An Analysis on the Inversion of Polynomials

ENSENANZA˜ REVISTA MEXICANA DE FISICA´ E52 (2) 163–171 DICIEMBRE 2006

An analysis on the inversion of polynomials

M.F. Gonzalez-Cardel´ and R. D´ıaz-Uribe Centro de Ciencias Aplicadas y el Desarrollo Tecnologico,´ Universidad Nacional Autonoma´ de Mexico,´ Apartado Postal 70-186, 04510 Mexico´ D.F., Mexico,´ e-mail: [email protected], ruﬁ[email protected] Recibido el 15 de diciembre de 2005; aceptado el 29 de marzo de 2006

In this work the application and the intervals of validity of an inverse polynomial, according to the method proposed by Arfken [1] for the inversioni of series, is analyzed. It is shown that, for the inverse polynomial there exists a restricted domain whose longitude depends on the magnitude of the acceptable error when the inverse polynomial is used to approximate the inverse function of the original polynomial. A method for calculating the error of the approximation and its use in determining the restricted domain is described and is fully developed up to the third order. In addition, ﬁve examples are presented where the inversion of a polynomial is applied in solving different problems encountered in basic courses on physics and mathematics. Furthermore, expressions for the eighth and ninth coefﬁcients of a ninth-degree inverse polynomial, which are not encountered explicitly in other known references, are deduced.

Keywords: Invertion of polynomial; equation solving; intervals of validity.

En este trabajo se analiza la aplicacion´ y los intervalos de validez de un polinomio inverso, segun´ el metodo´ propuesto por Arfken [1] para la inversion´ de series. Se muestra que existe un dominio restringido cuya longitud depende de la magnitud del error aceptable; esto se ejemplifica por simplicidad con un polinomio de tercer grado, aunque el procedimiento es aplicable a polinomios de cualquier grado. Se deduce una expresion´ para determinar el error del polinomio inverso; as´ı mismo, se presentan cinco ejemplos, con diferentes grados de dificultad, donde se aplica la inversion´ polinomial para resolver diversos problemas que pueden presentarse en f´ısica y matematicas.´ Se deducen las expresiones para los coeficientes octavo y noveno de un polinomio inverso de grado nueve, las cuales no se encuentran expl´ıcitamente en otras referencias conocidas.

Descriptores: Inversion´ polinomial; solucion´ de ecuaciones; intervalo de validez.

PACS: 01.40.-D; 02.30.Mv; 02.60.Cb

1. Introduction of functions, can be defined; undoubtedly this new polynomial has the same characteristics as any inverse function [2] Some physical and mathematical problems can be solved eas- (at least within the order of the approximation given by the ily if the functions involved are approximated, to some level polynomial). This is important, because inverse functions are of accuracy, by simpler and continuous functions, such as al- useful for solving many mathematical and physical problems; gebraic functions, better known as polynomials. Probably the the simplest process for solving an equation is an algebraic best known approximating polynomials are those due to Tay- solution; this process is one of the simpler applications of the lor, which are widely used to approximate any differentiable inverse function. function and are commonly represented as: For example, first degree equations have a simple method Xn of solution by solving the independent variable to obtain the i f(x) = aix inverse function. If we define i=0 where y = f(x) = mx + b, ¯ n ¯ 1 d f(x)¯ an = n ¯ , then we have: n! dx x=0 y − b when f(x) is expanded around x = 0; more general expres- x = = f −1(y), sions can be found in most of the calculus texts [2]. m The main advantage to this approach lies in the fact that a solution that corresponds to the inverse function of f(x). when adding, subtracting, multiplying, deriving and integrat- For second degree equations ing polynomials, other simple and also continuous polynomial functions are obtained. Some authors even deduce a Ax2 + Bx + C = 0, polynomial that is the multiplicative inverse [3] of another polynomial; this inverse polynomial transforms a quotient of a classic solution, given by the well-known formula of Fer- polynomials into a product of them, giving another polyno- rari [4]: mial as a result. √ −B ± B2 − 4AC The most interesting fact, however, is that another in- x = , verse polynomial [1], but now in the sense of a composition 2A 164 M.F. GONZALEZ-CARDEL´ AND R. DIAZ-URIBE´ is obtained by completing the square of the second degree 2. Iverting a polynomial equation and extracting its square root. The last operation is the inverse function of raising to the second power. To obtain the inverse of a polynomial: There exists a procedure for solving third degree equations of a kind given by the following expression: XN i P (x) = aix = y (1) 3 2 x + P x + Qx + R = 0. i=1

