WORKSHOP #6Linear Functions

Abstract. Students often identify a function with the ``rule" which describes it. Thus functions become purely algebraic objects. The concept of a function as a mapping of one set onto another or onto itself is often not well understood by our students. The mapping properties of linear functions are particularly easy to describe geometrically and hence particularly easy to work with. The goals of this workshop are: to develop a deeper understanding of linear functions as functions and to strengthen the skill of geometric reasoning. This geometric understanding of linear functions will then be used to investigate linear difference equations and their applications.

Note. There is enough material here for a half-day workshop. The material covered must be cut back for shorter workshop lengths. Another possible format would be to divide the material into three or four units of a minicourse. Some of the worksheets could then become homework assignments. A teacher who has had this workshop and is comfortable with this material can easily adapt it to an enrichment unit for high school or even middle school students. Most of the computations necessary for the first part workshop are carried out by hand. However, a scientific calculator with a table-making feature would be very helpful.

Format. We start with a description of the workshop material as we presented it formatted as a single narrative.  In an actual presentation, some material was simply present verbally, some in the form of projected transparencies (or projected computer screen) and some, particularly the problems, was distributed in handouts. The main advantage of having the material in Word is that the presenter can decide the best way to divide up the material for any given audience, producing their own slides and worksheets, choosing how little or how much you want to present. The worksheets that we used are attached to the end of this file. They too can be easily altered to fit in any redesign workshop. 

The Geometry of Linear Functions Consider the linear function f(x) = x + 4. Instead of the usual geometric method of visualizing this function, the traditional graph of the function in a Cartesian coordinate system, we will view this function as map of the Euclidean line onto itself:

It is clear that the function f(x) = x + 4 simply shifts each point (number on the number line) 4 units to the right (in the positive direction). Picturing f(x) = x – 5.5, we see that it simply shifts each point 5.5 units to the left (in the negative direction):

Describe the function f(x) = x + 400 geometrically. It is the translation that shifts each point 400 units to the right.

Describe the function f(x) = x−195

geometrically. It is the translation that shifts

each point −195

of a unit to the left.

Describe the function f(x) = x geometrically. It maps each point onto itself or it leaves each point fixed or it is the identity map.

Conclusion: The linear functions of the form f(x) = x + a translate the line a units to the right ifa > 0 ; a units to the left if a < 0

Now consider the linear function f(x) = -x. How does this function as map the Euclidean line onto itself? It interchanges each number with its negative: it is the reflection through the point 0.

What about the function f(x) = -x + 4? It is the reflection through the point 2.

What about f(x) = -x -6? f(x) = -x +13? After a few examples, the following seems clear:

Conclusion: The linear functions of the form f(x) = -x + b reflects the line about

the point b2

.

However, this does require a proof. Given the linear function f(x) = -x + b and an

arbitrary point y on the number line, we can write y in the form y=b2+ c. If c is

positive y is c units to the right of b2

and c units to its left when c is negative.

Applying f to y in this form: f (y)=f(b2+ c)=−(

b2+ c)+ b =

b2−c, we see that f

interchanges y and b2−c, its image reflected through

b2

.

Next consider the linear function f (x)=2x . How does this function as map the Euclidean line onto itself? It moves each point away from 0 doubling it distance from 0. We call this the dilation with multiplicative factor 2 centered at 0.

What about f (x)=12x? f(x) = 1.5x? f(x) = -2x? After a few examples, the following

seems clear:

Conclusion: The linear functions of the form f(x) = ax dilates the line about 0 by a factor of a, when a > 0, and it dilates the line about 0 by a factor of a , when a < 0. .

Finally we consider the general linear function f(x) = ax + b. How does this function map the Euclidean line onto itself? Let’s consider a specific example f(x) = 1.5x + 2.

This appears to be a dilation about the point -4 with magnification 1.5. We easily check that -4 is fixed: f (−4)=1.5 × (−4)+ 2 =−6 + 2 =−4. We also see that, for any number x, f (−4 + x)=−4 +1.5x . Thus for any point y =−4 + x, the distance from f(y), the image of y, to -4 is 1.5 times the distance from y to -4.

So what can we say about the general linear function f(x) = ax + b (a ≠0) ?

Given the linear function f(x) = ax + b (a ≠0) :1. all segment lengths are multiplied by a ; specifically,

f (x)−f(y) = a x−y ;

2. if a ≠1 , there is a unique fixed point, or center, c =b

(1−a) .

3. If a > 0 , orientation is preserved; if a < 0 , orientation is reversed.

All of these features of linear functions are easily verified: 1. f (x)−f(y) = (ax + b)−(ay+b) = ax−ay=a(x−y) = a x−y ;

2. if a ≠1 , b

(1−a) is defined and

f (b

(1−a))=

ab(1−a)

+ b=ab

(1−a)+(1−a)b(1−a)

=b

(1−a) .

