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POLYNOME - D.S.G POLLOCK TOPICS IN TIME-SERIES ANALYSIS...

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D.S.G. POLLOCK: TOPICS IN TIME-SERIES ANALYSIS ALGEBRAIC POLYNOMIALS Consider the equation φ 0 + φ 1 z + φ 2 z 2 = 0. Once the equation has been divided by φ 2 , it can be factorised as ( z λ 1 )( z λ 2 ) where λ 1 , λ 2 are the roots or zeros of the equation which are given by the formula (1) λ = φ 1 ± φ 2 1 4 φ 2 φ 0 2 φ 2 . If φ 2 1 4 φ 2 φ 0 , then the roots λ 1 , λ 2 are real. If φ 2 1 = 4 φ 2 φ 0 , then λ 1 = λ 2 . If φ 2 1 < 4 φ 2 φ 0 , then the roots are the conjugate complex numbers λ = α + , λ = α , where i = 1. There are three alternative ways of representing the conjugate complex numbers λ and λ : (2) λ = α + = ρ (cos θ + i sin θ ) = ρe , λ = α = ρ (cos θ i sin θ ) = ρe , where (3) ρ = α 2 + β 2 and θ = tan 1 β α . These are called, respectively, the Cartesian form, the trigonometrical form and the exponential form. The Cartesian and trigonometrical representations are understood by con- sidering the Argand diagram: ρ α β θ −θ λ λ * Re Im Figure 1. The Argand Diagram showing a complex number λ = α + and its conjugate λ = α . 1
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ALGEBRAIC POLYNOMIALS The exponential form is understood by considering the following series expansions of cos θ and i sin θ about the point θ = 0: (4) cos θ = 1 θ 2 2! + θ 4 4! θ 6 6! + · · · , i sin θ = 3 3! + 5 5! 7 7! + · · · . Adding these gives (5) cos θ + i sin θ = 1 + θ 2 2! 3 3! + θ 4 4! + · · · = e . Likewise, by subtraction, we get (6) cos θ i sin θ = 1 θ 2 2! + 3 3! + θ 4 4! − · · · = e . These are Euler’s equations. It follows from adding (5) and (6) that (7) cos θ = e + e 2 . Subtracting (6) from (5) gives (8) sin θ = i 2 ( e e ) = 1 2 i ( e e ) .
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