Although this procedure is not general, the roots of according to one of the procedures mentioned by Arfken [1], some of these equations can be found. The method due a polynomial solution of the same degree is proposed as: to Tartaglia [4,5] consists in making a change of variable (x = u + Q/3) to transform the complete third degree equation XN j into a reduced third degree equation: Q(y) = bjy = x (2) u3 + qu + r = 0. j=1 The roots of the reduced equation are given by: Substituting (1) in (2), the composition Q(P (x)) is car- r r ried out: r √ r √ 3 3 N u1 = + D + − D, P i 2 2 Q(y) = bi [P (x)] √ µr r ¶ i=1 · ¸ 1 3 r √ r √ PN PN j u = − u + 3 + D − 3 − D i, i 2 2 1 2 2 2 = bj aix (3) j=1 i=1 √ µr r ¶ 2 √ √ NP 1 3 3 r 3 r k u3 = − u1 − + D − − D i, = ckx = x, 2 2 2 2 k=1 where i is the imaginary unit, and the quantity D (called the where Discriminant), is given by: ³ ´ ³ ´ q 3 r 2 ck = ck(ai, bj). D = + . 3 2 This solution is valid only if D ≥ 0. Finally, the roots of the Equating the coefficients for each power on both sides of complete equation are obtained as the equation, we have Q xj = uj + / , for j = 1, 2, 3. 3 c1 = 1, ck = 0, for k = 2, 3, ... (4) For fourth degree equations, Ferrari [4,5] developed a 2 procedure similar to the above. The procedure, however, is a system of N equations for N unknown coefficients bj; too involved to be described here in detail. thus, the system is over determined. In order to find a consis- For fifth degree equations and higher, there is no specific tent solution, we need only consider the first Nequations and procedure for solving them; furthermore, Evariste Galois [6] neglect the remaining ones. Solving the resulting system, we showed almost two hundred years ago that these equations obtain the coefficients of the inverse polynomial. are not solvable by radicals. Thus, numerical methods are a As a particular case, let us consider a third degree poly- useful option. nomial, such as This introduction shows that the approximate solution of 2 3 even third degree polynomials and higher is well worth while. P (x) = a1x + a2x + a3x = y. (5) In this paper, we propose an approximate solution by finding the inverse polynomial through the procedure described by Then the proposed inverse polynomial denoted by Q(y) Arfken [1]; this is included in Sec. 2. It is shown that an im- will have the form portant advantage of this method over the numerical methods is that the inversion procedure gives analytical solutions, in- 2 3 Q(y) = b1y + b2y + b3y = x. (6) stead of only numerical data. In addition, in Sec. 3 we show that the inverse polynomial is valid only over a restricted do- The composition Q(P (x)) is carried out by substitut- main defined by the accepted error. Finally, in Sec. 4, addi- ing (5) in (6), so we get tional expressions for obtaining the inverse polynomial up to the ninth degree are given. We also include five examples to Q(P (x)) = b (a x + a x2 + a x3) show how the method is used in practice. All of them are 1 1 2 3 2 3 2 conducted only to the third degree, to avoid lengthy deduc- + b2(a1x + a2x + a3x ) tions and expressions. The last two examples show how to 2 3 3 solve simple undergraduate problems. + b3(a1x + a2x + a3x ) = x. (7)

