Walton's JMU Math Blog

Proof by Induction and Summations

2011-09-14T21:07:00.000-07:00

The two previous blog entries introduced the idea of the Principle of Mathematical Induction followed by a discussion of a typical example of a proof by induction. Be sure you read those two entries before you look at this one. This blog post extends these ideas by talking about how proof by induction applies to summations.

(I apologize right now about the formatting of summation notation, or sigma notation. I do not know how to get the math to look like math in a blog setting.)

Recall that the second condition of the PMI is that an arbitrary statement S(n) in the chain of statements being proved will guarantee that the next statement S(n+1) in the chain is also true. In a proof by induction, this step always involves using a recursive relation between something in the sentence S(n) with a corresponding object in the sentence S(n+1).

For summations, this recursive relation is always that a summation for statement S(n+1) exactly corresponds to the summation appearing in statement S(n) with some additional term(s).

For example, think back to our motivating example of a chain of statements:
S(1):    1 = 1(2)/2
S(2):    1+2 = 2(3)/2
S(3):    1+2+3 = 3(4)/2
S(4):    1+2+3+4 = 4(5)/2
S(5):    1+2+3+4+5 = 5(6)/2
...
Notice that the sum appearing on the left hand side of any given sentence appears in the sum on the very next sentence, but one more term has been added. Using parentheses to emphasize where the sum from the previous sentence appears in the new sentence, here is the same chain:

S(1):    1 = 1(2)/2
S(2):    (1)+2 = 2(3)/2
S(3):    (1+2)+3 = 3(4)/2
S(4):    (1+2+3)+4 = 4(5)/2
S(5):    (1+2+3+4)+5 = 5(6)/2
...

Summation (Sigma) notation gives us a handy way to describe sequences of sums like we see in the chain above. The terms of the sum follow a simple pattern. In this example, the pattern is the sequence of terms (1, 2, 3, 4, 5, ...). This sequence can be explicitly described as a_k = k for k=1, 2, 3, .... In the symbolism (notation) of summation notation, we write:

Σ¹_k=1(k) = 1
Σ²_k=1(k) = 1+2
Σ³_k=1(k) = 1+2+3
or
Σⁿ_k=1(k) = 1+2+3+...+n
(Notice that the formula in the parentheses (k) is the explicit formula of the sequence of terms in terms of the index which is listed at the bottom of Σ along with the first index of the sequence used in the sum. The index of the last term in the sum appears at the top of Σ.)

The pattern that we described above to illustrate the recursive relation between the sums of consecutive sentences in the chain of statements we wish to prove can also be written in terms of summation notation:
Σⁿ⁺¹_k=1(k) = Σⁿ_k=1(k) + (n+1)
When writing a proof by induction, we use this recursive relation.

Example: Prove that Σⁿ_k=1(k) = n(n+1)/2 for n=1, 2, 3, ...

In the proof, we will write S(n) to represent the sentence: Σⁿ_k=1(k) = n(n+1)/2.

Proof:
(1) We first need to think about the sentence where n is replaced by 1. That is, we need to prove that Σ¹_k=1(k) = 1(1+1)/2.
Σ¹_k=1(k) = 1 (Interpretation of summation notation)
(We are creating a true statement involving the summation symbol that appears on the left side of the equation we are proving. The sequence of terms is 1, 2, 3, 4, 5, ..., and the sum starts and ends with the 1st term.)
1(1+1)/2 = 1(2)/2 = 1 (Algebra)
(We now create an equation involving the formula on the right side of the equation we are proving, and we just showed it has the same value as the summation symbol.)

So Σ¹_k=1(k) = 1(1+1)/2.
(We obtained evidence above that the summation and the formula both represented the same value, so we conclude that they are equal.)

(2) We now need to show that S(n) implies S(n+1) for n=1, 2, 3, .... We will assume S(n), i.e. Σⁿ_k=1(k) = n(n+1)/2. Using the recursion connecting the sums, we will show S(n+1), i.e., Σⁿ⁺¹_k=1(k) = (n+1)((n+1)+1)/2.

Assume    Σⁿ_k=1(k) = n(n+1)/2    for some n=1, 2, 3, ...
Σⁿ⁺¹_k=1(k) = Σⁿ_k=1(k) + (n+1). (Recursive relation on summation of terms a_k = k)
Σⁿ⁺¹_k=1(k) = n(n+1)/2 + (n+1)     (Substitution: summation replaced by formula)
Σⁿ⁺¹_k=1(k) = n(n+1)/2 + 2(n+1)/2     (Find common denominator)
Σⁿ⁺¹_k=1(k) = (n²+n+2n+2)/2     (Distribution and adding fractions)
Σⁿ⁺¹_k=1(k) = (n²+3n+2)/2     (Combining terms)
(Notice that each equation was based on the previous equation. The key step was when we replaced Σⁿ_k=1(k) by the formula n(n+1)/2 that was provided by the assumed hypothesis. At this stage of the proof, we have a formula representing the value of the summation symbol with n+1 that is the left hand side of the equation in the sentence S(n+1). We now need to check the formula that appears in the right hand side of our sentence.)
(n+1)((n+1)+1)/2 = (n+1)(n+2)/2    (Arithmetic: 1+1=2)
(n+1)((n+1)+1)/2 = [n(n+2)+1(n+2)]/2     (Distributive law)
(n+1)((n+1)+1)/2 = [n²+2n + n+2]/2     (Distributive law again.)
(n+1)((n+1)+1)/2 = [n²+3n+2]/2     (Combining terms)
(Notice that each equation was based on the previous equation. Notice that the second and third steps (distributive law) are the formal steps in the idea commonly taught as FOIL when multiplying to binomial expressions together. The key is that we took the formula with n replaced by n+1 of the right hand side and discovered that it equals the same formula that we found for the summation above.)
So we have Σⁿ⁺¹_k=1(k) = (n+1)((n+1)+1)/2.
Consequently, S(n) implies S(n+1) for n=1, 2, 3, ....

By the PMI (since S(1) is true and S(n) implies S(n+1)), we have S(n) is true for n=1, 2, 3, ...
♦

Mathematical Induction (part 2)

2011-09-14T20:17:00.000-07:00

My previous blog post introduced the basic idea of what mathematical induction is about. This post focuses more on the mechanics of writing a proof.

A proof is the mathematical form of argument or persuasion. A proof consists of a sequence of logical statements, each of which is shown to be a true sentence based only on information previous to that sentence. Examples of sentences that might appear in a proof are equations or inequalities for which there is clear reason that it is true, and never based on what we hope is true or will later show is true.

For example, suppose I needed to prove (x+1)²=x²+2x+1. In a proof, I can not write down this equation as the first statement because it is not something I know (yet). Instead, I can write down equations known to be true based on basic principles:

(x+1)² = (x+1)(x+1) (meaning of power 2)
(x+1)(x+1) = x(x+1) + 1(x+1) (distributive law)
x(x+1) + 1(x+1) = x²+x + x+1 (distributive law again)
x²+x + x+1 = x²+2x+1 (collecting like terms)

Notice that each of the sentences is a statement of equality. Although they to suggest that the "=" sign is being used in a computational sense (i.e., it looks like it is being used to say apply a rule), we really are seeing these sentences as declaring that the result of applying a rule demonstrates that the sentence is actually true. (This is a subtle distinction that you need to fight your mind until it sinks in.) Technically, our proof is not yet complete. Each sentence on its own is a complete, true sentence. However, we need to end by stating that the sentence we were trying to prove is actually true:

(x+1)²=x²+2x+1 (equivalence of equal quantities, or substitution)

Now, you should read the above paragraphs as illustrating the ideas of a proof in that it illustrates how sentences (equations) are listed as statements that are demonstrated to be true. But so far, we have not dealt with the idea of implication.

Recall that the Principle of Mathematical Induction (PMI) involves verifying the two conditions:

Show that the first statement in the chain, which we call S(1), is true.
Show that if any single statement in the chain, which we call S(n), is true, then this implies the next statement, which will be S(n+1), is also true.

So the structure of every proof by induction involves two subproofs (showing S(1) is true; showing S(n) implies S(n+1)) followed by an application of the PMI. The subproofs are the method we verify that the two conditions of the PMI have been satisfied.

Here is what to look for in the subproofs.

S(1) is almost always very easy to show. However, you still need to be clear that you follow the pattern of a proof described above.
Showing S(n) implies S(n+1) will consist of assuming that S(n) is true for some n (don't forget that this is a specific but unspecified value, so you must leave n or k, depending on the label being used, as a symbol and not an actual number). Then the proof will almost always rely on some type of recursive relation between a quantity involved in the statement S(n) and a similar quantity in the statement S(n+1).
After the subproofs are complete, you must invoke the PMI and then declare that the statements are true for every n=1, 2, 3, ....

Example: Given a sequence defined by x1=3 and the recursive relation x_k+1=x_k+2 for k=1, 2, 3, ..., prove that x_k=1+2k for k=1, 2, 3, ....

Notice what the chain of statements we are trying to prove will be. The sentence S(k) is the statement x_k=1+2k, where x_k is the value of the sequence defined recursively, and 1+2k is just a formula involving k. It is very important to remember that the "=" sign in this equation is not defining the value of the sequence, but is simply stating that the two quantities (the sequence value and the formula value) happen to always be equal.

Everything in italics is not part of the proof.

