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The population of Osprey Eagles at a certain lake is dying out. Each year the new population is only $\\simplify{100*{a}}$% of the previous year's population. However, a new species of Trout appears in the lake. Conservationists find that the populations satisfy the following system of difference equations, where $x_t$ is the number of Osprey in year $t$ and $y_t$ is the number of trout in year $t$:

$\\boldsymbol{\\mathrm{x}_{t+1}} = A \\boldsymbol{\\mathrm{x}_{t}}$ where $\\boldsymbol{\\mathrm{x}_{t}} = \\begin{pmatrix} x_t \\\\ y_t \\end{pmatrix}$ , $A = \\var[fractionNumbers]{{M}}$ , and $\\boldsymbol{\\mathrm{x}_{0}} = \\var{{v}}$.

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Which of the following describes how the populations depend on the previous year's populations?

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[]

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The Osprey population is $\\var[fractionNumbers]{{a}}$ of previous year's Osprey population, plus $\\var[fractionNumbers]{{b}}$ of the previous year's Trout population; and the Trout population is $\\var[fractionNumbers]{{d}}$ of previous year's Trout population, minus $\\simplify[fractionNumbers]{-{c}}$ of the previous year's Osprey population.

", "

The Osprey population is $\\var[fractionNumbers]{{d}}$ of previous year's Trout population, minus $\\simplify[fractionNumbers]{-{c}}$ of the previous year's Osprey population; and the Trout population is $\\var[fractionNumbers]{{a}}$ of previous year's Osprey population, plus $\\var[fractionNumbers]{{b}}$ of the previous year's Trout population.

", "

The Osprey population is $\\var[fractionNumbers]{{a}}$ of previous year's Osprey population, minus $\\simplify[fractionNumbers]{-{c}}$ of the previous year's Trout population; and the Trout population is $\\var[fractionNumbers]{{b}}$ of previous year's Osprey population, plus $\\var[fractionNumbers]{{d}}$ of the previous year's Trout population.

", "

The Osprey population is $\\var[fractionNumbers]{{b}}$ of previous year's Osprey population, plus $\\var[fractionNumbers]{{d}}$ of the previous year's Trout population; and the Trout population is $\\var[fractionNumbers]{{a}}$ of previous year's Osprey population, minus $\\simplify[fractionNumbers]{-{c}}$ of the previous year's Trout population.

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Diagonalise the matrix $A$. That is, find a diagonal matrix $D$ and an invertible matrix $P$ such that $A = P D P^{-1}$.

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(For the diagonal matrix $D$, please ensure that the first diagonal entry is smaller than the second diagonal entry. For the matrix $P$, please ensure that the $(2,1)$ and $(2,2)$ entries are both equal to 1. We require these conditions as the system is not yet capable of easily accepting all possibilities of $D$ and $P$.)

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$D =$ [] ;

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$P =$ [] .

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By inspection the general solution is
$\\boldsymbol{\\mathrm{x}_1} = A \\boldsymbol{\\mathrm{x}_0}$ , $\\boldsymbol{\\mathrm{x}_2} = A \\boldsymbol{\\mathrm{x}_1} = A^2 \\boldsymbol{\\mathrm{x}_0}$ , ... , $\\boldsymbol{\\mathrm{x}_t} = A^t \\boldsymbol{\\mathrm{x}_0}$.

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If we now use the fact that $A = P D P^{-1}$, we get $\\boldsymbol{\\mathrm{x}_t} = P D^t P^{-1}\\boldsymbol{\\mathrm{x}_0}$. Indeed, if we expand $A^t = (P D P^{-1})^t$ we can see that nearly all of the $P$ and $P^{-1}$ matrices cancel out. The benefit of doing this is that we are left with $D^t$ which is much easier to calculate (since $D$ is a diagonal matrix) than $P^t$.

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If we expand $\\boldsymbol{\\mathrm{x}_t} = P D^t P^{-1}\\boldsymbol{\\mathrm{x}_0}$ we will be able to obtain an expression for $x_t$ and $y_t$.

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Now solve the system, subject to the given initial conditions. Press the \"Show steps\" button below if you require a hint.
(If you wish to write something like $\\frac{1}{2} (\\frac{7}{10})^t$ then please write (1/2)*((7/10)^t) .)

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$x_t =$ [] ;

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$y_t =$ [] .

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Which of the following describes how the populations of Osprey and of Trout will behave as $t \\longrightarrow \\infty$?

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[]

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As $t \\longrightarrow \\infty$ the populations of Osprey and of Trout both tend to infinity.

", "

As $t \\longrightarrow \\infty$ the populations of Osprey and of Trout both tend to zero.

", "

As $t \\longrightarrow \\infty$ the population of Osprey tends to infinity, and the population of Trout tends to zero.

", "

As $t \\longrightarrow \\infty$ the population of Osprey tends to zero, and the population of Trout tends to infinity.

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For the remainder of the question we will look at a larger system of difference equations. Some of the matrix calculations that you will perform could be quite tedious, and so we strongly recomend that you use an online matrix-multiplication calculator; however, please ensure that you can calculate matrix-multiplication and matrix-inverses if needed in the future.

