class 051103

If you have $1000 to invest at 100% annual interest:

How much would you have at the end of a year?

How much would have have at the end of a year if you could compound your interest 2 times during the year?

How much would have have at the end of a year if you could compound your interest 360 times during the year?

How much would have have at the end of a year if you could compound your interest continuously?

Evaluate (1 + 1/n)^n for n = 1, n = 10 and n = 100.

What do your results have to do with the number e?

What do your results have to do with compound interest?

Simplify the following expressions according to the laws of exponents:

( (1 + .10 / 1000) ^ 1000 ) ^ t

( (1 + .10 / n) ^ n ) ^ t

( (1 + r / n) ^ n) ^ t

What is the t = 0 value of the function P = 1000 * (1.08)^t?

Write an equation setting the value of the function equal to double its t = 0 value.

Simplify the equation as much as possible.

If we know that y = A * b^t, and we know that at clock times t = 2 and t = 5 the values of y are respectively 7 and 30, then what are the values of our parameters A and b?

A certain function has equation y = 50 * 2^(kx), and the same function also has the equation y = 50 * b^x.  Write an equation we could solve to find the value of b.

 

 

You need to understand the general scheme of the functions we use in this course. 

The general quadratic function is y = a t^2 + b t + c; another general form is y = A ( x – h ) ^ 2 + k.  The parameters of the first form are a, b, and c.  The parameters of the second form are A, h and k. 

The general linear function is y = m x + b.  Its parameters are m and b.   

The general power function is y = A ( x – h ) ^ p + c.  If the power p is known, its parameters are A, h and c.  If p is not known then its parameters are A, h, p and c. 

The general exponential function has several forms, including  

·         y = A * 2^(k t) + c with parameters A, k and c.

·         y = A * b^t + c with parameters A, b and c.

·         y = A * e^(k t) + c with parameters A, k and c. 

Once you know the values of the parameters you plug these into the form of the function and you get the equation of the function. 

If you have enough points on the graph of the function, you can set up a system of simultaneous equations for the parameters.  If you can then solve the system you substitute the parameters into the form to get the equation of the function. 

If the function has three parameters and you don’t know the value of any of the parameters, then you should plug in the coordinates of 3 points in order to get 3 simultaneous equations for the parameters.  This is what we did early in the course with the y = a t^2 + b t + c form.  We plugged in y and t coordinates to get three simultaneous linear equations with unknowns a, b and c.  We solved by adding multiples of some equations to others, using the standard techniques of simultaneous linear equations. 

If the function has two unknown parameters you would plug in the coordinates of two graph points.  The general linear function has two parameters m and b, so if you know two points on the graph you can do get two equations to solve for m and b.   

The general form of the power function can have up to 4 parameters, and sometimes you don’t know any of them.  That would give you 4 simultaneous equations in 4 unknowns, and it is in general very difficult to solve all those equations.  However in many cases you know the power p so you can fill that in, and you would only need 3 points on the graph.  In many other cases you know that there is no horizontal or vertical shift, so h and k would both be 0.  This would leave you only 2 unknown parameters and you would need only two points on the graph. 

We solve the simultaneous equations using the techniques appropriate to the system.  The goal is generally to eliminate variables one at a time.  Sometimes adding multiples of one equation to another will eliminate variables.  Sometimes dividing one equation by another will eliminate variables.  We often have to multiply or divide both sides of an equation by the same quantity or by equal quantities, or add or subtract the same or equal quantities on both sides.  Sometimes we have to take powers of both sides of the equations, being careful to use the laws of exponents.  Sometimes, as we will soon see, we have to take logarithms of both sides.

Once we have a model we can ask several questions.  If we represent the model by the equation

y = f(t)

then some of the most common questions are:

On a graph we would locate y on the vertical axis, project horizontally to the graph, then project vertically to the t axis to find our value(s) of t.

In function notation this would be written ( f(t2) - f(t1) ) / ( t2 - t1).

e is the limiting value of (1 + 1/n) ^ n, as n approaches infinity.  The following sequence of assigned problems demonstrates this:

12. Use your calculator to evaluate (1 + 1/n) ^ n for n = 2, 4, 10, 100, 1000, and 10000. For each value of n, write down the difference between 2.71828 and your result.   Make a reasonable estimate of what the differences would be for n = 100,000 and for n = 1,000,000.

13. As n continues to increase, (1 + 1/n) ^ n continues to approach 2.71828. However, your calculator will eventually begin to malfunction as you attempt to use larger and larger numbers for n. Most calculators will begin giving smaller and smaller results, and will finally give just 1. This is a result of the approximate nature of the calculator's binary approximation to base-10 arithmetic, and to the limits of its precision.

Determine the approximate value of n at which your calculator begins to give you bad answers. Suggestion: use n = 100,000, then 1,000,000, etc. (just add another 0 each time).

14. Use DERIVE to determine the approximate number n required to obtain the value 2.71828.

However this question came from the second half of the worksheet and you've taken the test before the final due date for this worksheet, so any time before tomorrow's class I'd omit this problem.

A recurrence relation can be used to generate a sequence of numbers.  The sequence of numbers can then be analyzed to see if successive difference sequences eventually become constant, or if some combination of ratio and difference sequences become constant.

The recurrence relation s(n) = s(n-1) + 3 n, with s(1) = -2, yields s(2) = 4, s(3) = 13, s(4) = 25, s(5) = 40, s(6) = 58, etc..  Writing these numbers as a sequence we get

-2, 4, 13, 25, 50, 58, ... .  The differences of this sequence are

6, 9, 12, 15, 18, ... and the differences of this sequence are

3, 3, 3, 3, 3, ... .

Since the second-difference sequence is constant, the original sequence is quadratic in nature.  A graph of s(n) vs. n would be modeled exactly by a quadratic function.  This graph would contain the points (1, -2), (2, 4), (3, 13), (4, 25), (5, 40) and (6, 58).

If we substitute these results into the form s = a n^2 + b n + c, using the points (1, -2), (3, 13) and (6, 58), we would get the equations

-2 = a * 1^2 + b * 1 + c

13 = a * 3^2 + b * 3 + c

58 = a * 6^2 + b * 6 + c.

If we solve these equations of a, b and c we will get a = 3/2, b = 3/2, c = -5.  So our model is

s = 3/2 n^2 + 3/2 n - 5.

Using function notation s(n) we could also write this as

s(n) = 3/2 n^2 + 3/2 n - 5.

If we evaluate s(n) for n values 1, 2, 3, ..., 10 we get -2, 4, 13, 25, 40, 58, 79, 103, 130, 160.

If we want to know the 100th member of the sequence, we just plug in 100.  We would get s(100) = 3/2 * 100^2 + 3/2 * 100 - 5 = 15,145.

This sequence not only agrees with the results we obtained from the recurrence relation, it extends them and allows us to find the value of s(n) for any n without having to find all the values in between.