Accurate algebraic formula for the quintic & Solution by iteration of radicals
aa r X i v : . [ m a t h . G M ] J a n Accurate algebraic formula for the quintic&Solution by iteration of radicals
Abdel Missa, Chrif YoussfiJanuary 15, 2021 ontents x + x + . = . . . . . . . . . . . . . . . . . . 327.1.1 Trigonometric algorithm . . . . . . . . . . . . . . . . . 327.1.2 Iteration of radicals algorithm . . . . . . . . . . . . . . 337.2 Example 2: x + x + ( . + . ) = . . . . . . . . . . . . 337.2.1 Trigonometric algorithm . . . . . . . . . . . . . . . . . 337.2.2 Iteration of radicals algorithm . . . . . . . . . . . . . . 34 bstract According to the Abel-Ruffini theorem [1] and Galois theory [2], there is nosolution in finite radicals to the general quintic equation. This article takesa different approach and proposes a new method to solve the quintic by it-eration of radicals. But, the most intriguing result is an accurate algebraicformula for absolute and relative root approximation: | formula- root | < . × − and | formula/root -1 | < . × − . We thenexpand some of the geometric properties discussed to construct a trigono-metric algorithm that derives all roots. Solving polynomial equations occupies a special place in the history of math-ematics. The first algebraic solution to the quadratic equation is attributedto Al-Khwarizmi [3] in the 9th century. In the 16th century, a solution to thecubic was found by Del Ferro, Tartaglia and Cardano [4] and was arguablythe first significant discovery of the European Renaissance. Soon thereafter,Ferrari created a method to solve the general quartic equation [5]. Thesuccess achieved in finding formulas to lower degree equations stimulated asearch to achieve the same ambition for the quintic. Despite efforts by Euler,Gauss, Lagrange and others, no general solutions were found. Finally, al-most 300 years after solving the cubic, Abel provided a complete proof of theimpossibility of solving general quintic polynomials by finite combination ofarithmetic operations and radicals. Subsequently, Galois went a step furtherand provided an exact criterion that characterizes solvable equations: theirGalois group must be solvable.Naturally, the quintic equation required more tools. The first significant de-velopment was the discovery of the Bring-Jerrard normal form x + ux + v = 0by using respectively quadratic and quartic Tschirnhaus tranformations [6].In 1858, Hermite [7] used a simplified normal form to provide the first knownsolution to general quintic equations in terms of elliptic modular functions.Shortly afterwards, Brioschi [8] and Kronecker [9] published equivalent solu-tions. In 1860, Cockle [10] and Harley [11] developed a method for solvingthe quintic using differential equations, leading to a solution in hypergeo-metric functions. And in 1884, Klein [12] provided an icosahedral solution of3he quintic. More recently, Doyle and McMullen [13] solved the quintic usingan iterative method, while Glasser [14] provided a derivation to trinomialequations that leads to the same hypergeometric solution for the quintic asCockle and Harley.In this article, we explore a new method that solves the quintic by itera-tion of radicals. Perhaps the most astonishing finding is the ability to findan accurate global approximation in radicals to one the five roots from thefirst iteration: | | < .
32 10 − . We then proceed by provingthe speed of convergence of the proposed iterative algorithm.Next, we discuss Bring radicals. Recall that the Bring radical of a com-plex number a is any of the five roots of: x + x + a = 0. More importantly,quintic equations can be solved in closed form using radicals and Bring rad-icals. One might think that a simple iterative algorithm x k +1 = − √ a + x k can solve the Bring equation. However, this is not the case as we show acounter example (when a is real) where the algorithm does not converge re-gardless of the starting point’s choice outside of the real root itself. On otherhand, we prove that our proposed iterative algorithm converges globally forany complex number a . Furthermore, we provide a radical approximation toone of the Bring radicals: | BR ( z )–approximation | < .
90 10 − .Finally, we expand some of the properties used in the proposed algorithm toprovide the location of the five roots in the complex plane. This leads to atrigonometric bisection algorithm that solves general quintic equations. The general quintic equation can be transformed to the Bring–Jerrard normalform: v + d v + d = 0From hereon, we define the n-th root of a complex number as the one withits argument in [ − π/n, π/n [.Excluding the trivial cases of d = 0 and d = 0, a change of variable4 = v/ √ d can further simplify the equation to Form 1: x + x + a = 0 where a = d p d ( a = 0) (1)Another change of variable z = a/x leads to Form 2: z + z λ where λ = − a y = uz , (2) becomes: y + uy u λ Choose u to satisfy: u = | λ | λ or u = (cid:18) | λ | λ (cid:19) = e iθ θ ∈ [ − π , π . This leads to Form 3: y + uy ξ ξ = | λ | ∈ R + ∗ (3)Think of Form 3 as a simple rotation of Form 2 with the benefit of solvingthe equation close to the real axis.Notice that s is solution of (3) if and only if its conjugate ¯ s is a solutionof y + ¯ uy ξ Therefore we can assume from hereon that 0 ≤ θ ≤ π/ Let us consider Equation (3) y + uy ξ ξ ∈ R + ∗ (3)5 = e iθ θ ∈ h , π i It can be exploited in two ways. First, by factorizing: y = s ξu + y (4)Second, by completing the quintic: (cid:16) y + u (cid:17) = 2 ξ + 2 u y + 2 u y + u y + u y in the right-hand side of (5): y = G ( ξ, y ) (6)with: G ( ξ, y ) = r ξ + 2 u t + 2 u t + u t + u − u t = s ξu + y (8)This inspires the following fixed point algorithm with the following sequence:( y k ) k ∈ N by:Starting point y = (cid:18) ξα (cid:19) with α = s √
24 (9)Iterative process y k +1 = G ( ξ, y k ) (10)To prove the convergence and the accuracy of this algorithm, we proceedwith the following steps: • Section 3.2 shows that there exists a unique root y ∗ for Equation (3)near the positive real axis. The intuition behind α is discussed below. Section 3.3 shows that ∀ ξ ∈ R + ∗ and ∀ θ ∈ [0 , π/ | y − y ∗ | < .
