Math 155: Calculus for the Biological Sciences

Eric Hanson

August 19, 2010

AbstractLecture notes on Calculus for the Biological Sciences based on Modeling the Dynam-

ics of Life the Second Edition by Frederick R. Adler

Contents

1 Introduction to Discrete-Time Dynamical Systems 31.1 Biology and Dynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.1.1 Models of Malaria . . . . . . . . . . . . . . . . . . . . . . . . . . 41.1.2 Dynamical Systems . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.2 Variables, Parameters, and Functions in Biology . . . . . . . . . . . . . 61.2.1 Variables and Parameters . . . . . . . . . . . . . . . . . . . . . . 61.2.2 Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.2.3 Combining Functions: Sum, Product, and Composition . . . . . 81.2.4 Inverse Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

1.3 The Units and Dimensions of Measurements and Functions . . . . . . . 121.3.1 Converting Units . . . . . . . . . . . . . . . . . . . . . . . . . . . 121.3.2 Dimensions vs. Units . . . . . . . . . . . . . . . . . . . . . . . . 13

1.4 Linear Functions and Their Graphs . . . . . . . . . . . . . . . . . . . . . 141.4.1 Proportional Relations . . . . . . . . . . . . . . . . . . . . . . . . 141.4.2 Slope . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141.4.3 Equations of Lines . . . . . . . . . . . . . . . . . . . . . . . . . . 15

1.5 Discrete-Time Dynamical Systems (DTDS) . . . . . . . . . . . . . . . . 161.5.1 DTDS and Updating Functions . . . . . . . . . . . . . . . . . . . 171.5.2 A Word On Updating Functions . . . . . . . . . . . . . . . . . . 181.5.3 DTDS and Units/Dimensioins . . . . . . . . . . . . . . . . . . . 181.5.4 Equations for Solutions to DTDS . . . . . . . . . . . . . . . . . . 19

1.6 DTDS Solutions and Exponential Functions (Adler 1.7) . . . . . . . . . 211.6.1 Generalizing the Bacterial Population Growth DTDS . . . . . . . 21

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Bibliography 22

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Day 1 Preliminaries: Go over the course information sheet.• Note the exam dates and times (Thursday evenings) and state that no make-

up exams are given, with a few exceptions if an alternate exam time request form issubmitted at least a week in advance.• Tutoring and office hours are held in the Great Hall of the TILT building. Students

are encouraged to meet there to work on practice problems.

1 Introduction to Discrete-Time Dynamical Sys-

tems

In this chapter, we review linear, trigonometric power, logarithmic, and exponentialfunctions, so that we can study functions and modeling. Functions and modeling arethe main tools needed to use mathematics to study biology. Modeling is the art oftaking a description of a biological phenomenon and converting it into mathematicalform. Our goal is to quantify the dynamics (change) of living things with functions.

Note: Please read Sections 1.1 to 1.4 in Adler. Each section should be entirelyreview of topics from the prerequisites for this course. We will cover them quickly, sohave them read by tomorrow and be ready to ask any questions that you may have.

1.1 Biology and Dynamics

Note: Please read Section 1.1 there may be a question on the quiz related to thematerial.

Living systems change; they are dynamic. What is life, anyway? One thing thatit involves is change. Taking a dynamical approach to studying biology is necessarilymathematical, because describing dynamics requires quantifying measurements. An-swering questions such as: What is changing? How fast is it changing? What is itchanging into? To answer these questions we will develop the tools of mathematics anduse the language of mathematics.

Step Definition

Quantify the measurements The numerical values that describe the systemDescribe a dynamical rule A description of how the basic measurements change

Develop a model A mathematical translation of the observationsFind a solution Use of mathematical methods to predict behavior

Write a Simulation Use a computer to predict behavior

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1.1.1 Models of Malaria

Early in the 20th century, it was discovered that malaria is transmitted by mosquitoes.(It is only female mosquitoes, by the way, that transmit the disease.)

Question 1. Can the spread of disease be controlled by killing mosquitoes, even if notevery mosquito is killed?

Some argued that it couldn’t. (You may have them discuss if it could or not and maybesee why it’s not so obvious.)

Sir Ronald Ross believed that malaria could be controlled in this way. The followingpoem Sir Ronald Ross wrote as he was researching malaria:

In this, O Nature, yield to me.

I pace and pace, and think and think, and take

with fever’d hands, and note down all I see.

That some distant light may haply break.

