Linear recurrence relations of order 1

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Created this as a copy of PV workLinear recurrence relations of order 1.

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Question statement

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

Give any introductory information the student needs.

			Name	Type	Generated Value

			Name	Type	Generated Value

dpct	integer	9
s	integer	2
lt	integer	5
verschil	integer	5
b	integer	10
d	rational	9/100
g	integer	10
Gpct	integer	110
tgrens	decimal	dec("25")
lt1	integer	10
d1	rational	1/5
d1pct	rational	20
opld	decimal	dec("0.23")
Te	decimal	dec("-2.302585092994045684017991454684364207601")
Ne	decimal	dec("-9.431067947124132687714272436023008075985e-2")
Be	decimal	dec("2.441489241625266409353703017300515625235e+1")
chem	string	thiotomoline
person	dict	Dictionary with 3 entries
oplf	decimal	dec("20")

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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).

Advice

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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This question is used in the following exams:

Exam Linear Recurrence by Paul Verheyen in PV English.
KostenExam by Paul Verheyen in Paul's workspace.
Test Exam by Clare Lundon in Clare Complex Numbers.

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Paul Verheyen 5 years, 6 months ago

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Linear recurrence relations of order 1

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Taxonomy: mathcentre

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Paul Verheyen 5 years, 6 months ago

Paul Verheyen 5 years, 6 months ago

Paul Verheyen 5 years, 6 months ago

Error in variable testing condition

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Generated value: integer

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Generated value: `integer`