32 10 − In other words, the closed form formula: y = r ξ + 2 u t + 2 u t + u t + u − u t = vuut ξu + (cid:0) ξα (cid:1) provides an accurate absolute approximation of y ∗ • Section 3.4 shows that the relative error of the formula is also small: ∀ ξ ∈ R + and ∀ θ ∈ [0 , π/ (cid:12)(cid:12)(cid:12)(cid:12) y y ∗ − (cid:12)(cid:12)(cid:12)(cid:12) < .
58 10 − • Finally, section 3.5 shows the sequence ( y k ) k ∈ N converges toward y ∗ and: ∀ k ∈ N | y k +1 − y ∗ | < K | y k − y ∗ | K ≈ . Consider equation (3). When θ = 0, u = 1 and it is obvious that there is aunique real solution that satisfies (3). Now assume θ >
0. We will show inthis section that there exists a unique root y of Equation (3) near the realaxis: y = re iσ with r > σ ∈ (cid:20) − θ , (cid:20) (11)The real and the imaginary parts of the equation are: r cos (5 σ ) + r cos ( θ + 4 σ ) = 2 ξ (12)and r sin (5 σ ) + r sin ( θ + 4 σ ) = 0 (13)7rom (13): r = − sin ( θ + 4 σ )sin (5 σ ) (14)Substituting r in (12): f ( σ ) = sin ( θ + 4 σ ) sin ( σ − θ )sin (5 σ ) = 2 ξ (15)Let’s write: f ( σ ) = g ( σ ) g ( σ )Where g ( σ ) = sin ( θ + 4 σ )sin (5 σ ) and g ( σ ) = sin ( σ − θ )sin (5 σ ) g is positive and stricly increasing in [ − θ , g is positive andstricly increasing since: g ′ ( σ ) = sin(4 σ + θ ) + 4 sin( θ − σ ) cos(5 σ )sin (5 σ ) > f is continuous and strictly increasing. In addition: f (cid:18) − θ (cid:19) = 0 and lim σ → − f ( σ ) = + ∞ (16)Consequently, there exists a unique σ ∈ [ − θ ,
0[ satisfying (15) or a uniqueroot y ∗ such as: − θ ≤ Arg ( y ∗ ) < In this section, we will prove the absolute error theorem: ∀ ξ ∈ R + ∗ and ∀ θ ∈ [0 , π/ | y − y ∗ | < .
32 10 − We will start by discussing the unique properties of the selected startingpoint. 8 .3.1 Starting point properties • P.1 | y ∗ | ≤ | y | (18) • P.2 | u + y ∗ | ≤ | u + y | (19) • P.3 Define h = r u + y u + y ∗ | Arg ( h ) | ≤ π
20 (20)
Proof: • P.1: | y ∗ | ≤ | y | Using (3) | y ∗ | | u + y ∗ | = 4 ξ Therefore | y ∗ | (cid:12)(cid:12) | y ∗ | + 2 | y ∗ | cos( θ − σ ) (cid:12)(cid:12) = 4 ξ Since 1 + | y ∗ | ≥ | y ∗ | : | y ∗ | cos (cid:18) θ − σ (cid:19) ≤ ξ or | y ∗ | ≤ ξ cos (cid:0) θ − σ (cid:1) ! Also: 0 ≤ θ ≤ π − θ ≤ σ ≤ ≤ θ − σ ≤ π (cid:12)(cid:12)(cid:12)(cid:12) cos (cid:18) θ − σ (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) ≥ cos (cid:16) π (cid:17) = s √
24 = α It follows that: | y ∗ | ≤ (cid:18) ξα (cid:19) = | y | P.2: | u + y ∗ | ≤ | u + y | Since | u | = 1 and the angle between u and y ∗ is θ − σ : | u + y ∗ | = 1 + | y ∗ | + 2 | y ∗ | cos ( θ − σ )Likewise, since the angle between u and y is θ : | u + y | = 1 + | y | + 2 | y | cos ( θ )From P.1: | y ∗ | ≤ | y | . Also 0 ≤ cos ( θ ) ≤ cos ( θ − σ ). Therefore: | u + y ∗ | ≤ | u + y | • P.3: h = r u + y u + y ∗ | Arg ( h ) | ≤ π Notice u + y = ( y + cos( θ )) + i sin( θ ). Since y > ≤ sin( θ ) y + cos( θ ) ≤ sin( θ )cos( θ )Therefore 0 ≤ Arg ( u + y ) ≤ θ ( P. . y ∗ ( u + y ∗ ) = 2 ξ ∈ R + and Arg ( y ∗ ) = σ : Arg ( u + y ∗ ) = − σ Since: − θ ≤ σ ≤ ≤ Arg ( u + y ∗ ) ≤ θ ( P. . − θ ≤ Arg ( u + y ) − Arg ( u + y ∗ ) ≤ θ (cid:12)(cid:12)(cid:12)(cid:12) Arg (cid:18) u + y u + y ∗ (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) ≤ θ ≤ π Arg ( h ) = 14 Arg (cid:18) u + y u + y ∗ (cid:19) | Arg ( h ) | ≤ π δ = | y − y ∗ | < C where C = .