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For $t \\in \\{ 0,1,2, \\ldots \\}$ consider the system of difference equations
$x_{t+1} = \\simplify[fractionNumbers]{{M1} * sub(x,t) + {M1} * sub(y,t) + {M1} * sub(z,t)}$
$y_{t+1} = \\simplify[fractionNumbers]{{M1} * sub(x,t) +{M1} * sub(y,t) +{M1} * sub(z,t)}$
$z_{t+1} = \\simplify[fractionNumbers]{{M1} * sub(x,t) +{M1} * sub(y,t) +{M1} * sub(z,t)}$

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subject to the initial conditions $x_0 = \\var{{i1}}$ , $y_0 = \\var{{i2}}$ , $z_0 = \\var{{i3}}$

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We can express this system as $\\boldsymbol{\\mathrm{x}_{t+1}} = A \\boldsymbol{\\mathrm{x}_{t}}$ and $\\boldsymbol{\\mathrm{x}_{0}} = \\begin{pmatrix} \\var{{i1}} \\\\ \\var{{i2}} \\\\ \\var{{i3}} \\end{pmatrix}$, where $\\boldsymbol{\\mathrm{x}_{t}} = \\begin{pmatrix} x_t \\\\ y_t \\\\ z_t \\end{pmatrix}$ and
$A =$ [] .

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Infact, similarly to parts a to d, we can express this system as $\\boldsymbol{\\mathrm{x}_{t}} = P D^t P^{-1} \\boldsymbol{v}$, where $P = \\var{{P1}}$,
$D =$ [] ,
and $\\boldsymbol{v} =$ [].

Similar to part c, solve the system subject to the initial conditions. Hint: Note that the columns of $A$ are orthogonal (although, they are not normalised).

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$x_t =$ [] ;

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$y_t =$ [] ;

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$z_t =$ [] .

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This helps to form one of the initial conditions.

", "definition": "random(1..5)", "group": "Variables for parts e and f"}, "j": {"name": "j", "templateType": "anything", "description": "

This is used in our definitions of v1 , v2 , and v3.

", "definition": "random(1..5)", "group": "Variables for parts e and f"}, "v": {"name": "v", "templateType": "anything", "description": "

This is the vector which we use for our initial conditions.

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This is our first eigenvector with eigenvalue a2. Note that our definitions of j,k,l ensure that this is not the zero vector.

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This is the first diagonal entry of the matrix Diag, and the eigenvalue for the eigenvector e1.

", "definition": "(a +3*d)/4", "group": "Variables for parts a to d"}, "P_inv": {"name": "P_inv", "templateType": "anything", "description": "

This is the inverse of P.

", "definition": "(3/(2*f)) * matrix([1 , -f/3] , [-1 , f])", "group": "Variables for parts a to d"}, "i3": {"name": "i3", "templateType": "anything", "description": "

This helps to form one of the initial conditions.

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This is the eigenvector with eigenvalue d1.

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This is the matrix with eigenvalues a1 , a2 , a3 and corresponsing eigenvecors v1 , v2 , v3.

", "definition": "P1 * Diag1 * P1_inv", "group": "Variables for parts e and f"}, "a2": {"name": "a2", "templateType": "anything", "description": "

This is the eigenvector of v2.

", "definition": "random(2..7 except(a1))", "group": "Variables for parts e and f"}, "c": {"name": "c", "templateType": "anything", "description": "

This is the 2,1 entry of the matrix M.

", "definition": "3*(a-d)/(4*f)", "group": "Variables for parts a to d"}, "a1": {"name": "a1", "templateType": "anything", "description": "

This is the eigenvector of v1.

", "definition": "random(2..6)", "group": "Variables for parts e and f"}, "P1": {"name": "P1", "templateType": "anything", "description": "

This is the matrix with columns equal to v1 , v2 , and v3.

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This helps to form one of the initial conditions.

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This is the first diagonal entry of the matrix Diag, and the eigenvalue for the eigenvector e1.

", "definition": "(3*a + d)/4", "group": "Variables for parts a to d"}, "a": {"name": "a", "templateType": "anything", "description": "

This is the 1,1 entry of the matrix M.

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This is the matrix describing the populations of Osprey and Trout. Note that by definition of a,b,c,d, the 1,1 entry of M is less than 0 showing that the Osprey population will decrease without the Trout. The 2,1 entry is negative as the Osprey eat the Trout. The 2,2, entry is greater than 1, which is necessary for the trout population to increase. The 1,2 entry is positive as the Osprey reproduce more when they eat the Trout. Note that the eigenvalues of M, which are d1 aand d2, are always greater than 1 (by definition of d). This ensures that the populations of both animals increases.

", "definition": "matrix([a , b] , [c , d])", "group": "Variables for parts a to d"}, "v2_abs_squared": {"name": "v2_abs_squared", "templateType": "anything", "description": "

This is the absolute value squared of v2.

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This is the absolute value squared of v1.

