Universidad Nacional de Luján Buenos Aires Argentina jaimczu@mailunlueduar In memory of my sister Fedra Marina Jaimczu ( Abstract We consider the functions /f and h/f, and study some properties of the successive derivatives of these functions We also study certain integer sequences connected with these successive derivatives Introduction In the next section we shall need the generalization of the chain rule of calculus to higher derivatives, that is, the Faá di Bruno s formula For the compact establishment of the Faá di Bruno s formula in the next section we need certain compact notation on partitions and multinomial coefficients This compact notation is due to Vella [2] We now explain this notation A partition π of a positive integer n is a representation of n as a sum of positive integers, n p +p 2 + +p m, called summands or parts of the partition The order of the summands is irrelevant The number of partitions of n we shall denote p(n The number of parts of the partition π we shall denote l(π Consequently l(π m For example, let us consider the following partition π of 55, We have l(π 5 A partition can be written more simply in the form π {p,p 2,,p m } For example our former partition π of 55 can be written in the form π {,,,,,, 2, 2, 2, 6, 6, 7, 7, 7, 0}

2 For each i ( i n, the number of times that i appears as a part of the partition π of n is denoted π i and is called the multiplicity of the part i in π For example, in our former partition π of 55 we have π 6, π 2 3 and π 3 0 Note that l(π n i π i The standard notation for partitions is π [ π, 2 π 2,,n πn ] with parts of multiplicity zero omitted For example, in standard notation, our partition π of 55 is π [ 6, 2 3, 6 2, 7 3, 0 ] Note that {π,π 2,,π n } is a partition of l(π We shall call to this partition the derived partition of π, and denote it δ(π For example, the derived partition δ(π of our partition π of 55 is the following partition of l(π 5 δ(π {, 2, 3, 3, 6} [, 2, 3 2, 6 ] Let us consider the partition π {p,p 2,,p m } of n We shall use the notation π! m i (p i! and we shall use the notation ( n π for the multinomial coefficient ( n p,p 2,,p m ( n p,p 2,,p m n! m i (p i! n! π π! In the next section (using this compact notation we establish the Faá di Bruno s formula and use this formula in the study of the successive derivatives of the function Furthermore, we introduce an integer sequence connected with these successive derivatives f(x In the third section we study the successive derivatives of the function h(x and generalize f(x the quotient rule to n th derivatives Finally, in the fourth section, we study some asymptotic properties of the integer sequence 2 Successive Derivatives of the Function f(x In the following theorem we establish in a compact and useful way the Faá di Bruno s formula This establishment is due to Vella [2] Theorem Suppose y g(u and u f(x are differentiable n times Then the composite function y (g f(x is also differentiable n times and (g f (n (x ( n n π δ(π! g(l(π f(x [f (i (x] π i, ( π Ω n where Ω n denotes the set of all partitions of n i 2

3 If f f(x we shall use the convention f f (0 Now, consider the function f(x f We have the following general theorem ( (0 (2 f Theorem 2 The successive derivatives of the function (2 satisfy the following formula, ( (n P n (n 0, (3 f f n+ where P n is a polynomial of integer coefficients in the variables f,f (,,f (n If n 0 then P 0, (4 and if n then P n ( l(π n ( l(π f(x n l(π [f (i (x] π i π δ(π π Ω n i ( l(π n ( l(π f n l(π [f (i ] π i (5 π δ(π π Ω n Proof Let g(u u u be Note that Substituting equation (6 into equation ( we find that ( (n (g f (n (x f ( n π π Ω(n π Ω(n i g (n (u ( n n! u n+ (6 π Ω(n ( n π δ(π! ( l(π l(π! f(x l(π+ n [f (i (x] π i i ( l(π l(π! n f(x n l(π [f (i (x] π i δ(π! f(x n+ i ( l(π ( l(π n π δ(π f(x n+ f(xn l(π [f (i (x] π i i ( l(π f(x n l(π n i [f(i (x] π i π Ω n ( l(π( n π P n f n+ δ(π f(x n+ (7 3