3. If a > 0 and x > y, then ax > ay and f (x)=ax+b > ay+b= f(y). So orientation is preserved. If, on the other hand, a < 0 and x > y, then ax < ay and f (x)=ax+b < ay+b = f(y). So orientation is reversed.

Any function from any Euclidean space onto itself that satisfies condition 1 above is called a similarity and, if the multiplicative factor is 1, it is called a congruence. As

we have seen the only congruences are translations and reflections about a point. All similarities that are not congruences have a fixed point and are either dilations from that point or dilations from that point composed with the reflection through that point. In the applications to follow this geometric point of view will be very useful.

Worksheet #2

Compute the magnification, the orientation and the center for the linear function f (x)=−2x + 3 . Then check your answers by plotting the trajectories of a few points.

The magnification is 2 and f is orientation reversing. The center is 3

(1−(−2))=1 .

Compute the magnification, the orientation and the center for the linear function

f (x)=−12x + 3 . Then check your answers by plotting the trajectories of a few

points.

The magnification is 12

and f is orientation reversing. The center is 3

(1−(−12))=2

.

Determining Linear Functions. The next worksheet asks us to derive the formula for a linear function given some information about it. One method for doing this is to write the formula with undetermined coefficients, f (x)=ax+ b , and then use the information that you have to solve for a and b. For example, if you know that f(3) = 7 while f(-2) = 12, you get the system of equations: 3a + b = 7; -2a + b = 12. Solving this system yields the values a = -1, b = 10 and the formula f(x) = -x + 10.

Given an arbitrary unknown linear function, f(x) = ax + b. If you know any two items from this list {the slope a, the constant b, the fixed point c, one function

value, a second function value} you can set up a system of equations and solve for both a and b.

Worksheet #3

Problem 1. Consider a thermometer that reads both Fahrenheit and Centigrade.

1. Give the temperature in Fahrenheit as a linear function of the temperature in Centigrade.

F(C)=95C + 32

2. Find the fixed point of this linear function and explain its physical meaning.

c=32

1−95

=−40; -40 degrees is the one temperature

that reads the same in Fahrenheit and Centigrade.

3. Compute the inverse function. What does it tell us?

C(F)=59F−17

79

. This gives the temperature in

Centigrade as a linear function of the temperature in Fahrenheit.

Problem 2. Suppose that your average in a course just before the final exam is 80% and that the final exam will account for 40% of your grade. Consider your final average as a (linear) function f(x) of the score x on your final exam.

1. What is the fixed point of this function?

80: if you get an 80 on the final exam you will get an 80 in the course.

2. What is the magnification of this function?.4: your grade on the final exam will be multiplied by .4

3. What is the equation for this function?f (x)=.4x + .6(80)=.4x + 48

4. What are the lowest and highest scores you could get in the course?f(0) = 48 (the lowest); f(100) = 88 (the highest).

5. What is the inverse of this function and what does it mean?f −1(x)=2.5x−120 It gives the score that you need to get on the final to get a course

average of x. For example to pass the course with a 60 average, you need to score 2.5 × 60 −120 =30on the final exam.

Problem 3. Consider the meterstick that can be found on any classroom. It is designed to read from left to right in centimeters on one side and from left to right in inches on the other.

1. Give the linear function that, for any centimeter measure on this stick, gives the inches measure on the opposite side. (Note: this is not the usual centimeters-to-inches conversion function.) I (C)=−0.3937C + 39.37 : we have I(0) = 39.37 and I(100) = 0.

2. Find the fixed point of this linear function and explain its physical meaning.39.371.3937

=28.25 . The point on the stick labeled 28.25cm is also labeled 28.25in.

3. Compute the inverse function. C(I )=−10039.37

I +100 : C(0) = 100; C(39.37) = 0.

Worksheet #4Some properties of linear functions:

Problem 1. Consider the linear function f (x)=ax+ b and compute the equation for the inverse function f −1(x) . Write the function in the form y =ax +b and solve for x

to get x =1ay−

ba. Thus f −1(x)=a−1x−a−1b.

Problem 2. Consider the linear functions f (x)=ax+ b and g(x)=rx + s. Find the conditions on f and g so that they commute: f (g(x))=g(f(x)).We have: f (g(x))=a(rx + s)+ b =arx + as+ b and g( f (x))=r(ax + b)+ s=rax + rb+ s.So f and g will commute if and only if as +b=rb+ s. Reorganizing, the condition becomes (1−r)b =(1−a)s. Both sides can equal 0, in several ways:

(1−r)=(1−a)=0 , f and g are both translations;(1−r)=s=0 , g is the identity function;b =(1−a)=0 , f is the identity function;b =s=0 , f and g are both dilations centered at 0.

In the case both sides are not zero we may divide both sides by (1−r)(1−a) to get:b1−a

=s

1−r.

