Article Explicit Construction of the Inverse of an Analytic Real : Some Applications

Joaquín Moreno 1, Miguel A. López 2,* and Raquel Martínez 2

1 Department of Applied Mathematics, Superior Technical School of Building Engineering, Polytechnic University of Valencia, 46022 Valencia, Spain; jmfl[email protected] 2 SIDIS Research Group, Department of Mathematics and Institute of Applied Mathematics in Science and Engineering (IMACI), Polytechnic School of Cuenca, University of Castilla-La Mancha, 16071 Cuenca, Spain; [email protected] * Correspondence: [email protected]

 Received: 12 October 2020; Accepted: 27 November 2020; Published: 3 December 2020 

Abstract: In this paper, we introduce a general procedure to construct the Taylor series development of the inverse of an analytical function; in other words, given y = f (x), we provide the that defines its inverse x = h f (y). We apply the obtained results to solve nonlinear in an analytic way, and generalize Catalan and Fuss–Catalan numbers.

Keywords: inverse functions; Taylor series; Taylor Remainder; nonlinear equations; Catalan numbers; Fuss–Catalan numbers

MSC: 40E99; 26A99; 30B10; 05C90

1. Introduction In this paper, we have taken as a basis the previous works [1,2], in which the inverse of a function is constructed, with the aim of generalizing the methods developed there to any . That is to say: given an analytic function around the point x0:

∞ (p) f (x0) p f (x) = f (x0) + ∑ (x − x0) p=1 p! where f (p)(x) is the p-th of f (x), throughout these lines, we construction the function, x = h f (y), with y = f (x) and, therefore, x0 = h f (y0) and y0 = f (x0), such that:

(p) ∞ h (y ) f 0 p h f (y) = h f (y0) + ∑ (y − y0) (1) p=1 p!

To accomplish this task we have organized this article as follows:

• In Section2, background theory is presented.

• In Section3, the successive of h f are computed in an explicit way. (p) • In Section4, a bound for |h f (y0)| (see (1)) is found. • In Section5, we study the convergence and establish the radius of convergence and Taylor Remainder of series (1). • In the next two sections, we introduce some applications for solving nonlinear equations in an analytic way, and to generalize the Catalan and Fuss–Catalan numbers.

Mathematics 2020, 8, 2154; doi:10.3390/math8122154 www.mdpi.com/journal/mathematics Mathematics 2020, 8, 2154 2 of 23

• In the last section, we present our conclusions. We recall that Catalan numbers is defined as:

1 2n (2n)! C = 1; C = = ; n = 1, 2, ··· (2) 0 n n + 1 n n! (n + 1)! and that they satisfy the recursive formula:

C = C C n ≥ n ∑ i1 i2 ; 1 . (3) i1+i2=n−1

Catalan numbers appeared for the first time in the book Quick Methods for Accurate Values of Circle Segments, by Ming Antu (1692–1763), a Chinese mathematician. In this book, he provides some trigonometric equalities and power series, in which Catalan numbers are involved. Nicolas Fuss (1755–1826) introduced, in his paper of 1791 (see [3]), the Fuss–Catalan numbers, as:

1 m n (m n)! Cm = 1; Cm = = (4) 0 n (m − 1)n + 1 n ((m − 1)n + 1)! n!

fixed m ≥ 2; with n = 1, 2, ··· . Notice that, for m = 2, they coincide with Catalan numbers. Furthermore, he provided a generalization of (3):

Cm = Cm ··· Cm ; fixed m ≥ 2; ∀n ≥ 1 . (5) n ∑ i1 im i1+···+im=n−1

From that time forward, throughout mathematical history, Catalan numbers have made important contributions. We list some of them that we have chosen in an arbitrary manner, by way of illustration: √ • Development in power series of the function f (x) = 1 − 4x (Euler (1707–1783)). • The ballot problem (Statistics and Probability), introduced for the first time in 1887 by Joseph Bertrand (1822–1900), see [4]. • In [5], the reader can find more than 200 practical combinatorial interpretations of Catalan numbers. • Binary trees (Graph Theory), see [6]. • path theory (Graph Theory), see [7]. Throughout this paper, all the necessary computational tasks have been performed with the program Wolfram Mathematica 11.2.0.0.

2. Some Recent Results In this section, we review some outcomes, previously published by the authors, (see [1,2]), which will be used in the following sections. Such a summary has been written in detail for the sake of clarity and the self-developed reading of these lines. In fact, for our goal, we will only need formulas (19)–(21). We posed the functional :

p 2 Q(x2 ··· , xp, h(x2, ··· , xp)) = xph (x) + ··· + x2h (x) − h(x) + 1 = 0 (6)

p−1 p−1 where p > 1, x = (x2, x3, ··· , xp) ∈ R and h : R → R is the unknown to solve. We proved that, if h(x) is a solution of (6), then it satisfies the first order partial derivative equation:

2 1 + (2x2 − 1)h(x) + (3x3 + 4x2 − x2)hx2 (x) + (4x4 + 6x2x3 − 2x3)hx3 (x) + ··· (7) + (pxp + 2(p − 1)x2xp−1 − (p − 2)xp−1)hxp−1 (x) + (2px2xp − (p − 1)xp)hxp (x) = 0

∂Q for all x, such that (x, h(x)) 6= 0. ∂h Mathematics 2020, 8, 2154 3 of 23

From (7), we showed that, if h(x) is a solution of Equation (6), then the equality:

(2q2 + 3q3 + ··· + pqp)! h(q2,··· ,qp)(0, ··· , 0) = (8) (q2 + 2q3 + ··· + (p − 1)qp + 1)!

(q ,··· ,qp) holds, with q2 + ··· + qp = n and q2, ..., qp non-negative , where h 2 (0, ··· , 0) is the n-th partial derivative of h, q2 times with respect to x2, ...., qp times with respect to xp. Therefore, we can express the function h(x) as:

∞ ( ) = q2 ··· qp h x ∑ ∑ Cq2···qp x2 xp (9) n=0 q2+···+qp=n

where (2 q2 + 3 q3 + ··· + p qp)! Cq2...qp = (10) (q2 + 2 q3 + ··· + (p − 1)qp + 1)! q2! q3! ··· qp! Consider the polynomial function of degree p, given by:

p y = P(x) = a0 + a1x + ··· + apx (a1, ap 6= 0) (11) with x, ai ∈ R, 0 ≤ i ≤ p, and the functions Xi, 2 ≤ i ≤ p:

(a − y)i−1a ( ) = 0 i Xi y i (12) (−a1) then, the series: a0 − y fP(y) = h(X2(y), ··· , Xp(y)) (13) −a1 is the inverse function of P(x), if it makes sense. Taking into account (9), by substituting functions (12) in (13), we obtain:

qp ∞  q2 p−1 ! a0 − y (a0 − y)a2 (a0 − y) ap f (y) = Cq q ... P −a ∑ ∑ 2... p (−a )2 (−a )p 1 n=0 q2+...+qp=n 1 1 (14) − ∞  q2  a qp y a0 q +...+(p−1) qp a2 p q +...+(p−1) qp = (−1) 2 Cq q ... (y − a ) 2 . a ∑ ∑ 2... p (−a )2 (−a )p 0 1 n=0 q2+...+qp=n 1 1

Next, by making the subscripts and superscripts change: n1 = q2 + 2q3 + ... + (p − 1) qp, the terms of series (14) are rearranged in the form:   ∞  q2  qp y − a a ap ( ) = 0 (− )n1 2 ( − )n1 fP y ∑  ∑ 1 Cq2...qp 2 ... p  y a0 . (15) a1 (−a1) (−a1) n1=0 q2+...+(p−1) qp=n1

Series (14) and (15) are absolutely convergent in a neighborhood of a0, Va0 , given by:

( 2 p p−1 ) p (a0 − y)a2 p (a0 − y) ap Va = y ∈ R; + ··· + < 1 (16) 0 p − 1 2 ( − )p−1 p a1 p 1 a1 that is, obviously, not empty. From now on, we will denote the inverse function of y = P(x) as x = hP(y). Therefore, if the inequality:

2 3 2 p p−1 p a a p a a p a ap 0 2 + 0 3 + ··· + 0 < 1 (17) p − 1 2 (p − 1)2 3 ( − )p−1 p a1 a1 p 1 a1 Mathematics 2020, 8, 2154 4 of 23 holds then, the series:

