A simplified expression for the solution of cubic polynomial equations using function evaluation
aa r X i v : . [ m a t h . G M ] F e b A simplified expression for the solution of cubic polynomial equations usingfunction evaluation
Ababu T. Tiruneh, University of Eswatini, Department of Env. Health Science, Swaziland.
Email: a [email protected] Abstract
This paper presents a simplified method of expressing the solution to cubic equations interms of function evaluation only. The method eliminates the need to manipulate the orig-inal coefficients of the cubic polynomial and makes the solution free from such coefficients.In addition, the usual substitution needed to reduce the cubics is implicit in that the finalsolution is expressed in terms only of the function and derivative values of the given cubicpolynomial at a single point. The proposed methodology simplifies the solution to cubicequations making them easy to remember and solve.
Keywords:
Polynomial equations, algebra, cubic equations, solution of equations, cubicpolynomials, mathematics
Polynomials of higher degree arise often in problems in science and engineering. According tothe fundamental theorem of algebra, a polynomial equation of degree n has at most n distinctsolutions [1]. The history of symbolic manipulation for solving polynomial equation notes thework of Luca Pacioli in 1494 [2] in which the basis was laid for solving linear and quadraticequations while he stated the cubic equation as impossible to solve and asking the Italian math-ematical community to take the challenge. Mathematicians seemed to have felt hitting a wallafter solving quadratic equations as it took quite a while for the solution to cubic equation to befound [3]. The solution to the cubic in the depressed form x + bx + c was discovered by del Ferroinitially but he passed it to his student instead of publishing it. Eventually Caradano in 1540got hold of the solution and published the results, crediting Del Ferro for the solution. Cardanoalso published Ferrari’s solution of the quartic equation.The general cubic of the form x + bx + cx + d can be reduced to a depressed form by a suitable substitution involving a new variable[4]. Ac-cordingly, the substitution takes the form of: x = y − b y + py + q Where p and q are expressed in the forms: p = c − b and q = d − bc b y = z − p z will reduce the depressed cubic after simplification to the form z + qz − p
27 = 0Solving this as a quadratic for z gives: z = − q ± √ R R = (cid:16) p (cid:17) + (cid:16) q (cid:17) Considering the three cube roots of unity [5] namely;1 , ω = −
12 + 12 √ i and ω = − − √ i It can then be easily verified [5] that the three solutions of the depressed cubic equations aregiven by: y = A + B, y = ωA + ω B and y = ωB + ω A The value of A and B in the above equation are given by: A = r − q √ R and B = r − q − √ R Other similar methods also arrived essentially at the same formula state above. For examplesee the work of Mukundan [6] that transforms the original depressed cubic equation into two cubicequivalents that are solved by simply taking the cube roots. A more comprehensive coverage ofthe formulae for the solution of the cubic equation is found at the Wolfram Mathworld website [7]in which some of the formula used in the examples used in this paper are referred to. Lagrange in1770-1771, as part of his study of higher degree equation used Fourier transform and subsequentinversion [8]. He hoped to extend his methods to arbitrarily high degree polynomials. However,Lagrange noted that the resulting resolvent polynomial for a five degree polynomial equationwas actually a six degree polynomial, prompting a possible hint to him that such equations may2ot be solvable.The solution to cubic equation has laid a basis for the methods developed for solving quarticequations. For example, Leonard Euler (1707-1783) appreciated the central role of the resolventcubic in the solution of the quartic polynomial equations [9]. Euler [10], through his solution tothe depressed (in which the x term is zero) quartic equations, showed that each root is expressedas the sum of three square roots that are solutions of the resolvent cubic equation. Even thefirst solution of the quartic equation by Del Ferro that was published together with the solutionof the cubic equation by