Consider a discrete-time dynamical system given by \\[P_{n+1}=-\\sqrt{P_{n}+2}\\] with $P_0=a\\geq-2$. Find $a$ such that the sequence $(P_n)_{n\\geq0}$ is constant.

", "preamble": {"js": "", "css": ""}, "tags": [], "rulesets": {}, "ungrouped_variables": [], "metadata": {"description": "", "licence": "None specified"}, "advice": "Let the sequence $(P_n)_{n\\geq0}$ be constant, i.e.

\\[

P_n=P_0=a

\\]

for all $n\\geq0$. Then we get the equation

\\[

a=-\\sqrt{a+2}.

\\]

Note that then $a\\geq-2$ (to ensure that $a+2\\geq0$), but also $a\\leq 0$ been equal to 'minus square root of...'. Then we get the equation \\[

a^2=a+2

\\]

that has two roots $a=2$ and $a=-1$, but the first one does not satisfy the condition $a\\leq0$.

Consider the Malthus equation

\\[

x'(t)=r x(t), \\qquad x(t_0)=x_0. \\tag{M}

\\]

What should be instead of $A$ and $B$ in the expression for the function

\\[

x(t)=A e^{rt-rB}

\\]

to ensure that this function solves the Malthus equation (M)?

The number $x=x(t)$ of fishes in a lake at a moment of time $t$ satisfies the equation \\[ x'(t)=1.1 x(t).\\] Initially, at the moment of time $t_0=\\var{year3}$ there were $\\var{fish}$ fishes. How many fishes does one expect to have in $3$ years?

"}, {"shuffleChoices": true, "marks": 0, "type": "1_n_2", "showCorrectAnswer": true, "distractors": ["Since the rate $\\var{rate}$ is negative, the population will always decrease, hence it could not become larger than it was. ", "Since the rate $\\var{rate}$ is negative, the population will always decrease, hence it could not become larger than it was. ", "Since the rate $\\var{rate}$ is negative, the population will always decrease, hence it could not become larger than it was.", "Correct. Since the rate $\\var{rate}$ is negative, the population will always decrease, hence it could not become larger than it was. "], "choices": ["In {year2}", "It has not reached yet, but it will in few years", "Just recently, this year", "It has not reached yet and will not reach ever."], "variableReplacements": [], "matrix": [0, 0, 0, "1"], "minMarks": 0, "extendBaseMarkingAlgorithm": true, "displayType": "radiogroup", "maxMarks": 0, "unitTests": [], "showFeedbackIcon": true, "scripts": {}, "displayColumns": "1", "variableReplacementStrategy": "originalfirst", "customMarkingAlgorithm": "", "prompt": "Let population $x=x(t)$ of a city evolve in time according to the equation \\[ x'(t)= \\var{rate} x(t) .\\] Suppose that in {year1} the population was {population1} people. When has the population reached the level of {population2} people?

"}], "preamble": {"js": "", "css": ""}, "metadata": {"licence": "None specified", "description": ""}, "ungrouped_variables": ["population1", "population2", "year1", "year2", "rate", "year3", "fish", "fishes"], "rulesets": {}, "variablesTest": {"maxRuns": 100, "condition": ""}, "functions": {}, "tags": [], "variables": {"fishes": {"templateType": "anything", "description": "", "name": "fishes", "definition": "3 fish", "group": "Ungrouped variables"}, "year1": {"templateType": "anything", "description": "", "name": "year1", "definition": "2000+random(1..16)", "group": "Ungrouped variables"}, "rate": {"templateType": "anything", "description": "", "name": "rate", "definition": "random(-10..-2)", "group": "Ungrouped variables"}, "population2": {"templateType": "anything", "description": "", "name": "population2", "definition": "random(population1+5000..population1+50000)", "group": "Ungrouped variables"}, "year3": {"templateType": "anything", "description": "", "name": "year3", "definition": "random(2001..2017)", "group": "Ungrouped variables"}, "fish": {"templateType": "anything", "description": "", "name": "fish", "definition": "random(101..115)", "group": "Ungrouped variables"}, "year2": {"templateType": "anything", "description": "", "name": "year2", "definition": "random(year1+1..2017)", "group": "Ungrouped variables"}, "population1": {"templateType": "anything", "description": "", "name": "population1", "definition": "random(100000..120000)", "group": "Ungrouped variables"}}, "variable_groups": [], "advice": "b) Note that the answer does not depend on $t_0$, since 'in three years' means that $t-t_0=3$ that appears at the formula.