Conclusion: two linear functions will commute if and only if at least one is the identity or both are translations or they have the same center.Iterating Linear Functions. In the last part of this workshop we plan to apply what we have learned about linear functions to get a better understanding of linear difference equations. For this application, we will need to understand what happens when we iterate a linear function. Given the linear function f (x)=ax +band an initial value x0, we wish to investigate the set of points x0 , x1,L xi,L , where xi =f(xi−1) for all positive integers i. The best way to get an understanding of these sets is to compute several examples.

Using the TI84 computing examples is easy. Consider f (x)=ax+ band an initial value x0:

type x0 and press ENTER (this puts x0 in the ANS memory); type a(ANS) + b and press ENTER (this computes x1 puts it in the ANS memory); repeatedly pressing ENTER will now give the sequence of iterates.

Consider the following two problems as a group:

Problem 1. Consider f (x)=x+ 2 ; plot the first few iterates starting with 1, with -8:

Give a formula for xi in terms of x0. xi =x0 + i × 2

Problem 2. Consider f (x)=−x+ 2 ; plot the first few iterates starting with 8, with -2.

Give a formula for xi in terms of x0. xi =1+ (−1)i(x0 −1)

Conclusions: The iterates of the translation f(x) = x + b are equally spaced points starting with x0 and adding b to each successive iterate. The iterates of a reflection f(x) = -x + b alternate between x0, (when i even) and f(x0) (when i odd).

Worksheet #5

Problem 1. Consider f (x)=1.5x ; plot the first few iterates starting with 1, with -2:

Problem 2. Consider f (x)=1.5x +1; plot the first few iterates starting with 0, with -2, with -3:

Problem 3. Consider f (x)=−1.5x + 5 ; plot the first few iterates starting with 0, with 4:

Problem 4. Consider f (x)=0.8x + 0.4 ; plot the first few iterates starting with -8 & 7:

Conclusions?? See below.

It is natural to ask if, given f (x)=ax+band an initial value x0, can we derive a formula for xi. The answer is yes and we have already done so when f is a congruence:

If f is a translation (a=1), then xi =x0 + i × b .

If f is a reflection (a=-1), then xi =x0 if i is even and xi =b−x0 when i is odd.

When f is not a congruence (a ≠1anda ≠−1) , it is clear that the position of the center plays an important role in the structure of the set of iterates. So we start by

rewriting the formula for f in terms of its center. Let c =b

(1−a)denote the center of

f. Then, we have f (x)=ax+b=ax + (1−a)c=a(x−c)+ c . Hence, x1 =f(x0 )=a(x0 −c)+ c or x1 −c=a(x0 −c)and, in general, xi −c=a(xi−1 −c). But then (x2 −c)=a(x1 −c)=a(a(x0 −c))=a

2(x0 −c) and in general: (xi −c)=ai(x0 −c)

or

xi =ai(x0 −c)+ c.

From this general formula for xi, we may verify the conclusions that we drew from the examples:

If a is negative, the iterations with alternate being above or below c; if a is positive, the iterations will all be on the same side of c.

If a < 0, successive iterations get closer and closer to c; if a > 0, successive iterations get further and further from c.

For example, f (x)=1.5x +1 has center c = -2 and xi =1.5i(x0 + 2)−2 . Giving:

x0 x1 x2 x3 x4 x5 x6

0 1 2.5 4.75 8.125 13.1875 20.78125-2 -2 -2 -2 -2 -2 -2-3 -3.5 -4.25 -5.375 -7.0625 -9.59375 -

13.390625

For example, f (x)=−1.5x + 5 has center c = 2 and xi =(−1.5)i(x0 −2)+ 2 . Giving:

x0 x1 x2 x3 x4 x5 x6

0 5 2.5 4.75 8.125 13.1875 20.781252 2 2 2 2 2 2-3 -3.5 -4.25 -5.375 -7.0625 -9.59375 -

13.390625

For example, f (x)=0.8x + 0.4 has center c = 2 and xi =(0.8)i(x0 −2)+ 2 . Giving:

x0 x1 x2 x3 x4 x5 x6

10 8.4 7.12 6.096 5.2768 4.62144 4.0971522 2 2 2 2 2 2-5 -3.6 -2.48 -1.584 -0.8672 -0.29376 0.164992

Worksheet #6

Suppose that you own a piece of land with a pond that you would like to stock with trout. Trout will not reproduce in this pond but roughly two-thirds of the fish population will survive from one year to the next. Suppose that you initially stock the pond with 5000 fish and then add 2000 in each successive year.

Problem 1. Give the linear difference equation for the number of trout in the pond

in successive years. tn =23tn−1 + 2000 with t0 =5000.

Problem 2. Consider the following questions1. At what level will the fish population stabilize? The fixed point is 6000

trout: 2000

1−23

=6000.

2. How important was the size of the initial stock? Not too important: the trout population will stabilize at 6000 no mater what the initial stock size. However, the closer you start to 6000 the sooner it will stabilize.

3. How important was the size of the annual stock? Very important: the size of the stable population will be 3 times the size of the annual stock.