Rev. Mex. F´ıs. E 52 (2) (2006) 163–171 AN ANALYSIS ON THE INVERSION OF POLYNOMIALS 165

After expanding and grouping, we obtain any higher degree than the third, we obtain the identity function; i.e. Q(P (x)) = x, showing consistency in the solution, 2 2 Q(P (x)) = a1b1x + (a2b1 + a1b2)x within the limits of the proposed approximation. 3 3 In order to get better insight into the consequences of this + (a3b1 + 2a1a2b2 + a1b3)x result, let us analyze numerically the next example in partic- 2 2 4 + (a2b2 + 2a1a3b2 + 3a1a2b3)x ular. 2 Example 1. According to Eqs. (6) and (10), the inverse poly- + (2a2a3b2 + 3a1a b3 2 nomial of + 3a2a b )x5 + (a2b + 6a a a b + a3b )x6 1 3 3 3 2 1 2 3 3 2 3 P (x) = x + x2 + x3 (11) 2 2 7 + (3a2a3b3 + 3a1a3b3)x is 2 8 3 9 + 3a2a b3x + a b3x 3 3 Q(x) = x − x2 + x3. (12) XN 2 i = cix = x (8) Evaluating P (x) in x = 0 [see Eq. (11)], we have i=1 P (0) = 0, and when evaluating Q(P (x)) = 0 = x, then P (x) Equating the coefficients of the same power in x from and Q(x) are inverse polynomials in x = 0. If now we eval- both sides of Eq. (8), it is required that all coefficients be uate P (1) = 3, substituting in Eq. (12) we get Q(3) = 21, so identical to zero, except the first. That is, Q(y) is not the inverse of P (x), or at least not in x = 1. Thus the solution is not a general one; there exists a region where

c1 = a1b1 = 1 the solution can be considered acceptable. 2 c2 = a1b2 + a2b1 = 0 3 c3 = a3b1 + 2a2b1b2 + a1b3 = 0 2 2 c4 = a2b2 + 2a1a3b2 + 3a1a2b3 = 0 2 2 c5 = 2a2a3b2 + 3a1a2b3 + 3a1a3b3 = 0 2 3 c6 = a3b2 + 6a1a2a3b3 + a2b3 = 0 2 2 c7 = 3a2a3b3 + 3a1a3b3 = 0 2 c8 = 3a2a3b3 = 0 3 c9 = a3b3 = 0 (9)

With the first three relationships, we form a system of three equations and three unknowns. The rest of the equations are ignored, otherwise we would have a system with more equations than unknowns whose solution is overdeter- mined. The system can be solved to obtain the coefficients bi as functions of the coefficients ai of the original polynomial; the solution to the system of three equations is 1 b1 = a1 a2 b2 = − 3 a1 2 2a2 − a3a1 b3 = 5 . (10) a1 It is well worth noting that, according to this solution, if the polynomial does not have a linear term, the inverse polynomial would not exist; this conclusion is valid for any degree polynomial. FIGURE 1. The two polynomials P(x) and Q(x) are shown. a) in Now if we make the composition of Q and P , with the the interval of x ∈ [0,16] symmetry near zero is not appreciated. bi coefficients given by Eqs. (10) and neglecting the terms of b) x ∈ [0,0.3] symmetry is observed.