The painful faces ask, can we not cure?

We answer, No, not yet; we seek the laws.

O reveal through all this thy obscure

The unseen, small, million-murdering cause.

Sir Ross built his mathematical model based on the following assumptions:

1. An infected person becomes infected when bitten by an infected mosquito and

2. an uninfected mosquito becomes infected after biting an infected person

Diagram 1. The dynamics of malaria

Based on this model Sir Ronald Ross, thought about how population sizes changewith time, and showed that the disease could be eradicated even without killing everymosquito. Evidence of this is seen in the United States, where malaria has been virtuallyeliminated even thought the mosquitoes capable of transmitting the disease persist inmany areas.

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1.1.2 Dynamical Systems

Dynamical systems encode how things change.

1. Discrete-time dynamical systems arise if measurements are made at equally spacedintervals (such as every second, every minute, every year...). Examples:

• The population (of, for example, bacteria or fish or tigers,...) in a givenyear (or second,...) as a function of the population in the previous year (orsecond,...)

• The number of mutant alleles present in one generation as a function of thenumber of mutant alleles present in the previous generation

2. Continuous-time dynamical systems arise if measurements are made continuously

Example: If you walk from one end of the board to the other, your position p(t)depends on time t, where t is now a continuous variable, any positive real number.The position function can be determined from i) your initial position p(0) and ii)your speed, your rate of change. Your speed may be positive or negative, constantor variable. Knowing your initial position and speed, you can sketch your positionas a function of continuous time.

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1.2 Variables, Parameters, and Functions in Biology

1.2.1 Variables and Parameters

Definition 1. A variable is a symbol that represents a measurement (numerical quan-tity) that can change during the course of an experiment

Example 1. Bacterial population growth

Time(t) Bacteria(b)0 1.01 2.52 5.53 10.0

Definition 2. A parameter is a symbol that represents a measurement (numericalquantity) that does not change during an experiment.

Example 2. Variables( b) and parameters( T )

t b when T = 5 b when T = 100 1.0 1.01 2.5 3.02 5.5 73 10.0 13

Recall the Cartesian Coordinate plane and its parts: origin, axes, and labels.

1.2.2 Functions

Functions describe relations between objects.

Definition 3. A function is a mathematical object that takes an input, performs anoperation on it, and returns a unique new object as an output. The input is called theargument or independent variable and the out put is called the value or dependentvariable.

The sets of objects that a function ”relates” are defined as follows:

Definition 4. The set of all possible inputs a function accepts is called the domain. Theset of all possible things a function can return as as an output is called the codomain,and the set of all things that the function does return as outputs is called the range.

Note: The subtle differences between the definitions.

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Diagram 2. ”Venn diagram for functions”

Example 3. A simple function for population growth over 3 weeks.

t Population (P )0 101 92 63 1

Formula for the function: P (t) = 10 − x2. The inputs are in weeks, in this case onlypositive values of time make sense, so the domain is t ≥ 0. The range is P ≤ 10, butthe codomain is all real numbers (R).

Graph of the function (relation):

Recall The vertical line test is a graphical method for recognizing relations thatare not functions. If some vertical line crosses the graph two or more time, the relationfails the vertical line test and is not a function.

Example 4. A relation that fails the vertical line test:

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1.2.3 Combining Functions: Sum, Product, and Composition

Definition 5. The sum of f + g of the functions f, g is the function defined by

(f + g)(x) = f(x) + g(x)

Example 5. Let f(x) = 2x and g(x) = 3x2 + x then

(f + g)(x) = f(x) + g(x)

= 2x+ 3x2 + x

= 3x2 + 3x

Evaluating for x = 2 we have

(f + g)(2) = 3(2)2 + 3(2) = 3(4) + 6 = 18

.Note that f(2) = 2(2) = 4 and g(2) = 3(2)2 + (2) = 12 + 2 = 14. Consider that

f(2) + g(2) = 4 + 14

= 18

gives us the same answer.

Remark 1. Graphically the sum of two function is a vertical sum across its inputs.

Definition 6. The product f · g of the functions f, g is the function defined by

(f · g)(x) = f(x) · g(x)

Example 6. Let f(x) = 2x and g(x) = 3x2 + x then

(f · g)(x) = f(x) · g(x)

= 2x(3x2 + x)

= 6x3 + 2x2

Evaluation for x = 2 we have

(f · g)(2) = 6(2)3 + 2(2)2

= 48 + 8

= 56

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Note that f(2) = 2(2) = 4 and g(2) = 3(2)2 + (2) = 12 + 2 = 14. Consider that

f(2) · g(2) = 4 · 14

= 56

gives us the same answer.