− qp ∞  q2 p 1 ! a a a a ap r = 0 C 0 2 ... 0 (18) −a ∑ ∑ q2...qp (−a )2 (−a )p 1 n=0 q2+...+qp=n 1 1 is absolutely convergent to the root of P(x), r, closest to the coordinate origin. 0 Finally, we consider the Taylor series of the polynomial, P(x), around the point x0, with P (x0) 6= 0, which is:

P00(x ) P(p)(x ) y = P(x) = P(x ) + P0(x )(x − x ) + 0 (x − x )2 + ... + 0 (x − x )p (19) 0 0 0 2! 0 p! 0 with y0 = P(x0). Then, one gets:  y − y ∞  P00(x ) q2 ( ) = + 0 (− )n1 0 hP y x0 0 ∑  ∑ 1 Cq2...qp 0 2 ... P (x0) 2! (−P (x0)) n1=0 q2+2 q3+...+(p−1) qp=n1 q (p) ! p # P (x0) n1 ··· 0 p y − y0 p! (−P (x0))  (20) 1 ∞  P00(x ) q2 = + (− )n1 0 x0 0 ∑  ∑ 1 Cq2...qp 0 2 ... P (x0) 2! (−P (x0)) n1=0 q2+2 q3+...+(p−1) qp=n1 !qp # P(p)(x ) 0 n1+1 ··· 0 p (y − y0) p! (−P (x0))

hP(y) being the inverse function of (19) in Vy0 :

( 2 00 p p−1 (p) ) p (y0 − y)P (x0) p (y0 − y) P (x0) V = y ∈ + ··· + < y0 R; 0 2 p−1 0 p 1 (21) p − 1 2! P (x0) (p − 1) p! P (x0) according to (16), with y0 = P(x0). Once we have seen the background on which this paper is based, we introduce the original contributions of this article in the coming sections.

3. Definition of the Function h f and Calculation of Its Derivatives

Theorem 1. Let ΩR ⊂ R be an open . Consider the function f : ΩR → R. Assume that f is analytic in the 0 open ball, BR(x0, R) ⊂ ΩR, R being its radius of convergence, and f (x0) 6= 0. Let z = P(x) be the Taylor Polynomial of degree p of the function f , that is to say:

00 (p) f (x ) 2 f (x ) p z = P(x) = f (x ) + f 0(x )x − x  + 0 x − x  + ··· + 0 x − x  . (22) 0 0 0 2! 0 p! 0

If we express the inverse functions of f (x) and P(x), around the point y0 = f (x0) = P(x0) as h f (y) and hP(z), respectively, then the next equality holds ∀ n ≥ 1 (n ≤ p):

(n) (n) h f (y0) = hP (y0) . (23)

Proof. By hypothesis, the series:

00 (p) f (x ) 2 f (x ) p y = f (x) = f (x ) + f 0(x )x − x  + 0 x − x  + ... + 0 x − x  + ... (24) 0 0 0 2! 0 p! 0 Mathematics 2020, 8, 2154 5 of 23 is absolutely convergent in the closed ball:

BR(x0, ρ) = {x ∈ R; |x − x0| ≤ ρ < R}; with ρ > 0

0 0 0 and f (x0) = P (x0), with f (x0) 6= 0 so, from the , there is a closed neighborhood of y0 = P(x0) = f (x0):

BR(y0, ρ1) = {y ∈ R; |y − y0| ≤ ρ1}; with ρ1 > 0 where h f (y) and hP(y) exist and they are . In order to simplify the notation, we suppose, without loss of generality, that ρ and ρ1 satisfy the properties BR(y0, ρ1) ⊂ f (BR(x0, ρ)) and BR(y0, ρ1) ⊂ P (BR(x0, ρ)). By substituting in (24) x by h f (y), where y is the only y ∈ BR(y0, ρ1), such that there is an only x ∈ BR(x0, ρ) with y = f (x), then (24) turns out to be:

00 (p) f (x ) 2 f (x ) p y = f (x ) + f 0(x )h (y) − x  + 0 h (y) − x  + ··· + 0 h (y) − x  + ··· (25) 0 0 f 0 2! f 0 p! f 0

In a similar way, by substituting in (22) x by hP(z), where z is the only z ∈ BR(y0, ρ1), such that there is an only x ∈ BR(x0, ρ) with z = P(x), then (22) becomes:

(p) f (x ) p z = f (x ) + f 0(x )h (z) − x  + ··· + 0 h (z) − x  . (26) 0 0 P 0 p! P 0

To prove (23), we use the induction method. For n = 1, on the one hand, we consider the Taylor Polynomial (26) of degree 1: 0  z = f (x0) + f (x0) hP(z) − x0 . (27)

We compute the first derivative of both sides of (27) with respect to z at the point z = y0, obtaining:

0 0 0 1 1 = f (x0)hP(y0) =⇒ hP(y0) = 0 . f (x0)

On the other hand, we compute the first derivative of both sides of (25) with respect to y at the point y = y0, obtaining:

 00 f (x ) 2 y0 = f (x ) + f 0(x )h (y) − x  + 0 h (y) − x  + ··· 0 0 f 0 2! f 0 0 (p) # (28) f (x ) p + 0 h (y) − x  + ··· . p! f 0 y=y0

Note that: (k) " (p) # f (x0) p h (y) − x = 0 (29) p! f 0

y=y0 if p > k. Indeed:

h pi(k) p! h (0)iq0 h (1)iq1 h f (y) − x0 = ∑ h f (y) − x0 h f (y) − x0 ··· q0! q1! ··· qk! 0q0+1 q1+2q2+···+kqk=k q0+q1+···+qk=p (30) h (k)iqk ··· h f (y) − x0

where: Mathematics 2020, 8, 2154 6 of 23

 • q0 is the number of times that h f (y) − x0 is repeated as a factor in its corresponding summands of (30). (1) • q1 plays the same role with respect to the expression h f (y) − x0 . ··· (k) • The same thing can be said with respect to qk, related to the expression h f (y) − x0 . Each term of (30) satisfies:

q1 + q2 + ··· + qk ≤ q1 + 2q2 + ··· + kqk = k

q0 + q1 + q2 + ··· + qk = p > k

q0 so q0 > 0 in all of them. Therefore, the factor h f (y) − x0 appears in each term of (30). As h f (y) − q0 x0 equals zero at the point y = y0,(29) follows. Hence, (28) comes to be:

0 0 0 1 1 = f (x0)h f (y0) =⇒ h f (y0) = 0 f (x0) and the result is true for n = 1. Suppose now that the result is true for n = r − 1. We consider the polynomial (26) with p = r. On the one side, we compute the r-th derivative of both sides of (26) with respect to z at the point z = y0, obtaining:

(r) " (r) # f (x ) r 0 = f (x ) + f 0(x )h (z) − x  + ··· + 0 h (z) − x  . (31) 0 0 P 0 r! P 0 z=y0

0 By applying to (31), we get a polynomial, P1, in the variables f (x0), ··· , (r) 0 (r) f (x0), hP(y0), ··· , hP (y0), that is to say:

 0 (r) 0 (r)  0 = P1 f (x0), ··· , f (x0), hP(y0), ··· , hP (y0) . (32)

(r) Observe that, due to differentiation rules (30), the exponent of hP (y0) as a variable of the polynomial function, P1, is equal to 1 and its coefficient is different to zero. Indeed, the coefficients and exponents corresponding to r-th derivative of h f are: 0 (r) • From [ f (x0)] , 0.