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TABLE I. Numerical data for P(x) and Q(x) from Example 1 and the error of the composition P(Q(x)) for –1 ≤ x ≤ 1 . 3 2 6 (a2b3 + a3b2 + 6a1a2a3b3)x x P (x) Q(P (x)) Error (%) 2a5 + 11a a3a − 7a2a a2 -1 -1 -3 200 = 2 1 2 3 1 2 3 x6 ≡ T x6 a5 3 -0.9 -0.819 -2.03911426 126.568251 1 2 2 7 -0.8 -0.672 -1.42704845 78.381056 (3a2a3b3 + 3a1a3b3)x -0.7 -0.553 -1.02792138 46.845911 4 2 2 2 3 6a2a3 + 3a1a2a3 − 3a1a3 7 7 = 5 x ≡ T4x -0.6 -0.456 -0.75875482 26.459136 a1 -0.5 -0.375 -0.56835938 13.671875 3 2 3 ¡ 2 ¢ 8 6a2a3 − 3a1a2a3 8 8 -0.4 -0.304 -0.42451046 6.127616 3a2a3b3 x = 5 x ≡ T5x a1 -0.3 -0.237 -0.30648105 2.160351 2 3 4 2a a − a1a -0.2 -0.168 -0.20096563 0.482816 a3b x9 = 2 3 3 x9 ≡ T x9 (13) 3 3 a5 6 -0.1 -0.091 -0.10003457 0.034571 1 0 0 0 0 0.1 0.111 0.10004663 0.046631 These are, in an approximate way, the six neglected terms 0.2 0.248 0.20174899 0.874496 that can be calculated with a knowledge of the particular values of the coefficients of the polynomial to invert; clearly they 0.3 0.417 0.31562271 5.207571 are dependent on the particular point x of the polynomial do- 0.4 0.624 0.47759462 19.398656 main. 0.5 0.875 0.77929687 55.859375 These terms are negligible provided that a1 is large; or 0.6 1.176 1.41940378 136.567296 that the domain is restricted, for instance to |x| < 1, be- 0.7 1.533 2.78559744 297.942491 cause xn, for n = 4, 5, 6, 7, 8, 9, is small. 0.8 1.952 5.57940941 597.426176 To answer the second question, we build Table 1, where 0.9 2.439 10.9992095 1122.13439 the errors are presented for different values of x. 1 3 21 2000 We can use the data in Table I to see that, for –0.2 ≤ x ≤ 0.2, the error in the approximate inverse polyno- In Fig. 1, the polynomials P (x) and Q(x) are shown mial is less than 1%; for –0.3 ≤ x ≤ 0.3, the error is not too graphically. In Fig. 1a the curves do not correspond to two much higher than 5%; but, for the interval –0.5 ≤ x ≤ 0.5, functions with one the inverse of the other, because if they the error is greater than 10%. For an error smaller than 1%, did so, the graphs should be symmetrical with respect to the by interpolation we find that x must be within the interval identity function. Certainly they are not. In Fig. 1b certain (-0.294, 0.241). In Fig. 2, the plots of both polynomials symmetry is appreciated, at least in a small interval. are shown in this interval; it is easy to observe the symmetry regarding the identity function (central line), thus, it can 3. Error Analysis be assumed that P (x) and Q(x) are inverse polynomials in this interval. Considering the previous example, it is convenient to ask: If we evaluate the neglected terms, we find that their sum how big are the neglected terms? When is the inverse poly- is exactly the error of P (Q(x)). As can be appreciated from nomial acceptable, and what is the error? To answer the first Eqs. (13), the error diminishes, increasing the a1 value, but of these questions, we evaluate the terms neglected in the in- it increases when a2 and a3 are increased. version of a third degree polynomial. By considering all the Another way to state the problem is as follows: if one terms in equation (8), and by substituting the expressions for wants a error smaller than or similar to e% of x, then b1, b2 and b3, obtained in Eqs. (10), we get ¯ ¯ 2 2 4 9 (a b2 + 2a1a3b2 + 3a a2b3)x ¯ 4 5 6 7 8 ¯ 2 1 ¯T1x +T2x +T3x +T4x +T5x +T6x ¯ e ¯ ¯ ≤ (14) 3 ¯ x ¯ 100 5a2 − 5a1a2a3 4 4 = 3 x ≡ T1x a1 2 2 5 or 2a2a3b2 + 3a1a2b3 + 3a1a3b3)x 4 2 2 2 6a + a1a a3 − 3a a ¯ ¯ = 2 2 1 3 x5 ≡ T x5 3 4 5 6 7 8 e 4 2 ¯T1x +T2x +T3x +T4x +T5x +T6x ¯ ≤ (15) a1 100