4. If the pond could accommodate a stable population of 7500 trout, how could you achieve this? Add 2500 each year instead of 2000.

All of the fish that you have been using to stock your pond come from a friend who owns a pond in which trout do reproduce. Each year the trout population in his pond increases by 25\% until the maximum population of 12,000 trout is reached. Suppose that you stock your pond with 5000 fish the first year and 2000 a year thereafter. Assume that when you initially stocked your pond his pond was at its capacity of 12,000.

Problem 3. 1. Give the linear difference equation that governs the population size in your

friend's pond. Tn =1.25Tn−1 −2000 with T0=7000.

2. Assume that you continue to stock your pond with 2000 trout each year from your friend's pond. What will the trout population be in your friend's pond after 10 years?

n 1 2 3 4 5 6 7 8 9 10Tn 6750 6438 6047 5559 4984 4185 3231 2039 549 0

3. Could this have been avoided and still stock your pond? Yes! The fixed point of this difference equation is 8000. As long as the population is above 8000 it will grow by more than 2000 each year and the population will eventually grow back to its full capacity of 12000. Hence the initial stocking of your pond should be with fewer than 4000 trout.

Worksheet #7

Another interesting and useful application of linear difference equations is to weight control. Lets consider Jack. He weighs 80 kilograms and, due to a high blood-pressure problem, his doctor has suggested that he reduce his weight by 10% to 72 kilograms. Jack maintains his weight of 80kg on an average of 2480 calories per day. We see that, with his present level of physical activity, Jack needs 31 calories per kilogram per day to maintain his weight.So, Jack's first reaction was: "This isn't too difficult; all I have to do is to reduce my daily intake by 10% to 2232 calories per day." The doctor acknowledged that Jack's calculations were correct but pointed out that, while a diet of 2232 calories per day would maintain a weight of 72kg, it would literally take years for Jack to get his weight down to that level on a diet of 2232 calories per day. In making his calculation, Jack was unaware of another important constant: the number of calories he needed to ``burn off" in order to lose 1 kilogram in weight. Given Jack's life style and metabolism the doctor believes that

to lose 1kg in his weight, Jack must reduce his over all intake a total of 7000cal. The doctor and Jack agree that Jack will try to get down to the 72kg over the next 90 days. Jack, who is very good at arithmetic, calculated:

8 × 7000

90=622 and 2480−622 =1858.

He concluded that, if he reduced his daily intake to 1858 calories per day, he would reach his target weight in the 90 days. While Jack made his computations, the doctor consulted a table and informed Jack that he must go on a 1725cal per day diet! Jack was surprised and confused. He showed the doctor his calculations and asked why they were in error. The doctor responded that the calculations were correct but the argument was wrong.

Problem 1. Explain the error in Jack's argument.

It is true that, when Jack weighs 80kg, a 1858cal diet would give a daily deficit of 622cal which would go toward weight loss. But the deficit when Jack's weight has drops to say 78kg is only 31× 78 −1858 =560 calories. And as his weight drops, the deficits get smaller. Hence the deficits don’t add up to the required 56000 calories. When Jack finally understood the error in his argument, he asked the doctor where the 1725cal figure came from. The doctor showed Jack the table and explained that the entries in the table were computed from a linear difference equation! Jack went home, learned all about difference equations and finally came up with the following (correct) analysis.

Let d denote Jack's daily caloric intake (2480 now); let x0 denote Jack's weight today (80kg) and let xn denote Jack's weight n days from now. We wish to set d so that x90 will equal 72kg. Recall there are two constants:

31cal per day are needed to maintain each kg of Jack's weight and 7000cal below the maintenance level is needed for a 1kg weight

reduction.

For the nth day, we make the following computation: xn =xn−1 +d −31xn−17000

. Which

we explain by observing that d −31xn−1 is the excess (when positive) or deficit (when negative) in calories on the nth day and noting that this will result in a d −31xn−17000

kilogram weight change (increase when positive, decrease when

negative) on the nth day. Thus Jack's daily weight is given by the linear difference equation:

xn =69697000

xn−1 +d

7000.

We easily see the fixed point to be d31

. So, if d=2232 (Jack's first solution), Jack's

weight will eventually approach the fixed-point 72kg. Using the formula for the ith iterate, we can see just how this diet would work out.

xn =69697000

⎛⎝⎜

⎞⎠⎟n

(80 −72)+ 72

Problem 2. Fill in the following table of Jack’s weight if he were to go on a 2232cal diet:

n 90 180 365 730 1095

xn 77.4 75.6 73.6 72.3 72.06

Problem 3. Set up the difference equation for a diet of 1860 calories per day (Jack's second computation) and track Jack’s weight at 30-day intervals until it reaches 72kg.

The fixed point is now 186031

=60 and xn =69697000

⎛⎝⎜

⎞⎠⎟n

(80 −60)+ 60 is the new

difference equation and the new values are listed here.