", "definition": "j^2 + k^2 + l^2", "group": "Variables for parts e and f"}, "Diag": {"name": "Diag", "templateType": "anything", "description": "

This is the diagonal matrix with diagonal entries d1 and d2.

", "definition": "matrix([d1 , 0] , [0 , d2])", "group": "Variables for parts a to d"}, "a3": {"name": "a3", "templateType": "anything", "description": "

This is the eigenvector of v3.

", "definition": "random(2..8 except(a1) except(a2))", "group": "Variables for parts e and f"}, "g": {"name": "g", "templateType": "anything", "description": "

This is used in the definition of the vector v.

", "definition": "random(1..9) * 10", "group": "Variables for parts a to d"}, "k": {"name": "k", "templateType": "anything", "description": "

This is used in our definitions of v1 , v2 , and v3.

", "definition": "random(1..6 except(j))", "group": "Variables for parts e and f"}, "P": {"name": "P", "templateType": "anything", "description": "

This is the matrix with columns equal to e1 and e2.

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This is used in our definitions of v1 , v2 , and v3.

", "definition": "random(1..7 except(j) except(k))", "group": "Variables for parts e and f"}, "v3_abs_squared": {"name": "v3_abs_squared", "templateType": "anything", "description": "

This is the absolute value squared of v3.

", "definition": "2*j^4 + 2*k^4 + 2*l^4 + 4*(j^2)*(k^2) + 4*(j^2)*(l^2) + 4*(k^2)*(l^2) -2*(j^3)*k - 2*(j^3)*l - 2*j*(k^3) -2*(k^3)*l - 2*j*(l^3) - 2*k*(l^3) - 2*(j^2)*k*l - 2*j*(k^2)*l - 2*j*k*(l^2)", "group": "Variables for parts e and f"}, "v3": {"name": "v3", "templateType": "anything", "description": "

This is our first eigenvector with eigenvalue a3.

", "definition": "vector((k+l)*j - (k^2 + l^2) , (j+l)*k - (j^2 + l^2) , (j+k)*l - (j^2 + k^2))", "group": "Variables for parts e and f"}, "f": {"name": "f", "templateType": "anything", "description": "

This is used in the definition of b and c

", "definition": "random(3..7)/2", "group": "Variables for parts a to d"}, "e2": {"name": "e2", "templateType": "anything", "description": "

This is the eigenvector with eigenvalue d2.

", "definition": "vector(f/3 , 1)", "group": "Variables for parts a to d"}, "Diag1": {"name": "Diag1", "templateType": "anything", "description": "

This is the diagonal matrix with entries equal to the eigenvalues.

", "definition": "matrix([a1 , 0 , 0], [0 , a2 , 0], [0 , 0 , a3])", "group": "Variables for parts e and f"}, "b": {"name": "b", "templateType": "anything", "description": "

This is the 1,2 entry of the matrix M.

", "definition": "-f*(a-d)/4", "group": "Variables for parts a to d"}, "v1": {"name": "v1", "templateType": "anything", "description": "

This is our first eigenvector with eigenvalue a1.

", "definition": "vector(j,k,l)", "group": "Variables for parts e and f"}, "h": {"name": "h", "templateType": "anything", "description": "

This is used in the definition of the vector v.

", "definition": "random(1..9) * 1000", "group": "Variables for parts a to d"}, "P1_inv": {"name": "P1_inv", "templateType": "anything", "description": "

This is the inverse of P1

", "definition": "matrix([1/v1_abs_squared , 0 , 0], [0 , 1/v2_abs_squared , 0], [0 , 0 , 1/v3_abs_squared]) * transpose(p1)", "group": "Variables for parts e and f"}, "d": {"name": "d", "templateType": "anything", "description": "

This is the 2,2 entry of the matrix M. Note that d is greater than a.

", "definition": "4 - 2*a + random(1..5)/10", "group": "Variables for parts a to d"}}, "advice": "", "variablesTest": {"condition": "", "maxRuns": 100}, "name": "MA100 LT Week 10", "preamble": {"js": "", "css": ""}, "statement": "

Lent Term Week 10 (lectures 39): In this question you will look at systems of difference equations.

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As with all questions, there may be parts where you can choose to \"Show steps\". This may give a hint, or it may present sub-parts which will help you to solve that part of the question. Furthermore, remember to always press the \"Show feedback\" button at the end of each part. Sometimes, helpful feedback will be given here, and often it will depend on how correctly you have answered and will link to other parts of the question. Hence, always retry the parts until you obtain full marks, and then look at the feedback again.

Keep in mind that in order to see the feedback for a particular part of a question, you must provide a full (but not necessarily correct) answer to that part. Do not worry though, as you can look at the feedback and then ammend your answer accordingly.

Furthermore, as with all questions, choosing to reveal the answers will only show you the answers which change every time the question is loaded (i.e. answers to randomised questions); the fixed answers will not be revealed.

", "tags": [], "rulesets": {}, "ungrouped_variables": [], "metadata": {"description": "

This is the question for Lent Term week 10 of the MA100 course at the LSE. It looks at material from chapters 39.

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