Proof:
(1) We first write a subproof that S(1) is true: x₁=1+2(1). To do this, we need to create a sequence of equations that we know are true based on the symbols themselves, and not the equation above.

x₁=3 (Given)
(We looked at the given information and saw this was provided.)
1+2(1) = 1+2 = 3 (Arithmetic)
(We needed an equation involving 1+2(1) so we wrote down an equation that was based on the rules of evaluating this formula.)
So x₁=1+2(1).   (Equivalence)
(Earlier, we showed x₁=3 and then we showed 1+2(1)=3, so this means the two quantities are equal. This ends the subproof since we just finished writing the statement corresponding to S(1) being true.)

(2) We next write a subproof that S(k) implies S(k+1) for k=1,2,3,... To do this, we will start by assuming S(k) is true for some unspecified k. Using the recursion to compute x_k+1 we will demonstrate that S(k+1) is also true. Note: S(k+1) is the statement x_k+1=1+2(k+1).

Assume x_k=1+2k is true for some k=1, 2, 3, ....
x_k+1=x_k+2 (Given recursive definition of sequence)
(The statement we are trying to prove involves the symbol x_k+1 so we need to create equations based on that symbol.)
x_k+1=(1+2k)+2 (Substitution: x_k=1+2k from assumed hypothesis)
(This is the key step: we use the recursive connection between x_k+1 and x_k to establish a formula for x_k+1.)
x_k+1=3+2k    (Algebra)
(We now have a simple formula for x_k+1 and now we turn our attention to the other half of the statement S(k+1), namely the formula 1+2(k+1).)
1+2(k+1) = 1 + 2k + 2   (Distributive law)
(We are creating a true equation involving the formula by considering the results of applying mathematical laws.)
1+2(k+1) = 3+2k    (Algebra from above: 1+2=3)
(Our target should always be that the formula with k+1 in place of k will match whatever we found by the recursion to compute a value for x_k+1.)
So x_k+1=1+2(k+1)    (Equivalence or Substitution)
That is, S(k) implies S(k+1) for k=1, 2, 3,....
(This ends the second subproof because we just finished writing the statement we were trying to prove in this part. We are now ready to invoke the PMI.)

Since S(1) is true and S(k) implies S(k+1) for k=1, 2, 3, ..., the Principle of Mathematical Induction guarantees S(k) is true for every k=1, 2, 3, .... That is, x_k=1+2k for k=1, 2, 3, ...
♦

Mathematical Induction

2011-09-14T19:14:00.000-07:00

I actually posted on this very topic a few years ago. However, I have refined how I think about doing proofs by mathematical induction. And so I am writing one more time.

First, you might find it interesting to look at the Wikipedia article on mathematical induction.

The Principle of Mathematical Induction (PMI) is an axiom that describes the set of natural numbers, which is N = {1, 2, 3, 4, ...}. From one point of view, the PMI says that if S is a set that (1) contains the number 1 and (2) guarantees that whenever any number n is in the set, then n+1 is also in the set, then the set S contains all of N. From the perspective of proofs, however, we are really interested in showing that an infinite chain of logical statements is true.

Before we proceed with the idea of the PMI, let me be clear about a logical statement. A logical statement is a well-constructed sentence that is definitively true or false. Two well-known but easily misunderstood examples of logical statements are equations and inequalities. An equation "A=B" is a logical statement that declares two quantities (A and B) are equal (have the same value). An inequality "Afalse, but the statement itself is logically complete. (Here logical does not mean `makes sense' but it means `can be decided between True or False'.)

Related to this issue is a common misunderstanding by students that "=" is like an operation that means "has the value" or "find the answer" as it is often used on a calculator. This is especially important in a proof, where each statement (usually an equation) must be clearly true rather than a statement of what you hope or what you are trying to calculate.

Okay, back to the Principle of Mathematical Induction. This principle is about dealing with an infinite chain of logical sentences. For example, consider the following chain of statements. S(1) is the label for the first sentence, S(2) is the label for the second sentence, and so on:

S(1):     1 = 1(2)/2
S(2):     1+2 = 2(3)/2
S(3):     1+2+3 = 3(4)/2
S(4):     1+2+3+4 = 4(5)/2
S(5):     1+2+3+4+5 = 5(6)/2
S(6):     1+2+3+4+5+6 = 6(7)/2
...
(the pattern continues so that if n is any number n=1,2,3,..., we have):
S(n):     1+2+3+...+n = n(n+1)/2

On the left of each sentence is a summation. On the right of each sentence is a simple formula involving n. Using summation notation, we would have written this sentence:
S(n):    Σⁿ_k=1 k = n(n+1)/2

The pattern described by S(n) looks like a single sentence, but it is important to remember that it is describing the entire infinitely long chain of logical sentences. If you add up the values on the left side of any single sentence and compare it to the value of the formula on the right, you will see that the answers are the same. But this only verifies the formulas that you actually check.

The Principle of Mathematical Induction is a tool to prove that the entire chain of sentences are all true. The PMI is much like an infinite chain of dominoes (see the earlier linked wikipedia article). To knock over a chain, if you knock them down one at a time, you'll never finish. But if you show that knocking any single domino down will guarantee the next domino also falls and you show that you can knock down the first domino, then this is enough to guarantee that the entire chain will fall down.

So here is the PMI:

Suppose S(n), n=1, 2, 3, ..., is a chain of logical statements and that (1) S(1) is true and (2) S(n) implies S(n+1) for any n=1, 2, 3, ..., then S(n) is true for every n=1, 2, 3, ....

Notice that there are two conditions to use PMI:

We must verify that S(1) (the first sentence) is true.
We must show that the sentence S(n+1) is true whenever the previous sentence S(n) is true.

A proof by induction is the process whereby we verify these two conditions and then apply the PMI to conclude that the entire chain is true.

To learn more about writing the proofs and an example with commentary, please continue by reading the next blog post.

How We Learn Mathematics

2011-03-08T09:44:00.000-08:00

I was reading the following paper: D. Breidenbach, E. Dubinsky, J. Hawks, and D. Nichols, "Development of the Process Conception of Function," Educational Studies in Mathematics, 23: 247-285, 1992.

Quote Dubinsky (1989): "A person's mathematical knowledge is her or his tendency to respond to certain kinds of perceived problem situations by constructing, reconstructing and organizing mental processes and objects to use in dealing with the situations."

"Applying this point of view to mathematics (or any other subject) consists of determining the nature of the specific processes and objects that are constructed and how they are organized when one studies mathematics"

Ways of thinking about functions:

prefunction - does not understand any real ways of using function concepts
action - repeatable mental or physical manipulation (e.g., plug in numbers and calculate); static; one step at a time
process - think of function as a single dynamic transformation

I then found another article: A. Sfard and L. Linchevski, "The Gains and the Pitfalls of Reification: The Case of Algebra," Educational Studies in Mathematics, 26 (2/3), 191-228, 1994 [Learning Mathematics: Constructivist and Interactionist Theories of Mathematical Development]

This article proceeds with the view that in mathematics, there is a duality in mathematical constructs being a process or an object. That is, conceive of things operationally (process) or structurally (object). Historical examples include the expansion of number systems: positive to negative (operational: subtraction as adding a negative to structural: negative numbers as objects), and real to complex (i=sqrt(-1) as an operational convenience to an actual object)

An included reference suggests finding another article: Kieran, C.: 1992, 'The learning and teaching of school algebra', in D. A. Grouws (ed.), The Handbook of Research on Mathematics Teaching and Learning, Macmillan, New York, pp. 390-419. I'll have to see if I can find this one, as it is cited for the sentence, "[reification] was also used to introduce some order into the quickly growing bulk of findings about algebraic thinking."

Interesting phrase: "the ability to grasp the structural aspect is not easy to achieve" and "those crucial junctions in the development of mathematics where a transition from one level to another takes place are the most problematic."

Another interesting way to think about how mathematics is organized: (1) Logical, or the way it fits together; (2) Historical, or the way in which it was developed; and (3) Cognitive, or the processes in which people learn.

Modes of Algebra
1.1) Algebra as Generalized Arithmetic: The Operational Phase
-- solve for the unknown, but not using symbols (grade school algebraic thinking)
-- rhetoric algebra
-- principally reversing processes
1.2) Algebra as Generalized Arithmetic: The Structural Phase
1.2.1) algebra of a fixed value (unknown)
-- Notational convenience, but treat variable as a fixed value
:::: becomes a mental challenge to think of formula as both a process and result
:::: example given: 2+3 represents process, 5 represents result. But x+3 represents both, no separate "result"
:::: compare to the challenges of new number types required to think about division, subtraction, and extracting square roots
** Nice comment: "Once we manage to overcome this difficulty, it is quickly forgotten. ... Our eyes are easily blinded by habit and by our own ontological beliefs. Nevertheless, much evidence for the difficulty of reification may also be found in today's classroom, provided those who listen to the students are open-minded enough to grasp the ontological gap between themselves and the less experienced learners."
1.2.2) Functional algebra (of a variable)
:::: View formula as object
:::: Parameters represented as symbols not numbers.
2) Abstract Algebra

Give examples of interview questions. Students at early stages of thinking think about formulas as recipes for computations (process) but do not perceive them as valid objects. "The equality sign is interpreted as a 'do something signal' (Behr et al 1976; Kieran 1981)"

Here's something I see all the time in calculus classes: "It [the = symbol] serves here as a 'run' command. When treated in this way, the equality symbol looses [sic] the basic characteristics of an equivalence predicate: it stops being symmetrical or transitive. Indeed, young children seem to have no qualms about solving word problems with the help of a chain of non-transitive equalities. For instance, when asked 'How many marbles do you have after you win 4 marbles 3 times and 2 marbles 5 times?', the child would often write: 3*4=12+5*2=12+10=22."

