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The first question is whether or not all fixed points live in the bottoms of bowls in the dynamical landscape
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This is absolutely not the case
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For example, we saw in lecture that fixed points can not only occur on the bottoms of bowls, but also on the top of bowls that are upside down, and at saddle points
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So this question is false
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Problem 2 asks us to identify which of the following fixed points of the undamped pendulum are saddle points in the dynamical landscape
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The simple way to solve this problem is to remember that even multiples of pi, like 0, 2 pi, 4 pi, etc. are elliptic fixed points, not saddle points
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So this one is not a saddle point; its elliptic
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And this one is not a saddle point; its also elliptic
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However, odd multiples of pi in the theta coordinate are saddle points
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So this is a saddle point; this is, and this is as well
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As the hint suggested, there is a slightly more sophisticated way of solving for this
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For this, well need to use a little bit of mathematics
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The solution Im about to present is a little bit outside the scope of this course, but for those students with a little bit more mathematical expertise, I thought this might be an interesting solution
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Recall that this is the system of differential equations that describes the undamped pendulum
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To analyze which fixed points are saddle points, we need to linearize this system about the fixed point
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That is, we need to calculate the Jacobian, evaluate it at a fixed point
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The next step is to find the eigenvalues of this matrix at a particular theta*, which we know to be a fixed point
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If we solve for the zeros of the characteristic polynomial, we see that the possible eigenvalues are plus or minus the square root of g/l*cos(theta*)
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We now simply need to evaluate this equation at the fixed points of this dynamical system
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Looking back at the dynamical system, we can see that these fixed points are omega equals 0, from the first equation, and any multiple of pi from the second equation
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This gives us two cases
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If theta* is an even multiple of pi, for example 0, we have the cosine of 0 (or any even multiple of pi) equals 1
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And we get that the associated eigenvalues of these fixed points are plus or minus i times the square root of a positive number
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So we have two imaginary eigenvalues, with zero real part, so the fixed points with theta that have even multiples of pi are elliptic fixed points
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If we instead choose theta to be an odd multiple of pi, for example pi, we see that cosine of pi is equal to -1, and we get the possible eigenvalues are plus or minus the square root of g divided by l
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Since g and l are both positive, we have one positive eigenvalue and one negative eigenvalue
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This means that the fixed points with odd multiples of pi for theta, and zero for omega, are hyperbolic fixed points
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Remember, this discussion is somewhat outside the scope of this course, and is intended for experts
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Now lets get back to the quiz
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For this next problem, we need to calculate the eigenvalues of this two by two matrix
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The first step in doing this is to subtract s off of the diagonal, and then take the determinant
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The result of this is known as the characteristic polynomial
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If we find the roots of the characteristic polynomial, this will give us our eigenvalues
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If we do this, we get that our eigenvalues are 2 and 5
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This is the answer to question 3
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For problem 4, all you need to know is that the shape of a matrix is defined as the row times the column
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Since theres 2 rows, and 1, 2, 3, 4, 5 columns, this is a 2x5 matrix
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For question 5, we just need to know how many eigenvalues a square matrix has
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A matrix A, which is 2x2, has 2 eigenvalues, one for each dimension of the matrix
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These two eigenvalues may be repeated that is, you may have an eigenvalue of multiplicity
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They may also be real and they may be complex
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If A is a matrix that describes the evolution of a dynamical system, then any point along an eigenvector must stay on that eigenvector
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This fact was stated in lecture, but heres an easy way to see that this is the case
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If the matrix A describes the action of the dynamical system, we can represent the action of the dynamical system on a point w as the multiplication of A times w
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If w lies along an eigenvector of this system A, then we can rewrite w as a constant times that eigenvector, where that constant is some real number
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But from the definition of eigenvalue and eigenvector, we know that if s is the eigenvalue associated with the eigenvector v, then we can rewrite the multiplication of A times v as s times v
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We can then combine the two constants s, the eigenvalue, and the constant k, into another constant beta, and we see that we get beta times v
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But, by similar logic, we know that beta times v is just another vector which points along the eigenvector v
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This shows that if any point is along an eigenvector of A, it must stay on that eigenvector
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If A captures the action of a dynamical system and the eigenvalues of A are negative, then we must have a fixed-point dynamic, as we are shrinking in both directions
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In general, an eigenvalue defines how much the action of a matrix on a vector stretches or shrinks that vector along the corresponding eigenvector
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In dynamical systems terminology, this corresponds to how fast a trajectory moves along a corresponding eigenvector
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For question 9, in the linear case, a matrix can absolutely give you an accurate representation of the evolution of a dynamical system
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So this question is true
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In fact, translating from a linear dynamical system to a matrix is simply notation
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Question 10, however, is slightly more complicated
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For a nonlinear dynamical system, a matrix can only give you an accurate local representation
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This word local is key here
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This idea goes back to everything is flat if you look closely enough
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Imagine a ball for example: if you zoom in and zoom in and zoom in to a ball, eventually youll see a flat surface
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That is, at this very local scale, representing a ball by a plane, or equivalently a matrix, may be a fairly accurate representation
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This is obviously an analogy, but hopefully one that will help you understand this
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Question 11, however, is false
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A matrix, or a linearization, cannot give you an accurate global representation of the evolution of a nonlinear dynamical system
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Going back to the ball analogy, this would be like trying to fit a plane to a ball
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Aside from the tangent point where the plane touched the ball, moving away from this point the plane representation, or the linear representation, would very quickly lose the global representation of the ball
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Hopefully this analogy gives you insight into why global representation using linear methods of a nonlinear dynamical system is dangerous, the same thing as representing a ball with a plane