Problem 4. Set up the difference equation for a diet of 1725 calories per day (from the doctor’s table) and track Jack’s weight at 30-day intervals until it reaches 72kg.

The fixed point is now 172531

=55.65 and xn =69697000

⎛⎝⎜

⎞⎠⎟n

(80 −55.65)+ 55.65 is the

new difference equation and the new values are listed here.

Worksheet #8

n 30 60 90 120

xn 77.5 75.3 73.4 71.7

n 30 60 90

xn 77 74.3 72

We close with an application of linear difference equations to the mathematics of personal finance. We start with a quick look at compound interest. Interest rates are always stated as annual rates and given as a percent. For example, your bank may offer a CD at 5.3% and your credit card may charge you 11.7% interest on your unpaid balance. In both cases it is understood that these are annual rates. Interest rates are always stated as annual percentages; for computations they are converted to decimals and prorated for the actual time period. For example the interest on

$1,000 at 6% for one month is given by1000 ×.0612

=5 . The quotient .0612

=.005 is

called the periodic rate and is usually denoted by i. If you were to deposit $1,000 in a bank account paying 6% for 3 years compounded monthly, you would have (1.005)36 ×1000 =1196.68 dollars in the account at the end of the 3 years:

at the end of the first month you have1000 + i ×1000 =(1+ i)1000 ; at the end of the second month, [(1+ i)1000] + i[(1=i)1000] =(1+ i)21000 ; at the end of the nth month the amount has grown to (1+i)n1000.

So we can think of compound interest in terms of a particularly simple linear difference equation: xn =(1+ i)xn−1 ; the amount at the end of the nth period is 1 plus the periodic rate times the amount at the end of the previous period.

Now consider a loan. Suppose you borrow $10,000 at 6 percent with regular monthly payments of $200. How much will you still owe at the end of one year? How long will it take to payoff the loan? At the end of the first month, just before you make your first payment, the bank will add that month’s interest to your balance due 1.005 ×10000 and then subtract off your first payment to get your new balance of $985. This process is then repeated each month: xn =1.005 × xn−1 −200 . You can easily track your loan on the TI84: type 10000 and press ENTER; type (1.005)×ANS−200 and press ENTER; press ENTER repeatedly to get each successive month’s balance. At the end of 57 months the debt will be reduced to $135.46. Of course, the payments on most loans are arranged to come out to exactly 0 at the end of a chosen time period. If this were a 5-year loan, the monthly payments would be $193.33. [Check this on the TI84.] So, it is natural to ask just how is the payment size computed? To answer that we must get a better understanding of this difference equation

Let’s consider a loan of L dollars at a periodic rate i and payments of size P. The linear difference equation that describes this loan is xn =(1+ i)xn−1 −Pwhere x0=L.

The first thing we do is to compute the fixed point: c =−P

1−(1+ i)=−P−i

=Pi . Since the

periodic rate is so small, the fixed point is a large positive number: c=200.005

=40000

in our example. Since the multiplicative factor 1+i is greater than 1, successive iterations move away from the fixed point; since the multiplicative factor is so close to 1, they move away rather slowly. It is also clear that, if L =x0 < c , the balance due xn will cross the 0 line for large enough n. The condition L < cis equivalent to the

condition iL < Pwhich simply states that the payments are larger than the monthly interest.

Above we have plotted the course of this loan with $200 payments. If we reduce the monthly payments on this loan to $150 it will obviously take longer to pay off. But how much longer; on the other hand increasing the payments to $350 will shorten the time pay off the loan.

Problem 1. Plot the trajectory of this $10,000 loan payment sizes $150 and $350.

Suppose that we wish to pay off this loan in 4 years. Start with the formula for xn re

written in terms of or loan symbols: xn =(1+ i)n(L−

Pi)+

Pi. If we wish to pay the

loan off in exactly N payments, we set xN =0 and then solve for P:

(1+ i)N (L−Pi)+

Pi=0

Multiplying through by i and regrouping: i(1+ i)NL−[(1+ i)N −1]P=0Finally, solving for P:

P =i(1+ i)NL(1+ i)N −1

=iL

1−(1+ i)−N

In our case, L=10,000 and i=.005; so we may write P as a function of N:

P =50

1−1.005−N .

If we wish to pay of the loan in 5 years: P =50

1−1.005−60 =193.33 ;

if we wish to pay of the loan in 3 years: P =50

1−1.005−36 =304.22 .

Problem 2. Compute the payment size if this were a 10-year loan. For each of these cases 3-year, 5-year and a 10-year loan, compute the total amount of interest that you will pay.

For the 10-year loan P =50

1−1.005−120 =111.02 .

Total interest = total payments – 10,000. For the 3-year loan: 36 × 304.22 −10,000 =951.92 ;

for the 5-year loan: 60 ×193.33−10,000 =1,599.80 ;for the 10-year loan: 120 ×111.02 −10,000 =3,346.4 .

