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The first question is whether or not conservative systems have attractors
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And the answer is no, conservative systems do not have attractors
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However, even though conservative systems do not have attractors, they can be chaotic.
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For example, the motion of Pluto is chaotic, but our solar system, on human time scales, is a conservative system
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So conservative systems can be chaotic
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Question 3 asks us which one of these would let us know that were at a fixed point of a dynamical system
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The last option, If the system is at that state and you perturb it, that perturbation will shrink, is a characterization of stability
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While stability conditions are often associated with fixed points, that is not the definition of a fixed point, so this condition is not sufficient to let us know that were at a fixed point of the dynamical system
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The trajectory starting there is also not good enough; the trajectory must stay there
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The definition of a fixed point is that the dynamics of f take the fixed point x* back to x*
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That is, that once a trajectory has reached that state, a trajectory will never leave that state
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So, the answer to question 3 is that trajectories never leave that state is a sufficient condition to tell us that were at a fixed point of a dynamical system
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The next question asks which of these characterizes a stable fixed point
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We know from the previous question that the second answer characterizes a fixed point, but it doesnt talk about stability
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We also know that the first answer wasnt even enough to characterize a fixed point, let alone a stable fixed point
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The first part of the last option characterizes stability; that is, if a system is in the state and you slightly perturb it, that perturbation does not grow
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That characterizes the stability portion
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And trajectories never leave that state without external influence characterizes the fixed point portion
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For these reasons, this last option is correct
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In the last problem, we need to characterize each of these points of the damped pendulum as stable, unstable, or not a fixed point
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Part a corresponds to this pendulum, where theta is equal to zero and its not moving
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As we saw in the lecture, this is a stable fixed point
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Part b is a little bit trickier
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So theta equals zero, so it does correspond to same pendulum; however, omega, which is the velocity, is pi
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So while the pendulum is in this bottom position, it is moving at a velocity of pi
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Because of its angular velocity, even though the pendulum is in this bottom position, at the next time step the pendulum will continue to move
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And so f(x*) will not be x*
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So, this is not a fixed point
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In this next problem, the pendulum is now in this state
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As you saw in the lecture, this was a fixed point the pendulum will balance here but its very unstable; even blowing on the pendulum will cause it to fall
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For parts d and e, all we need to know is all even multiples of pi with no angular velocity are stable fixed points
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And all odd multiples of pi with no angular velocity are unstable fixed points