Equations of the form 2x-3 = 11 can be interpreted as a formula whose result is 11 (which can be solved by inverse operations); equations of the form 2x-3=5x-9 appear to be two different formulas, and inverse operations do not make sense.

Graph from a graph of f ' (x)

2010-04-28T19:57:00.001-07:00

First, pay attention: the graph provided on the assignment is the graph of the derivative f '(x) and not the graph of f. So you can't look at the picture and say that because the graph you are looking at is increasing that f '(x) is positive; if the graph is increasing, then that means f '(x) is increasing, and not f(x). (This is useful information, but you just need to think about what it does say.)

Second, the number line sign analysis summaries will help identify the shape of the graph. Imagine taking the unit circle and breaking it up according to quadrants. The signs of f '(x) and f ''(x) determine which of these four basic shapes the graph is most like.

f '(x) = + and f ''(x) = + means f(x) looks like Quadrant IV (incr, conc. up)
f '(x) = - and f ''(x) = + means f(x) looks like Quadrant III (decr, conc. up)
f '(x) = + and f ''(x) = - means f(x) looks like Quadrant II (incr, conc. down)
f '(x) = - and f ''(x) = - means f(x) looks like Quadrant I (decr, conc. down)

The graph is just formed by taking these shapes and putting them end-to-end. You wouldn't actually use the entire portion of the unit circle because we probably don't want vertical tangents like the unit circle has. The circle just helps us remember the basic shape. The points where we join the shapes together will probably be inflection points (concavity changes) or extreme values.

However, sign analysis does not tell us the heights of any points. The problem gives only one point: f(0) = 1. The rest of the points of interest (especially the local extreme values) can be found by thinking about the information relating to the areas of the graph of f '(x). (Again, think about the Fundamental Theorem of Calculus).

Sums of Geometric Sequences

2010-04-28T19:47:00.001-07:00

The second problem on the project introduces a new closed form for a sum:

∑_k=1ⁿ [A ρ^k] = A ρ (ρⁿ-1)/(ρ-1).

Unfortunately, too many of you are still intimidated simply by the symbols that are used.

The formula for the geometric sequence, A ρ^k, is like an exponential, except the power is an integer variable rather than a continuous variable like x. For example, if A=2 and ρ=1/3, we have terms that are increasing powers of (1/3) times 2:
2/3 (k=1), 2/9 (k=2), 2/27 (k=3), 2/81 (k=4), etc.

The summation is simply the sum of these values:
∑_k=1ⁿ [2 (1/3)^k] = 2/3 + 2/9 + 2/27 + 2/81 + ... + 2/3ⁿ.
The closed form gives a formula answering the value of this sum:
2(1/3)[(1/3)ⁿ-1]/[(1/3) - 1] = (2/3)*[(1/3)ⁿ-1]/(-2/3) = 1 - (1/3)ⁿ.

So when you write down the Riemann sum for the integral in question, you need to look at using the properties of exponentials so that the Riemann sum looks just like a sum of a geometric sequence. You should identify a factor that does not involve k, and this is A. You should identify the other factor as some number raised to the k power. Then you can use the closed form.

Populations, Birth Rates, and Death Rates

2010-04-28T19:25:00.000-07:00

This entry is a general assist for my class working on a project. Suppose you knew the rate at which births are occurring (call it a function of time, b(t)) and you knew the rate at which deaths are occurring (a function d(t)). If the only way the population changes is through births and deaths, then if P(t) is the function describing the size of the population in time, then P'(t) = b(t) - d(t). (It is still your job to explain why this makes biological sense.)

Okay, now for the general principle. Anytime you know the rate of change of a quantity, you can always get back to the original quantity through a definite integral (assuming the rate of change is continuous, anyway). This is the heart of the 2nd Fundamental Theorem of Calculus. Not using P and t as variables (so that you have at least something to translate), here is the basic idea.

Suppose you know f '(x). Then A(x) = ∫₀^x f '(z) dz is an antiderivative of f '. But so is f(x) since that is where f '(x) comes from. That is f(x) = A(x) + C for some constant. In particular, A(0) = 0, so C=f(0). That is, f(x) = f(0) + ∫₀^x f '(z) dz.

This will always work, even if I don't start the integral at 0: f(x) = f(a) + ∫_a^x f '(z) dz. Written another way, it looks like the first Fundamental Theorem of Calculus: f(x) - f(a) = ∫_a^x f '(z) dz.
In other words, the 2nd FTC implies that every function is its starting value plus the integral of its rate of change.

Now, for our population problem, we don't actually know the rate of change completely; we only know the value at specific points. So instead of computing an integral (to get an exact value), we will approximate the integral using a Riemann sum. We are restricted to using the table data, so Δt=2 is forced upon us. For example, ∫₀² b(t) dt can only be estimated with a single rectangle while ∫₀⁴ b(t) dt would involve two rectangles. The idea of the Riemann sum is that we choose b(t_k*) as one of our data points (either on the left or right).

More specifically, on the interval [0,2] (k=1), we can either use t₁*=0 so that b(t₁*)=100 or use t₁*=2 so that b(t₁*)=135. In the first case, the rectangle for k=1 contributes b(t₁*)Δt = 200, while the second case leads to a contribution of b(t₁*)Δt = 270. The average (midpoint) of these two values (200+270)/2 = 235 is the estimate that would come from using the trapezoid sum. We do this for each of the 8 intervals between our data points, for both the births and the deaths.

By considering our estimates for the number of births and deaths in each of the intervals ([0,2], [2,4], [4,6], etc.), we can produce an estimate of the new population at each of the times (2, 4, 6, 8, etc.). By thinking about what estimates lead to the largest predicted population, we get an upper limit (i.e. bound) for our estimate --- no population consistent with this data can ever go above that value. Similarly, we can choose those estimates to create a lower bound. The true population will be somewhere in between.

Exponential Functions

2010-02-05T05:32:00.000-08:00

I'm getting feedback that exponential functions are giving you extra trouble. I'd appreciate getting feedback to help know how I can clarify the concepts. Here is a summary of some of the key concepts that I'm wanting you to understand:

b^x is not just a formula "b to the power x" but is a new function, which I'm asking you to call exp_b(x).
The properties of exponents like b^x+y=b^x b^y and (b^x)^y=b^xy become properties of the exponential functions.
exp_b(x+y)=exp_b(x)*exp_b(y)
exp_b(xy)= [exp_b(x)]^y

Logarithms are the inverse functions of exponential functions.
exp_b(log_b(x))=x
log_b(exp_b(x))=x

In formula representation, these are written as follows:
b^log_b(x)=x
log_b(b^x)=x
Whenever you see a formula with an exponential, say b^3x-2, you should be able to think in both formula and function modes interchangeably.
b^3x-2 = b^3x b^-2 = (b³)^x b^-2
b^3x-2 = exp_b(3x-2)
The first mode allows us to recognize that b^3x-2t is actually of the form Aq^x where A=b^-2 and q=b³. The second mode is useful to remind us that we really have a composition when we need to compute a derivative or when dealing with inverse functions.
There is a special base e that is most important because the corresponding exponential function is its own derivative.
exp'_e(x)=exp_e(x)
d/dx[e^x]=e^x

The natural exponential is written without a base. The corresponding inverse function is called the natural logarithm, ln(x).
exp(ln(x))=x
ln(exp(x))=x

In formula representation, these are written as follows:
e^ln(x)=x
ln(e^x)=x
The x in these formulas, as always, is a placeholder. So any number or formula could be used in place of x.
Any exponential can be written in terms of the natural exponential. The key is to use the properties of exponents.
b = e^ln(b)
b^x = [e^ln(b)]^x = e^ln(b)^x
Another way to think of this is using composition of inverse functions: exp(ln( ))
b^x = exp(ln(b^x))
ln(b^x) = x ln(b)
b^x = exp(x ln(b)) = e^{x ln(b)}

That is, by replacing b^x by e^ln(b)^x, we can write any exponential A b^x in the form A e^kx where k is the number ln(b).
Derivatives of exponentials use the basic property that exp'(x) = exp(x), or d/dx[e^x] = e^x. Usually, this also requires the chain rule: d/dx[e^u] = e^u u'.
d/dx[e^2x] = e^2x (2) = 2e^2x (u = 2x)
d/dx[e^-3x] = e^-3x (-3) = -3e^-3x (u = -3x)
d/dt[e^(t²-5t)] = e^(t²-5t) (2t-5) = (2t-5)e^(t²-5t) (u = t²-5t)

So this was a moderately long list. Perhaps just re-reading it helped you understand something better. Or perhaps you realize something is still confusing. Please post comments explaining why you are finding problems challenging or perhaps explaining what helped you suddenly understand what you had missed earlier.

Concerns about Derivatives

2009-11-17T12:11:00.001-08:00

Okay, I have just finished grading HW #4. This assignment had some differentiation rules. But many of you are not comprehending the purpose of a derivative rule.

For example, we had the special rule for squares of functions:

d/dx[f²(x)] = 2 f(x) f '(x)

This means that any time there is a formula squared and you need to take its derivative, you can apply this rule.

(2x+3)² corresponds to the function f(x)=2x+3 being squared. So since f '(x) = 2:

d/dx[(2x+3)²] = 2 (2x+3) (2) = 4(2x+3)

Similarly, (x²-4x+5)² corresponds to the function f(x) = x²-4x+5 being squared, and f '(x)=2x-4

d/dx[(x²-4x+5)²] = 2 (x²-4x+5) (2x-4)

You need to be an expert at identifying the form of an expression in order to apply appropriate rules of differentiation, and next semester, rules of integration (anti-differentiation).