It should be clear form these last examples that short-term and long-term loans behave very differently with respect to the amount of interest paid over the life of the loan. Most people encounter short-term car loans and long-term mortgages in their lifetime. In the last two problems we track one of each.

Problem 3. 1. Suppose you want to buy the new 2010 Civic Hybrid for $23,800 and you finance the full amount with a 5-year loan at 5.4%. What is the periodic rate? How much are your monthly payments?

i =0.05412

=0.0045 ; P =0.0045 × 238001−1.0045−60 =453.51

2. Given the information in part a), how much do you end up paying for the car at the end of five years? How much of that was interest?

Total payment $27,210.60 (453.51x60); total interest $3,410.60 (27210.60-23800)

3. Suppose you prefer the 2010 Civic Sedan with some extra features for $21,850. The auto dealer gives you two options: offers 0% financing for five years or $3,000 cash back and a 5.4% 5-year loan for the rest. Which is the better deal?

If you choose 0% financing, your monthly payments will be $364.17. (21,850/60)

If you choose cash back and use the $3,000 to reduce the amount of your loan to,

your monthly payments will be P =

0.0045 ×188501−1.0045−60 =359.19.

4. You promised your daughter that if she went to a public school for her undergraduate degree you would buy her a new car at graduation. Currently you can only afford an extra $400 a month towards the car and the best loan available is a 5 year loan at 5.4% APR. What price range should you be looking at?

Solving our payment formula for L =(1−1.0045−60 )P

iand evaluating at P = 400, we

see that we should be looking for a car that costs near $23,989.14.

Problem 4. As an example of a long-term loan, consider a $250,000, 25-year mortgage at 5.4%.

1. What are your monthly payments? P =0.0045 × 2500001−1.0045−300 =1520.33

2. How much do you end up paying for your house? How much is interest?

$1520.33 x 300 = $456,099 and $456,099 - $250,000 = $206,099 in interest.

3. How much of your first payment was interest and how much went to reduce your debt?

The interest accrued in the first month is .0045 x 250,000 = 1125. So $1,125 of you first payment is interest and $1,520.33 – $1,125 = $395.33 is the amount by which your debt is reduced.

4. How much do you still owe after 10 years?

By our basic formula x120 =1.0045120(250000 −

1520.33.0045

)+1520.33.0045

=187280.51

5. In order to reduce your payments you are considering a 30-year mortgage at he same rate. What would the payments be? How much would the house eventually cost?

Your new payments would be $1403.83 and the total amount that you would pay is $505,378.80

Worksheet #1

Worksheet #2

Compute the magnification, the orientation and the center for the linear function f (x)=−2x + 3 . Then check your answers by plotting the trajectories of a few points.

Compute the magnification, the orientation and the center for the linear function

f (x)=−12x + 3 . Then check your answers by plotting the trajectories of a few

points.

Worksheet #3

Problem 1. Consider a thermometer that reads both Fahrenheit and Centigrade.

1. Give the temperature in Fahrenheit as a linear function of the temperature in Centigrade.

2. Find the fixed point of this linear function and explain its physical meaning.

3. Compute the inverse function. What does it tell us?

Problem 2. Suppose that your average in a course just before the final exam is 80\% and that the final exam will account for 40\% of your grade. Consider your final average as a (linear) function f(x) of the score x on your final exam.

1. What is the fixed point of this function?

2. What is the magnification of this function?

3. What is the equation for this function?

4. What are the lowest and highest scores you could get in the course?

5. What is the inverse of this function and what does it mean?

Problem 3. Consider the meterstick that can be found on any classroom. It is designed to read from left to right in centimeters on one side and from left to right in inches on the other.

1. Give the linear function that, for any centimeter measure on this stick, gives the inches measure on the opposite side. (Note: this is not the usual centimeters-to-inches conversion function.)

2. Find the fixed point of this linear function and explain its physical meaning.

3. Compute the inverse function. What does it tell us?

Worksheet #4Some properties of linear functions:

Problem 1. Consider the linear function f (x)=ax+ b and compute the equation for the inverse function f −1(x) .

Problem 2. Consider the linear functions f (x)=ax+ b and g(x)=rx + s. Find the conditions on f and g so that they commute: f (g(x))=g(f(x)).

Worksheet #5

Problem 1. Consider f (x)=1.5x ; plot the first few iterates starting with 1, with -2:

Problem 2. Consider f (x)=1.5x +1; plot the first few iterates starting with 0, with -2, with -3:

Problem 3. Consider f (x)=−1.5x + 5 ; plot the first few iterates starting with 0, with 4:

Problem 4. Consider f (x)=0.8x + 0.4 ; plot the first few iterates starting with -8, with 7:

Conclusions??

Worksheet #6

Suppose that you own a piece of land with a pond that you would like to stock with trout. Trout will not reproduce in this pond but roughly two-thirds of the fish population will survive from one year to the next. Suppose that you initially stock the pond with 5000 fish and then add 2000 in each successive year.

