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In this assessment you always have to enter fractions (no decimals), unless the number is an integer!

a) Since the daily decrease percentage of {chem} is \$$\\var{dpct}\$$ %, and since the daily use of 1 ml medication daily  adds $\\var{s}$ mg of {chem}, we have
\$y_{t+1} = y_t - \\var{d} \\cdot y_t + \\var{s} = \\var{1-d} \\cdot y_t + \\var{s} \$
or without using decimal forms
\$y_{t+1} = y_t - \\simplify[all,fractionNumbers]{{dpct}/100} \\cdot y_t + \\var{s} = \\simplify[all,fractionNumbers]{{100-dpct}/100} \\cdot y_t + \\var{s} \$

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b) You obtain the general solution to a first order recurrence equation by adding the general solution to the associated homogeneous recurrence equation and a \"particular\" solution to the original recurrence equation.

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The associated homogeneous recurrence equation \$$y_{t+1} = \\simplify[all,fractionNumbers]{{100-dpct}/100} \\cdot y_t \$$ has as general solution
\$y_t^{H} = C \\cdot \\left( \\simplify[all,fractionNumbers]{{100-dpct}/100} \\right)^t ,\$
with $C$ a real number.

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To find a particular solution to the original recurrence equation, you can imitate the right-hand side of
\$y_{t+1} - \\simplify[all,fractionNumbers]{{100-dpct}/100} \\cdot y_t = \\var{s} .\$
Assume \$$y_t = \\alpha\$$ with \$$\\alpha \$$ a real number, then \$$y_{t+1} = \\alpha\$$, implying
\$\\alpha - \\simplify[all,fractionNumbers]{{100-dpct}/100} \\cdot \\alpha = \\var{s} \$
\$\\alpha = \\simplify[all,fractionNumbers]{{100*s}/{dpct}} .\$

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Consequently, the general solution is \$y_t = C \\cdot \\left( \\simplify[all,fractionNumbers]{{100-dpct}/100} \\right)^t + \\simplify[all,fractionNumbers]{{100*s}/{dpct}} .\$

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Since \$$y_0 = \\var{b} \$$, you can solve  \$$C \$$ from \$\\var{b}= C + \\simplify[all,fractionNumbers]{{100*s}/{dpct}} ,\$ i.e.
\$C = \\simplify[all,fractionNumbers]{{b-100*s/{dpct}}} .\$

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The final solution to the recurrence equation with the given initial condition is
\$y_t = \\simplify[all,fractionNumbers]{{b-100*s/{dpct}}} \\cdot \\left( \\simplify[all,fractionNumbers]{{100-dpct}/100} \\right)^t + \\simplify[all,fractionNumbers]{{100*s}/{dpct}} .\$

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c) The long run concentration (in mg) of {chem} in {person['name']}'s blood is the limit value for  \$$t\$$ going to \$$+\\infty \$$ of the solution \$$y_t \$$ supra:
\$\\simplify[all,fractionNumbers]{{100*s}/{dpct}} .\$

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d) Suppose that {person['name']} daily uses \$$x \$$ ml of medication, then linearity implies a daily increase in the concentration of {chem} in {if(person['gender']='female','her','his')} blood of \$$\\var{s} \\cdot x \$$ mg.

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The new recurrence equation

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\$y_{t+1} = \\simplify[all,fractionNumbers]{{100-dpct}/100} \\cdot y_t + \\var{s} \\cdot x \$

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has as solution

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\$y_t = C \\cdot \\left( \\simplify[all,fractionNumbers]{{100-dpct}/100} \\right)^t + \\simplify[all,fractionNumbers]{{100*s}/{dpct}} \\cdot x \$

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with resulting long run concentration

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\$\\simplify[all,fractionNumbers]{{100*s}/{dpct}} \\cdot x .\$

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In order to reach a long run equilibrium of \$$\\var{lt} \$$ mg of {chem} in the blood,

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\$\\simplify[all,fractionNumbers]{{100*s}/{dpct}} \\cdot x = \\var{lt} ,\$

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i.e.

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\$x = \\simplify[all,fractionNumbers]{{lt*dpct}/{100*s}} ,\$

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such that {person['name']} needs to use (approximately) \$$\\var{opld} \$$ ml of medication on a daily basis.