Cardano involved converting the original quartic equation in to twocomplete squares. During the process of solving such equation a resolvent cubic equations arisesthat must be solved first [11]. Several of the recent methods for solving quartic equations alsodo require solving the resolvent cubic. For example see the works of Saghe [12] Fathi [13] andthat of Kulkarni [14].Solution to five degree polynomials and greater were met with limited success as it latertranspired to mathematicians that the solutions could not be expressed in terms of radicals. Thefirst clear proof came from Abel in 1824 who proved that the general polynomial of degree fivecan not be solved in terms of radicals. Abel later revised his proof by verifying that certainforms of the quintic equation (that were later termed as ’Abelian Galois group’) can be solvedby radicals. Galois in 1832 gave a proof that a five degree polynomial equation can be solved byradicals if and only if its Galois group is solvable [11].Certain polynomials equations of higherdegree also require solving a resolvent cubic. Example is the so called palindromic polynomialswith symmetric coefficients. A six degree polynomial that is palindromic results in a resolventcubic through substitution of the fom y = x + x [11]Some authors noted the relation of the transformation variables to the depressed cubic andquartic equations with derivative values. For example, Das, [15], noted that the transformationto the depressed form of the quadratic, cubic and quartic equations correspond to the equationsthat set to zero the first derivative, second and third derivative respectively of the originalequations. However, beyond this the transformation of the other coefficients of the cubic equationto functions and derivative values is not explained. Consider the cubic polynomial equation of the form ax + bx + cx + d = 0 Without loss ofgenerality assume a=1 in which the equation reduces to x + bx + cx + d = 0 The case in which a = 1 will be dealt with a simple revision of the solution at a later stage.The methodology of arriving at the final simplified expression in terms of function evaluation3ollows the well known procedure of reducing the cubic to the deflated form and using Viete’smagic substitution to reduce the equation to quadratic form. It should be known that in the endboth procedures become implicit in the solution whereby the final expression consists of functionvalue of the original cubic and its derivatives.Define new variable t and constant z such that x = z + t (1)Substituting this new expression from eq.1 into the original cubic equation x + bx + cx + d = 0results in the following: x + bx + cx + d = ( z + t ) + b ( z + t ) + c ( z + t ) + d = 0After suitable simplification the above expression takes the form: t + ( b + 3 z ) t + (3 z + 2 bz + c ) t + ( z + bz + cz + d ) = 0 (2)It is clear that eq.2 contains expressions in the brackets that are functional and derivativevalues of the constant z defined in eq.1. Therefore, eq.2 can be expressed as: t + (cid:18) f ′′ ( z )2 (cid:19) t + ( f ′ ( z )) t + f ( z ) = 0 (3). The deflated form is easily obtained by setting f ′′ ( z ) = 6 z + 2 b = 0 in eq.3 which gives z = − b/
3. After this eq.3 will reduce to the reduced form given below. t + ( f ′ ( z )) t + f ( z ) = 0 (4)Now the famous Viete’s substitution can be made in eq.4 by defining a variable s and aconstant α such that: t = s + αs (5)Substituting the expression in eq.5 into eq.4 and further simplification results in the followingequation: s + (3 α + f ′ ( z )) s + (cid:0) α + f ′ ( z ) α (cid:1) s + f ( z ) + α s = 0 (6)4t is apparent that the expression containing s and 1/s in eq.6 will vanish if Viete’s substitu-tion, i.e., α = − f ′ ( z )3 is made. After this substitution, eq.6 reduces to: s + f ( z ) s − f ′ ( z )
27 = 0 (7)Substituting r = s in eq.7 will reduce it to the quadratic form: r + f ( z ) r − f ′ ( z )