(r) h i1 (r) • From [ f 0(x )(h (y) − x )] , f 0(x ) (h (y) − x )(r) = f 0(x )h (y ). 0 P 0 0 P 0 0 P 0 y=y0  00 (r) f (x0) h i1 h i1 • From (h (y) − x )2 , f 00(x ) (h (y) − x )(r) (h (y) − x )(0) = 0. 2! P 0 0 P 0 P 0 y=y0  000 (r) 000 f (x0) f (x0) h ( )i1 h ( )i2 • From (h (y) − x )3 , (h (y) − x ) r (h (y) − x ) 0 = 0. 3! P 0 2 P 0 P 0 y=y0 • ··· " #(r) (r)( ) (r)( ) h i1 h ir−1 f x0 r f x0 (r) (0) • From (hP(y) − x0) , (hP(y) − x0) (hP(y) − x0) = 0. r! (r − 1)! y=y0 (r) 0 (r) Therefore, the coefficient of hP (y0) is f (x0) 6= 0 and its exponent, 1. Thus, solving for hP (y0) in (32) makes sense and we arrive at the only solution:

(r)  0 (r) 0 (r−1)  hP (y0) = R1 f (x0), ··· , f (x0), hP(y0), ··· , hP (y0) where R1 is a . Mathematics 2020, 8, 2154 7 of 23

On the other hand, we compute the r-th derivative of both sides of (25) with respect to y at the point y = y0, obtaining:

(r) " (r) # f (x ) r 0 = f (x ) + f 0(x )h (y) − x  + ··· + 0 h (y) − x  0 0 f 0 r! f 0 y=y 0 (33) (r) " (r+1) # f (x ) r+1 + 0 h (y) − x  + ··· . (r + 1)! f 0 y=y0

From the (r + 1)-th term of the second side of (33) onward, the r-th derivative at the point y = y0 equals zero, due to (29). By applying again differentiation rules to the first r terms of the second side of (33), we obtain the same polynomial function as in (32), in other words:

 0 (r) 0 (r)  0 = P1 f (x0), ··· , f (x0), h f (y0), ··· , h f (y0) by induction hypothesis, we conclude that:

 0 (r) 0 (r−1) (r)  0 = P1 f (x0), ··· , f (x0), hP(y0), ··· , hP (y0), h f (y0) .

Therefore: (r)  0 (r) 0 (r−1)  h f (y0) = R1 f (x0), ··· , f (x0), hP(y0), ··· , hP (y0) and the result follows.

Remark 1. Theorem1 may have been approached by applying Bell and the Faà di Bruno formula (see [8,9]), in the following way:

Bell polynomials are introduced as:

n Yn(x1, ..., xn−k+1) = ∑ Bn,k(x1, ..., xn−k+1); ∀n ≥ 1 k=1 where Bn,k(x1, ..., xn−k+1) are the partial Bell polynomials, given by:

q q n!  x1  1  x2  2 Bn,k(x1, ..., xn−k+1) = ∑ ··· q1! ... qn! 1! 2! q1+2q2+3q3+...=n q1+q2+q3+...=k with q1, q2,... non-negative numbers. As q1 + 2q2 + 3q3 + ... = n ⇒ qi = 0; ∀ i > n, and   q1 + 2q2 + 3q3 + ... + nqn = n q2 + 2q3 + ... + (n − 1)qn = n − k ⇒ ⇒ qn−k+1 = ... = qn = 0 q1 + q2 + q3 + ... + qn = k q1 + q2 + q3 + ... + qn = k then, we can rewrite:

q q q n!  x1  1  x2  2  xn  n Bn,k(x1, ..., xn) = ∑ ··· (34) q1! ... qn! 1! 2! n! q1+2q2+3q3+...+qnn=n q1+q2+q3+...+qn=k taking 00 = 1, if necessary. ( ) For n = 1, h 1 = 1 , where h is the inverse function of f . f f (1) f Mathematics 2020, 8, 2154 8 of 23

For n > 1, we use Faà di Bruno formula, obtaining:

n = (k) ( (1) (2) (n)) 0 ∑ h f Bn,k f , f , ..., f k=1

(1) (2) (n) where Bn,k( f , f , ..., f ) are the partial Bell polynomials, with k = 1, 2, ..., n. Substituting Bn,k, according to (34), we arrive at:

!q1 !q2 !qn n n! f (1) f (2) f (n) = (k) ··· 0 ∑ h f ∑ . (35) q1! ... qn! 1! 2! n! k=1 q1+2q2+...+nqn=n q1+q2+...+qn=k

We remove the last term, Bn,n, of the sum (35). As:

{(q1, q2, ..., qn); q1 + 2q2 + ... + nqn = n and q1 + q2 + ... + qn = n} = {(n, 0, ..., 0)}

we obtain:

!q1 !q2 !qn n−1 n! f (1) f (2) f (n) = (k) ··· + (n)( (1))n 0 ∑ h f ∑ h f f . q1! ... qn! 1! 2! n! k=1 q1+2q2+...+nqn=n q1+q2+...+qn=k

(n) In , solving for h f , we get:

q q n−1 (2) ! 2 (n) ! n (n) (k) n! f f h = − h ··· . (36) f ∑ f ∑ (1) (1) n q1! ... qn! 2!( f )2 n!( f ) k=1 q1+2q2+...+nqn=n q1+q2+...+qn=k

(n) (k) The calculation of h f requires the reiterative substitution of h f ; k = 1, ··· , n − 1 in formula (36). (k) Moreover, the general expression of each h f has the same structure and complexity as (36) itself. For this reason, we were facing great difficulties and we opted for the process described in Section2, which we considered more affordable, providing, furthermore, a new alternative.

Corollary 1. Let ΩR ⊂ R be an open set. Consider the function f : ΩR → R. Assume that f is analytic in 0 a neighborhood of each point x ∈ ΩR, with f (x) 6= 0. If h f (y) is the inverse function of f around the point y = f (x), then the (p + 1)-th derivative of h f at the point, y, divided by (p + 1)! is given by:

 00 q2  000 q3 1 p f (x) f (x) Ap(x) = (−1) Cq q f 0(x) ∑ 2... p+1 2! (− f 0(x))2 3! (− f 0(x))3 q2+2 q3+···+p qp+1=p (37) !qp+1 f (p+1)(x) ··· . (p + 1)! (− f 0(x))p+1

Proof. The result follows from Theorem1 and formulas (19)–(21). Indeed:

(p+1)  00 q2  000 q3 f (x) 1 p f (x) f (x) = (−1) Cq q ··· (p + 1)! f 0(x) ∑ 2... p+1 2! (− f 0(x))2 3! (− f 0(x))3 q2+2 q3+···+p qp+1=p !qp+1 f (p+1)(x) ··· = Ap(x) . (p + 1)! (− f 0(x))p+1 Mathematics 2020, 8, 2154 9 of 23

In other words: (p+1) f (x) = (p + 1)! Ap(x) .

Remark 2. Consider the set, Sp, of subscripts and superscripts of the definition of Ap, see (37), given by:

Sp = {{q2, q3, ··· , qp+1}; q2, q3, ..., qp+1; non-negative integers; q2 + 2q3 + ··· + pqp+1 = p} .

As the number p increases, so too does the run time of the computer code for calculating Ap, disproportionately. As an illustration, we show them until p = 10:

S1 = {1}

S2 = {{2, 0}, {0, 1}}

S3 = {{0, 0, 1}, {1, 1, 0}, {3, 0, 0}}

S4 = {{0, 0, 0, 1}, {0, 2, 0, 0}, {1, 0, 1, 0}, {2, 1, 0, 0}, {4, 0, 0, 0}} ....

S10 = {{0, 0, 0, 0, 0, 0, 0, 0, 0, 1}, {0, 0, 0, 0, 2, 0, 0, 0, 0, 0}, {0, 0, 0, 1, 0, 1, 0, 0, 0, 0}, {0, 0, 1, 0, 0, 0, 1, 0, 0, 0}, {0, 0, 2, 1, 0, 0, 0, 0, 0, 0}, {0, 1, 0, 0, 0, 0, 0, 1, 0, 0}, {0, 1, 0, 2, 0, 0, 0, 0, 0, 0}, {0, 1, 1, 0, 1, 0, 0, 0, 0, 0}, {0, 2, 0, 0, 0, 1, 0, 0, 0, 0}, {0, 2, 2, 0, 0, 0, 0, 0, 0, 0}, {0, 3, 0, 1, 0, 0, 0, 0, 0, 0}, {0, 5, 0, 0, 0, 0, 0, 0, 0, 0}, {1, 0, 0, 0, 0, 0, 0, 0, 1, 0}, {1, 0, 0, 1, 1, 0, 0, 0, 0, 0}, {1, 0, 1, 0, 0, 1, 0, 0, 0, 0}, {1, 0, 3, 0, 0, 0, 0, 0, 0, 0}, {1, 1, 0, 0, 0, 0, 1, 0, 0, 0}, {1, 1, 1, 1, 0, 0, 0, 0, 0, 0}, {1, 2, 0, 0, 1, 0, 0, 0, 0, 0}, {1, 3, 1, 0, 0, 0, 0, 0, 0, 0}, {2, 0, 0, 0, 0, 0, 0, 1, 0, 0}, {2, 0, 0, 2, 0, 0, 0, 0, 0, 0}, {2, 0, 1, 0, 1, 0, 0, 0, 0, 0}, {2, 1, 0, 0, 0, 1, 0, 0, 0, 0}, {2, 1, 2, 0, 0, 0, 0, 0, 0, 0}, {2, 2, 0, 1, 0, 0, 0, 0, 0, 0}, {2, 4, 0, 0, 0, 0, 0, 0, 0, 0}, {3, 0, 0, 0, 0, 0, 1, 0, 0, 0}, {3, 0, 1, 1, 0, 0, 0, 0, 0, 0}, {3, 1, 0, 0, 1, 0, 0, 0, 0, 0}, {3, 2, 1, 0, 0, 0, 0, 0, 0, 0}, {4, 0, 0, 0, 0, 1, 0, 0, 0, 0}, {4, 0, 2, 0, 0, 0, 0, 0, 0, 0}, {4, 1, 0, 1, 0, 0, 0, 0, 0, 0}, {4, 3, 0, 0, 0, 0, 0, 0, 0, 0}, {5, 0, 0, 0, 1, 0, 0, 0, 0, 0}, {5, 1, 1, 0, 0, 0, 0, 0, 0, 0}, {6, 0, 0, 1, 0, 0, 0, 0, 0, 0}, {6, 2, 0, 0, 0, 0, 0, 0, 0, 0}, {7, 0, 1, 0, 0, 0, 0, 0, 0, 0}, {8, 1, 0, 0, 0, 0, 0, 0, 0, 0}, {10, 0, 0, 0, 0, 0, 0, 0, 0, 0}} .