Problem 1. Give the linear difference equation for the number of trout in the pond in successive years.

Problem 2. Consider the following questions

1. At what level will the fish population stabilize?

2. How important was the size of the initial stock?

3. How important was the size of the annual stock?

4. If the pond could accommodate a stable population of 7500 trout, how could you achieve this?

All of the fish that you have been using to stock your pond come from a friend who owns a pond in which trout do reproduce. Each year the trout population in his pond increases by 25\% until the maximum population of 12,000 trout is reached. Suppose that you stock your pond with 5000 fish the first year and 2000 a year thereafter. Assume that when you initially stocked your pond his pond was at its capacity of 12,000.

Problem 3. 1. Give the linear difference equation that governs the population size in

your friend's pond.

2. Assume that you continue to stock your pond with 2000 trout each year from your friend's pond. What will the trout population be in your friend's pond after 10 years?

3. Could this have been avoided and still stock your pond?

Worksheet #7

Another interesting and useful application of linear difference equations is to weight control. Lets consider Jack. He weighs 80 kilograms and, due to a high blood-pressure problem, his doctor has suggested that he reduce his weight by 10% to 72 kilograms. Jack maintains his weight of 80kg on an average of 2480 calories per day. We see that, with his present level of physical activity, Jack needs 31 calories per kilogram per day to maintain his weight.So, Jack's first reaction was: "This isn't too difficult; all I have to do is to reduce my daily intake by 10% to 2232 calories per day." The doctor acknowledged that Jack's calculations were correct but pointed out that, while a diet of 2232 calories per day would maintain a weight of 72kg, it would literally take years for Jack to get his weight down to that level on a diet of 2232 calories per day. In making his calculation, Jack was unaware of another important constant: the number of calories he needed to ``burn off" in order to lose 1 kilogram in weight. Given Jack's life style and metabolism the doctor believes that to lose 1kg in his weight, Jack must reduce his over all intake a total of 7000cal. The doctor and Jack agree that Jack will try to get down to the 72kg over the next 90 days. Jack, who is very good at arithmetic, calculated:

8 × 7000

90=622 and 2480−622 =1858.

He concluded that, if he reduced his daily intake to 1858 calories per day, he would reach his target weight in the 90 days. While Jack made his computations, the

doctor consulted a table and informed Jack that he must go on a 1725cal per day diet! Jack was surprised and confused. He showed the doctor his calculations and asked why they were in error. The doctor responded that the calculations were correct but the argument was wrong.

Problem 1. Explain the error in Jack's argument.

When Jack finally understood the error in his argument, he asked the doctor where the 1725cal figure came from. The doctor showed Jack the table and explained that the entries in the table were computed from a linear difference equation! Jack went home, learned all about difference equations and finally came up with the following (correct) analysis.

Let d denote Jack's daily caloric intake (2480 now); let x0 denote Jack's weight today (80kg) and let xn denote Jack's weight n days from now. We wish to set d so that x90 will equal 72kg. Recall there are two constants:

31cal per day are needed to maintain each kg of Jack's weight and 7000cal below the maintenance level is needed for a 1kg weight

reduction.

For the nth day, we make the following computation: xn =xn−1 +d −31xn−17000

. Which

we explain by observing that d −31xn−1 is the excess (when positive) or deficit (when negative) in calories on the nth day and noting that this will result in a d −31xn−17000

kilogram weight change (increase when positive, decrease when

negative) on the nth day. Thus Jack's daily weight is given by the linear difference equation:

xn =69697000

xn−1 +d

7000.

We easily see the fixed point to be d31

. So, if d=2232 (Jack's first solution), Jack's

weight will eventually approach the fixed-point 72kg. Using the formula for the ith iterate, we can see just how this diet would work out.

xn =69697000

⎛⎝⎜

⎞⎠⎟n

(80 −72)+ 72

Problem 2. Fill in the following table of Jack’s weight if he were to go on a 2232cal diet:

n 90 180 365 730 1095

xn

Problem 3. Set up the difference equation for a diet of 1860 calories per day (Jack's second computation) and track Jack’s weight at 30-day intervals until it reaches 72kg.

Problem 4. Set up the difference equation for a diet of 1725 calories per day (from the doctor’s table) and track Jack’s weight at 30-day intervals until it reaches 72kg.

Worksheet #8

We close with an application of linear difference equations to the mathematics of personal finance. We start with a quick look at compound interest. Interest rates are always stated as annual rates and given as a percent. For example, your bank may offer a CD at 5.3% and your credit card may charge you 11.7% interest on your unpaid balance. In both cases it is understood that these are annual rates. Interest rates are always stated as annual percentages; for computations they are converted to decimals and prorated for the actual time period. For example the interest on

$1,000 at 6% for one month is given by1000 ×.0612

=5 . The quotient .0612

=.005 is

called the periodic rate and is usually denoted by i. If you were to deposit $1,000 in a bank account paying 6% for 3 years compounded monthly, you would have (1.005)36 ×1000 =1196.68 dollars in the account at the end of the 3 years:

at the end of the first month you have1000 + i ×1000 =(1+ i)1000 ; at the end of the second month, [(1+ i)1000] + i[(1=i)1000] =(1+ i)21000 ; at the end of the nth month the amount has grown to (1+i)n1000.