32 10 − (21)For k ∈ N define: t k = s ξu + y k , h k = y ∗ t k = r u + y k u + y ∗ , δ k = | y k − y ∗ | (22)where ( y k ) k ∈ N is defined in (9) and (10) (section 3.1).Also for t ∈ C define: P ( t ) = 2 t + 2 u t + u t (23)Recall that y ∗ = G ( ξ, y ∗ ): y ∗ = r ξ + 2 u y ∗ + 2 u y ∗ + u y ∗ + u − u y ∗ + u r ξ + u u P ( y ∗ )Or: 2 ξ + u (cid:16) y ∗ + u (cid:17) − u P ( y ∗ ) (24)Likewise y = G ( ξ, y ) leads to: y + u r ξ + u u P ( t ) = r(cid:16) y ∗ + u (cid:17) + u P ( t ) − P ( y ∗ ))11herefore: y + u (cid:16) y ∗ + u (cid:17) √ ǫ where ǫ = u P ( t ) − P ( y ∗ ) (cid:0) y ∗ + u (cid:1) (25)In the following two lemmas we will show that | ǫ | < . Lemma 1: Upper bound for | ǫ | Define: ω = α (cid:18) y ∗ u (cid:19) + (cid:18) uy ∗ (cid:19) ! r = | y ∗ | d = . = α B = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) u ω − ω (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) C = + r + r + r (cid:12)(cid:12) y ∗ + u5 (cid:12)(cid:12) ǫ satisfies: | ǫ | < ABCProof: P ( t ) − P ( y ∗ ) = ( t − y ∗ ) (cid:18) t + 2 t y ∗ + 2 y ∗ + 25 ( t + y ∗ ) + u (cid:19) (26)From P.2, | u + y ∗ | ≤ | u + y | therefore: | h | = (cid:12)(cid:12)(cid:12)(cid:12) y ∗ t (cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12) r u + y u + y ∗ (cid:12)(cid:12)(cid:12)(cid:12) ≥ | t | ≤ r = | y ∗ | This combined with (26) leads to: | P ( t ) − P ( y ∗ ) | ≤ | t − y ∗ | (cid:18) r + 45 r + 125 (cid:19) (27)On the other hand: t − y ∗ = t − y ∗ t + t y ∗ + t y ∗ + y ∗ (28)12ince t = 2 ξu + y and y ∗ = 2 ξu + y ∗ t − y ∗ = 2 ξu + y ∗ y ∗ − y u + y = y ∗ y ∗ − y u + y It follows that: t − y ∗ = y ∗ y ∗ − y ( u + y )( t + t y ∗ + t y ∗ + y ∗ )Recall from (22): t = y ∗ h and u + y = h ( u + y ∗ )Therefore: t − y ∗ = y ∗ ( y ∗ − y )( u + y ∗ ) (cid:0) h + h + h + h (cid:1) (29)Since | h | ≥ | Arg ( h ) | ≤ π/
20 and the modulus of a complex number isalways greater than its real part : | h || h + h + h | ≥ X k =0 cos (cid:16) k π (cid:17) ≥ d = 3 . | t − y ∗ | ≤ | y ∗ | | y ∗ − y | d | u + y ∗ | = (cid:12)(cid:12)(cid:12)(cid:12) y y ∗ − (cid:12)(cid:12)(cid:12)(cid:12) | y ∗ | d | u + y ∗ | Therefore | t − y ∗ | ≤ (cid:12)(cid:12)(cid:12)(cid:12) y y ∗ − (cid:12)(cid:12)(cid:12)(cid:12) r d | u + y ∗ | (31)Let’s now turn our attention to the denominator of ǫ in equation (25). Sincecos( u, y ∗ ) ≥ cos( π/ (cid:12)(cid:12)(cid:12) y ∗ + u (cid:12)(cid:12)(cid:12) ≥ s r + 125 + r √
25 (32) This is a good approximation due to the benefit of working near the real axis. | ǫ | ≤ d | y y ∗ − || u + y ∗ | r + r + r (cid:12)(cid:12) y ∗ + u (cid:12)(cid:12) (33)Going back to equation (3): y ∗ ( u + y ∗ ) = 2 ξ = 2 αy By factorizing: u y ∗ (cid:18) uy ∗ (cid:19) + (cid:18) y ∗ u (cid:19) ! = 2 αy Since ω = 12 α (cid:18) y ∗ u (cid:19) + (cid:18) uy ∗ (cid:19) ! αy = u y ∗ (cid:18) y ∗ u (cid:19) + (cid:18) uy ∗ (cid:19) ! = 2 αu y ∗ ω Therefore (cid:18) y y ∗ (cid:19) = u ω Since | Arg (cid:16) y y ∗ (cid:17) | ≤ π y y ∗ = u ω (34)From (33) and (34): | ǫ | ≤ d (cid:12)(cid:12)(cid:12) u ω − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (1 + y ∗ u ) (cid:12)(cid:12) r + r + r (cid:12)(cid:12) y ∗ + u (cid:12)(cid:12) Multiplying the numerator and the denominator by | u/y ∗ || ǫ | ≤ d (cid:12)(cid:12)(cid:12) u ω − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (1 + y ∗ u )(1 + uy ∗ ) (cid:12)(cid:12)(cid:12) r + r + r (cid:12)(cid:12) y ∗ + u (cid:12)(cid:12) | uy ∗ | (35)14ince ω = 14 α (cid:18) y ∗ u (cid:19) (cid:18) uy ∗ (cid:19) | ǫ | ≤ d α (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) u ω − ω (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) r + r + r (cid:12)(cid:12) y ∗ + u (cid:12)(cid:12) | uy ∗ | (36)Recall that | uy ∗ | ≤ (1 + 1 /r ), then: (cid:18) r + 45 r + 125 r (cid:19) (cid:12)(cid:12)(cid:12)(cid:12) uy ∗ (cid:12)(cid:12)(cid:12)(cid:12) ≤ r + 345 r + 2125 r + 125 r This combined with (36) leads to: | ǫ | ≤ ABC (37)With A = 120 d α B = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) u ω − ω (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) C = 6 r + r + r + r (cid:12)(cid:12) y ∗ + u (cid:12)(cid:12) Lemma 2: Upper bounds of A, B, C and ǫ • P.4 A ≤ . • P.5 B ≤ . • P.6 C ≤ . • P.7 ǫ ≤ . roof: • P.4 A ≤ . Recall: A = 120 d α Since: d = 3 . α = s √ A ≤ . (38) • P.5 B ≤ . Define v = u ω = y /y ∗ Using P.1 | v | ≥
1. Also since y ∈ R + β = arg ( v ) = − arg ( y ∗ ) = − σ Therefore 0 ≤ β ≤ θ/ ≤ π . B = (cid:12)(cid:12)(cid:12)(cid:12) v − v (cid:12)(cid:12)(cid:12)(cid:12) = s | v | + 1 − | v | cos( β ) | v | (39)This expression can be rewritten as: B = s m ( m + cos( β )) + sin ( β ) | v | with m = | v | − cos( β ) ≥ B ≤ q ( m + cos( β ) m − ) − + sin ( β ) (41)16he maximum of the right-hand side is reached when the derivativewith respect to m is zero at m = cos( β ) /
8. This leads to: B ≤ s (cid:18)
89 cos( β ) (cid:19) + sin ( β ) (42)As a result: B ≤ . (43) • P.6 C ≤ . Recall that: C = 6 r + r + r + r (cid:12)(cid:12) y ∗ + u (cid:12)(cid:12) Using (32) C ≤ r + r + r + r (cid:16) r + + r √ (cid:17) Therefore: C ≤ I + 345 I + 2125 I + 125 I (44)Where I k = r k (cid:16) r + + r √ (cid:17) k = 1 , , , I k = r − k + 125 r − k + r − k √ ! − By deriving with respect to r , the upper bound for I k is reached at: r k = (2 k − √ √ − k + 40 k − k Which leads to: C ≤ . (45)17 P.7 ǫ ≤ . Using (37), (38), (43) and (45): | ǫ | ≤ . (46) Lemma 3: Upper bound of | y − y ∗ | Define: d = . ′ = + r + r + r (cid:12)(cid:12) y ∗ + u5 (cid:12)(cid:12) • P.8 | y − y ∗ | ≤ A . B . C ′ • P.9 C ′ ≤ . • P.10 | y − y ∗ | ≤ . • P.8 | y − y ∗ | ≤ A . B . C ′ From (25): | y − y ∗ | = (cid:12)(cid:12)(cid:12)(cid:16) y ∗ + u (cid:17) ( √ ǫ − (cid:12)(cid:12)(cid:12) (47)Using the maximum of the modulus of the derivative: | ( √ ǫ − | ≤
15 (1 − | ǫ | ) | ǫ | Since | ǫ | ≤ . − | ǫ | ) ≤ d where d = 4 . | ( √ ǫ − | ≤ d | ǫ | (48)and | y − y ∗ | ≤ d (cid:12)(cid:12)(cid:12) y ∗ + u (cid:12)(cid:12)(cid:12) | ǫ | (49)Recall from Lemma 1 that | ǫ | ≤ ABC , therefore: | y − y ∗ | ≤ d A.B.C ′ (50)With C ′ = (cid:12)(cid:12)(cid:12) y ∗ + u (cid:12)(cid:12)(cid:12) C = 6 r + r + r + r (cid:12)(cid:12) y ∗ + u (cid:12)(cid:12) (51) • P.9 C ′ ≤ . Using (32) C ′ ≤ r + r + r + r (cid:16) r + + r √ (cid:17) ≤ a J + a J + a J + a J Where a = 34 − √ a = − a = 5 − √ a = − J k = r k (cid:16) r + + r √ (cid:17) = r + 125 r − + r − √ ! − Since a , a and a are negative C ′ ≤ a J Using the derivative, J k ’s maximum is reached at: r k = ( k − √ √ k − k − k Therefore when k = 3: J ≤ . C ′ ≤ . (52)19 P.10 | y − y ∗ | ≤ . From (37), (38), (43) and (52) | y − y ∗ | ≤ . (53) Theorem 2: (cid:12)(cid:12)(cid:12)(cid:12) y y ∗ − (cid:12)(cid:12)(cid:12)(cid:12) ≤ C where C = .
