New PDF release: Instructor's Solutions Manual to An Introduction to

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14 does not hold, since it requires that pk (n) = 0 for all n ≥ n0 . 13. 6, one may choose two solutions x1 (n), x2 (n) such that x1 (0) = −1, x1 (1) = 0 and x2 (0) = 0, x2 (1) = 1. −1 0 Then C(0) = = −1. Therefore C(n) = −1. 0 1 u1 (n)∆u2 (n) − u2 (n)∆u1 (n) u2 (n) = . u1 (n) u1 (n)u1 (n + 1) If we use the hint, we obtain 14. 18) follows directly. 15. 10) to evaluate W (n). 2n ⇒ u1 (n) = n! (a) 2n+2 (n + 3) 2n+1 2 2n − · + · (n + 2)! (n + 2) (n + 1) n + 2 n! 2n+1 n+3 2 = − +1 =0 (n + 2)n! 10) n−1 C(n) = i=0 2 i+2 C(0) = 2n .

28 Dynamics of First Order Difference Equations x∗1 = 0, x∗2 = 1, x∗3 , x∗4 are fixed points of f . f (x) = −2x |f (x∗1 )f (x∗2 )| = 0 < 1, {0, 1} is asymptotically stable. 6. The 2-cycle can be found from x∗ = 5 − 6 . 5 − x6∗ So we get x∗ = 2, x∗ = 3. Since these are equilibrium points of the original equation, we conclude that this equation does not possess a 2-cycle. 7. f (x) = 3ax∗ − b |f (0)f (1)| = |(−b)(3a − b)| = |b2 − 3ab| < 1 8. The following figure shows Baker’s function. 1 1/2 1 Baker’s Map Let N be the number of fixed points of B n (x).

Therefore rn+2 cos(n + 2)θ + p1 rn+1 cos(n + 1)θ + p2 rn cos nθ + i[rn+2 sin(n + 2)θ + p1 rn+1 sin(n + 1)θ + p2 rn sin nθ] = 0. Thus rn+2 cos(n + 2)θ + p1 rn+1 cos(n + 1)θ + p2 rn cos nθ = 0 rn+2 sin(n + 2)θ + p1 rn+1 sin(n + 1)θ + p2 rn sin nθ = 0. 10. (a) The left side of the difference equation can be written as follows: π I= 0 H(θ, ϕ) dθ, where cos θ − cos ϕ H(θ, ϕ) = cos(n + 2)θ − cos(n + 2)ϕ − 2 cos ϕ[cos(n + 1)θ − cos(n + 1)ϕ] + cos nθ − cos nϕ. Using the identity cos(n + 2)u + cos nu = 2 cos(n + 1)u cos u the function H(θ, ϕ) can be written as follows H(θ, ϕ) = 2 cos(n + 1)θ cos θ − 2 cos(n + 1)ϕ cos ϕ− − 2cosϕ[cos(n + 1)θ − cos(n + 1)ϕ] = 2 cos(n + 1)θ[cos θ − cos ϕ].

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