Remark 2. Graphically the product of two function is a vertical product across itsinputs.

Definition 7. The composition f ◦ g of function f, g is the function defined by

(f ◦ g)(x) = f(g(x))

Example 7. Let f(x) = 2x and g(x) = 3x2 + x then

(f ◦ g)(x) = f(g(x))

= f(3x2 + x)

= 2(3x2 + x)

= 6x2 + 2x

Evaluation for x = 2 we have

(f ◦ g)(x) = 6(2)2 + 2(2)

= 6(4) + 4

= 28

Note that g(2) = 3(2)2 + (2) = 12 + 2 = 14 and then we have

f(g(2)) = f(14)

= 2(14)

= 28

which is the same answer as above.

Remark 3. Composition, f ◦ g, is the same as evaluating g at x and then evaluating fat the output. In this situation f is called the outer function and g is called the innerfunction.

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Diagram 3. ”Venn diagram” for composition of functions

Note that the domain and range of the functions in function composition must”match up” and that in general f ◦ g is different form g ◦ f .

Example 8. Composition of functions in BiologyLet P (T ) 1

10 ·T be the population of squirrels per square mile in a forest in terms of thenumber of trees per square mile, T , and assume the population of trees per square mileis related to the age of the forest by T (t) = 10, 000 − 2t2 + t where t is years beginningin 1990. According to the model, what is the population of squirrels this year.

Solution:

P (t) = P (T (t))

=1

10(10, 000− 2t2 + t)

= 1, 000− 1

5t2 +

1

10t

2010 corresponds to t = 20, so the number of squirrels is

P (20) = 1, 000− 1

5(20)2 +

1

10(20)

= 1, 000− 1

5(400) + 2

= 1, 000− 20 + 2

= 982

1.2.4 Inverse Functions

Inverse functions reverse the process that functions apply to inputs. That is given anoutput to a particular function, the inverse of the function determines the input thatproduced the output.

Diagram 4. ”Venn diagram” for inverse functions

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Definition 8. The function f−1 is the inverse of f if

f(f−1(x)) = x

andf−1(f(x)) = x

Algorithm 1. Finding the Inverse of a Function

1. Write the equation y = f(x)

2. Solve for x in terms on y

3. The inverse function is the operation done to y

Remark 4. In applied mathematics we do not switch the x and y. Even though this looksweird, in applied mathematics variables have meaning, that is they represent specificobjects. Thus switching ”names” can cause confusion.

Example 9. Let y = 56x then the inverse is x = 6

5y

Some functions do not have an inverse, consider the following:

Example 10. What is the inverse parabola, see graph below:

Any function that has the same output for more than one input does not have aninverse. The horizontal line test is a graphical method for determining if a functionhas an inverse. If the graph of a function intersects any horizontal line in two or morepoints then the function does not have an inverse.

However if a function does not have an inverse, it might have an inverse on part ofit’s domain.

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Example 11. Suppose a population of bacteria is modeled by b(t) = t2

10 + 55 usingAlgorithm 1 we find,

b =t2

10+ 55

b− 55 =t2

1010(b− 55) = t2√10(b− 55) = |t|

⇒ t = ±√

10(b− 55)

If we restrict the domain to positive values of t, then b(t) passes the horizontal line testand t = b−1(t) =

√10(b− 55)

Remark 5. The function f(x) = x5+x+1 passes the horizontal line test but we cannotsolve for the inverse algebraically. In fact French mathematician Evariste Galois provedthat there is no formula for the solution of a general polynomial with degree greater than4. (Galois was only 20 years old). However, in mathematical modeling just knowing theexistence of an inverse is important, and the the computation of an inverse can then bedone numerically.

1.3 The Units and Dimensions of Measurements and Func-tions

Question 2. Why are units and dimensions important?

Example 12. Consider 2 + 2 = 2. Is this wrong? Not if we have the proper labels forthe units, 2Na+ + 2Cl− = 2NaCl. This is now a standard chemistry formula.