Science and Mathematics

2009-09-03T08:31:00.000-07:00

The other day, I had my students respond to a question about how mathematics relates to science.

In class, I had pointed out that mathematical definitions are very precise while scientific measurements can be rather messy. A mathematician has a very precise meaning when they say that two variables are proportional or have a linear relation. But when we get real data, even if they do not satisfy these precise meanings, we still gain significant information about the relation and might even say that the measured variables are proportional or linear. Unfortunately, many students seemed to think I was looking for a repeated discussion of this point.

Science can be thought of as the study of the physical world through the scientific method. Essentially, we make observations on what happens in the world (whether that be physical, chemical or biological interactions) and want to understand why that is happening as well as to predict what will happen in the future. In order to do this, scientists propose various hypotheses based on their observations (and past accumulated scientific experience) and then test those hypotheses. Experiments quite often include quantitative measurements, and part of the prediction is often to propose relationships between independent variables (the variables in treatments and control) and the dependent variables (outcomes). Experience may support a hypothesis or falsify the hypothesis, but it never can prove a hypothesis.

Knowledge based on patterns that we predict will continue, but which we can support but never prove, is called inductive knowledge. Science is an example of inductive knowledge.

Mathematics can be thought of as the study of structures that satisfy very specific rules. We have properties of arithmetic, algebra, and calculus. We establish specific axioms that describe our basic assumptions about the structures and then use logical argument to deduce the behavior of more complicated constructions. We might look at examples to see what ideas might be true or false, and in this sense mathematics can also take advantage of inductive knowledge. However, the objective in mathematics is not just to suppose that a pattern will continue; the objective is to determine conclusively if it must continue. We seek for proofs (that it is true) or counterexamples (break the pattern).

Knowledge based on basic assumptions (axioms) and logical argument that determines conclusively what must follow from these assumptions is called deductive knowledge. mathematics is an example of deductive knowledge.

Models form a connection between mathematics and science. Data often appear to follow a general trend, even in the presence of the noise of messy observation. A mathematical model takes that messiness and forms an abstract clean relationship that mathematics can work with. Based on the deductive approach of mathematics, we can often establish consequences of the assumed model form. We then apply those consequences as hypotheses in our scientific framework. The predictions from the deductive approach provide the predictions that can be used to falsify these hypotheses.

First Reading Assignment

2009-08-24T10:36:00.000-07:00

Since I am not sure when Blackboard is going to be available for the class (I procrastinated asking for the two sections to be merged into a single section), here is the reading assignment and preparation for classwork for Wednesday.

Reading Assignments:

Online Calculus Textbook: Read Sections 1.1 and 1.2 (link below). These sections emphasize the idea that variables (which represent physical quantities) can be related, as independent and dependent variables. We want to think about related quantities throughout this semester. This text specifically asks for the web-browser Firefox version 3.0 or later (this is to render formulas correctly).

Come prepared to class having prepared answers for Problems 3, 4, 6, 7, 10 from Section 1.2 Problems. We will discuss these problems but will not turn them in.

How Students Learn: This book prepared by the National Academy of Sciences is actually written for teachers to focus on making courses better suited for students to learn. The first 12 pages introduce three concepts that students should be aware of in their own learning. The link below goes to the first page, and then follow the links to read through page 12.

How Students Learn, page 1 through page 12.

Why is math fun? Why is math hard?

2009-08-24T10:25:00.000-07:00

Today in class, I tried to help break the ice and reduce some of the anxiety related to taking a university mathematics course (Math 231). I asked students for examples of why they might find mathematics fun and why they might find mathematics hard. Here are some of the responses.

Why fun?

It's fun when you struggle with a concept and then it finally clicks and you understand.
It's fun to see mathematics actually being applied to a real problem.
It's fun when you are able to spot trends and make predictions based on data.
It's fun why you really understand why instead of just the "required" steps.
When you understand, it becomes easy.
It can be a lot like a game or solving a puzzle.
It's fun to develop things logically.

Why hard?

Later material builds on earlier material, so missing something early is permanent hardship.
It can be really hard when the teacher goes too fast.
It can be really hard when the teacher is unclear, especially if they can't give alternate ways of thinking about an idea.
It can be hard if the teaching style is very different from your learning style.
There are so many formulas, it can be overwhelming to try to memorize them.
The theorems, rules and definitions are full of little details.
It can be difficult to understand the many conceptual ideas that interact.
Learning related technology can be challenging.
Only one answer, so you can't fake it.
Very hard to cram for exams.
It can be really hard to find a little (stupid) mistake when proofing your work.
Lots of homework, and problems can take a lot of time.

I'd welcome more comments, including examples of when you found mathematics especially exciting or examples of how your relationship with mathematics soured. Feel free to post a comment.

A New Semester --- Two Classes

2009-01-13T12:55:00.001-08:00

This semester I am teaching a mathematical models in biology (Math/Bio 342) as well as the first semester of Calculus with Functions (Math 231). So I expect to have entries for both of these courses showing up.

Values, Equations, and Theorems

2008-12-03T08:08:00.000-08:00

There is some confusion about theorems. For example, consider the Mean Value Theorem: If f is continuous on [a,b] and differentiable on (a,b), then there is some value c∈(a,b) so that f'(c)=(f(b)-f(a))/(b-a).

Some students think that the ratio (f(b)-f(a))/(b-a) is the Mean Value Theorem. But it is not; it is just a value that is called the average rate of change of f between a and b. You can compute this value as long as both f(a) and f(b) exist. It has nothing to do with derivatives or continuity.

Other students think that the equation f'(c)=(f(b)-f(a))/(b-a). This is closer to the truth, but still is incorrect. First of all, what is c? Second, this statement may not be true. For example, suppose that f(x)=-1 if x<0 and f(x)=1 if x>0. Suppose that a=-2 and b=+2. Then the ratio (f(b)-f(a))/(b-a) is equal to 1/2. But f'(x)=0 everywhere except at x=0, where f'(0) does not exist.

Even closer is to say that f'(c)=(f(b)-f(a))/(b-a) for some c between a and b. This is actually the conclusion of the Mean Value Theorem. It requires the entire statement, particularly the statement that the equation is true for some c and that the value c must be between a and b.

But my example above provides an example where the conclusion of the Mean Value Theorem is false. That does not mean that the Mean Value Theorem itself is false. After all, it is a theorem, and that means that it has been proved to be true always. The part that is missing is the hypothesis for the theorem. The conclusion can only be guaranteed to be true using the theorem if the hypotheses are all satisfied. In this case, you must also check (or give a reason why) the function f is continuous and differentiable on the interval from a to b, including the endpoints for continuity.

Similarly, ∫_a^b f(x) dx/(b-a) computes the average value of a function f on an interval [a,b]. The value can be computed anytime the function is integrable over the interval [a,b]. The Mean Value Theorem for Integrals has nothing to do (in principle) with this calculation.

However, if f is continuous on [a,b], then
f(c)=∫_a^b f(x) dx/(b-a)
for some c∈(a,b). This entire statement comprises the Mean Value Theorem for Integrals. The hypothesis that must be verified to use the theorem is that f is continuous on [a,b]. The conclusion is that you are guaranteed that
f(c)=∫_a^b f(x) dx / (b-a)
for at least one value c between a and b.

Limit or Function Evaluation?

2008-12-02T10:19:00.000-08:00

I've noticed that some students are perplexed about when they use a limit or function evaluation. I presume that the cause of this confusion is that students have learned that you evaluate a limit by plugging in a value. But this is only because nearly all functions that they work with are continuous.

You use a limit evaluation when you need to know what the value of the function should be by using information from the side of the point of interest. When using a limit, you must use limit notation: lim_x→c f(x). Then you use the appropriate rules of limits to evaluate (and hopefully, the function is continuous).

You use a function evaluation when you need to know the value of the function at an actual point. There is no limit involved, just function evaluation. You just use function notation, say f(c), and compute the value defined by the function.

For example, suppose you are calculating an instantaneous rate of change as the limit of an average rate of change. The average rate of change only makes sense when the interval of interest includes two points (endpoints of an interval). The instantaneous rate of change is found by seeing what the value of the average rate of change does when the two points move closer to each other, or more particularly, as the second point approaches the first point.

On the other hand, suppose that you know the derivative, which is itself a function. Then the instantaneous rate of change is calculated by function evaluation using the derivative function. (The limit was already used to create this new function.)

In particular, I have noticed this problem when dealing with finding extreme values of a function. When the interval of interest is an open interval, we are acting as though the domain does not include the endpoints. So, with this restricted domain, evaluation of the function is not possible (since the points are not in the domain). So we must use the information about the function immediately adjacent to the endpoints, and this is with a limit. In this context, the value found in the limit is not achieved at the endpoint, although it might be achieved somewhere else in the domain.

On the other hand, if the interval is a closed interval, then the endpoints are included in the restricted domain. If the function is continuous at these points, then evaluating the function directly is appropriate, since the point is in the domain.

Final remark, if the function is discontinuous at some point in the interval, you must also check the limits at that point for consideration when looking for extreme values.

Differential Equations Project Tips

2008-11-17T17:59:00.000-08:00

As some questions are asked more regularly, I thought I'd provide some general discussion here.

(1) Start with the proposed form of X(t). Compute X'(t) and X''(t) based on that form. Then use those calculations to discover when X'' + k/m X = 0.

(2) The question "mean physically about the mass on a spring" is not asking you to think about the mass (as in measurement) but is asking you to think about what the statement X(0)=1 means about the state of the mass at time t=0 and what the statement X'(0)=0 means about the state of the mass at time t=0.

