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THE QURAN ON THE CEREBRUM

August 26, 2017

Méthode de Sotta : f(x) = a x³ + b x² + c x + d = 0 3 racines cubiques de l'unité, solutions de z³ - 1 = 0 j[k] = exp(2kiπ/3) avec i² = -1 et k = {0; 1; 2} j[0] = 1 j[1] = -0,5 + 0,86602540378 i = (- 1 + i √3) / 2 j[2] = -0,5 - 0,86602540378 i = (- 1 - i √3) / 2 A = b² - 3ac B = bc - 9ad C = c² - 3bd SI (A == 0) x[k] = - b / 3a + j[k] ∛(b / 3a)³ - d / a SINON { Δ_{_R} = B² - 4AC SI (Δ_{_R} == 0) { x[0] = x[1] = - B / 2A x[2] = - 2 x[0] - b / a } SINON { α = (- B + √Δ_{_R}) / 2A β = (- B - √Δ_{_R}) / 2A γ = ∛(b + 3αa) / (b + 3βa) x[k] = (α - βγj[k]) / (1 - γj[k]) } } Cas général avec A ≠ 0 et Δ_{_R} ≠ 0 α et β les 2 solutions de l'équation AX² + BX + C = 0 b + 3αa γ³ = ——————— b + 3βa α - βγj[k] x[k] = —————————— 1 - γj[k]

Méthode trigonométrique : g(z) = z³ + pz + q = 0 avec p ≠ 0 on pose z = ρ cos(θ) donc g(z) = ρ³ cos³(θ) + pρ cos(θ) + q = cos³(θ) + p/ρ² cos(θ) + q/ρ³ = cos(3θ)/4 + 3 cos(θ)/4 + p/ρ² cos(θ) + q/ρ³ = 0 = cos(3θ) + 3 cos(θ) + 4p/ρ² cos(θ) + 4q/ρ³ = cos(3θ) + 4q/ρ³ + (3 + 4p/ρ²) cos(θ) = 0 On pose 3 + 4p/ρ² = 0 et cos(3θ) + 4q/ρ³ = 0 si |4q/ρ³| ≤ 1 et ρ ≠ 0 3ρ² + 4p = 0 et cos(3θ) = - 4q/ρ³ donc ρ² = -4p/3 et ρ = 2√-p/3 et cos(3θ) = (3q/2p)√-3/p donc 3θ = acos((3q/2p)√-3/p) + 2kπ et θ = acos((3q/2p)√-3/p)/3 + 2kπ/3 et donc z = 2√-p/3 cos(θ) = 2√-p/3 cos(acos((3q/2p)√-3/p)/3 + 2kπ/3) et x = 2√-p/3 cos(acos((3q/2p)√-3/p)/3 + 2kπ/3) - b/3a

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You can use the overflow property when you want to have better control of the layout. The overflow property specifies what happens if content overflows an element's box.

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