For this reason, we provide the next Corollary as an alternative.

Corollary 2. As in Corollary1, the successive A p(x), p ≥ 0, with x = h f (y), are given by:

0 1 Ap−1(x) 1 A (x) = ; A (x) = ; p ≥ 1 . 0 f 0(x) p p + 1 f 0(x)

Proof. From Corollary1, we have that:

(p) h f (y) A − (x) = ; y = f (x); p ≥ 1 . p 1 p!

Thus:

(p+1) (p+1) h f (y) (p + 1) h f (y) A0 (x) = f 0(x) = f 0(x) =⇒ A0 (x) = (p + 1)A (x) f 0(x) p−1 p! (p + 1)! p−1 p and the result follows. Mathematics 2020, 8, 2154 10 of 23

4. Calculation of an Upper Bound for the Derivatives of h f Jacques Hadamard said: “The shortest path between two truths in the real domain passes through the complex domain”. We find this quote very valid for the ensuing paragraphs, in which we accomplish the task of bounding formula (37).

Remark 3. Mean Value Theorem is not true in general in the complex field. Nevertheless, it can be applied under some conditions that we next establish, in agreement with [10]. Let ΩC ⊂ C be and open and convex set and G : ΩC → C, an holomorphic function. Consider w1, w2 ∈ ΩC. Then, there exist a, b ∈ L, where L is the segment with endpoints w1 and w2, such that:

 G(w ) − G(w )   G(w ) − G(w )  Re(G0(a)) = Re 2 1 ; Im(G0(b)) = Im 2 1 . w2 − w1 w2 − w1

Hence:  0 0  G(w2) − G(w1) = Re(G (a)) + i Im(G (b)) (w2 − w1) . (38)

Theorem 2. Under the same hypothesis as Theorem1, let BC(x0, R) be an open ball in the complex field, whose restriction to R is just BR(x0, R) (introduced in Theorem1) and let F : BC(x0, R) → C be the complex function, given by:

00 (p) f (x ) 2 f (x ) p s = F(w) = f (x ) + f 0(x )w − x  + 0 w − x  + ... + 0 w − x  + ... 0 0 0 2! 0 p! 0

Then, there exists a M > 0 (defined below) such that:

( ) ( ) p! (|x | + ρ ) p ( ) = p ( ) ≤ 0 2 ∀ ≥ < < h f y0 hF y0 p , p 1; with 0 ρ2 R. (ρ2 M)

Remark 4. Observe that the restriction of F to R is equal to the function f , in agreement with (24).

Proof. Obviously, F is also an analytic function in the complex field, with the same radius of convergence as f , R. Consequently, F is also an holomorphic function. 0 0 F (x0) = f (x0) 6= 0; therefore, from the Inverse Function Theorem for holomorphic functions, we can find two closed balls, BC(x0, ρ2) ⊂ BC(x0, R) and BC(y0, ρ3) and a closed neighborhood, Vx0 , such that there exists the inverse function of F, hF and both functions:

F : Vx0 ⊂ BC(x0, ρ2) → BC(y0, ρ3); hF : BC(y0, ρ3) → Vx0 (39) are bijections.

As hF is an holomorphic function in the closed ball BC(y0, ρ3), we can apply the Cauchy Integral Formula, in the following form:

Z 1 hF(s) it hF(y0) = ds; γ = {s ∈ C; s = y0 + ρ3e ; t ∈ [0, 2π]} . (40) 2πi γ s − y0

By differentiating (40) with respect to y0, we obtain the Cauchy Differentiation Formula for the function hF, in the following manner:

( ) p! Z h (s) p ( ) = F hF y0 p+1 ds . (41) 2πi γ (s − y0) Mathematics 2020, 8, 2154 11 of 23

If s ∈ BC(y0, ρ3), then the next inequalities hold:

|hF(s)| − |hF(y0)| ≤ |hF(s) − hF(y0)| = |w − x0| ≤ ρ2 =⇒ |hF(s)| ≤ |x0| + ρ2 (42)

since w ∈ Vx0 and, in correspondence with (38), if s ∈ γ:

0 0 ρ3 = |s − y0| = |F(w) − F(x0)| = Re(F (a)) + i Im(F (b)) |w − x0| (43) for some a, b ∈ L, where L ⊂ BC(x0, ρ2) is the segment with endpoints w and x0. If we indicate w, a, b and F as w = c + id, a = a1 + i a2, b = b1 + i b2 and F(c + id) = u(c, d) + i v(c, d); u(c, d), v(c, d) being functions from R2 into R that represent the real and imaginary parts of F, respectively, then (43) turns out to be: r  2  2 = | − | = ∂ u ( ) + ∂ v ( )) | − | = ∂ u ( ) + ∂ v ( ) | − | ρ3 s y0 ∂ c a1, a2 i ∂ c b1, b2 w x0 ∂ c a1, a2 ∂ c b1, b2 w x0 . (44)

If: ( )  ∂ u 2 M ≤ min (c, d) ; c + id ∈ B (x , ρ ) 1 ∂ c C 0 2 ( ) (45)  ∂ v 2 M ≤ min (c, d) ; c + id ∈ B (x , ρ ) 2 ∂ c C 0 2 √ and if we take M such that 0 < M ≤ M1 + M2 then, in consonance with the hypothesis of the theorem, we can choose ρ3 between the values given by the inequalities:

ρ2 M ≤ ρ3 = |s − y0| ≤ ρ2 N (46) s    ∂ u 2  ∂ v 2  where, in agreement with (44), N = max (c,d) + (c, d) ; c + id ∈ BC(x0, ρ2) .  ∂ c ∂ c  As a consequence of all of this, using (41), (42) and (46), we obtain Cauchy’s Estimate for the successive derivatives of hF as follows:

( ) p! Z |h (s)| p! |x | + ρ h p (y ) ≤ F ds ≤ 0 2 F 0 p+1 p+1 2πρ3 2π γ |s − y0| 2π ρ 3 (47) p! (|x | + ρ ) p! (|x | + ρ ) = 0 2 ≤ 0 2 ; ∀ p ≥ 1 . p (ρ M)p ρ3 2

In addition, the result follows.

5. Taylor Series and Taylor Remainder of the Function h f

Theorem 3. If the same hypothesis as Theorems1 and2 hold, then h f (y) is an analytic function, around the point y = y0, with a radius of convergence of, at least, ρ2 M.

Proof. Once we have calculated and bounded the successive derivatives of h f in Theorems1 and2, respectively, we are now in position for introducing its corresponding Taylor series:

1 ∞ h (y) = x + (y − y ) + A (x )(y − y )p+1 . (48) f 0 0( ) 0 ∑ p 0 0 f x0 p=1 Mathematics 2020, 8, 2154 12 of 23

Hence: ∞ 1 p+1 h (y) ≤ |x | + |y − y | + A (x ) |y − y | . (49) f 0 0( ) 0 ∑ p 0 0 f x0 p=1 In concordance with (47), we have that:

( + ) h p 1 (y ) f 0 |x0| + ρ2 Ap(x0) = ≤ ; ∀ p ≥ 1 . (50) (p + 1)! (ρ M)p+1 2

By substituting (50) in (49), we arrive at:

∞ 1 |x0| + ρ2 p+1 h (y) ≤ |x | + |y − y | + |y − y | . f 0 0( ) 0 ∑ p+1 0 f x0 p=1 (ρ2 M)

Therefore:

∞ p+1 1 y − y0 h (y) ≤ |x | + |y − y | + (|x | + ρ ) . f 0 0( ) 0 0 2 ∑ f x0 p=1 ρ2 M

In addition, the result follows.