51 10 − (54) Proof:
Using (49): (cid:12)(cid:12)(cid:12)(cid:12) y y ∗ − (cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12) y − y ∗ y ∗ (cid:12)(cid:12)(cid:12)(cid:12) ≤ d (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (cid:0) y ∗ + u (cid:1) y ∗ ǫ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (55)Since ǫ ≤ ABC : (cid:12)(cid:12)(cid:12)(cid:12) y y ∗ − (cid:12)(cid:12)(cid:12)(cid:12) ≤ d ABC ′′ (56)Where C ′′ = C (cid:12)(cid:12) y ∗ + u (cid:12)(cid:12) r = 6 r + r + r + (cid:12)(cid:12) y ∗ + u (cid:12)(cid:12) (57)Using (32) C ′′ ≤ r + r + r + (cid:16) r + + r √ (cid:17) = 6 J + 295 J + 21 − √ J + 1 (cid:16) r + + r √ (cid:17) Since 1 r + + r √ ≤
25 (decreasing function with maximum at 0)and J ≤ . J ≤ . J ≤ . C ′′ ≤ . (cid:12)(cid:12)(cid:12)(cid:12) y y ∗ − (cid:12)(cid:12)(cid:12)(cid:12) ≤ . (59)20 .5 Speed of convergence Theorem 3: ∀ k ≥ | y k + − y ∗ | < | y k − y ∗ | and | y k + − y ∗ | < C K k (60) with K = . . Proof:
Recall that r = | y ∗ | and from (22): h k = y ∗ t k = r u + y k u + y ∗ and t k = s ξu + y k We will breakdown the proof into three lemmas:
Lemma 4:
For k ∈ N ∗ if | y k − y ∗ | ≤ C then for p = 1 , , , − q ≤ | h kp | ≤ q | t kp | ≤ (1 + q ) r p with q = 0 . Proof:
Since cos( u, y ∗ ) ≥ | u + y ∗ | ≥ | u | = 1 and (cid:12)(cid:12)(cid:12)(cid:12) y k − y ∗ u + y ∗ (cid:12)(cid:12)(cid:12)(cid:12) ≤ | y k − y ∗ | For k ≥ p = 1 , , ,
4: using the maximum of the modulus of thederivative for the function: x (1 + x ) p | h kp − | = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:18) y k − y ∗ u + y ∗ (cid:19) p − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ p | y k − y ∗ | (1 − | y k − y ∗ | ) − p ≤ | y k − y ∗ | − | y k − y ∗ | Which leads to: | h kp − | ≤ C − C ≤ q = 0 . − q ≤ | h kp | ≤ q (61)Also, since | t kp | = r p | h kp | | t kp | ≤ (1 + q ) r p In particular: (cid:12)(cid:12) h k + h k + h k + h k (cid:12)(cid:12) = (cid:12)(cid:12) h k −
1) + ( h k −
1) + ( h k −
1) + ( h k − (cid:12)(cid:12) Therefore (cid:12)(cid:12) h k + h k + h k + h k (cid:12)(cid:12) ≥ − | h k − | − (cid:12)(cid:12) h k − (cid:12)(cid:12) − (cid:12)(cid:12) h k − (cid:12)(cid:12) − (cid:12)(cid:12) h k − (cid:12)(cid:12) Using Lemma 4 (cid:12)(cid:12) h k + h k + h k + h k (cid:12)(cid:12) ≥ − q (62) Lemma 5: ∀ k ≥ | y k − y ∗ | ≤ C then | y k + − y ∗ | ≤ | y k − y ∗ | KProof:
Using exactly the same steps that led to (25): y k +1 − y ∗ = (cid:16) y ∗ + u (cid:17) (cid:16) p ǫ k +1 − (cid:17) (63)With: ǫ k +1 = u P ( t k ) − P ( y ∗ ) (cid:0) y ∗ + u (cid:1) Similarily, using the steps that led to (26) and (29), we obtain: ǫ k +1 = y ∗ u y ∗ − y k )(2 t k + 2 t k y ∗ + 2 y ∗ + ( t k + y ∗ ) + u )( u + y ∗ ) (cid:0) h k + h k + h k + h k (cid:1) (cid:0) y ∗ + u (cid:1) (64)Using Lemma 4: | ǫ k +1 | ≤ q − q ) (cid:18) I + 45 I + 125 I (cid:19) C ≤ . q | p ǫ k +1 − | ≤ − q ) | ǫ k +1 | ≤ d | ǫ k +1 | (65)Where d = 4 . | y k +1 − y ∗ | ≤ d (cid:12)(cid:12)(cid:12)(cid:16) y ∗ + u (cid:17) ǫ k +1 (cid:12)(cid:12)(cid:12) Or | y k +1 − y ∗ | ≤ d (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) y ∗ u y ∗ − y k )(2 t k + 2 t k y ∗ + 2 y ∗ + ( t k + y ∗ ) + u )( u + y ∗ ) (cid:0) h k + h k + h k + h k (cid:1) (cid:0) y ∗ + u (cid:1) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) | u + y ∗ | ≥ | y k +1 − y ∗ | ≤ | y k +1 − y ∗ | q d (1 − q ) 6 r + r + r (cid:16) r + + r √ (cid:17) Therefore | y k +1 − y ∗ | ≤ | y k +1 − y ∗ | q − q ) d (cid:18) J + 45 J + 125 J (cid:19) Since 1 + q − q ) d (cid:18) J + 45 J + 125 J (cid:19) ≤ K where K = 15 . | y k +1 − y ∗ | ≤ | y k − y ∗ | K Lemma 6: ∀ k ≥ | y k +1 − y ∗ | < K | y k − y ∗ | and | y k +1 − y ∗ | < C K k with K = 15 .
44. 23 roof:
By induction, since | y − y ∗ | < C and K > k ≥ | y k − y ∗ | < C and : | y k +1 − y ∗ | ≤ | y k − y ∗ | K and | y k +1 − y ∗ | ≤ C K k Consider equations in Form 2: z + z λ Recall from section 2 that a rotation y = uz leads to Form 3: y + uy ξ with ξ = | λ | and u = (cid:18) | λ | λ (cid:19) Therefore the same algorithm would apply to Form 2:1. Starting point z = (cid:18) | λ | α (cid:19) (cid:18) λ | λ | (cid:19) with α = s √
24 (66)2. Iteration z k +1 = r λ + 25 m k + 225 m k + 1125 m k + 13125 −
15 (67)Where m k = r λ z k Since | u | = 1 and z k = y k /u for all k ∈ N , the properties discussed in section(3) would apply: 24. Absolute error theorem: | z − z ∗ | < C where C = 4 .
32 10 −
2. Relative error theorem: (cid:12)(cid:12)(cid:12) z z ∗ − (cid:12)(cid:12)(cid:12) < C where C = 2 .
51 10 −
3. Speed of convergence theorem: ∀ k ≥ | z k +1 − z ∗ | < K | z k − z ∗ | and | z k +1 − z ∗ | < C K k with K = 15 . Arg ( z ∗ ) = Arg ( y ∗ ) − θ , | Arg ( y ∗ ) | ≤ π/
20 and | θ | ≤ π/ | Arg ( z ∗ ) | ≤ π Consider the equation x + x + a = 0 a = 0 . x = 0 and x k +1 = − √ a + x Which produces the following sequence (approximated to the fifth digit): x = 0, x = − . x = 0 . x − . x = 0 . x = − . x = 0 . x = − . x = 0 . x = − . x = 0 . x = − . x = 0 . x = − . x = 0 . x lead to a different outcome? Outside the real-root itself, we will show belowthat it’s not the case. 25ndeed assume that the sequence converges toward a real number x ∗ . Since x ∗ ( x ∗ + 1) = − a , | x ∗ | ≤ a .Use ǫ = 0 . a . There exists an integer k such that for all k > k : | x k − x ∗ | < ǫ Note that: x ∗ = − √ a + x ∗ x k +1 = − √ a + x k Using the mean value theorem for the function − √ a + x between x ∗ and x k ,there exist a real number c between x ∗ and x k such that: | x k +1 − x ∗ | = 15 | a + c | − | x k − x ∗ | Since | a + c | ≤ | a | + | c | ≤ | a | + | x ∗ | + ǫ ≤ a + ǫ ≤ . a | x k +1 − x ∗ | ≥ . | x k − x ∗ | Which proves that the algorithm is not stable.
Consider equation (1): x + x + a = 0 with a = 0As discussed in section 2, a change of variable z = a/x leads to: z + z = − a As proven in section 4, the proposed algorithm provides an accurate firstestimate z for one of the five roots z ∗ : | z − z ∗ | < C and (cid:12)(cid:12)(cid:12) z z ∗ − (cid:12)(cid:12)(cid:12) < C Consider x ∗ = a/z ∗ and the sequence x k = a/z k for k ∈ N emma 7:
1. Absolute error: | x − x ∗ | < C C = 2 .
90 10 −
2. Relative error: (cid:12)(cid:12)(cid:12) x x ∗ − (cid:12)(cid:12)(cid:12) < C ′ C ′ = 2 .
57 10 −
3. Speed of convergence: ∀ k ≥ | x k +1 − x ∗ | < K ′ | x k − x ∗ | and | x k +1 − x ∗ | < C K ′ k with K ′ = 14 . Proof :1.