(3) X(0) is a constant and has derivative of d/dt[X(0)] = 0. Recall that dX/dt > 0 implies that X is increasing, dX/dt < 0 implies that X is stationary (instantaneous rate = 0)

(4) An arbitrary quantity A is proportional to some other quantity B if it is always that case that A = k B for some constant value k (the constant of proportionality). Now interpret the statements to identify what pieces of the equation are proportional to what.

(5) Although you have studied e^x in precalculus, you will not use the logarithm at all in this work. Instead, I just want you to consider some function that has the special property that exp' = exp. (This sentence is analogous to sin'=cos and cos'=-sin.) However, you do need to think about the chain rule: X(t) = A exp(rt) (Since the argument is not simply t, you must use the chain rule.) This problem is exactly analogous to Step 1.

(6) You will get something like

dX/dt = "formula involving X and a, b, and m"

X is increasing when "formula" > 0, decreasing when "formula" < 0, and stationary when "formula" = 0. So use your skills with algebra (think sign analysis) to find conditions when these are the case.

(7) You need to understand the relationship between a rate of change and an actual change. To understand this as well as possible, see the section we skipped in Chapter 3 (last section). But what you essentially need is that we will follow the tangent line for the time increment Δt. How much change is there when the rate of change and the duration of time are both known?

(8) The new version of Excel has some unanticipated differences from what I had when I wrote the project. The labels are not assigned from a menu anymore. Instead of the 3-step process that is described, you just click in the label field in the header section of Excel and type in the new label and then hit enter.

The calculations you see in the first few lines should exactly match your hand calculations in part (7).

Do not print the spreadsheet (it takes WAY too many pages). That is why I ask you to submit your spreadsheet on Blackboard as part of the project.

(9) I must receive a print out of the graph -- hand drawn figures are not acceptable. Ideally, this entire project report would be typed (perhaps using Equation Editor for the equations), with the figures naturally fitting in.

(10) Make hypotheses and test your hypotheses.

Exponential Project Tips

2008-11-17T17:35:00.000-08:00

As some questions are asked more regularly, I thought I'd provide some general discussion here.

(1) exp is the name of the function, just as sin and cos are names of functions. From calculus, you learn that sin'=cos and cos'=-sin. This step shows that exp_b' = ln b * exp_b. (That is, it leaves the function alone except for a constant multiple. (But be careful where the chain rule is needed!)

(2) You do not need to use the limit definition (epsilons and deltas). Instead, for perhaps the easiest solution, you should think about how to finish the statement:

lim b^x = lim [(b^x-1)/x ... ]

That is, if you start with (b^x-1)/x, what do you do to that expression to leave only b^x. Then use elementary limit rules to compute your resulting limit.

(3) One method is to use the method of substitution for limits (change of variables) and then use an identity for the function so that the result of Step 2 is applied --- this method mimics what is done to show that sin x is continuous everywhere. A second method is to use a general theorem that makes continuity an obvious conclusion of the results from Step 1.

(4) You must start with a statement like:

ln (1/b) = lim_{x → 0} [(1/b)^x - 1]/x

There are two easy approaches: (1) Find a common denominator to rewrite this as a simple fraction before continuing or (2) Think of (1/b)^x as b to some appropriate power and then use a limit substitution.

(5) Since you do not know the derivative of ln x, it is incorrect to use the Mean Value Theorem applied to the logarithm. Instead, you should apply the MVT to the function exp_b(x) on an interval so that b^a and b^b are incredibly easy and where it is clear which value is larger (so that you know if the average rate of change is positive or negative). You may use the fact that b^x is positive for all values of x.

(6) The function fb(x) is a linear function. You should write it in slope-intercept form (e.g., mx+b).

(7) Do not attempt to solve the equation fb(x) = exp_b(x). There is one obvious solution from the definition: x=0. But the formulas themselves do not explain where there would not be more solutions. Instead, you should define a function (perhaps g) so that

g(x) = exp_b(x) - fb(x).

You know that g(0) = 0. You need to show that g(x)>0 for all x ≠ 0. My hint suggested Rolle's theorem, but I have since found that the Mean Value Theorem helps even more. Use the Mean Value Theorem to show that for x>0, the average rate of change between 0 and x must be positive. What about x<0? x="0?">

(8) You may not use a limit form of the type b^∞. You may take a limit of the function fb(x) because that is of a form we know how to work with. Then you should use the result of (7) to conclude what the limit of exp_b(x) must be.

(9) and (10) put all of the previous steps together to perform analysis similar to Sections 4.2 and 4.4 to understand the graph.

Spotlight: Math Games

2008-10-24T13:06:00.001-07:00

The other day before class, I introduced a little game called Sprouts. I found the rules summarized at the MAA website. There is also a nice discussion on the Science News website. I find it interesting that such a simple game can be analyzed using mathematical properties.

One of my favorite "math" games is a game called Eleusis. This game was invented Robert Abbott as an analogy of the scientific method. So perhaps we should call this a "science" game. The game is played with a deck (or multiple decks) of playing cards. I turn over the first card and then think of a pattern that would start with that card. Now, the remaining players take turns attempting to choose a card from their hands that they believe would be a valid next card in the pattern. If they are correct, I leave it there. If they are incorrect, I move the card out of the sequence and below the card they tried to follow (for future reference).

The goal for the players is to eventually arrive at a hypothesis that they believe explains the pattern. By playing cards, they attempt to critically assess whether their hypothesis is a complete explanation of the pattern. This mimics the scientific method because we see patterns in how nature functions, and through experiment we attempt to see if controlled efforts are consistent with or contradict our acting hypotheses.

Try it out? Let me know how the game goes.

Mathematical Induction

2008-10-10T08:58:00.000-07:00

The principle of mathematical induction is a topic that our textbook unfortunately skips over. It is used when we want to prove a rule that applies to positive integers. Often, it is the argument that is needed when you want to say, "See! It works the same way for this case and that case, so the pattern will just keep repeating." But to say that a pattern keeps repeating is exactly what we should attempt to make more precise.

The natural numbers are the set of all positive integers: 1, 2, 3, 4, .... It is the dot-dot-dot that creates the problem. Using "..." attempts to tell us that the pattern continues. But what exactly is the pattern? For the natural numbers, the pattern is that you just add one to the previous number. So here is one way of describing the natural numbers, and it is what motivates the principle of mathematical induction.

1 is in the set
For every number that is in the set, call it n, we also have n+1 in the set.

We could restate this using an implication:

1 is in the set
If n is in the set, then (n+1) is in the set.

And that is what we do for all applications of mathematical induction. We provide a starting point (such as 1 is in the set). Then we establish an implication that if a statement is true for one value (n is in the set), then it must also be that the statement is true for the next value (n+1 is in the set).

Here is an example from our past that should have used induction.

Theorem: xⁿ is continuous for n=1, 2, 3, ....

Scratchwork:
Before proving this statement, we should think how we might attempt this without induction. Well, f(x) is really a product xⁿ= x x x ... x, where there are n factors of x. (See how the "..." allows us to hand-wave our notation?) Well, we know that the limit of each factor x will just go to the value c, so the limit must be lim_{x → c} f(x) = cⁿ. That use of "..." keeps us from clearly stating how we used the limit of a product, other than again referring to a pattern: "Use the limit of a product n-1 times." The use of induction makes this precise.

Proof:
We prove by induction.
1) First, we show that f(x) = x¹ = x is continuous. (This is the starting point)
But this is already known: lim_{x → c} x = c.
So f is continuous at any point c.
The statement is true when n=1.
2) Second, we assume that f(x) = xⁿ is continuous and now show that this implies that g(x) = xⁿ⁺¹ is also continuous. (This is the inductive step)
So assume that f is continuous.
g(x) = x f(x) (Relate the new function in terms of what is assumed)
So using the limit of a product:
lim_{x → c} g(x) = lim_{x → c} x f(x) = c f(c) = c cⁿ = cⁿ⁺¹ = g(c)
So xⁿ⁺¹ is continuous whenever xⁿ is continuous.

So by induction, since the statement is true for n=1, and whenever the statement is true for one value n it is also true for the next value n+1, the statement is true for all integers starting with 1.
(End of Proof)

Sometimes induction is compared to reaching different rungs on a ladder. The first statement is what allows you to climb onto the first step. The inductive implication is what says that if you have already reached one rung, then you can move to the next rung. Putting the two together, you first climb on the ladder's first rung. Then you know that you can climb from the first to the second rung, from the second rung to the third rung, from the third to the fourth, and so on forever. The implication, in one fell swoop, justifies climbing each step from the previous. The principle of mathematical induction replaces the uncertainty in "..."

Intermediate Value Theorem

2008-10-06T07:37:00.000-07:00

A theorem is a statement that is always true because it has been proved. Theorems are usually stated as implications. That is, they usually are stated as "If [something is true], then [something else is true]." However, this does not mean that the hypothesis (what appears as [something is true]) is actually true. Nor does it mean that the conclusion (the statement instead of [something else is true]) is true. It means that you are guaranteed to know that the conclusion is true whenever the hypothesis is true.

When applying a theorem, it is your task to establish that the hypothesis is true. Then, by stating the theorem, you are allowed to state that the conclusion is also true.

Here is an example using the Intermediate Value Theorem. Recall that the theorem states that if you have a function f that is continuous on a closed interval [a,b] (where a and b can be any numbers with a < b), then for any y-value C between the values f(a) and f(b), you are guaranteed to be able to find a value x such that a < x < c and f(x) = C.