Rev. Mex. F´ıs. E 52 (2) (2006) 163–171 AN ANALYSIS ON THE INVERSION OF POLYNOMIALS 167

FIGURE 3. Plot of the inequality (17), where the interval solution is observed FIGURE 2. The polynomials P(x) and Q(x) are shown in an interval where certain symmetry is appreciated. Solving this polynomial by the method of the Regula Falsi [7] with an error smaller then 0.01%, the roots obtained The maximum value for the error should be given in xm, are: -0.2937164 and 0.2418198, therefore, inequality (18) is where xm is the greatest value of x in the restricted domain. satisfied when For example if xm = 0.1, then the error is −0.2937164 ≤ x ≤ 0.2418198, ¯ −3 −4 −5 −6 −7 which coincides with the interval found from Table I. ¯10 T1 + 10 T2 + 10 T3 + 10 T4 + 10 T5 ¯ Example 3. An application of inverting a polynomial is to −8 ¯ −2 + 10 T6 ≤ 10 e. (16) find roots of a polynomial. Let us suppose that we want to find a root of the polynomial 1 Knowing the coefficients Tk, with k = 1,. . . , 6, we can y = P (x) = x3 + x2 + x − = 0, (20) readily estimate the error of an inverse third degree polyno- 4 mial. which includes a zero order term. In general, (20) has the form Example 2. Let us consider again the polynomial (11) in 2 3 P (x) = a0 + a1x + a2x + a3x = y (21) example 1, we have that a1 = a2 = a3 = 1; then, using Eqs. (13) we find that the coefficients T are: In order to find the solution, it is convenient to rewrite Eq. (21) as 3 2 T1= 0, T2 = 4, T3 = 6, T4 = 6, T5 = 3 and T6 = 1. z = (y − a0) = a3x + a2x + a1x. (22) As we already know, the inverse polynomial is of the form So, if we allow a maximum percent error equal to 2%, x 2 3 must satisfy the following relationship: x = b1z + b2z + b3z . (23) Substituting the value of z, we have: 4 5 6 7 8 2 3 −0.02 ≤ 4x + 6x + 6x + 3x + x ≤ 0.02 (17) x = (y − a0) − (y − a0) + (y − a0) (24) Developing the binomials we get A plot of the inequality is shown in Fig. 3; it is not diffi- 3 2 x = b3y + (b2 − 3b3a0)y + (b1 − 2b2a0 cult to see that the left inequality is always satisfied, and the 2 + 2 3 right side inequality can be expressed as: + 3b3a0)y + (−b1a0 b2a0 − b3a0) (25) The third degree polynomial in Eq. (20) has coefficients 4 5 6 7 8 f(x) ≡ 4x + 6x + 6x + 3x + x − 0.02 ≤ 0. (18) a0= -1/4, a1= 1, a2= 1 and a3= 1, according to the relationships (10), the coefficents of the inverse polynomial are b1= 1, b2= -1 and b3= 1. In the same figure, it can be seen that the solution interval Substituting these values in (25), we have f(x) x is the region where the plot of is below the axis. That 1 11 13 interval is determined by the roots of x = y3 − y2 + y + (26) 4 16 64 One root of the polynomial is obtained by making y = 0 4 5 6 7 8 4x + 6x + 6x + 3x + x − 0.02 = 0. (19) in the inverted polynomial Eq. (26). Doing this, we get

Rev. Mex. F´ıs. E 52 (2) (2006) 163–171 168 M.F. GONZALEZ-CARDEL´ AND R. DIAZ-URIBE´ x = 13/64 ≈ 0.203125. On the other hand, numerically solv- Approximating the sine function with the first three terms ing this polynomial again with the method of the Reg- of its McLaurin series, ula Falsi [7], with a smaller error than 0.01%, the root is µ ¶ x3 x5 x7 x = 0.201392 ± 0.01%. The value obtained from the inverse sin(x) x− + − x2 x4 x6 = 3! 5! 7! =1− + − . (30) polynomial differs from that found numerically by less than x x 6 120 5040 1%, a value that corresponds to the evaluation of the error through the expressions in (14), and that may be acceptable Then, substituting Eq. (30) in to Eq. (29) µ ¶ depending on the specific problem. sin(x) 1 1 x2 x4 x6 Example 4. It is important to recognize that this method can −√ = 1−√ − + − =0, (31) x 2 2 6 120 5040 be applied to solve some common physical problems. For example, the Fraunhofer intensity pattern [8] of a single slit and defining the following polynomial: is given by µ ¶ 2 4 6 sin(x) 2 x x x I = I . (27) P (x) = − + − 0.2929, (32) m x 6 120 5040 Usually the width of the pattern is evaluated through the and making the following changes of variable: s = x2 and points where w = P (x)+ 0.2929, we get: sin(x) = 0 s s2 s3 x w = (P (x) + 0.2929) = − + = 0 (33) (see Fig 4); this is given by x = ± nπ with n an integer. 6 120 5040 The width of the pattern is found by making n = ± 1, and The coefficients of the inverse polynomial, according to by subtraction, ∆x = 2π. In many cases, however, the prob- the relationships in (10) are: lem is not so easy, and one of the important parameters is the