So we can think of compound interest in terms of a particularly simple linear difference equation: xn =(1+ i)xn−1 ; the amount at the end of the nth period is 1 plus the periodic rate times the amount at the end of the previous period.

Now consider a loan. Suppose you borrow $10,000 at 6 percent with regular monthly payments of $200. How much will you still owe at the end of one year? How long will it take to payoff the loan? At the end of the first month, just before you make your first payment, the bank will add that month’s interest to your balance due 1.005 ×10000 and then subtract off your first payment to get your new balance of $985. This process is then repeated each month: xn =1.005 × xn−1 −200 . You can easily track your loan on the TI84: type 10000 and press ENTER; type (1.005)×ANS−200 and press ENTER; press ENTER repeatedly to get each successive month’s balance. At the end of 57 months the debt will be reduced to $135.46. Of course, the payments on most loans are arranged to come out to exactly 0 at the end of a chosen time period. If this were a 5-year loan, the monthly payments would be $193.33. [Check this on the TI84.] So, it is natural to ask just how is the payment size computed? To answer that we must get a better understanding of this difference equation

Let’s consider a loan of L dollars at a periodic rate i and payments of size P. The linear difference equation that describes this loan is xn =(1+ i)xn−1 −Pwhere x0=L.

The first thing we do is to compute the fixed point: c =−P

1−(1+ i)=−P−i

=Pi . Since the

periodic rate is so small, the fixed point is a large positive number: c=200.005

=40000

in our example. Since the multiplicative factor 1+i is greater than 1, successive iterations move away from the fixed point; since the multiplicative factor is so close to 1, they move away rather slowly. It is also clear that, if L =x0 < c , the balance due xn will cross the 0 line for large enough n. The condition L < cis equivalent to the condition iL < Pwhich simply states that the payments are larger than the monthly interest.

Above we have plotted the course of this loan with $200 payments. If we reduce the monthly payments on this loan to $150 it will obviously take longer to pay off. But how much longer; on the other hand increasing the payments to $350 will shorten the time pay off the loan.

Problem 1. Plot the trajectory of this $10,000 loan payment sizes $150 and $350.

Suppose that we wish to pay off this loan in 4 years. Start with the formula for xn re

written in terms of or loan symbols: xn =(1+ i)n(L−

Pi)+

Pi. If we wish to pay the

loan off in exactly N payments, we set xN =0 and then solve for P:

(1+ i)N (L−Pi)+

Pi=0

Multiplying through by i and regrouping: i(1+ i)NL−[(1+ i)N −1]P=0Finally, solving for P:

P =i(1+ i)NL(1+ i)N −1

=iL

1−(1+ i)−N

In our case, L=10,000 and i=.005; so we may write P as a function of N:

P =50

1−1.005−N .

If we wish to pay of the loan in 5 years: P =50

1−1.005−60 =193.33 ;

if we wish to pay of the loan in 3 years: P =50

1−1.005−36 =304.22 .

Problem 2. Compute the payment size if this were a 10-year loan. For each of these cases 3-year, 5-year and a 10-year loan, compute the total amount of interest that you will pay.

It should be clear form these last examples that short-term and long-term loans behave very differently with respect to the amount of interest paid over the life of the loan. Most people encounter short-term car loans and long-term mortgages in their lifetime. In the last two problems we track one of each.

Problem 3.

1. Suppose you want to buy the new 2010 Civic Hybrid for $23,800 and you finance the full amount with a 5-year loan at 5.4%. What is the periodic rate? How much are your monthly payments?

2. How much do you end up paying for the car at the end of five years? How much of that was interest?

3. Suppose you prefer the 2010 Civic Sedan with some extra features for $21,850. The auto dealer gives you two options: offers 0% financing for five years or $3,000 cash back and a 5.4% 5-year loan for the rest. Which is the better deal?

4. You promised your daughter that if she went to a public school for her undergraduate degree you would buy her a new car at graduation. Currently you can only afford an extra $400 a month towards the car and the best loan available is a 5 year loan at 5.4% APR. What price range should you be looking at?

Problem 4. As an example of a long-term loan, consider a $250,000, 25-year mortgage at 5.4%.

1. What are your monthly payments?

2. How much do you end up paying for your house? How much is interest?

3. How much of your first payment was interest and how much went to reduce your debt?

4. How much do you still owe after 10 years?

5. In order to reduce your payments you are considering a 30-year mortgage at he same rate. What would the payments be? How much would the house eventually cost?