Here is a hypothetical situation. My car holds 12 gallons of gasoline. (That is not the hypothetical part -- I have actually filled the tank :-) I have installed an automated gas-tank tracking system that records the amount of gas as a function of the car's mileage. (Yep, that's the hypothetical part) If you ask me how much gas I had when the car was at 97,034 miles, then I can tell you it had exactly 5.93 gallons of gasoline.

Last week, I filled up my tank when the car was at 98,012 miles. This morning, I checked my car and it now records the tank as having 1.45 gallons and 98,143 miles. (All figures are also hypothetical, including mileage) So here is a question: will I actually be able to identify a mileage on the car when between that last fill up and today when the car contained exactly 4.7 gallons?

Hmm. Let's see. Imagine that we use the variable x to represent the mileage on the car. Also, let f be a function that measures the gallons in the car f(x) when the mileage is x. We know that f(98,012) = 12 and f(98,143) = 1.45. So C=4.7 is between f(98,012) and f(98,143). What does the Intermediate Value Theorem say?

Now, before you go on, I need to tell you a story. On Friday, I needed to mow the lawn. My backyard is pretty large, so it takes a while. Funny thing! I ran out of gas. I knew I had recently filled the car, so I found my gas siphon and pumped a gallon out of the car's tank and into my gas can. Phew! Glad that was available! Finished the lawn with nary a problem.

So what did you answer?

(Extra credit toward quiz grade if you answer correctly this week by e-mail: waltondb at jmu dot edu)

Common Limit Issues

2008-10-02T12:59:00.000-07:00

Part of the challenge of mathematics is learning the language of mathematics. Mathematics is meant to be spoken, kind of like poetry. But in my class, I'm not wanting a cinquain. Scattered thoughts that are related, but not directly connected in sentences, do not a coherent message provide.

So, here are the top ten problems in limits:

1. Dropping "lim" suddenly. If you have two expressions f(x) = g(x) where g(x) is a simplified version of f(x) (cancelled something), you should write lim f(x) = lim g(x) and NOT lim f(x)=g(x). That "lim" doesn't apply to both sides of the equal sign.

2. Writing "lim" too many times. Just because you don't want to forget to write "lim" doesn't mean you write it in front of everything. You keep writing "lim" while you massage the formula into a form where you can decide the limit. As soon as you are allowed to "plug-in" the value, you have just "taken the limit" and you should stop writing "lim".

Example: f(x) = (x^2-4)/(x-2) and g(x) = x+2. We know that f(x)=g(x) for x ≠ 2. So

lim_{x → 2}(x^2-4)/(x-2) = lim_{x → 2}(x+2) = 2+2 = 4.

3. Lonely "lim". "lim" is not simply an abbreviation for the word "the limit". It is an operator wanting to do something to a formula. It needs a formula next to it at all times. It is without a formula. So when students write "lim = 3", clearly intending to say, "the limit is 3", they are really saying, "the limit of is 3". Of what? And that is the problem. The limit is lonely and has nothing to act on.

4. Stopping at a limit form. Just because you see a zero (0) in the denominator in the limit form does not mean you are done. If the limit has form 0/0, you must try to factor and cancel. If the limit has form L/0, you must identify the sign of the function to decide whether it is going to +&infty; or -&infty;.

5. Writing "=" for undefined values. (Don't do that!) Use a limit form notation to indicate that the denominator is 0 or terms go to infinity.

6. Piecewise using x=a. A piecewise function that has a formula when x=a is a distractor for limits. Remember, a limit always determines what the function would predict if you came from the sides. So a limit never checks at x=a.

7. Writing f(a) instead of lim f(x). This is another piecewise function issue. To check the value predicted by the two sides, you need to say you are checking the sides (meaning limit). So you must write that it is a limit.

8. Using rules for x going to infinity at a. When x goes to infinity, we can ignore any terms that look like 1/x since those terms go to zero as x goes to infinity. However, when x goes to a, those terms are still numbers other than zero. Don't just forget about them (and don't even factor out the dominant terms).

9. Step-by-step when not required. Unless I explicitly ask you to show that the limit has a value using the elementary limit rules, you should just compute the limit. You don't need to spend time showing the step-by-step justification unless asked.

10. No work at all. Often you can see the limit from a graph (say on a calculator). But you need to show a reason on the paper based on mathematics that gives your answer. At the very least, say you looked at a calculator to motivate your answer.

Epsilons, Deltas and Limits... Oh My!

2008-10-02T05:23:00.000-07:00

Yes, Toto, writing proofs of limits can be as scary as the wicked witch from the east! But do not fear, with the right direction, we can squash those problems with ease.

The first step is to realize that we are proving a limit based on its definition. Suppose we need to prove the statement written in its general form:

lim_{x → a} f(x) = L

Notice that this really is saying that when x is a value close to a, the value of f(x) is close to L. The mathematical statement of this says:

"For any ε > 0, there exists a value δ > 0 so that if 0 < |x-a| < δ then |f(x)-a| < ε."

Now just because the Scarecrow is flapping in the breeze, we don't need to be afraid of this complicated looking formula. Our task is to be able to find a formula for δ in terms of ε so that once you know that the value of x is within δ of the value a (δ says how close), then the value f(x) is within ε of the value L.

To reach that fabled wizard of mathematics called a proof, we just need to follow the yellow brick road outlined below. The proof will always take a form involving four steps corresponding to the four parts of the definition:

1. For any ε > 0: We need to create a proof that works for any ε > 0. So that we have a value to work with, we start with any ε with the requirement ε > 0. So the first statement of the proof is something like, "Given ε > 0" or "Suppose ε > 0" or "Let ε > 0."

2. ... there exists δ > 0 such that: The second step is that we need to provide a recipe for how to provide δ > 0 that will make the rest of the statement true. Unfortunately, by the time we reach this step of the proof, we don't yet know what the right recipe is. Personally, I just write, "Let δ=____" and leave enough space to fill in later.

3. ... if 0 < |x-a| < δ ...: We are starting to prove an implication (if...then... statement). We are successful if we can show the conclusion is true whenever the hypothesis is true. So, to accomplish this, we assume that the hypothesis is true and see what happens. I write, "Assume 0 < |x-a| < δ."

4. ... then |f(x)-L| < ε: This is the conclusion of the implication. And this is also the hardest part of the proof. In the middle of completing the work in this proof, we will discover the recipe needed for δ. At that time, we can go back and fill in the missing pieces.

So the 4th step is the hard one. Don't be cowardly like the lion and give up; there is a method to this as well. For the polynomials that we work with, the value of |f(x)-L| will always factor into something of the form:

|f(x)-L| = |x-a| |"stuff"|

We know from step 3 that |x-a|< δ. We want |"stuff"| to be less or equal to a number, which for convenience in discussion we'll call k. Once we find that number k, then we know:

|f(x)-L| < δ k

So part of our recipe will be to make sure that δ≤ε/k. If the recipe requires no other parts, we can even just use δ=ε/k. With this knowledge, we will have found:

|f(x)-L| < δ k ≤ (ε/k)k = ε

All would be well, except that the wicked professor of the west hasn't yet told you how to find k. Let's step back a moment to the |"stuff"| factor. If you don't, I'll send my winged monkeys to bring you back :-). When we're lucky (and f(x)=mx+b), the factor is already a number. But for any other problem, there will be a formula that still involves x. In these cases, without knowing more about x, we won't know how big the extra "stuff" can become. In order to keep a handle on this "stuff" we are going to require for our recipe that δ itself never gets too large. For the simplest cases, we can require δ ≤ 1. And this means that we can take advantage of knowing that x will be between a-1 and a+1.

In a general problem, we could find the largest value for |"stuff"| using values of x between a-1 and a+1. This might take a bit of work. But for f(x) that is quadratic, "stuff" is going to be another linear looking term. We want that "stuff" to involve x-a, so use x = (x-a) + a.

For example, when a=2, the term x+1 can be rewritten x+1 = (x-2)+2+1 = (x-2)+3. The awesome Triangle Inequality then tells us:

|x+1| = |(x-2)+3| ≤ |x-2| + 3

But we know that |x-2|<δ and we required δ ≤ 1 for our recipe. So |x+1|< 4.

For another example, suppose that f(x)=x²-x, a=3, and L=6. We assume 0<|x-3|<δ and rewrite

|f(x)-L| = |x²-x-6| = |x-3||x+2|

We know |x-3|<δ and we need to find a number k so that |x+2|≤k. Since x is in "stuff", we require δ ≤ 1 and use the triangle inequality:

|x+2| = |(x-3)+3+2| ≤ |(x-3)|+5 < δ+5 ≤6

So |x+2| < 6 (This is our value k=6). Thus we also want to use δ = ε/6 in our recipe. Both requirements are taken care of by the formula δ = min(1, ε/6). So now we know

|x²-x-6| = |x+2||x-3| < 6 δ ≤ 6(ε/6) = ε

We have arrived at the emerald city of our desire and proved the limit statement

lim_{x → 3} x²-x = 6.

But the proof needs to be in the right order:
Given ε>0.
Let δ = min(1, ε/6).
Assume 0 < |x-3| < δ. |x²-x - 6| = |x-3||x+2|
|x+2| = |x-3 + 5|
|x+2| ≤ |x-3| + 5
|x+2| < δ + 5 and δ ≤ 1 so |x+2| < 6
|x²-x-6| = |x-3||x+2| < 6δ and δ ≤ ε/6
So |x²-x-6| < ε
Thus, for all ε>0, there exists δ>0 so that if 0 < |x-3| < δ, then |(x²-x)-6| < ε.
Therefore, lim_{x → 3} x²-x = 6

Proofs and Even/Odd Functions

2008-09-15T12:43:00.001-07:00

Here is the second part of my chat discussion. I have edited this some, but I hope the essence of the questions and answers.