4 Using (5 we obtain the first polynomials P n P f ( (x f (, (8 P 2 f(x f (2 (x + 2f ( (xf ( (x f f (2 + 2f ( f (, (9 P 3 fff (3 + 6ff ( f (2 6f ( f ( f (, (0 P 4 ffff (4 + 8fff ( f (3 + 6fff (2 f (2 36ff ( f ( f (2 ( + 24f ( f ( f ( f (, P 5 fffff (5 + 0ffff ( f (4 60fff ( f ( f (3 + 20ffff (2 f (3 90fff ( f (2 f ( ff ( f ( f ( f (2 20f ( f ( f ( f ( f ( (2 In the following theorem we establish some general properties of the polynomials P n Theorem 3 The polynomial P n (n has the following properties a Each term (monomial in the polynomial P n has n factors and the sum of superscripts is n the monomials are of the form f (i f (i 2 f (in where i + i i n n (f f (0 Consequently we can establish a correspondence between each monomial and a partition of n The number of terms (monomials in the polynomial P n is p(n, that is, the number of partitions of n b If n is even then the sum of the coefficients is On the other hand, if n is odd the sum of the coefficients is the sum of the coefficients of the polynomial P n is ( n c If n is even the monomials with an even number of f have positive coefficient and the monomials with a odd number of f have negative coefficient If n is odd the monomials with an even number of f have negative coefficient and the monomials with a odd number of f have positive coefficient d If is the sum of the absolute values of the coefficients in the polynomial P n then the following formula holds (A 0 : q(x 2 e x 0 A! x (3 Therefore, A ( 0 is the th derivative of the function q(x at x 0 (A 2 e x q ( (0 The radius of convergence of (3 is R log 2 Furthermore, ( l(π (n (4 π δ(π π Ω n e The coefficient of the monomial f ff (n is The coefficient of the monomial f ( f ( is ( n n! Consequently n! (see part (d 4

5 Proof Proof of part (a It is an immediate consequence of equation (5 Part (a is proved Proof of part (b We apply (7 in the case f(x e x Note that f (i (0 for all i 0 Therefore the left side of (7 at x 0 becomes ( e x (n (0 dn dx n(e x x0 ( n On the other hand the right side of (7 at x 0 becomes π Ω n ( l(π( ( n l(π π δ(π f(0 n l(π n i [f(i (0] π i f(0 n+ π Ω n ( l(π( ( n l(π π δ(π n l(π n i (π i n+ ( l(π ( l(π, π δ(π π Ω n which is the sum of the coefficients of the polynomial P n Part (b is proved Proof of part (c It is an immediate consequence of equation (5 Part (c is proved Proof of part (d We apply (7 in the case f(x 2 e x Note that f(0 and f (i (0 for all i > 0 Consequently, the right side of (7 at x 0 becomes π Ω n ( l(π( ( n l(π π δ(π f(0 n l(π n i [f(i (0] π i f(0 n+ π Ω n ( l(π( ( n l(π π δ(π n l(π n i ( π i n+ ( l(π ( l(π ( P n i π i π δ(π π Ω n ( l(π ( l(π ( l(π π δ(π π Ω n ( l(π (5 π δ(π π Ω n Note that (5 (see the numerator of (7 is the sum of the absolute values of the coefficients in the polynomial P n On the other hand, the left side of (7 at x 0 becomes Therefore (g f (n (0 q (n (0 q(x (g f(x 2 e x 0 A! x Note that the function of the complex variable q(z 2 e z is analytical in the dis z < log 2 and consequently the radius of convergence of (3 is R log 2 Note also that in terms of 5