Half Width of the Full Maximum, commonly abbreviated as b1 = 6, b2 = 1.8 and b3 = 0.8232; HWFM. This is obtained by taking therefore, the inverse polynomial is ± 1 I(x ) = Im. (28) 2 s = 6w + 1.8w2 + 0.8232w3 (34) Substituting (28) in (27) one obtains or, using (33), µ ¶ sin(x±) 2 1 = . (29) s = 6 (P (x) + 0.2929) x± 2 + 1.8 (P (x)+0.2929)2 +0.8232 (P (x)+0.2929)3 (35) In this equation, x± cannot be directly solved analytically; we can, however, use polynomial inversion approach, for ex- Developing, ample, starting from the McLaurin the polynomial expansion and then inverting this polynomial. Then it is solved follow- s = 1.9369 + 7.2662P (x) ing the procedure of the previous examples. + 2.5233P (x)2 + 0.8232P (x)3, (36)

and evaluating in P (x) = 0, we have s=1.9369 or x± = x = ±1.3917. Evaluating the error through the expressions in (14) applied to the polynomial (33) we obtain an error smaller than 0.3%. Solving Eq. (29) for the method of the Regula Falsi, we ﬁnd x = 1.3915, a value that differs from the one obtained for the inverse polynomial by less than 0.02%. Example 5. In the study [10] of the radiation emitted by a black body, it is often necessary to ﬁnd the wavelength λ, for which the density of energy radiated by a black body has a maximum at a given temperature. As is well known, the density of energy (see Fig. 5) is given by:

8πk5T 5 x5 E(λ) = 4 4 x , sin(x) c h e − 1 FIGURE 4. Plot of the function f(x) = x , it is not difﬁcult to see that f(x) = 0, in x =± nπ, with n integer. where: x = hc/λkT .

Rev. Mex. F´ıs. E 52 (2) (2006) 163–171 AN ANALYSIS ON THE INVERSION OF POLYNOMIALS 169

highly probable that the resulting solution will have a large error. So, in order to ﬁnd what is the best a value in such a way that the solution for x has a low error, we ﬁrst make some estimations of error for several values of a. After several trials, we found that a = 8 is a good choice. By substituting this value in Eq. (37), we have

y = −5.591 × 10−5x3 + 0.00150958x2 + 0.18624603x − 0.95761989

and:

z = y+0.95761989 FIGURE 5. Plot of density of energy emitted by a black body. =−5.591×10−5x3+0.00150958x2+0.18624603x. To find the maximum, it is necessary to find the roots of: The coefficients of the inverse polynomial, according to dE = 0. the relationships (10), are: dx Expanding the last expression, the problem is reduced to b1 = −1.25, b2 = 0.9765625 and b3 = −1.11897786 solving the following equation: x e−x + − 1 = 0 (37) Therefore: 5 This is a transcendental equation, not easily solved by x = 0.06680501z3 − 0.23366602z2 + 5.3692419z elementary algebra. The solution can be approximated, however, by expanding in Taylor’s series, around x = a.We ob- or tain to the third degree x 3 y(x) = e−x + − 1 x = 0.06680501(y + 0.95761989) 5 ½µ ¶ ¾ − 0.23366602(y + 0.95761989)2 e−a 1 a ≈ − x3+ + e−a x2 6 2 2 + 5.3692419(y + 0.95761989) ½µ ¶ ¾ a2 1 − 1+a+ e−a− x 2 5 Expanding and evaluating in y = 0, we get the root of (38) ½µ ¶ ¾ as a2 a3 + 1+a+ + e−a−1 , 2 6 x = 4.9860789. defining: This is a very good approximation to the root of the function (37), because if we solve numerically through the e−a method of the Regula Falsi [7], with a smaller error than a = − ; 3 6 0.01%, the root is: 4.965332 ± 0.01%. Thus, the value ob- ½µ ¶ ¾ 1 a tained by inverting the polynomial differs from the numerical a = + e−a ; 2 2 2 result in less than 0.5%. Significantly, we found that, for a values between 6 and ½µ 2 ¶ ¾ a −a 1 10, the result is very similar to the above. a1 = − 1 + a + e − ; 2 5 These examples have shown how to estimate the valid- ½µ ¶ ¾ a2 a3 ity of the solutions of the method of polynomial inversion, a = 1 + a + + e−a − 1 0 2 6 for an approximating polynomial of third degree with a error smaller than 2%, but these conditions can be varied depend- y(x) can be expressed approximately as: ing on the problem that is to be solved. 2 3 Next, some expressions are given to determine the coef- y(x) = a0 + a1x + a2x + a3x . (38) ficients of an inverse polynomial of a higher degree, which At this point, the value of a is arbitrary. If we try to in- allows a reduction of the error and an increased of the inter- vert the polynomial (38) for a particular choice of a, it is very val of validity of the solution.

