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Economics Dynamics Problems 145

# Economics Dynamics Problems 145 - Discrete dynamic systems...

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Unformatted text preview: Discrete dynamic systems ∗ Take ﬁrst x1 = 1, then ∗ f (x1 ) = 1 ∗ ∗ f (x1 ) = 2(4x1 − 3)−1/2 = 2 Hence xt = 1 + 2(xt−1 − 1) = −1 + 2xt−1 ∗ which is unstable since f (x1 ) = 2 > 1. ∗ Next consider x2 = 3 ∗ f (x2 ) = 3 ∗ ∗ f (x2 ) = 2(4x2 − 3)−1/2 = 2 3 Hence xt = 3 + 2 (xt−1 − 3) 3 2 = 1 + xt−1 3 ∗ which is stable since f (x2 ) = 2/3 < 1. Example 3.19 yt+1 = f ( yt ) = 3.2yt − 0.8y2 t Letting yt = y∗ for all t we can readily establish two equilibria: y∗ = 0 and y∗ = 1 2 2.75. Considering the nonzero equilibrium, then f ( y∗ ) = 2.75 2 f ( y∗ ) = 3.2 − 1.6y∗ = −1.2 2 2 Hence, the linear approximation is yt+1 = 2.75 − 1.2( yt − 2.75) = 6.05 − 1.2yt The situation is shown in ﬁgure 3.20. The solution to this model is yt+1 = 2.75 + (−1.2)t ( y0 − 2.75) which is oscillatory and explosive. Although the linear approximation leads to an explosive oscillatory equilibrium, the system in its nonlinear form exhibits a two-cycle with values 2.0522 and 3.1978.12 What the linear approximation reveals is the movement away from y∗ = 2.75. What it cannot show is that it will converge on a two-cycle. This example, therefore, illustrates the care required in interpreting the stability of nonlinear difference equations using their linear approximations. 12 This can be established quite readily with a spreadsheet or as explained in appendices 3.1 and 3.2. 129 ...
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