6 generating functions (see reference [3] we have proved that q(x 2 e x is the exponential generating function of the sequence Part (d is proved Proof of part (e It is an immediate consequence of equation (5 Part (e is proved Note that the first values of the integer sequence are (see either (4, (8, (9, (0, ( and (2 or use directly equation (4: A 0, A, A 2 3, A 3 3, A 4 75, A 5 54 In the next section we shall prove the following recurrence combinatorial formula n 0 A 0, A simple consequence of this formula is the inequality A (n (6 + > (n 3 Successive Derivatives of the Function h(x f(x We shall need the well-nown binomial formula n (a + b n a n b, (7 0 where ( n n! and 0! (n!! We also shall need the following well-nown result of Leibniz on the successive derivatives of the function fg f(xg(x Lemma 4 Let us consider the function fg The following formula holds: where f (0 f and g (0 g (fg (n n 0 Theorem 5 Let us consider the function The following formula holds ( (n h Q n f f n+ f (n g ( (n 0, (8 h(x f(x h f n 0 ( n h (n f n P (n 0 (9 f n+ 6

7 Proof We have Lemma 4 and equation (3 give ( (n h f n ( n 0 h f h f h (n ( f ( n 0 h (n P f + n ( n 0 h (n f n P f n+ equation (9 Using (9, (4, (8, (9 and (0 we obtain the first polynomials Q n Q 0 h, Q h ( f hf (, Q 2 h (2 ff 2h ( ff ( hff (2 + 2hf ( f (, Q 3 h (3 fff 3h (2 fff ( 3h ( fff (2 + 6h ( ff ( f ( hfff (3 + 6hff ( f (2 6hf ( f ( f ( In the following theorem we obtain some information on the polynomial Q n Theorem 6 Let us consider the polynomial Q(n (n 0 a The sum of the coefficients of the polynomial Q(n (n is zero b If C n is the sum of the absolute values of the coefficients in the polynomial Q n then the following formula holds: C n n 0 A (n 0 (20 c If n is even the monomials with an even number of f have positive coefficient and the monomials with a odd number of f have negative coefficient If n is odd the monomials with an even number of f have negative coefficient and the monomials with a odd number of f have positive coefficient d If C n is the sum of the absolute values of the coefficients in the polynomial Q n then the following formula holds: p(x ex 2 e x 0 C! x (2 Therefore, C ( 0 is the th derivative of the function p(x p ( (0 The radius of convergence of (2 is R log 2 ex at x 0 (C 2 e x 7

8 e We have C 0 A 0, C n 2 (n f The following combinatorial formula holds: n A (n 0 g Each term (monomial in the polynomial Q n has (n + factors and the sum of superscripts is n the monomials are of the form h (i f (i 2 f (i (n+, where i + i i (n+ n and h h (0, f f (0 The number of term (monomials in the polynomial Q n is n 0 p( where p( is the number of partitions of (p(0 Proof Proof of part (a The sum of the coefficients in the polynomial P is ( ( see part (b of Theorem 3 Consequently the sum of the coefficients in the polynomial Q n will be (see (9 and (7 n ( ( n 0 0 Part (a is proved Proof of part (b The sum of the absolute values of the coefficients in the polynomial P is A (see part (d of Theorem 3 Consequently the sum of the absolute values of the coefficients in the polynomial Q n will be (20 (see (9 Part (b is proved Proof of part (c It is an immediate consequence of (9 and part (c of Theorem 3 Part (c is proved Proof of part (d Let us consider the function f(x f e x 2 We have f(0 and since f (n (x f (n e x (n we have f (n (0 On the other hand consider the function h(x h e x We have h(0 and since h (n (x h (n e x (n we have h (n (0 Therefore part (c and equation (9 imply that the function p(x h(x f(x ex satisfies e x 2 p(n (0 C n (n 0 On the other hand the function of the complex variable p(z h(z ez is analytical in the dis z < log 2 and consequently the radius of f(z e z 2 convergence of (2 is R log 2 Part (d is proved Proof of part (e We have (see part (d of Theorem 3 ( p(x ex 2 e q(x x 2 e x Consequently C 0 p(0 q(0 A 0 8