27 = 0 (8)The solution to eq.8 will be r = − f ( z ) ± s f ( z ) + 4 f ′ ( z ) s = r / in the above expression results in the solution in terms of s, i.e., s = − f ( z ) ± s f ( z ) + 4 f ′ ( z ) / (9)Finally, substituting the expression in eq.9 in eq.5 and using the value of the constant zto reduce the cubic to the defalted form in eq.1 will give the solution of the cubic polynomialequation which will be in the form given below after simplification: x = f ′′− (0) + − f ( z ) + s f ( z ) + 4 f ′ ( z ) / + − f ( z ) − s f ( z ) + 4 f ′ ( z ) / (10)It is shown, therefore, that the solution of the general cubic equation x + bx + cx + d = 0is expressed in eq.1 in terms only of the the functional and derivative values at the constantz. The z value is found from the equation f ′′ ( z ) = 0 and this value is given by z = f ′′− (0)in eq.10. It is apparent that eq.10 is free of the original coefficient of the cubic equation which5ight make it difficult to remember the solution in terms of these coefficients. The case of thegeneral polynomial ax + bx + cx + d = 0 is handled by dividing both the functional value f(z)and the derivative f’(z) in eq.10 by the coefficient a which will eventually give the solution.Equation (10) can be reduced further with the following usual substitution of cubic equations: R = − f ( z )2 and Q = f ′ ( z )3With the above definition of R and Q, Equation (10) will be rewritten as: x = f ′′− (0) + h R + p R + Q i / + h R − p R + Q i / (11)The application of this simplified expression will be illustrated with the three examples givenin the following section. Example 1 f ( x ) = x − x + 11 x − f ′′ ( z ) = 6 z −
12 = 0 means z = = 2 f ′ ( z ) = 3 z − z ) + 11 = 3(2) − − f ( z ) = 2 − ) + 11(2) − Q = f ′ ( z )3 = − and R = − f ( z )2 = 0The discriminant D is evaluated as: D = Q + R = ( − ) + (0) = − < <
0. The three roots are then computed as follows: θ = cos − R q ( − Q ) = cos − q ( ) = cos − (0) = π x = 2 p ( − Q ) cos ( θ z = 2 r ( 13 ) cos ( π x = 2 p ( − Q ) cos ( θ + 2 π z = 2 r ( 13 ) cos ( 5 π − x = 2 p ( − Q ) cos ( θ + 4 π z = 2 r ( 13 ) cos ( 3 π x = { , , } Example 2 f ( x ) = x − x − f ′′ ( z ) = 6 z = 0 means z = 0 f ′ ( z ) = 3 z −
15 = 3(0) −
15 = − f ( z ) = (0) − − − Q = f ′ ( z )3 = −
153 = − and R = − f ( z )2 = − (cid:18) − (cid:19) = 2The discriminant D is evaluated as: D = Q + R = ( − + (2) = − −
125 + 4 = − < <
0. The three roots are thencomputed as follows: θ = cos − R q ( − Q ) = cos − p (125) ! = 1 . x = 2 p ( − Q ) cos ( θ z = 2 p (5) cos ( θ x = 2 p ( − Q ) cos ( θ + 2 π z = − − √ x = 2 p ( − Q ) cos ( θ + 4 π z = − √ x = { , − − √ , − √ } Example 3 f ( x ) = x − x + 9 x − ′′ ( z ) = 6 z −
10 = 0 means z = = f ′ ( z ) = 3 z − z ) + 9 = 3 (cid:18) (cid:19) − (cid:18) (cid:19) + 9 = 23 f ( z ) = (cid:18) (cid:19) − (cid:18) (cid:19) + 9 (cid:18) (cid:19) − − Q = f ′ ( z )3 = and R = − f ( z )2 = − (cid:18) − (cid:19) = 4427The discriminant D is evaluated as: D = Q + R = (cid:0) (cid:1) + (cid:0) (cid:1) > x = z + B where B is given by: B = h R + p R + Q i / + h R − p R + Q i / (12)Substituting the computed values of R and Q in equation (12) gives B = 4 / x = z + B = + = = 3The complex conjugate roots are obtained from the following equation: x , = z − (cid:18) (cid:19) B ± √ ! i p B + 4 Q (13)Substituting the computed values of z = B = and Q = in equation (13) gives: X , = 1 ± √ i The three roots of the given cubic equation are, therefore, x = { , √ i, − √ i } The formula for the solution to cubic equations are a bit complex to work with. However,this paper demonstrated that it is possible to simplify the expression for the solutions of thecubic equations further through a method that requires only evaluation of the functional value8nd the derivative at x value which makes the second derivative of the cubic equation zero.When the solution is expressed essentially in terms of these two functional values, it would beeasy to remember the formula for the solution as given by equation (10). In other words, theoriginal general cubic equation containing the coefficients is solved in terms of functional valuesevaluated at a single point. There is no need to remember a formula containing the coefficientsof the original cubic equation which can be complex for a general cubic. From educationalperspective, this paper also provides an alternative mechanism for understanding and solvingcubic equations using function evaluation. Such mechanism can also be extended to higherdegree polynomial equations such as four degree polynomial and solvable quintics. The solutionto quadratic equation using functional evaluation is, of course, trivially simple and likewiseapplicable.
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