Rev. Mex. F´ıs. E 52 (2) (2006) 163–171 170 M.F. GONZALEZ-CARDEL´ AND R. DIAZ-URIBE´

4. Coefficients for inverse polynomials of As for the error terms, we only present the error terms for higher degree. a third degree polynomial, and they are shown in Eq. (14); the error terms for higher degree can be calculated starting from Using the method as described above, the coefficients of in- the composition of polynomials, as was done in section 3; a verse, higher degree polynomials can be calculated; some of detailed computation of the error terms would be too lengthy them are listed below. Expressions for b4 to b7 are the same for this paper. as those reported by Dwight [9], whereas b8 and b9 are not reported in any other work we know ii . Let y(x) be given by 5. Conclusions 2 3 4 y = a1x + a2x + a3x + a4x In this present work, we have evaluated the error of the in- + a x5 + a x6 + a x7 + ..., a = 0. 5 6 7 with 1 (39) verse polynomial obtained through a method proposed by Ar- Then, its inverse polynomial is given by fken [1]. In addition, accepting a particular amount of error, we have shown how to obtain the interval of validity for the x = b y+b y2+b y3+b y4+b y5+b y6+b y7+..., (40) 1 2 3 4 5 6 7 inverse approximating polynomial. We have shown how to where coefficients b1,b2 and b3 are given by the relationships apply the polynomial inversion method up to the third de- in (10), and [9] gree for solving five different problems not easily solved at the undergraduate level in physics and engineering. The first 1 ¡ ¢ shows only how to do the inversion of the polynomial, and b = 5a a a − a2a − 5a3 , 4 7 1 2 3 1 4 2 the second explains the procedure for finding the interval of a1 validity for the inversion made in problem 1. The third ex- 1 ¡ ¢ 2 2 2 4 3 2 ample shows how to find one root of a third-degree equation. b5 = 9 6a1a2a4 + 3a1a3 + 14a2 − a1a5 − 21a1a2a3 , a1 The fourth and fifth examples find the solutions for two basic 1 physics problems where transcendental equations are found: b = (7a3a a + 7a3a a + 84a a3a − a4a ) 6 11 1 2 5 1 3 4 1 2 3 1 6 one is for obtaining the semi width of a single-slit Fraunhofer a1 diffraction pattern. The other is the calculation of the wave- 2 2 2 2 5 − 28a1a2a4 − 28a1a2a3 − 42a2, length at which the radiation emitted by a black body has a