9 On the other hand we have Therefore p (n (x 2q (n (x (n C n p (n (0 2q (n (0 2 (n Part (e is proved Proof of part (f It is an immediate consequence of part (b and part (e Part (f is proved Proof of part (g It is an immediate consequence of (9 and part (a of Theorem 3 Part (g is proved 4 Asymptotic Results on the Sequence Lemma 7 Let us consider the power series (3, that is 0 A! x The ratio test for power series is valid, that is the following limit holds: Proof We have Consequently and lim n 0 Using (24 repeatedly we obtain A! A (! A lim A log 2 (22 n 2 A 0 A 0 A 0 A An 2 A A 2 A A A 2 An 2 A 2 A 3 2 A 2 A 2 A 3 An 2 A 3 A 4 3 A + n n, (23 n (24 2 (n 0! (n! 3 (n! (n! 4 (n 2! (n! 9

10 A This last equation and (6 give,! (n! (0 n n ( n 0 n n Finally, (23 and (25 give! A n 0 n n!! (n!! (n! n 0 (n! n(e (25 n (e (26 Suppose that the following inequality holds from a certain n n + We obtain, A Equations (6 and (29 give, n 0! (n! A n n n! hn (27 hn (28 (n n (29 h n 0 A + (! C + (n!! (n! n n ( C + n h C + n! n n h ( h where C n ( n A 0 Equation (30 gives the inequality (from a certain n A n C + n! n n ( h n (n! C + n e h, (30 h e h + ǫ (ǫ > 0 (3 n h 0

11 Suppose that the following inequality holds from a certain n n + We obtain A Equations (6 and (34 give, n 0 n! (n! A n n! n n n pn (32 pn (33 (n n (34 p n 0! (n!! (n! ( p! A n + n n n p n A ( n p (n! (35 Equation (35 gives (from a certain n n n n ( p! ( p! λ e p p λ (λ > 0 Note that the function e p λ (λ > 0 (36 n p f(x ex x is positive and strictly increasing in the interval (, since its derivative is positive in the interval (, On the other hand its image is the interval (0, Furthermore, f(0 and f( e Let us consider the inequality (26 x! h e0 0 e n e p (37 This inequality is our first inequality

12 The inequality in the right side (that is, equation (32 give us the inequality (see (36 Note that since p > 0 we have the inequality Let λ 2 be the least number between and Consequently h 2 e p p λ 2 n (38 e p p > h /2 e p h p 2 h 2 > h (39 Inequality (38 (that is, equation (27 give us (see equation (3 the inequality Note that since (see (39 h 2 > we have the inequality Let ǫ 2 be the least number between and Consequently e h 2 + ǫ 2 p 2 (40 n h 2 e h 2 < e h 2 /2 p e h2 2 h 2 e p p 2 < p e (4 On the other hand, inequalities (38 and (40 give us the following second inequality h 2 e p λ 2 e h 2 + ǫ 2 p 2 (42 p n h 2 Inequality (40 (that is, equation (32 give us the inequality (see (36 h 3 e p 2 p 2 λ 3 n (43 2

13 Equation (4 implies that Let λ 3 be the least number between and Consequently e p 2 p 2 e p 2 p 2 > e p /3 e 2 p p p h 3 > h 2 > h (44 Inequality (43 (that is, equation (27 give us (see equation (3 the inequality Equation (44 implies that Let ǫ 3 be the least number between and Consequently e h 3 + ǫ 3 p 3 (45 n h 3 e h 3 h 3 < e h 2 /3 e h 2 e h 3 h 2 h 3 2 h 2 p 3 < p 2 < p e (46 On the other hand, inequalities (43 and (45 give us the third inequality h 3 e p 2 λ 3 e h 3 + ǫ 3 p 3 (47 p 2 n h 3 In this form we build the following sequence of inequalities (see (37, (42, (47 h e0 0 e n e p h 2 e p λ 2 e h 2 + ǫ 2 p 2 p n h 2 h 3 e p 2 λ 3 e h 3 + ǫ 3 p 3 p 2 n h 3 3