[Student]: ok, so im still iffy on how to structure proofs, ill know the assumption (obviosly) and the conclusion, but I'm unsure how to structure the premises and justify them
[Professor]: Well, basic ideas include starting the proof with your assumptions. These are the basic statements you know are true.
[Student]: right
[Professor]: The very last line of the proof should match the conclusion. (and it shouldn't appear earlier) The hard part is what comes in between :-)
[Professor]: But seriously, usually, you take a look at your conclusion and see what type of statement it requires.
[Professor]: If it is an equation that is needed, then you can usually start with one side of the equation and then see what you can put on the other side that you know must be true. (Usually using a definition or algebra)
[Professor]: Then you see how you can use your assumptions to create a statement that leads to your conclusion.

After turning to an actual problem, we started to get into more specifics. The problem was stated as: Suppose that f1 and f2 are odd functions. Prove that f1*f2 is an even function.

[Professor]: What is given knowledge?
[Student]: f1 and f2 are both odd and g1 and g2 are even
[Professor]: What is desired? (conclusion)
[Student]: f1 * f2 is even
[Professor]: What does it mean that f1 is odd?
[Student]: that f(-x) = -f(x)
[Professor]: Make the f into f1, and correct.
[Student]: ah ok ya
[Professor]: So in the actual proof, one line would be...
f1 is an odd function
[Professor]: The next line would interpret...
f1(-x) = -f1(x) for all x in domain of f1
[Professor]: Repeat these two lines for f2.
[Professor]: So far, we have simply restated facts based on the assumptions and our knowledge of odd functions.
[Student]: and that is all stuff that is under the given part? rite
[Professor]: In the proof, it doesn't belong to a "given" part per se, but the reason on the right hand side (if in tabular form) would be that the statement was "Given"
[Professor]: So our proof now has 4 statements.

[Professor]: Now, (scratch work) look at what you need to show: f1*f2 is even. What does that mean?
[Student]: ok ya i have no idea, i feel like it would involve one of the multiplicative properties involving f1 and f2 but i dont really know where to start
[Professor]: What does it mean if I said that G is an even function? g(x) = g(-x) My name was G, so it would be G(x)=G(-x).
[Professor]: But the name could be Brian: Brian(-x)=Brian(x). Voila! I [Brian] am even.
[Professor]: f1*f2 is the name of a function. And for f1*f2 to be even, you need: f1*f2(-x) = f1*f2(x)
[Professor]: It just happens that the name looks like a formula.
[Student]: theres noit a step in between there?
[Student]: to justify how multiplying the functions is ok
[Professor]: Well, there are still steps. But we need to know what we are aiming for.
[Student]: ah
[Professor]: You know what f1 and f2 are. They are odd functions. But exactly what is this new function that we call f1*f2?
[Student]: idk f1(x) * f2(x)
[Professor]: Exactly!
[Student]: or f1(-x)
[Professor]: No
[Student]: * f2(-x)
[Student]: no ok
[Professor]: So we know f1*f2(x) = f1(x)*f2(x)
[Student]: yep
[Professor]: And it looks like you were in the middle of the thought: f1*f2(-x) = f1(-x)*f2(-x)
[Student]: ya
[Professor]: Okay, things are starting to come together.
[Student]: sort of
[Professor]: We want to show f1*f2(-x)=f1*f2(x)
[Student]: substitution?
[Professor]: Yep. Back to the proof. We need to use substitution to get our final result. But we do it one step at a time.

[Professor]: I'll type the left hand side, you type the right hand side.
[Professor]: f1*f2(-x) = ?
[Student]: f1(-x) * f2(-x)
[Professor]: Good. But f1(-x) = ? and f2(-x) = ?
[Student]: f1(x) * f2(x)
[Student]: oh
[Professor]: And finally, f1(x)*f2(x) = ?
[Student]: f1*f2(x)
[Professor]: Precisely. Putting those together, we now know f1*f2(-x) = f1*f2(x).
[Professor]: And that means ...
[Professor]: f1 * f2 is even
[Professor]: Q.E.D. :-)
[Student]: sooo i guess i then assemble all of those steps or just some of them
[Professor]: You can leave out the parts that I said belonged as scratch work.
[Professor]: There were four lines associated with the given information. Then I said "back to the proof" and there were probably four more lines. That is the proof.
[Student]: ic ok
[Student]: ugh, this is def not my fav stuff, i cant believe i haven't learned this before

In summary, here is our proof:

f1 is an odd function
f1(-x) = -f1(x) for x in domain
f2 is an odd function
f2(-x) = -f2(x) for x in domain
f1*f2(-x) = f1(-x) * f2(-x)
= (-f1(x))*(-f2(x))
= f1(x)*f2(x)
= f1*f2(x)
So f1*f2(-x) = f1*f2(x) for x in domain
f1*f2 is an even function

Now, back to the discussion. Another problem dealt with composition. That is, students were to prove: If f1 is an odd function and g1 is an even function, then g1•f1 is an even function. Here is where part of that discussion went.

[Student]: so how do the g1(f1(x)) differ, i guess proving that f1(x) is odd, and then proving then that g1(x) is odd because of f1(x)?
[Professor]: One key point is that the definition of odd or even --- f(-x) = -f(x) and g(-x)=g(x) --- is that the x is simply a place-holder and the same statement would be true regardless of what is in the place of the x.
[Professor]: So for example f(-(x+2)) = -f(x+2), where the "x" was actually the formula x+2.
[Student]: right
[Professor]: Now, be more specific on your question related to composition.
[Student]: so once i prove that f1 is even then it would be the same justification for g1
[Professor]: Careful! f1 is odd (from the given information), so you can't prove it is even.
[Student]: well, opposite justification
[Student]: I'm still not sure of the question.
[Professor]: g1 alone is known to be even. There is nothing to prove about that.
[Student]: ok
[Student]: timeout, so all id have to do is justify g1 as being even regardless of f1?
[Professor]: Why? What are your steps?
[Professor]: And what are you trying to prove?
[Student]: so its given that g1 is even and f1 is odd and we're trying to prove that g1(f1(x)) is even
[Professor]: Technically, g1•f1 is even.
[Professor]: g1(f1(x)) is a value, not a function.
[Professor]: So you need to show g1•f1(-x) = g1•f1(x).
[Student]: wait im confused
[Student]: ok
[Student]: i get it
[Student]: couldnt you just subsitute -x for f1(-x)
[Professor]: No for what you said. f1(-x) and -x are different, so they don't substitute.
[Professor]: But I don't think that is what you were thinking. Try to restate.
[Student]: you could justify g1(f1(x)) = g1(f1(x)) with g(x)=g(-x)
[Student]: -f1(x)*
[Professor]: I think you're on the right track. But be careful that you go step by step.
[Professor]: What are all the steps?
[Professor]: I'll start it off... g1•f1(-x) = ?
[Student]: thats where im confused, do you need a a step for that?
[Professor]: Yes. You must relate the function (g1•f1) to the rest of the formulas.
[Student]: g1(f1(-x))= g1(f1(x) )
[Professor]: What justifies that? The -x belongs to f1, not g1
[Student]: g1(-f1(x)) def of odd fcn
[Professor]: So don't skip that step
[Student]: so whats after that then?
[Professor]: Well, why don't you summarize the statements so far. Start again with: g1•f1(-x) = ...
[Student]: g1(f1(-x)) = g1(-f1(x)) b/c of the def of odd fcns
[Professor]: Very good. Now, what does g1 do when you have g1(-[anything])?
[Student]: g1(-x)=-g1(x) ? but that would make it odd
[Professor]: So use the fact that g1 is even. g1(-x)=?
[Student]: g1(x)
[Professor]: So g1(-U) = g1(U) or g1(-f1(x)) = g1(f1(x))
[Student]: k
[Professor]: It doesn't matter what appears.
[Professor]: g1(-[stuff]) = g1([stuff])
[Professor]: So you left off at: g1 o f1(-x) = g1(f1(-x)) = g1(-f1(x)) = ... (finish it off)
[Student]: g1(-f1(x))=g1(f1(x))
[Professor]: And how does that relate to g1•f1?
[Student]: g1(f1(x)) = even
[Professor]: not equal even. g1(f1(x)) = g1• f1(x).
[Professor]: Recall, you are working with the function g1•f1. You need to show g1•f1(-x) = g1•f1(x).
[Student]: i just got a little confused
[Student]: whats the diff between g1(f1(x)) = g1•f1(x)
[Student]: ah i c nev mind
[Professor]: same value but only g•f can be called the name of the function
[Student]: ok
[Student]: so g1•f1(x) = g1•f1(-x)
[Professor]: That would be the final line to show that g1•f1 is an even function.
[Student]: ok that makes sense

And that leads us to our second proof:

f1 is an odd function
f1(-x) = -f1(x) for x in domain
g1 is an even function
g1(-x) = g1(x) for x in domain
g1•f1(-x) = g1(f1(-x))
= g1(-f1(x))
= g1(f1(x)
= g1•f1(x)
So g1•f1(-x) = g1•f1(x) for x in domain
g1•f1 is an even function

Do you know how to do the justifications of each line now?