Absolute error: | x − x ∗ | < C C = .
90 10 − Notice: | x − x ∗ | = (cid:12)(cid:12)(cid:12)(cid:12) az z ∗ (cid:12)(cid:12)(cid:12)(cid:12) | z − z ∗ | . (68) z ∗ (1 + z ∗ ) = − a Since, φ the argument z ∗ is between [ − π/ , π/ | z ∗ | = 1 + | z ∗ | + 2 | z ∗ | cos( φ ) ≥ | z ∗ | ≥ | z ∗ | | z ∗ | = | a | | z ∗ | ≤ | a | Or r = | z ∗ | ≤ | a | . (69)On the other hand, equation (2) implies: | z ∗ | + | z ∗ | ≥ | a | . (70)We distinguish two cases: 27 | a | ≤
1: Using (69), r ≤
1. From (70): | a | ≤ | z ∗ | + | z ∗ | ≤ | z ∗ | therefore | a || z ∗ | ≤ Since | z /z ∗ − | < C (relative error theorem in section 4)11 + C ≤ | z ∗ /z | < − C (71) | a || z | ≤ | a || z ∗ | | z ∗ || z | ≤ − C It follows that: | x − x ∗ | = (cid:12)(cid:12)(cid:12)(cid:12) az z ∗ (cid:12)(cid:12)(cid:12)(cid:12) | z − z ∗ | = (cid:12)(cid:12)(cid:12)(cid:12) az (cid:12)(cid:12)(cid:12)(cid:12) (cid:12)(cid:12)(cid:12) z z ∗ − (cid:12)(cid:12)(cid:12) Therefore: | x − x ∗ | ≤ C − C ≤ .
90 10 − (72) • | a | >
1: Using (70), | z ∗ | > C = 0 .
856 (function | z ∗ | + | z ∗ | isincreasing in | z ∗ | and C + C < | a | ≤ | z ∗ | (cid:18) | z ∗ | (cid:19) ≤ | z ∗ | (cid:18) C (cid:19) Therefore | z ∗ | ≥ | a |
45 5 r C C (73)Using (71): | z | ≥ | a | (1 − C ) r C C (74)From (73) and (74): | a || z z ∗ | ≤ | a | − (1 − C ) − (cid:18) C C (cid:19) − < . | x − x ∗ | ≤ . ≤ .
90 10 − (75)28. Relative error: (cid:12)(cid:12)(cid:12) x x ∗ − (cid:12)(cid:12)(cid:12) < C ′ C ′ = .
57 10 − Since x ∗ = a/z ∗ and x = a/x : (cid:12)(cid:12)(cid:12) x x ∗ − (cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12) z ∗ − z z (cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12) z ∗ z (cid:12)(cid:12)(cid:12)(cid:12) (cid:12)(cid:12)(cid:12) z z ∗ − (cid:12)(cid:12)(cid:12) Using the relative error theorem for Form 2 and (71): (cid:12)(cid:12)(cid:12) z z ∗ − (cid:12)(cid:12)(cid:12) < C (cid:12)(cid:12)(cid:12)(cid:12) z ∗ − z z (cid:12)(cid:12)(cid:12)(cid:12) < − C Therefore (cid:12)(cid:12)(cid:12) x x ∗ − (cid:12)(cid:12)(cid:12) < C − C < C ′ = 2 .
57 10 − Speed of convergence: | x k + − x ∗ | < ′ | x k − x ∗ | with K ′ = . Since x = a/z : x k +1 − x ∗ = az k +1 − az ∗ = a ( z ∗ − z k +1 ) z k +1 z ∗ and x k − x ∗ = az k − az ∗ = a ( z ∗ − z k ) z k z ∗ Which leads to: (cid:12)(cid:12)(cid:12)(cid:12) x k +1 − x ∗ x k − x ∗ (cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12) z k z ∗ (cid:12)(cid:12)(cid:12) (cid:12)(cid:12)(cid:12)(cid:12) z ∗ z k +1 (cid:12)(cid:12)(cid:12)(cid:12) (cid:12)(cid:12)(cid:12)(cid:12) z k +1 − z ∗ z k − z ∗ (cid:12)(cid:12)(cid:12)(cid:12) (5 . (cid:12)(cid:12)(cid:12)(cid:12) z k +1 − z ∗ z k − z ∗ (cid:12)(cid:12)(cid:12)(cid:12) ≤ K Second, (cid:12)(cid:12)(cid:12) z k z ∗ (cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12) z k − z ∗ z ∗ (cid:12)(cid:12)(cid:12)(cid:12) ≤ | z k − z ∗ || z ∗ | ≤ | z − z ∗ || z ∗ | ≤ C (cid:12)(cid:12)(cid:12) z k z ∗ (cid:12)(cid:12)(cid:12) ≤ C (5 . (cid:12)(cid:12)(cid:12) z k +1 z ∗ (cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12) z k − z ∗ z ∗ (cid:12)(cid:12)(cid:12)(cid:12) ≥ − | z k − z ∗ || z ∗ | ≥ − | z − z ∗ || z ∗ | ≥ − C Therefore: (cid:12)(cid:12)(cid:12)(cid:12) z ∗ z k +1 (cid:12)(cid:12)(cid:12)(cid:12) ≤ − C (5 . (cid:12)(cid:12)(cid:12)(cid:12) x k +1 − x ∗ x k − x ∗ (cid:12)(cid:12)(cid:12)(cid:12) ≤ K C − C ≤ K ′ with K ′ = 14 . The logic used in section 3.2 can be expanded to establish the geometriclocation of the five roots in the complex plane.Case 1: 0 < θ < π First, recall that y ∗ = re iσ is a root of (3) if