\nc) The solution to the given equation has the form

\\[

u(t)=u(t_0)e^{\\var{rate}t},

\\]

where $t_0$ is the initial moment of time, i.e. $t_0=\\var{year1}$. Since $\\var{rate}<0$, $u(t)$ is decreasing, i.e. $u(t)<u(t_0)$ for all $t>t_0$. Since $\\var{population2}>\\var{population1}$, the population would not become larger than it was, and will not start to grow ever.

Consider a dynamical system of the form

\\[

x'=f(x). \\tag{DS}

\\]

Fixed points are solutions to the equation

\\[

f(x)=0

\\]

If the sign of $f$ by passing a fixed point changes from $+$ to $-$ it means that the point is an attractor. If th sign changes from $-$ to $+$, the point is a repeller.

Let in (DS)

\\[

f(x)=\\var{a}x^2-\\var{b}. \\tag{F}

\\]

How many fixed points this dynamical system does have?

Match choices with answers

", "showFeedbackIcon": true, "answers": ["solutions to the equation $f(x)=0$.", "attractor or sink.", "repeller or source."], "unitTests": [], "layout": {"type": "all", "expression": ""}, "customMarkingAlgorithm": "", "type": "m_n_x", "variableReplacements": [], "matrix": [["1", 0, 0], [0, "1", 0], [0, 0, "1"]], "displayType": "radiogroup", "shuffleAnswers": true}, {"showFeedbackIcon": true, "scripts": {}, "minMarks": 0, "shuffleChoices": true, "variableReplacementStrategy": "originalfirst", "customMarkingAlgorithm": "", "maxMarks": 0, "displayColumns": "1", "choices": ["$-\\var{c}$", "$\\var{c}$", "This dynamical system does not have an attractor.", "$\\pm\\var{c}$"], "showCorrectAnswer": true, "unitTests": [], "type": "1_n_2", "extendBaseMarkingAlgorithm": true, "variableReplacements": [], "distractors": ["Yes, since $f(x)$ is positive in a left neighbourhood of $-\\var{c}$ and is negative in a right neighbourhood.", "No, since $f(x)$ is negative in a left neighbourhood of $\\var{c}$ and is positive in a right neighbourhood.", "Since $f(x)$ is positive in a left neighbourhood of $-\\var{c}$ and is negative in a right neighbourhood, $x=-\\var{c}$ is a stable fixed point.", "No, since $f(x)$ is negative in a left neighbourhood of $\\var{c}$ and is positive in a right neighbourhood."], "marks": 0, "matrix": ["1", 0, 0, 0], "displayType": "radiogroup", "prompt": "Consider the dynamical system (DS) with $f(x)$ given by (F). Find the attractor of this dynamical system.

"}, {"showFeedbackIcon": true, "scripts": {}, "minMarks": 0, "shuffleChoices": true, "variableReplacementStrategy": "originalfirst", "customMarkingAlgorithm": "", "maxMarks": 0, "displayColumns": "1", "choices": ["is a sink.", "is a source.", "is not here a fixed point at all.", "is a fixed point, but it is neither sink nor source."], "showCorrectAnswer": true, "unitTests": [], "type": "1_n_2", "extendBaseMarkingAlgorithm": true, "variableReplacements": [], "distractors": ["No, the sign of $f(x)$ near $x=\\pi$ changes from $-$ to $+$.", "Yes, the sign of $f(x)$ near $x=\\pi$ changes from $-$ to $+$.", "No, $f(\\pi)=\\sin 2\\pi=0$.", "No, the sign of $f(x)$ near $x=\\pi$ changes from $-$ to $+$."], "marks": 0, "matrix": [0, "1", 0, 0], "displayType": "radiogroup", "prompt": "Consider the dynamical system (DS) with

\\[

f(x)=\\sin 2x.

\\]

Using the phase portrait of this system check that $x=\\pi$...

This is an **unassessed** homework. Note that, however, it compulsory to pass this homework to proceed further.