1.3.1 Converting Units

Example 13. Converting miles to centimeters. We use the basic identities:

5280ft = 1mile

12in = 1ft

2.54cm = 1in

these imply

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5280ft

mile= 1

12in

ft= 1

2.54cm

in= 1

so we have

1mile = 1mile · 1 · 1 · 1

= 1mile · 5280ft

mile· 12

in

ft· 2.54

cm

in= 160, 934.4cm

Example 14. Convert 1 C (cup) to ml given that 1 pint = 0.473176 liters and that 1pint = 2 C. solution: 1 C = 236.588 ml

Note: There is an algorithm for the conversion procedure on page 26 of Adler.

1.3.2 Dimensions vs. Units

Definition 9. Dimensions describe underlying quantities that can be measured indifferent units. Units are a particular standard of measure that describe dimensions.

Quantity Dimensions Sample Units

length length meter, micron, inchduration time second, minute, day

mass mass gram, kilogramarea length2 square meter, acre

volume length3 liter, cubic meter, gallonspeed length/time meters/second, mph

acceleration length/time2 meters/second2

force mass × length/time2 dynes, pounds

density mass/length3 grams/liter

Fundamental relations are formulas for translation between dimensions, some arestated below:

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Volume of a Sphere V = 4π3 r

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Surface Area of a Sphere S = 4πr2

Area of a Circle A = πr2

Perimeter of a circle P = 2πrVolume, Area, and Thickness V = AT

Total Number and Mass m = µbMass, Density, and Volume M = ρV

Note: Please review scaling and shifting of functions, pages 29-35. There may be aquiz question on this topic.

1.4 Linear Functions and Their Graphs

1.4.1 Proportional Relations

Definition 10. A proportional relation is a relation where the output is proportionalto the output, that is the ratio of the output to the input is a constant. The generalformula for a proportional relations is:

f(x) = ax

where a is some constant value, called the constant of proportionality. Clearly,

a =ax

x=

output

input

for x 6= 0.

Example 15. bout = 3bin clearly a = 3. Suppose bin = 10 then bout = 30, now considerboutbin

= 3010 = 3 = a

It is easy to see that the constant of proportionality is the slope of the graph of aproportional relation. Now we recall how to find the slope of general linear relation.

1.4.2 Slope

Definition 11. Most generally slope is

∆output

∆input

. The slope of a line is the change between two data points. Suppose we call the points(x1, y1) and (x2, y2) then the slope, m, is given by:

m =∆output

∆input=y2 − y1x2 − x1

.

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Example 16. Find the slope between (1, 14) and (20, 25). Then

m =y2 − y1x2 − x1

=25− 1

20− 14

=24

6= 4

Note: Notice the relationship between slope of linear relations and slope of proportionalrelations (i.e. constant of proportionality).

Example 17. f(x) = x+ 4 is not a proportional relation. Can you see why?

1.4.3 Equations of Lines

Definition 12. A line passing through the point (x0, y0) with slope m has formula:

1. Point-Slope Form of a Line

y = m(x− x0) + y0

2. Slope-Intercept Form for a Line

y = mx+ b

where b is the y-intercept.

Example 18. Finding line equation examples:

1. m = 2 and (0, 4), then we have line y = 2x+ 4. Why?

2. (0, 4) and (4, 2), then we have m = 4−20−4 = 2

−4 = −12 and the line equation is

y = −12x+ 4

3. (2, 4) and (4, 2), then we have m = 4−22−4 = 2

−2 = −1 and the line equation isy = −1(x− 2) + 4

Example 19. Does the following data line on a line.

time Mites on a lizard y0 101 152 203 254 305 40

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In order for the data to lie on a line the slope must be the same between all datapoints (each row is a data point). Clearly the data does on line on a line since the slopebetween t = 0 and t = 1 (m = 5) is different from the slope between t = 4 and t = 5(m = 10).

Now suppose we have the following data,

time Mites on a lizard, y0 101 152 203 254 30

Clearly this data lies on a line with equation y = 5t+ 10. We can then interpolate,that is estimate for untested inputs. Suppose we want to estimate for t = 1.5 and t = 5,then we find there are an estimated 17.5 and 35 mites respectively.

Recall that when solving for the intersection of two lines, parallel lines do not inter-sect and identical lines have infinitely many solutions. Read pages 48 -50 for examplesof word problems solved using linear equations.

1.5 Discrete-Time Dynamical Systems (DTDS)

In a Discrete-Time Dynamical Systems (DTDS) a function is applied to an input toobtain an output. In biology we often use DTDS’s to examine populations, so with apopulation as an input we can obtain an output that is the new population say a yearlater. DTDS provide a way to predict what will happen to a population in the long run.However, if the DTDS has an output that is the population value one year later, thatsame system cannot tell us what the value is half a year later. This is the concept ofdiscrete vs. continuous time.