Theorem 4. Under the same hypothesis as Theorems1 and2, then the (p + 1)-th Taylor Remainder of the series h f (y) is given by: p+1 y − y0 1 R ≤ (|x | + ρ ) p+1 0 2 − . (51) ρ2 M 1 − y y0 ρ2 M

Proof. The error in evaluating h f (y) by its Taylor Polynomial of degree p is: ! p−1 ∞ n+1 n+1 Rp+1 = h f (y) − x0 + ∑ An(x0)(y − y0) = ∑ An(x0)(y − y0) . (52) n=0 n=p

Taking into account (50), then (52) becomes:

∞ |x | + ρ ∞ y − y n+1 ≤ 0 2 | − |n+1 = (| | + ) 0 Rp+1 ∑ n+1 y y0 x0 ρ2 ∑ . n=p (ρ2 M) n=p ρ2 M

Therefore: p+1 y − y0 1 R ≤ (|x | + ρ ) p+1 0 2 − . ρ2 M 1 − y y0 ρ2 M The result follows.

In the following paragraphs, we introduce an example, in detail, of the construction of the inverse function, h f . In oder to ensure the veracity of the inequalities, we have rounded up the numerical values when estimating an upper bound, and we have rounded down for a lower bound.

x Example 1. Find the Taylor series of the inverse function of y = f (x) = e + x + 1, h f (y), around the point y0 = f (0) = 2, compute the power series that defines h f (1.9), in an analytic way and, finally, approximate its numerical value with an error lower or equal to 0.001, using the Taylor Remainder. Mathematics 2020, 8, 2154 13 of 23

Step 1: Definition of F(w) and F0(w), according to Theorem2:

f (x) = ex + x + 1; F(w) = ew + w + 1; with w = c + id F(w) = u(c, d) + i v(c, d) = ec cos(d) + c + 1 + i (ec sin(d) + d) ∂ u  ∂ v  F0(w) = (c, d) + i (c, d) = 1 + ec cos(d) + i ec sin(d). ∂ c ∂ c

Step 2: Construction of the closed ball BC(x0, ρ2). We look for a square Sq = [x0 − ρ2, x0 + ρ2] × [−ρ2, ρ2] to contain BC(x0, ρ2) and in which either c c 1 + e cos(d) or e sin(d) does not change their signs. If we take x0 = 0 and ρ2 = 3, due to the fact that in the interval [−3, 3], ec is increasing, cos(d) reaches its minimum and maximum values at the points d = 3 and d = 0, respectively, and sin(d) at the points d = −π/2 and d = π/2, respectively, we can say that: 0.950711 ≤ 1 + e−3 cos(3) ≤ 1 + ec cos(d) ≤ 1 + e3 cos(0) ≤ 21.085537 (53) 0 and, therefore, F (w) 6= 0 for all w ∈ BC(x0, ρ2) ⊂ Sq.

Step 3: Calculation of M1, M2, and M. For the same reasons as in Step 2, we arrive at:

 π   π  − 20.085537 ≤ e3 sin − ≤ ec sin(d) ≤ e3 sin ≤ 20.085537 . (54) 2 2

According to (45) and in agreement with (53) and (54), M1, M2, and M are given by:

2 p M1 = 0.950711 ; M2 = 0; M1 + M2 = 0.950711 ≥ M = 0.95.

Step 4: Calculation of the radius of convergence and the Taylor series of h f around the point y0 = f (x0) = 2. As a consequence of all of this, we conclude that the inverse function of f according to (48) is:

∞ p+1 h f (y) = 0 + ∑ Ap(0) (y − 2) p=0 and has a radius of convergence of, at least, ρ2 M = 2.85. The exact value of h f (1.9) is given by the power series:

∞ p+1 h f (1.9) = 0 + ∑ Ap(0) (1.9 − 2) p=0 that is well defined, since: 2 − 2.85 ≤ 1.9 ≤ 2 + 2.85

Step 5: Numerical approximation of h f (1.9). For approximating the numerical value of the irrational number h f (1.9), with an error lower or equal to 0.001, we need to give numerical values to the number p in (51). In Table1, we resume the performed calculations with such a purpose. Its first row indicates the values of p, the second one shows the corresponding values of Rp+1.

Table 1. Values of p and Rp+1.

p 1 2 3 4

Rp+1 0.003827 0.000135 0.000004 0.00000016 Mathematics 2020, 8, 2154 14 of 23

For 0 ≤ p ≤ 1, Rp+1 does not attain the required value. For p = 2, we obtain R3 ≤ 0.000135 < 0.001. For p ≥ 3, the error decreases more and more. Therefore: 1 p+1 h f (1.9) ≈ 0 + ∑ Ap(0) (1.9 − 2) ≈ −0.050625 p=0 and: −0.05076 ≤ −0.050625 − 0.000135 ≤ h f (1.9) ≤ −0.050625 + 0.000135 ≤ −0.05049 .

6. Applications I: Resolution of Nonlinear Equations Corollary 3. Under the same hypothesis as Theorem3, consider the closed interval: h i  h f (y0 − ρ2 M), h f (y0 + ρ2 M) ; i f h f (y0 − ρ2 M) < h f (y0 + ρ2 M) [a, b] = h i  h f (y0 + ρ2 M), h f (y0 − ρ2 M) ; i f h f (y0 − ρ2 M) > h f (y0 + ρ2 M)

Then, there is one root and only one root, r ∈ [a, b] of the function y = f(x) if, and only if, 0 ∈ [y0 − ρ2 M, y0 + ρ2 M].

Proof. Throughout the proof of Theorem2, formula (39) brings to light that the function F : Vx0 ⊂ BC(x0, ρ2) → BC(y0, ρ3) is a : As f is the restriction of F to the real numbers then, in agreement with (46), ρ3 > ρ2 M, so f : [a, b] → [y0 − ρ2 M, y0 + ρ2 M] is a bijection too. Therefore, if r ∈ [a, b], then f (r) = 0 ∈ [y0 − ρ2 M, y0 + ρ2 M]. Conversely, if 0 ∈ [y0 − ρ2 M, y0 + ρ2 M], then h f (0) = r ∈ [a, b], and the result follows.

x Example 2. Find a root of the function y = f (x) = e + x + 1, in an analytic way, through the series h f (0), and approximate its numerical value with an error lower or equal to 0.001.

With the same values as in the previous example for the parameters x0, y0, M, and ρ2, the series h f (0) makes sense, since 2 − 2.85 ≤ 0 ≤ 2 + 2.85, so the exact value of the required root is given by:

∞ p+1 h f (0) = 0 + ∑ Ap(0)(−2) . p=0

With these values of the parameters, we need to take p = 26, at least, to approximate h f (0) with the required precision, in consonance with Table2. For p ≥ 27, the error decreases more and more.

Table 2. Values of p and Rp+1.

p 25 26 27 28

Rp+1 0.001008 0.000708 0.000496 0.000348

Then, the radius of convergence of the series h f around the point y0 is, at least, ρ2 M = 2.85, so h f (0) is well defined and R27 ≤ 0.000708 < 0.001. Hence:

25 p+1 h f (0) ≈ 0 + ∑ Ap(0)(−2) ≈ −1.278464 p=0 with:

−1.279172 ≤ −1.278464 − 0.000708 ≤ h f (0) ≤ −1.278464 + 0.000708 ≤ −1.277756 . Mathematics 2020, 8, 2154 15 of 23

Remark 5. In the previous example, the parameters x0, M, and ρ2 needed to find the required root are given in the solution of Example1, but it may be questioned, in general, how to get such values. In the following example, we attempt to find a possible solution to this problem.

1 Example 3. Calculate the parameters x , M, and ρ in order to find a root of y = f (x) = 2 cos(x) − x in 0 2 2 the interval [0, π], if it exists.

We solve the problem in the following steps: Step 1: Definition of F(w) and F0(w).