and only if two conditionsare met: f ( σ ) = sin ( θ + 4 σ ) sin ( σ − θ )sin (5 σ ) = 2 ξ (76) r = − sin ( θ + 4 σ )sin (5 σ ) > y ∗ (cid:18) uy ∗ (cid:19) = 2 ξ σ + Arg (cid:18) uy (cid:19) = 0 mod 2 π Let σ k be the argument of the root y ∗ k :5 σ k + Arg (cid:18) uy ∗ k (cid:19) = 2 kπ − ≤ k ≤ ξ tends to 0 + , either y ∗ tends to y ∗− = − u with the argument − π + θ or y ∗ tends to 0 with the argument for that tends to:2 kπ k = − , , , ξ tends + ∞ , | y ∗ k | tends to + ∞ with an argument σ k that tends to:2 kπ − θ − ≤ k ≤ I − = (cid:20) − π − θ , − π (cid:20) I = (cid:20) − θ , − (cid:20) I = (cid:20) π , π − θ (cid:20) I = (cid:20) π , π − θ (cid:20) and I − = (cid:20) − π + θ, − π (cid:20) f is continuous in each of these intervals. f is zero at the lower bound of I k for k = − , , I k for k = 1 , f also tends to+ ∞ on the upper bound of I k for k = − , , I k for k = 1 ,
2. Notice also that if σ ∈ I k then (77) is satisfied ( r > σ k in each of the fivedisjoint intervals. Since f is strictly monotonic in each interval, we can definea bisection method to approximate σ k and the associated modulus from (77),leading to the five roots of equation (3).Case 2: θ = 0 or θ = π The same algorithm described in the previous case would apply to iden-tify four roots, while the fifth can be obtained from Vieta’s formulas. When θ = 0, we use a bisection in the intervals I − , I − , I and I . For θ = π , weuse the bisection in the intervals I − , I , I and I . All numbers presented in this section are approximated to the 10th decimaldigit. 31 .1 Example 1: x + x + . = Using the transformation to Form (3) ξ = 0 . θ = π y ∗ k for k = − , , , x ∗ k using the transformation x ∗ k = auy ∗ k .The fifth root ( k = −
2) is obtained from Vieta’s formula. k y ∗ k x ∗ k -2 − . − . i − . − . − . i − . . i . − . i . . i . . i . − . i − . . i − . − . i ℜℑ x ∗− x ∗− x ∗ x ∗ x ∗ Figure 1: Roots of the equation x + x + 0 .
01 = 032 .1.2 Iteration of radicals algorithm
Using the algorithm described in section 3 to estimate one root (coincideswith k = 0 from trigonometric algorithm): y ∗ = 0 . − . i Iteration y n | y n − y ∗ | (cid:12)(cid:12)(cid:12) y n y ∗ − (cid:12)(cid:12)(cid:12) . − . i . × − . × − . − . i . × − . × − . − . i . × − . × − Since x = auy : x ∗ = 0 . . i and the iterative esti-mates are:Iteration x n | x n − x ∗ | (cid:12)(cid:12)(cid:12) x n x ∗ − (cid:12)(cid:12)(cid:12) . . i . × − . × − . . i . × − . × − . . i . × − . × − + x + ( . + . ) = Using the transformation to Form (3) ξ ≈ . θ ≈ . y ∗ k of equation3 and consequently to roots x ∗ k using the transformation x ∗ k = auy ∗ k . k y ∗ k x ∗ k -2 − . − . i − . − . i -1 0 . − . i − . . i . − . i . . i . . i . − . i − . . i − . − . i ℑ x ∗− x ∗− x ∗ x ∗ x ∗ Figure 2: Roots of the equation x + x + (3 .
08 + 1 . i ) = 0 Using the algorithm described in section 3 to estimate one root (coincideswith k = 0 from trigonometric algorithm): y ∗ = 2 . − . i Iteration y n | y n − y ∗ | (cid:12)(cid:12)(cid:12) y n y ∗ − (cid:12)(cid:12)(cid:12) . − . i . × − . × − . − . i . × − . × − . − . i . × − . × − Since x = auy , x ∗ = 1 . . i and the iterative esti-mates are:Iteration x n | x n − x ∗ | (cid:12)(cid:12)(cid:12) x n x ∗ − (cid:12)(cid:12)(cid:12) . . i . × − . × − . . i . × − . × − . . i . × − . × − Future direction of work
It would be interesting to study the feasibility of expanding the proposediteration of radicals algorithm to identify all roots by selecting appropriatestarting points and the right combination of the 5th and 4th unity roots .