Question 3. What is the difference between discrete and continuous time?

Later we will learn a way to solve DTDS so that we can obtain solutions that arenot restricted to the discrete time step.

Diagram 5. Venn diagram for a DTDS.

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1.5.1 DTDS and Updating Functions

We can think of a DTDS as describing the relation between a quantity measured atthe beginning and the end of an experiment or a time interval. We now more formallydefine a DTDS:

Let m be a measurement, then we denote mt as the measurement at the beginningof a time interval. We think of t as the current time and then t+ 1 is the time one stepinto the future. Then we can describe the relation between the initial measurement mt

and the next measurement, mt+1, by a DTDS:

Definition 13. A DTDS takes a measurement mt and applies an updating functionf to determine the value of the measurement one time step later, mt+1. In mathematicalnotation we have

mt+1 = f(mt)

Example 20. Suppose a population of bacteria triples every month and that the pop-ulation is initially 1000 bacteria. Clearly we know how to calculate the population onemonth later,

1000(3) = 3000

.In the notation of discrete time dynamical systems we have, b0 = 1000 as an initial

condition and the population the next year is given by the relation

bt+1 = 3bt

.We can now iterate the DTDS to determine the population values for the next 5

months.

Month Population1 b1 = 3b0 = 3(1000) = 30002 b2 = 3b1 = 3(3000) = 90003 b3 = 3b2 = 3(9000) = 270004 b4 = 3b3 = 3(27000) = 810005 b5 = 3b4 = 3(81000) = 243000

Example 21. Suppose it is spring time for a male (bull) moose, so his antlers are begin-ning to grow. Moose antlers grow to full size in 3 to 5 months, making them the fastestgrowing animal organ. Suppose that Marvin the moose’s antlers grow 6 inches a month.How big will they be in 5 months? (information obtained from http://en.wikipedia.org/wiki/Moose)

Initial condition: a0 = 0DTDS: at+1 = at + 6

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Month Antler Length1 a1 = a0 + 6 = 0 + 6 = 62 a2 = a1 + 6 = 6 + 6 = 123 a3 = a2 + 6 = 12 + 6 = 184 a4 = a3 + 6 = 18 + 6 = 245 a5 = a4 + 6 = 24 + 6 = 30

Thus Marvin’s antlers grow to 30 inches.

Example 22. Suppose we know that each day a patient uses up half of the medicationin her bloodstream. However he is given a new dose sufficient to raise the concentrationin the bloodstream by 1.0 milligrams per liter. Then we have the DTDS

Mt+1 = 0.5Mt + 1

. If the initial condition for the system is M0 = 0 can you graph the outputs of thesystem. (Note the updating function is linear).

1.5.2 A Word On Updating Functions

Updating functions are FUNCTIONS, this means that any of the operations that we canperform on function we can also perform on updating functions. We can sum (subtract),multiply (divide**?), compose updating functions, and find inverses (if they exist).

Example 23. Consider the composition of the bacteria growth updating function, P (bt) =3bt.

(P ◦ P )(bt) = P (P (bt))

= P (3bt)

= 3(3bt)

= 9bt

Now consider the initial condition b0 = 1000 under our new updating function. Thenour next value is 9000 which is b2. If we continue with more iterations it become clearthat the composition of the updating function corresponds to a two-step updating func-tion, that is we have

bt+2 = 9bt

.

Question 4. Can you find the inverse of the updating function in Example 22?

1.5.3 DTDS and Units/Dimensioins

Please read pages 57-58 on units and dimensions.

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1.5.4 Equations for Solutions to DTDS

Definition 14. The sequence of values mt for t = 1, 2, 3, . . . is called the solution tothe DTDS mt+1 = f(mt) starting with initial condition m0

The graph of a solution is the discrete set of points

(0,m0), (1,m1), (2,m2), . . . (t,mt)

It is possible to find a formula for the solution to simple DTDS’s, but not in manycomplicated cases. However, if we can find this formula it is very useful, since with itwe can easily calculate values for large values of t. Also we can interpolate for all valuesof t not just for values in the discrete set of solutions (i.e. t = 1.5).