14 h 4 e p 3 λ 4 e h 4 + ǫ 4 p 4 p 3 n h 4 In this sequence of inequalities we have the strictly increasing and bounded sequence h n and the strictly decreasing and bounded sequence p n Consequently h n has limit l and p n has limit l 2 Next, we shall prove that l l 2 l Therefore we obtain the desired log 2 limit (22 lim n n log 2 Note that the sequence λ n has limit zero since λ n and the sequence ǫ n n has also limit zero since ǫ n n The sequence h n satisfies (see the sequence of inequalities the following recurrence relation h n e p n λ n e p n e h n h n +ǫ n h e n +ǫ n h n λ n Consequently if we tae limits in both sides we obtain that l satisfies the equation l e e l l (48 e The sequence p n satisfies (see the sequence of inequalities the following recurrence relation p n e hn + ǫ n e h n l l e p n p n λn p e n λ n p n + ǫ n Consequently if we tae limits in both sides we obtain that l 2 satisfies the equation l 2 e e l 2 l 2 (49 e l 2 l 2 4

15 l and l 2 satisfy the same equation (see (48 and (49 This equation has the solution l e e l l e l l log 2 l (50 We shall prove that this solution is the unique positive solution to equation (50 Consequently l l 2 l log 2 Equation (50 is, l e! e l l e l «l e l e ( e l e! e l l! e l l! e e l l l e e l e e! e l l e! e l! e l l! e e l l l e 0 ( l e l Let us write x l and consider the function f(x e x e x ( e x We have to prove that the unique positive solution to the equation f(x e x e x ( e x 5

16 is x log 2 Note that f(0 0 and if x > 0 then f(x > 0 On the other hand If x > 0 the derivative of f(x is Let us consider the function We have Its derivative is f (x e x e x lim f(x x x + 2 2e x e x g(x x + 2 2e x g(0 0, g(log 2 > 0, lim g(x x g (x 2e x On the interval [0, log 2 we have g (x > 0 In x log 2 we have g (x 0 On the interval (log 2, ] we have g (x < 0 Therefore there exists a > log 2 such that g(a 0 On the interval (0,a is g(x > 0 and on the interval (a, is g(x < 0 Hence f(x is strictly increasing in the interval [0,a and f(x is strictly decreasing and greater than in the interval [a, Consequently the unique positive solution to (50 is x log 2 Theorem 8 The following asymptotic formulae hold n log 2, (5 lim n + lim, (52 n (+ +, (53 n A A 2 A A 0 A n e, (54 + ea + n n, (55 log n log n ( + log log 2n + o(n, (56 6

17 log n log n, (57 n n (log 2 n e (+o(n (58 Proof Equation (5 is an immediate consequence of limit (22 Equations (54, (55, (56 and (58 can be proved from limit (22 (see reference [, Theorem 3] Equation (52 is an immediate consequence of equation (5 Equation (53 is an immediate consequence of equation (52 Equation (57 is an immediate consequence of equation (56 5 Acnowledgements The author would lie to than the anonymous referee for his/her valuable comments and suggestions for improving the original version of Theorem 2 and Theorem 3 The proofs of Theorem 2 and Theorem 3 in this article using the Faá di Bruno s formula pertain to the referee, and consequently the equations (5 and (4, are due to the referee The author is also very grateful to Universidad Nacional de Luján References [] R Jaimczu, ote on power series and the e number, Int Math Forum 6 (20, [2] D Vella, Explicit formulas for Bernoulli and Euler numbers, Integers 8 (2008, Paper #A0 [3] H Wilf, Generatingfunctionology, Academic Press, Mathematics Subject Classification: Primary 26A24; Secondary B75 Keywords: Successive derivative, integer sequence Received April 9 20; revised version received July 7 20 Published in Journal of Integer Sequences, September 4 20 Return to Journal of Integer Sequences home page 7

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