For the given map equations x n+1 = x n cos α − ( y n − x n 2 ) sin α y n+1 = x n sin α + ( y n − x n 2 ) cos α To verify the map is area preserving To find inverse mapping. To explore the map using the computer for various values of α Concept Introduction: ➢ Jacobian matrix is used for checking the area behavior of the system J = ( ∂ f ∂ x ∂ f ∂ y ∂ g ∂ x ∂ g ∂ y ) If the determinant of the Jacobian matrix is 1, then the area is preserved; if the determinant is less than 1, then the area is contracted; if the determinant is greater than 1, then the area expands. ➢ Hénon map is given by x n + 1 = y n + 1 - ax n 2 y n + 1 = bx n where a,b are adjustable parameters. ➢ Properties of Hénon map: It is invertible. It is dissipative. It has a trapping region for certain parameter values. Some orbits of the Hénon map escape to infinity.

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Chapter 12.1, Problem 8E

Interpretation Introduction

Interpretation:

For the given map equations

$xn+1= xncosα−(yn−xn2)sinαyn+1= xnsinα+ (yn−xn2)cosα$

To verify the map is area preserving

To find inverse mapping.

To explore the map using the computer for various values of $α$

Concept Introduction:

➢ Jacobian matrix is used for checking the area behavior of the system

$J=(∂f∂x∂f∂y∂g∂x∂g∂y)$

If the determinant of the Jacobian matrix is 1, then the area is preserved; if the determinant is less than 1, then the area is contracted; if the determinant is greater than 1, then the area expands.

➢ Hénon map is given by

$xn + 1 = yn + 1 - axn2$

$yn + 1 = bxn$

where $a,b$ are adjustable parameters.

➢ Properties of Hénon map:

It is invertible.

It is dissipative.

It has a trapping region for certain parameter values.

Some orbits of the Hénon map escape to infinity.

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