1 1 f (x) = 2 cos(x) − x; F(w) = 2 cos(w) − w; with w = c + id 2 2 1  1  F(w) = u(c, d) + i v(c, d) = 2 cos(c) cosh(d) − c + i − 2 sin(c) sinh(d) − d 2 2 ∂ u  ∂ v  1   F0(w) = (c, d) + i (c, d) = − − 2 cosh(d) sin(c) + i − 2 cos(c) sinh(d) . ∂ c ∂ c 2

Step 2: Setting of the initial parameters ρ2, x0, and M. As said in Example1, the main idea is to look for rectangles, Sq = [x0 − ρ2, x0 + ρ2] × [−ρ2, + ρ2], where the closed ball, BC(x0, ρ2), is contained, and, in addition, either −1/2 − 2 cosh(d) sin(c) or −2 cos(c) sinh(d) do not change their signs. With such a purpose, we fix ρ2 = π/2 as the half of the length of the interval and x0 as its middle point. Therefore, Sq is left as [0, π] × [−π/2, π/2] and x0 = π/2. We bound both partial derivatives in Sq, obtaining:

1 1 1 − − 2 cosh(d) sin(c) ≤ − − 2 cosh(0) sin(0) ≤ − . 2 2 2

It is easy to see that −2 cos(c) sinh(d) changes its sign in Sq, since −2 cos(c) sinh(−ρ2) and −2 cos(c) sinh(ρ2) take values with contrary signs. Thus, from now on, we are only going to pay 1 attention to the expression − − 2 cosh(d) sin(c). 2 Thus, the real part of the derivative of F(w) is not positive and, furthermore,

f (x0 − ρ2) · f (x0 + ρ2) < 0.

Then, there is only one root in (x0 − ρ2, x0 + ρ2). Under these conditions, we can compute M = 0.5, in agreement with (45), the radius of convergence of h f , ρ2 M = π/4 in consonance with Theorem3 and the interval [y0 − ρ2 M, y0 + ρ2 M]=[−1.570796, 0]. As 0 ∈/ (−1.570796, 0), we look for a better ρ2, M, and x0 (for practical purposes, we consider 0 ∈/ (−1.570796, 0) instead of 0 ∈ [−1.570796, 0], in order to improve the convergence of the series h f (0)).

Step 3: Improvement of ρ2, M, and x0. We fix ρ2 = π/3. In other words, the third of the interval [0, π], and we define the rectangles: h π π i  2π  h π π i π • S = [0, 0 + 2ρ ] × − , = 0, × − , , with x = . q1 2 3 3 3 3 3 0 3 h π π i h π i h π π i 2π • S = [0 + ρ , 0 + 3ρ ] × − , = , π × − , , with x = . q2 2 2 3 3 3 3 3 0 3 h π π i In such a way that S ∪ S = [0, π]× − , . q1 q2 3 3 We bound −1/2 − 2 cosh(d) sin(c) in Sq1 :

1 1 1 − − 2 cosh(d) sin(c) ≤ − − 2 cosh(0) sin(0) ≤ − . 2 2 2 Mathematics 2020, 8, 2154 16 of 23

Again, the real part of the derivative of F(w) is not positive and, furthermore,

f (x0 − ρ2) · f (x0 + ρ2) < 0.

Then, there is only one root in (x0 − ρ2, x0 + ρ2). Under these conditions, we can compute M = 0.5, the radius of convergence of h f , ρ2 M = π/6 and the interval [y0 − ρ2 M, y0 + ρ2 M]=[−0.047198, 1]. As 0 ∈ (−0.047198, 1), we take ρ2 = π/3, M = 0.5, and x0 = π/3. See a resume of this process in Table3.

Table 3. Search for ρ2, x0, and M in the interval [0, π], with E1 = x0 − ρ2 and E2 = x0 + ρ2.

∂u Sign of n Interval ρ2 Sign of M x0 y0 − ρ2 M y0 + ρ2 M ∂c f (E1) · f (E2)

2 [0, π] π/2 − − 0.5 π/2 −1.570796 0

 2π  3 0, π/3 − − 0.5 π/3 −0.047198 1 3

1 Example 4. Find a root in the interval [0, π] of the function y = f (x) = 2 cos(x) − x, in an analytic way, 2 through the series h f (0), and approximate its numerical value with an error lower or equal to 0.001.

With the same values as in the previous example for the parameters x0 = π/3, ρ2 = π/3 and M = 0.5, and y0 = f (x0) = 0.476401 the series h f (0) makes sense, since −0.047198 ≤ 0 ≤ 1, so the exact value of the required root is given by:

∞ π  π  p+1 h f (0) = + ∑ Ap (−0.476401) . 3 p=0 3

With these values of the parameters, we need to take p = 106, at least, to approximate h f (0) with the required precision. Thus, in order to get a better result, we repeat Step 3 of the Example3 for ρ2 = π/3 and ρ2 = π/4 (see the outcomes in Table4).

Table 4. Search for ρ2, x0, and M in the interval [0, π], with E1 = x0 − ρ2 and E2 = x0 + ρ2.

∂u Sign of n Interval ρ2 Sign of M x0 y0 − ρ2 M y0 + ρ2 M ∂c f (E1) · f (E2)

2 [0, π] π/2 − − 0.5 π/2 −1.570796 0

 2π  3 0, π/3 − − 0.5 π/3 −0.047198 1 3 h π i 3 , π π/3 − − 0.5 2π/3 −2.570796 −1.523598 3 h π i 4 0, π/4 − − 0.5 π/4 0.628815 1.414214 2  π 3π  4 , π/4 − − 1.91 π/2 −2.285509 0.714712 4 4 h π i 4 , π π/4 − + 0.5 3π/4 −2.98501 −2.199612 2

In consonance with Table4, we make a new choice: x0 = π/2, y0 = −0.785398, ρ2 = π/4, M = 1.91, and p = 13 (see Table5). For p ≥ 14, the error decreases more and more. Mathematics 2020, 8, 2154 17 of 23

Table 5. Values of p and Rp+1.

p 12 13 14 15

Rp+1 0.001098 0.000576 0.000301 0.000157

Then, the radius of convergence of the series h f around the point y0 is, at least, ρ2 M = 1.5, so h f (0) is well defined and R14 ≤ 0.000576 < 0.001. Hence:

12 π  π  p+1 h f (0) ≈ + ∑ Ap (−0.785398) ≈ 1.252353 2 p=0 2 with: 1.251777 ≤ 1.252353 − 0.000576 ≤ h f (0) ≤ 1.252353 + 0.000576 ≤ 1.252929 .

Example 5. Find a root in the interval [0.5, 1.3] of the function y = f (x) = cos(x) − x3, in an analytic way, through the series h f (0), and approximate its numerical value with an error lower or equal to 0.001.

Step 1: Definition of F(w) and F0(w), according to Theorem2.

F(w) = cos(w) − w3; w = c + id F(w) = cos(c) cosh(d) − c3 + 3cd2 + i (− sin(c) sinh(d) − 3c2d + d3) ∂ u  ∂ v  F0(w) = (c, d) + i (c, d) = − sin(c) cosh(d) − 3c2 + 3d2 + i (− cos(c) sinh(d) − 6cd). ∂ c ∂c

Step 2: Construction of the closed ball BC(x0, ρ2). Proceeding as in Example3, we take x0 = 0.9 and ρ2 = 0.4, due to the fact that in the interval [0.5, 1.3]: • The function cos(c) reaches its minimum and maximum values at the points c = 1.3 and c = 0.5, respectively. • The function sin(c) reaches its minimum and maximum values at the points c = 0.5 and c = 1.3, respectively. • The function sinh(d), at the points d = −0.4 and d = 0.4, respectively. • The function cosh(d), at the points d = 0 and d = 0.4, respectively. Then, we can say that:

− sin(c) cosh(d) − 3c2 + 3d2 ≤ − sin(0.5) cosh(0) − 3 · 0.52 + 3 · 0.42 ≤ −0.749426 (55)

0 for all w ∈ BC(0.9, 0.4) and, therefore, F (w) 6= 0 in BC(0.9, 0.4) as required.

Step 3: Calculation of M1, M2, and M. For the same reasons as in Step 2, we arrive at:

−3.480469 ≤ − cos(0.5) sinh(0.4) − 6 · 0.4 · 1.3 ≤ − cos(c) sinh(d) − 6dc (56) ≤ − cos(0.5) sinh(−0.4) − 6 (−0.4) 1.3 ≤ 3.480469 .

According to (45) and in agreement with (56) and (55), M1, M2, and M are given by:

2 p M1 = (−0.749426) ; M2 = 0; M1 + M2 = 0.749426 ≥ M = 0.74.