Example 24. A Solution to the Bacteria DTDS Suppose again that we have DTDSbt+1 = 3bt with initial condition b0 = 1000. Then we have the following solutions:

b1 = 3b0 = 3(1000)

b2 = 3b1 = 3(3b0) = 3 · 3 · 1000 = 32 · 1000

b3 = 3b2 = 3(3b1) = 3(3(3b0)) = 33 · 1000

...

bt = 3t · 1000

bt = 3t · b0 is called the closed form solution to the DTDS. Using this form wecan calculate the solution for any value of t.

Example 25. Now consider bt+1 = 3bt with a different initial condition b0 = 50. Thewe have the following solution:

b1 = 3b0 = 3(50)

b2 = 3b1 = 3(3b0) = 3 · 3 · 50 = 32 · 50

b3 = 3b2 = 3(3b1) = 3(3(3b0)) = 33 · 50

...

bt = 3t · 50

Clearly this is a different solution than what we found above, so it is apparent the thesolution to a DTDS is dependent on both the initial condition and the updating function.

Note: That while the closed form solution provides a continuous set the solution set tothe DTDS is still discrete and a graph of the solution should be a set of points. Unlesswe wish to connect the points for emphasis of a pattern.

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Example 26. Can we find a closed form solution to the DTDS for the length of Marvinthe Moose’s antlers?

Recall that Marvin’s antlers growth follows the DTDS, at+1 = at + 6 with initialcondition, a0 = 0 and that a moose’s antlers grow for at most 5 months. So we onlyhave the values of at for t = 0, 1, 2, 3, 4, 5, thus the domain of the closed form solutionis the set, dom(f) = {t|0 ≤ t ≤ 5}.

Then the solution to the DTDS is:

a1 = a0 + 6 = 0 + 6 = 6

a2 = a1 + 6 = (0 + 6) + 6 = 0 + 2 · 6a3 = a2 + 6 = ((0 + 6) + 6) + 6 = 0 + 3 · 6

...

at = 0 + 6t

What would the solution be if a0 = 1?

Example 27. Let’s find a solution to the the Medication DTDS. Recall mt+1 = 0.5mt+1and let the initial condition for the system be M0 = 5.0 The we have solutions:

m1 = 0.5(5) + 1 = 3.5

m2 = 0.5(3.25) + 1 = 2.75

m3 = 0.5(2.75) + 1 = 2.375

m4 = 0.5(2.375) + 1 = 2.1875

...

The solution values appear to be approaching 2. In fact each step the solutions ”movehalfway closer to 2.” Consider the following:

m0 − 2 = 5− 2 = 3

m1 − 2 = 3.25− 2 = 1.25 = 0.5 · 3m2 − 2 = 2.75− 2 = 0.75 = 0.5 · 1.5m3 − 2 = 2.375− 2 = 0.375 = 0.5 · 0.75

m4 − 2 = 2.1875− 2 = 0.1875 = 0.5 · 0.375

...

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These observations can lead us to a solution:

m0 = 2 + 3

m1 = 2 + 0.5 · 3m2 = 2 + 0.52 · 3m3 = 2 + 0.53 · 3m4 = 2 + 0.54 · 3

...

mt = 2 + 0.5t · 3

Question 5. Consider finding a closed form for the DTDS in the Example 27, usinginitial condition m0 = 1 (the solution is on page 63).

Remark 6. In general finding closed form solutions to a DTDS is not easy. Considerxt+1 = 2xt+30 with initial condition x0 = 10 and you will find that a pattern is not veryobvious in this case. In fact given a random DTDS it is likely a closed form solution isvery hard to find.

1.6 DTDS Solutions and Exponential Functions (Adler 1.7)

Recall the bacterial population growth DTDS, bt+1 = 3bt with initial condition b0 = 1has closed form solution

bt = 3t

This is an example of an exponential function. Suppose we want to solve for bt = 300then we must be able to solve

300 = 3t

Question 6. How do we do this? Answer: We use the natural logarithm.

1.6.1 Generalizing the Bacterial Population Growth DTDS

Thus far we have assumed that all bacteria in our model live through each cycle. Clearlythis is a big assumption, so lets now suppose that only a fraction of the new bacteriasurvive. We will call the fraction σ.

Let ω be the number of offspring per bacteria. Then in the DTDS bt+1 = 3bt, ω = 3.Let r be the per capita production, that is the number of surviving offspring per

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Bibliography

References

[1] F.R. Adler, Modeling the Dynamics of Life: Calculus and Probability for Life Sci-ences Second Edition, Thomson Brooks/Cole, Belmont CA, (2005)

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