1 4 4 4 2 2 3 maximum value. In addition, as a new result, we have ob- b7 = 13 (8a1a2a6 + 8a1a3a5 + 4a1a4 + 120a1a2a4 tained explicit expressions for two new coefficients for the a1 inverse polynomial not reported in other references. 2 2 2 6 5 3 2 + 180a1a2a3 + 132a2 − a1a7 − 36a1a2a5 Once the expressions for the coefficients of the inverse 3 3 3 4 polynomial and the expression of the error are obtained, the − 72a1a2a3a4 − 12a1a3 − 330a1a2a3). procedure of the inverted polynomial has the advantage of The two new coefficients are: being very easy to apply, since it reduces to a direct evalua- 1 5 5 5 3 3 tion, and in the case of extraction of roots, it also reduces to b8 = 15 (9a1a2a7 + 9a1a3a6 + 9a1a4a5 + 165a1a2a5 a1 an evaluation, having the disadvantage that it is useful only + 495a3a2a a + 165a3a a3 + 1287a a5a − a6a in a vicinity around the origin; the radius of this vicinity de- 1 2 3 4 1 2 3 1 2 3 1 8 pends on the magnitudes of the coefficients of the polynomial 7 4 2 4 2 4 − 429a2 − 45a1a2a6 − 45a1a2a4 − 90a1a2a3a5 to invert. A useful advantage that the inverse polynomial of- 4 2 2 4 2 3 2 fers is that it allows one to obtain an approximate analytic − 45a1a3a4 − 495a1a2a4 − 990a1a2a3), expression for transcendent inverse functions, which can be 1 easily evaluated. It can be concluded that the procedure of b = (10a6a a + 10a6a a + 10a6a a + 5a6a2 9 17 1 2 8 1 3 7 1 4 6 1 5 inversion of polynomials is a good first approach to the solu- a1 tions of some problems that involve polynomials or that can 4 4 4 3 4 2 2 4 2 + 55a1a3 + 220a1a2a6 + 330a1a2a4 + 660a1a2a3a5 be approximated with them. 4 2 2 4 2 2 5 8 + 660a1a2a3a4 + 5005a1a2a3 + 2002a1a2a4 + 1480a2 7 5 2 5 2 5 − a1a9 − 55a1a2a7 − 55a1a3a5 − 110a1a2a3a6 Acknowledgements 5 5 2 3 4 − 110a1a2a4a5 − 55a1a3a4 − 715a1a2a5 3 3 3 2 3 6 This investigation was sponsored by Consejo Nacional de − 2860a1a2a3a4 − 1430a1a2a3 − 5005a1a2a3), Ciencia y Tecnolog´ıa under the project 37077-E.

Rev. Mex. F´ıs. E 52 (2) (2006) 163–171 AN ANALYSIS ON THE INVERSION OF POLYNOMIALS 171 i. It is worth to recognize here that in many text books [1,9] the 5. M.A. Hall and B.A. Knight, Higher Algebra (Macmillan and term reversion is used instead; the meaning, however, is the Con., Ltd., Londres, UK, 1982). same as implied along this paper. 6. C. Coulston Gillispie, Dictionary of Scientific Biography ii. We tried to find the two additional references given by Dwight (Charles Scribner’s Sons, New York, USA, 1981) Vol. 5. on page 15 of Ref. [9]; however we were not able to find them. Our sources state that the Philosophical Magazine was not pub- 7. V.O. Panteleeva and C.M. Gonzalez,´ Metodos´ Numericos´ (Edi- lished until 1921. The other reference is a book not found in ciones Instituto de Investigacion´ de Tecnolog´ıa Educativa de la Mexico.´ In addition, MATHEMATICA [11] reports a general ex- Universidad Tecnologica´ de Mexico´ S. C., Mexico,´ 2002) p. 85, pression which is not completely defined because it is formally 177, 288. defined but not explicitly expressed; furthermore, MATHEMAT- 8. D. Halliday and R. Resnick, Physics (part II) (John Wiley & ICA does not account for the error of the inversion procedure. sons, inc., USA, 1980). 1. G. Arfken, Mathematical Methods for Physicists (Academic 9. B.H. Dwight, Tables of Integrals and other mathematical data, Press, New York, 1981). 4a. Edicion´ (The Macmillan Company, 1961, USA.) p.15, 136. 2. M. Spivak, Calculus, W.A. Benjamin, Inc. (New York 1978). 10. M. Alonso and E.J. Finn, Fundamental University Physics, 3. P. Henrici, Applied and computational complex analysis, (Ed. Volume III, Quantum and Statistical Physics (Addison-Wesley John Wiley, New York, 1974) p.12. Publishing Company, Massachusetts, E.U.A., 1968). 4. J.E. Thompson, Algebra for the practical man, D. Van Nostrand Company, Inc. (New York EUA, 1968). 11. Wolfram Research Inc., Mathematica V 5.0.

Rev. Mex. F´ıs. E 52 (2) (2006) 163–171