Step 4: Calculation of the radius of convergence, and the Taylor series of h f around the point y0 = f (x0) = −0.10739. Mathematics 2020, 8, 2154 18 of 23

As a consequence of all of this, we conclude that the inverse function of f is:

∞ p+1 h f (y) = 0.9 + ∑ Ap(0.9) (y + 0.10739) p=0 and has a radius of convergence of, at least, ρ2 M = 0.3. The exact value of h f (0) is given by the power series:

∞ p+1 h f (0) = 0.9 + ∑ Ap(0.9) (0.10739) p=0 that is well defined, since: −0.10739 − 0.3 ≤ 0 ≤ −0.10739 + 0.3

Step 5: Numerical approximation of h f (0). We choose p = 7, according to the Table6. For p ≥ 8, the error decreases more and more.

Table 6. Values of p and Rp+1.

p 6 7 8 9

Rp+1 0.001688 0.000613 0.000222 0.000081

Then, R8 ≤ 0.000613 < 0.001. Hence: 6 p+1 h f (0) ≈ 0.9 + ∑ Ap(0.9) (0.10739) ≈ 0.865474 p=0 with: 0.864861 ≤ 0.865474 − 0.000613 ≤ h f (0) ≤ 0.865474 + 0.000613 = 0.866087 .

7. Applications II: A Generalization of Catalan and Fuss–Catalan Numbers In this section, we are going to analyze the relations between Catalan and Fuss–Catalan numbers and the inverse functions of polynomials. 2 In agreement with (14), the inverse function of y = f (x) = a0 + a1x + a2x , with a1, a2 6= 0, is defined as: !q2 ∞ a 1 q2 2 q2+1 x = h (y) = (−1) Cq (y − a0) (57) f a ∑ 2 a2 1 q2=0 1 being: (2q2)! Cq2 = ; q2 = 0, 1, 2, ... (q2 + 1)! q2! the sequence of Catalan numbers, in concordance with (2).

Taking into account (16), series (57) is absolutely convergent if y ∈ Vao that in this case is defined as: ( ) (y − a )a = ∈ 0 2 < Va0 y R; 4 2 1 . a1

Choosing a0, a1, and a2 in such a way that 0 ∈ Va0 , from (57), we get:

2 −a0 a2 2 0 = a0 + a1h f (0) + a2h f (0) =⇒ h f (0) = − h f (0) . (58) a1 a1 Mathematics 2020, 8, 2154 19 of 23

As:

!q2 a ∞ (2q )! a a ( ) = − 0 2 0 2 h f 0 ∑ 2 a (q2 + 1)! q2! a 1 q2=0 1

!i1+i2 a2 ∞ (2i )! (2i )! a a 2 ( ) = 0 1 2 0 2 h f 0 2 ∑ ∑ 2 a i1! i2! (i1 + 1)! (i2 + 1)! a 1 p=0 i1+i2=p 1

2 i1, i2 being non-negative integer numbers; then, substituting h f (0) and h f (0) in the second formula of (58), we obtain:

!q2 a a ∞ (2q )! a a − 0 − 0 2 0 2 ∑ 2 a1 a1 q2! (q2 + 1)! a q2=1 1

!i1+i2 a a a2 ∞ (2i )! (2i )! a a = − 0 − 2 · 0 1 2 0 2 2 ∑ ∑ 2 . a1 a1 a i1! i2! (i1 + 1)! (i2 + 1)! a 1 p=0 i1+i2=p 1

We deduce from this that:

!q2 !i1+i2+1 ∞ (2q )! a a ∞ (2i )! (2i )! a a 2 0 2 = 1 2 0 2 ∑ 2 ∑ ∑ 2 . (59) q2! (q2 + 1)! a i1! i2! (i1 + 1)! (i2 + 1)! a q2=1 1 p=0 i1+i2=p 1

By equating the terms with the same degree of the series of both sides of (59) and making q2 = i1 + i2 + 1 ≥ 1; i1 + i2 = q2 − 1, we arrive at:

(2q2)! (2i1)! (2i2)! = ∑ ; ∀q2 ≥ 1 . q2! (q2 + 1)! i1! i2! (i1 + 1)! (i2 + 1)! i1+i2=q2−1

In this way, we have provided an alternative proof of the recursive relation, given by (3). m Equivalently, the inverse function of y = f (x) = a0 + a1x + amx , with a1, am 6= 0, m > 2, is:

1 ∞  a qm x = h (y) = (−1)(m−1)qm C m (y − a )(m−1)qm+1 (60) f a ∑ qm (−a )m 0 1 qm=0 1 where, for n = qm, we have that:

m (m qm)! Cn = Cqm = ; qm = 0, 1, 2, ... ((m − 1)qm + 1)! qm! that is, the Fuss–Catalan numbers sequence, in consonance with (4).

In agreement with (16), series (60) is absolutely convergent if y ∈ Va0 that in this case is defined as:

 mm (y − a )m−1a  = ∈ 0 m < Va0 y R; m−1 m 1 . (m − 1) a1

Choosing a0, a1, and am in such a way that 0 ∈ Va0 , from (60), we get:

m −a0 am m 0 = a0 + a1 h f (0) + am h f (0) =⇒ h f (0) = − h f (0) . (61) a1 a1 Mathematics 2020, 8, 2154 20 of 23

As:

q ∞ m−1 ! m a am a h (0) = − 0 C 0 f a ∑ qm (−a )m 1 qm=0 1 i +···+i  m ∞ m−1 ! 1 m a am a hm( ) = − 0 C C ··· C 0 f 0 ∑ ∑ i1 i2 im m a1 (−a1) p=0 i1+i2+...+im=p

m i1, i2, ..., im being non-negative integers numbers; then, substituting h f (0) and h f (0) in the second formula of (61), we arrive at:

q ∞ m−1 ! m a a am a − 0 − 0 C 0 a a ∑ qm (−a )m 1 1 qm=1 1 i +···+i  m ∞ m−1 ! 1 m a a a am a = − 0 − m 0 C ··· C 0 ∑ ∑ i1 im m . a1 a1 −a1 (−a1) p=0 i1+···+im=p

We deduce from this that:

q i +···+i + ∞ m−1 ! m ∞ m−1 ! 1 m 1 am a am a C 0 = C ··· C 0 ∑ qm m ∑ ∑ i1 im m . (62) (−a1) (−a1) qm=1 p=0 i1+···+im=p

By equating the terms with the same degree of the series of both sides of (62) and making qm = i1 + ··· + im + 1 ≥ 1; i1 + ··· im = qm − 1, we arrive at:

C = C ··· C ∀q ≥ qm ∑ i1 im ; m 1 . i1+···im=qm−1

That is the well known recursive formula (5), for which we have provided a new proof. In this framework, we provide an original theorem.

Theorem 5. The numbers Cq2...qp , introduced in (10), are a generalization of Catalan and Fuss–Catalan numbers and they hold the recursive sequence:

Cq2...qp = C 1 1 1 C 2 2 2 + C 1 1 1 C 2 2 2 C 3 3 3 ∑ i2,i3,...ip i2,i3,...ip ∑ i2,i3,...ip i2,i3,...ip i2,i3,...ip 1 2 1 2 3 i2+i2=q2−1 i2+i2+i2=q2 1 2 1 2 3 i +i =q3 i +i +i =q3−1 3 ···3 3 3 ···3 1 2 1 2 3 ip+ip=qq ip+ip+ip=qp (63) + ··· + C 1 1 1 C 2 2 2 ... C p p p , ∑ i2,i3,...ip i2,i3,...ip i2 ,i3 ,...ip 1 2 p i2+i2+...+i2 =q2 1 2 p i +i +...+i =q3 3 3 ··· 3 1 2 p ip+ip+...+ip=qp−1 which is a generalization of (3) and (5). Formula (63) is true for all n ≥ 1 and for all p ≥ 2, with q2 + q3 + ... + qp = n. If q2 = 0; then, the first summand of the second side of (63) does not exist, so we can take it as zero in (63); equally, if q3 = 0, then the second summand of the second side of (63) does not exist, so we can take it as zero; and so on for the following q4, ..., qp.