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\$\$

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e) In this case solving the recurrence relation

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\$y_{t+1} = \\simplify[all,fractionNumbers]{{100-dpct}/100} \\cdot y_t + \\simplify[all,fractionNumbers]{{lt*dpct}/{100}} \$

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\$y_t = \\var{b - lt} \\cdot \\left( \\simplify[all,fractionNumbers]{{100-dpct}/100} \\right)^t + \\var{lt} .\$

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In order to reach, for the first time, less than {Gpct}% of the ideally desired long run limit value of \$$\\var{lt} \$$ mg of {chem} in {person['name']}'s blood, you need

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\$\\var{b - lt} \\cdot \\left( \\simplify[all,fractionNumbers]{{100-dpct}/100} \\right)^t + \\var{lt} < \\simplify[all,fractionNumbers]{{Gpct*lt}/{100}} \$

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which means

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\$\\var{b - lt} \\cdot \\left( \\simplify[all,fractionNumbers]{{100-dpct}/100} \\right)^t < \\simplify[all,fractionNumbers]{{(Gpct-100)*lt}/{100}} \$

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or equivalently

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\$\\left( \\simplify[all,fractionNumbers]{{100-dpct}/100} \\right)^t < \\simplify[all,fractionNumbers]{{(Gpct-100)*lt}/{100*({b-lt})}} .\$

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Applying the natural logarithm to the left- and right-hand side gives:

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\$t \\cdot \\ln \\left( \\simplify[all,fractionNumbers]{{100-dpct}/100} \\right) < \\ln \\left( \\simplify[all,fractionNumbers]{{(Gpct-100)*lt}/{100*({b-lt})}}\\right) ,\$

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i.e.

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\$t \\cdot (\\var{Ne}) < \\var{Te} ,\$

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resulting in

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\$t > \\var{Be} .\$

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Rounding up leads to the number of  \$$\\var{ceil(Be)} \$$ days in order to reach for the first time less than {Gpct}% of the ideally desired long run limit value \$$\\var{lt} \$$ mg {chem} in the blood.

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\$\$

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f) Suppose {person['name']}'s body breaks down $x$% of {chem} daily and {person['name']} daily uses 1 ml of medication, where $0 < x < 100$. Then the corresponding recurrence equation
\$y_{t+1} = y_t - {\\frac{x} {100}} \\cdot y_t + \\var{s} ,\$
has as solution
\$y_t = \\left( {\\var{b}}-{\\frac {\\var{100*s}} {x}} \\right)\\cdot \\left( {1 - {\\frac{x} {100}}} \\right)^t + {\\frac{\\var{100*s}} {x}} ,\$
resulting in al long run equilibrium of
\${\\frac{\\var{100*s}} {x}} .\$
For a  long run  concentration of {lt1} mg of {chem} in the blood, this implies
\$x = \\simplify[all,fractionNumbers]{{100*s}/{lt1}} .\$
So on a daily basis {person['name']}'s body needs (approximately) to break down \$$\\var{oplf} \$$ % of {chem} in {if(person['gender']='female','her','his')} blood.

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## Problem

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#### This morning, at the beginning of the day, {person['name']} had {b} mg of {chem} (a chemical compund) in {if(person['gender']='female','her','his')} blood.Each day {if(person['gender']='female','she','he')} uses 1 ml of medication, which adds at the beginning of the following day {s} mg of {chem} to {if(person['gender']='female','her','his')} blood.Daily {if(person['gender']='female','her','his')} body breaks down {dpct}% of the concentration of {chem} that is present at the beginning of the day.Let $y_t$ be the amount in mg {chem} that will be in {if(person['gender']='female','her','his')} blood in $t$ days at the beginning of the day (consider the time $t$ as a discrete variable).

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Formulate a recurrence relation for the sequence $y$ in the form  $y_{t+1} = number1 \\cdot y_t + number2$ :

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$y_{t+1}$ = [[0]] $\\cdot y_t$ + [[1]]

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What is in the long run the amount of {chem} (in mg) in {if(person['gender']='female','her','his')} blood?

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[[0]]

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How many days will it take in order to reach, for the first time, less than {Gpct}% of the ideally desired long run limit value [as proposed in d)] in {person['name']}'s blood?

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[[0]]

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Suppose {person['name']}'s body does not break down {dpct}% daily, but another (constant) percentage.
What should that percentage be if {if(person['gender']='female','she','he')} is to end up having {lt1} mg of {chem} in the blood in the long run, when {if(person['gender']='female','she','he')} uses 1 ml of medication on a daily basis?

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[[0]]

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