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

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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} \\]

\\[\\]

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.

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.

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} \\]
This leads to
\\[ \\alpha = \\simplify[all,fractionNumbers]{{100*s}/{dpct}} .\\]

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

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}}} .\\]

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}} .\\]

\\[\\]

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}} .\\]

\\[\\]

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.

The new recurrence equation

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

has as solution

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

with resulting long run concentration

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

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

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

i.e.

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

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

\\[\\]

e) In this case solving the recurrence relation

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

leads to

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

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

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

which means

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

or equivalently

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

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

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

i.e.

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

resulting in

\\[ t > \\var{Be} .\\]

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.

\\[\\]

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

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 $y_{t+1} = number1 \\cdot y_t + number2$ :

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

Solve this recurrence equation with given initial condition, and write this solution in the form $y_t = number1 \\cdot basenumber^t + number2$.

$y_t$ = [[0]] $\\cdot$ [[1]] $+$ [[2]]

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

[[0]]

Ideally there should be {lt} mg of {chem} in {if(person['gender']='female','her','his')} blood in the long run. Determine the amount of medication (in ml) {if(person['gender']='female','she','he')} should use daily to reach this long run equilibrium. Suppose that the medication linearly increases the amount of {chem} added to the blood.

[[0]]

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?

[[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?

[[0]]