Proof. Without loss of generality, we prove the Theorem only for the case Cq2q3 , since its generalization follows exactly the same process and, in this way, the development of the reasoning gains enough clarity, due to the complexity of the superscripts and subscripts. Mathematics 2020, 8, 2154 21 of 23

From (10), it is easy to check that Cq20...0 = Cq2 (Catalan numbers) and that C0...0qm0...0 = Cqm (Fuss–Catalan numbers). 2 3 Taking into account (14), the inverse function of y = a0 + a1x + a2x + a3x (a1, a3 6= 0) is given by:

!q2 ∞ a  a q3 1 q2+2q3 2 3 q2+2q3+1 h (y) = (−1) Cq q (y − a0) (64) f a ∑ ∑ 2 3 a2 (−a )3 1 p=0 q2+q3=p 1 1

with (2q2 + 3q3)! Cq2q3 = q2! q3! (q2 + 2q3 + 1)! according to (10).

In agreement with (16), series (64) is absolutely convergent if y ∈ Va0 that in this case is defined as: ( ) 9 (y − a )a 27 (y − a )2a = ∈ 0 2 + 0 3 < Va0 y R; 2 3 1 . 2 a1 4 a1

Choosing a0, a1, a2, and a3 in such a way that 0 ∈ Va0 , from (64), we get:

2 3 −a0 a2 2 a3 3 0 = a0 + a1h f (0) + a2h f (0) + a3h f (0) =⇒ h f (0) = − h f (0) − h f (0) . (65) a1 a1 a1

We define: 2 a2 a0 a3 a0 q2 q3 m0 = ; n0 = and An = Cq q m n . a2 (−a )3 ∑ 2 3 0 0 1 1 q2+q3=n From (64), we can say that:

∞ ∞ a q q a h (0) = 0 C m 2 n 3 = 0 A f −a ∑ ∑ q2q3 0 0 −a ∑ p 1 p=0 q2+q3=p 1 p=0 2 ∞ 2 a0 h (0) = Ai Ai ; with i , i2 non-negative integers f a2 ∑ ∑ 1 2 1 1 p=0 i1+i2=p a3 ∞ h3 ( ) = 0 A A A j j j f 0 3 ∑ ∑ j1 j2 j3 ; with 1, 2, 3 non-negative integers (−a1) p=0 j1+j2+j3=p

2 3 then, substituting h f (0), h f (0), and h f (0) in the second equation of (65) and cancelling the common factor a0/(−a1), we obtain:

∞ ∞ ∞ A = + m A A + n A A A ∑ p 1 0 ∑ ∑ i1 i2 0 ∑ ∑ j1 j2 j3 . (66) p=0 p=0 i1+i2=p p=0 j1+j2+j3=p

The second summand of the second side of (66) becomes:

∞ ! ! r2 r3 s2 s3 m0 ∑ ∑ ∑ Cr2r3 m0 n0 ∑ Cs2s3 m0 n0 . p=0 i1+i2=p r2+r3=i1 s2+s3=i2

From Cauchy product rule:

∞ ∞ i0 + i0 r2 r3 s2 s3 = 1 1 2 m0 ∑ ∑ ∑ Cr2r3 m0 n0 Cs2s3 m0 n0 ∑ ∑ ∑ Cr2r3 Cs2s3 m0 n0 p=0 0 0 0 p=0 0 0 0 (67) i1+i2=p r2+s2=i1 i1+i2=p r2+s2=i1 0 0 r3+s3=i2 r3+s3=i2 Mathematics 2020, 8, 2154 22 of 23

0 0 with i1, i2 non-negative integers. The third summand of the second side of (66) turns out to be:

∞ ! ! ! t2 t3 u2 u3 v2 v3 n0 ∑ ∑ ∑ Ct2t3 m0 n0 ∑ Cu2u3 m0 n0 ∑ Cv2v3 m0 n0 . p=0 j1+j2+j3=p t2+t3=j1 u2+u3=j2 v2+v3=j3

Again, from Cauchy product rule:

∞ 0 0 i1 i2+1 ∑ ∑ ∑ Ct2t3 Cu2u3 Cv2v3 m0 n0 (68) p=0 0 0 0 i1+i2=p t2+u2+v2=i1 0 t3+u3+v3=i2

0 0 0 0 with i1, i2 non-negative integers. If we make the change i1 + 1 = q2 and i2 + 1 = q3, from (66)–(68), as m0 and n0 are arbitrary numbers, we have that:

= + Cq2q3 ∑ Cr2r3 Cs2s3 ∑ Ct2t3 Cu2u3 Cv2v3 (69) r2+s2=q2−1 t2+u2+v2=q2 r3+s3=q3 t3+u3+v3=q3−1

for Cq2q3 6= C00. 0 2 If q2 = 0, then i1 = −1 and its corresponding term in the development of h f (0) does not exist, in agreement with (67), so we can take the first summand of the second side of (69) as zero. The same 0 3 thing happens if q3 = 0, then i2 = −1 and its corresponding term of the development of h f (0) does not exist, in correspondence with (68), so we can take the second summand of the second side of (69) as zero too. Finally, it is easy to check that, for q3 = 0, (69) coincides with (3) and that for q2 = 0, with (5). The result follows.

To close this section, we provide a procedure to generate combinatorial identities by comparing the Taylor development of a function, f , and its inverse h f . To give an example, consider the function y = ex and its inverse, x = log(y). From (37), with x = 0 and y = e0 = 1, we have that:

(p+1) p log (1) = (−1) p! = Ap(0)(p + 1)!  q2  qp+1 p 1 1 = (−1) Cq ...q ··· . ∑ 2 p+1 2! (−1)2 (p + 1)! (−1)p+1 q2+2 q3+···+p qp+1=p

Taking into account that:

(−1)p (−1)2 q2+3 q3+···+(p+1) qp+1 = (−1)p (−1)p (−1)q2+...+qp+1 = (−1)q2+...+qp+1 we arrive at the amazing relation:

p (− ) Cq ...q + 1 q2+...+qp+1 2 p 1 = (−1) q . p + 1 ∑ (2!)q2 (3!)q3 ··· ((p + 1)!) p+1 q2+2 q3+···+p qp+1=p

8. Conclusions As known, due to the Inverse Function Theorem, given an analytic real function, f (x), and a point, 0 x0, with f (x0) 6= 0, there is a neighborhood of x0, Vx0 , in such a manner that the inverse function of

f (x) is well defined in f (Vx0 ). In other words, we know about its existence, but, with respect to its explicit formulation, in the general case, nothing has yet been established. In this context, throughout this paper, we have provided a general procedure to construct the inverse function, x = h f (y), of an arbitrary analytic real function, f (x). Mathematics 2020, 8, 2154 23 of 23

We have addressed this problem by developing the Taylor series of h f (y), in the same way as the most important functions in real analysis ( ex, log(x), sin(x), cos(x), ...) are defined. With this aim in mind, we have found a general formula to calculate the n-th derivative of x = h f (y) as a function of the derivatives of y = f (x). Just as it happens, for example, with e3 or sin(2), for practical purposes, the numerical values of the inverse function, h f (y), need to be expressed with a prefixed number of digits of accuracy. We have faced this question by elaborating a formulation of the Taylor Remainder, valid for any inverse function, h f (y). We have shown, through several examples, how the inverse function, x = h f (y), can be used to solve, in an analytic, not numeric, way the nonlinear equation f (x) = 0. Indeed, the series h f (0) gives an exact solution of the equation f (x) = 0, in an analytic manner and, providing a numeric value, with a predetermined accuracy, to the number h f (0) (the already obtained solution), it is not the same as finding it by numerical methods. Finally, we have obtained a new expansion of the Catalan and Fuss–Catalan numbers that we hope can be used in future research in the field of Graph Theory.

Author Contributions: Conceptualization, J.M.; Formal analysis, J.M., M.A.L., and R.M.; Funding acquisition, M.A.L. and R.M.; Investigation, J.M., M.A.L., and R.M.; Methodology, J.M., M.A.L., and R.M.; Software, J.M. and M.A.L.; Supervision, J.M., M.A.L., and R.M.; Validation, J.M., M.A.L., and R.M.; Visualization, J.M., M.A.L., and R.M.; Writing—original draft, J.M., M.A.L., and R.M.; Writing—review and editing, J.M., M.A.L., and R.M. All authors have read and agreed to the published version of the manuscript. Funding: This work has been partially supported by Fundación Séneca (Spain), grant 20783/PI/18, Ministerio de Ciencia, Innovación y Universidades (Spain), grant PGC2018-097198-B-I00, and FEDER OP2014-2020 of Castilla-La Mancha (Spain), grant 2020-GRIN-29225. Conflicts of Interest: The authors declare no conflict of interest.

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