Linear recurrence relations of order 1

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Question 1

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

Problem

This morning, at the beginning of the day, Casey had 6 mg of thiotemoline (a chemical compund) in her blood.
Each day she uses 1 ml of medication, which adds at the beginning of the following day 4 mg of thiotemoline to her blood.
Daily her body breaks down 16% of the concentration of thiotemoline that is present at the beginning of the day.
Let $y_t$ be the amount in mg thiotemoline that will be in her blood in $t$ days at the beginning of the day (consider the time $t$ as a discrete variable).

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a)

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

$y_{t+1}$ = $\cdot y_t$ + interpreted as $\displaystyle{}$ Expected answer:interpreted as $\displaystyle{4}$

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b)

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

$y_t$ = $\cdot$ $+$ interpreted as $\displaystyle{}$ Expected answer:interpreted as $\displaystyle{25}$

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c)

What is in the long run the amount of thiotemoline (in mg) in her blood?

interpreted as $\displaystyle{}$ Expected answer:interpreted as $\displaystyle{25}$

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d)

Ideally there should be 4 mg of thiotemoline in her blood in the long run. Determine the amount of medication (in ml) she should use daily to reach this long run equilibrium. Suppose that the medication linearly increases the amount of thiotemoline added to the blood.

interpreted as $\displaystyle{}$ Expected answer:interpreted as $\displaystyle{\frac{ 4 }{ 25 }}$

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e)

How many days will it take in order to reach, for the first time, less than 106% of the ideally desired long run limit value [as proposed in d)] in Casey's blood?

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f)

Suppose Casey's body does not break down 16% daily, but another (constant) percentage.
What should that percentage be if she is to end up having 9 mg of thiotemoline in the blood in the long run, when she uses 1 ml of medication on a daily basis?

interpreted as $\displaystyle{}$ Expected answer:interpreted as $\displaystyle{\frac{ 400 }{ 9 }}$

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Advice

a) Since the daily decrease percentage of thiotemoline is $\var{dpct}$ %, and since the daily use of 1 ml medication daily adds $\var{s}$ mg of thiotemoline, 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 thiotemoline in Casey'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 Casey daily uses $x$ ml of medication, then linearity implies a daily increase in the concentration of thiotemoline in her 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 thiotemoline 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 Casey 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 106% of the ideally desired long run limit value of $\var{lt}$ mg of thiotemoline in Casey'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 106% of the ideally desired long run limit value $\var{lt}$ mg thiotemoline in the blood.

f) Suppose Casey's body breaks down $x$ % of thiotemoline daily and Casey 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 9 mg of thiotemoline in the blood, this implies
$x = \simplify[all,fractionNumbers]{{100*s}/{lt1}} .$
So on a daily basis Casey's body needs (approximately) to break down $\var{oplf}$ % of thiotemoline in her blood.

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Numbas has failed

Linear recurrence relations of order 1

Problem

a)

b)

c)

d)

e)

f)

Advice