Domain and Codomain

2008-09-15T10:09:00.000-07:00

I had a nice chat this afternoon with one of my students. The first topic had to do with the notation f : D→S. Here is what we said:

hi prof walton, i have a quick question

Shoot (but don't hurt me.)

haha, ok so i am under the impression that when functions
say f: x → x that means that the domain and the codomain
are the same but the ranges can be different correct?

The codomain (after arrow) lists the type of numbers
that might be values in the output, while the domain
(before arrow) lists all of the numbers that are in the
list of potential inputs. The range is the list of numbers
that actually are outputs. Is this the question?

yep, so the range is a sub"field" of the codomain rite
it lists all possibles while the range is what numbers
are in the function ?

subset instead of sub"field". Otherwise, yes.

haha ya i was lookin for that word

The cheapest answer for codomain would be simply R
(all real numbers). If the codomain is listed as something
more specific, that helps us understand the function better,
but we still might skip some of the numbers in the set.

ic, so when im looking at f and g that have the same
codomain, that then would not imply that f(g) = g(f)
because the ranges could be different? or is
f(g) being = to g(f) relational to the domain only

Equality of functions requires that they have the same
domain and the same values at every point in that
domain. If the domain is different, then the functions
must not be equal. If the functions have different
values for any point, then the functions are not equal.

so d → s just means that the domain could produce
these outputs right?

That's right. The outputs must be somewhere in the
list known as S. And the inputs only make sense if they
are in D.

but s just means possible outputs because its the codomain rite

Yes, because it is the codomain (after arrow)

rite rite rite, gotcha, this stuff is weird

Fun with Functions

2008-09-08T08:37:00.000-07:00

Cryptography is an interesting application of functions. A cipher allows you to take text (for example) and encrypt it into a new form of information that can then be transmitted. Most children learn a particularly simple cipher that is called a substitution cipher. Such a cipher does a direct translation of letter for letter. Below is a simple example, motivated by the "stage-appearance" of the letters in the phrase "The quick brown fox jumps over the lazy dog."

Other simple ciphers include shifting the alphabet a fixed number of letters or reversing the alphabet.

If we think of encrypting a message as being a function from character sequences to new character sequences, then we might imagine applying two different encryptions one after the other. This is function composition. Or we might want to decrypt a message. This is applying the inverse function. In particular, note that for an encryption method to be useful, the inverse must exist. That is, the function must be one-to-one.

Here is a message that I constructed using the QUICKBROWN cipher (above) followed by a shift cipher where an A becomes an R. Have fun!

JLWH EL MBSJ KJ LPMGK VGLHSM ZOG DCSX TGKHL.

Absolute Value Inequalities

2008-08-27T12:03:00.000-07:00

Well, perhaps I muddied the water for some of you. Sorry about that. In terms of skills, when you solve |u|<a where u is any expression and a is another expression [I thought it was for constants, but it turns out to always work], you can solve by finding the intersection (and) of the solutions to u<a and to -u<a, which we write u<a and -u<a. When you solve |u|>a, you find the union (or) of the solutions to u>a and to -u>a, which we write u>a or -u>a.

My explanation in the supplemental handout was to motivate why this works. After all, the course is not just about skills, but it is also about justification. The absolute value is a piecewise-defined function. That is, there are different rules depending on the value of the expression being worked with. The skill-based method that works for absolute value does not work for other piecewise-defined functions. But thinking about each of the "pieces" separately and joining them properly will always work.

Since the handout was a first edition, I'm curious where you found the biggest issues.

Solving Equations Graphically

2008-08-25T11:30:00.000-07:00

So, in class today, some of you may have been wondering about the computer program that I was using. This utility is called Grapher and it is installed in any recent Mac OS computer. You'll find it in the Utilities folder within Applications.

I was wondering then whether there is a similar resource available for Windows computers as well. Doing a Google search on "Graphing Calculator" I found the following possibility: GraphCalc. I don't have immediate access to check this out, so I'd certainly welcome some comments here as to how well it works.

Now that you have something to work with (and a calculator will work as well, just a little slower), here is something interesting to notice. To solve ax=sin(x), we need to plot y=ax and y=sin(x). You also need to choose a value of a. In Grapher, you would add a New Equation (Cmd-Opt-N) like a=0.25. You now need to find where these graphs intersect.

In class, we learned that we can create new equations that have the same solutions by performing the same operation to both sides (other than division, where we worry about division by zero). So we could get a new equation like a=sin(x)/x. Now we plot y=a and y=sin(x)/x. If you add these as two new graphs instead of getting rid of the old plots, you can compare the two equations graphically. Here is the plot:
I used different colors to distinguish which intersections I was looking for.

For the original equation (ax=sin(x), shown in red), we see there are three intersections. For the new equation (a=sin(x)/x, shown in green), we only have two intersections. But those two intersections agree exactly with the original (see the highlighted intersection at the same value of x marked by circles), and the third corresponds to x=0 which disappeared because we divided both sides by x.

Calculus I -- Welcome Fall 2008

2008-08-25T08:20:00.000-07:00

Welcome to JMU and your first semester of calculus. I'm excited this semester to be teaching calculus again. I hope that you're excited as well.

So, while I was preparing for class, I came across MIT's open courseware program. This is a pretty amazing collection of knowledge that is freely accessible. Feel free to browse it. In particular, I found a text written by Gilbert Strang that will be an outstanding parallel reference for our course this year. I'm especially impressed with how he makes the text conversational in style rather than the more typical dry style of math textbooks.

We'll be pushing through the first chapter of our official textbook very quickly as it should be a review of mathematics that you have already taken.

See you in class!

Central Limit Theorem

2008-03-31T08:49:00.000-07:00

The central limit theorem allows to use a standard normal random variable as an approximating proxy for a properly centered and rescaled sample mean.

First, for any random variable Y with expected value E[Y]=μ_y and variance Var(Y)=σ_y², if we center and rescale as W=(Y-μ_y)/&sigma_y;, then W will always have expected value E[W]=0 and variance Var(W)=1. But it is not the case that W behaves like a standard normal distribution unless Y is somehow special. However, in special cases, when Y can be represented in terms of a sum of many independent and identically distributed random variables, then the central limit theorem can help us.

In particular, if Y is the sample mean of a random sample X₁, X₂, ..., X_n, then Y=(X₁+...+X_n)/n. Suppose that each X_i has expected value E[X_i]=μ_x and variance Var(X_i)=σ_x². Then we know that μ_y=μ_x and σ_y²=σ_x²/n. Our centered and rescaled version of Y can be expressed in terms of the parameters for X: W=(Y-μ_y)/σ_y = (Y-μ_x)/(σ_x/√n). Because Y can be defined in terms of a constant times the sum of i.i.d. random variables, W has an approximately standard normal distribution.

As a second example, suppose that Y is the sum of a random sample X₁, X₂, ..., X_n, with Y=X₁+...+X_n. Suppose that each X_i has expected value E[X_i]=μ_x and variance Var(X_i)=σ_x². Then we know that μ_y=nμ_x and σ_y²=nσ_x². Our centered and rescaled version of Y can be expressed in terms of the parameters for X: W=(Y-μ_y)/σ_y = (Y-nμ_x)/((√n)σ_x). Again, since Y can be defined in terms of a constant times the sum of i.i.d. random variables, W has an approximately standard normal distribution.

Quite a few of our known distributions can be described as a sum of i.i.d. random variables. The most famous is the binomial distribution. Suppose that Y has a binomial distribution with n trials and probability p. Then we can think of Y as the sum of n i.i.d. Bernoulli random variables X₁,...,X_n, each with parameter p. Then μ_x=E[X_i]=p and σ_x²=p(1-p), leading to μ_y=np and σ_y²=np(1-p). Consequently, W=(X-np)/√(np(1-p)) has an approximately normal distribution. In this example, we usually find that we need np≥5 and n(1-p)≥5 for the approximation to be good.

Other random variables that can be represented as a sum of simpler i.i.d. random variables. The Poisson distribution for large value λ can be rewritten as the sum of many smaller Poisson RVs with small values of λ (X₁,...,X_n each Poisson with rate λ/n). The Gamma distribution (including Chi-Square) can similarly be written as a sum of many simpler Gamma (or Chi-Square) random variables. In each of these cases, a centered and rescaled version of the random variable W=(Y-μ_y)/σ_y will be approximately a standard normal distribution.

Success with Random Variables

2008-02-09T18:33:00.000-08:00

So this afternoon, I finished grading a quiz focusing on random variables associated with a sequence of Bernoulli trials. Based on these quizzes, I'm trying to understand what makes a probability course so difficult for students to understand. The other night, I was at a department social and talking with a professor who has taught Math 318 many times. He came right out and stated (without me giving any prompting) that he is always surprised at how students have such a hard time with the course, even though the actual mathematics involved in the course are fairly straight-forward.

In the last entry, I noted that at least part of the challenge is that there are so many different types of functions that appear. But I think that a significant issue comes back to confusion about what random variables represent and how they relate to questions that are posed. Recall that a random variable summarizes some aspect of a random experiment as a single number. (We will soon generalize to the ability to summarize with multiple random variables.) Typical questions in probability focus on the probability of events and the average of certain quantities. Almost always, we answer such questions by identifying an appropriate random variable. We then decide how to characterize events in terms of that random variable, or how to express the quantity being averaged in terms of that random variable.

The examples of random variables related to a sequence of Bernoulli trials provide our first real example of this type of reasoning, and I think this transition is part of why the topic was difficult for many of you. First, we must remember that the random variable is not the same as the random experiment, but simply one of many different ways to summarize an aspect of that experiment. The experiment itself is characterized completely by the sequence of Bernoulli trials, each of which is going to be either a success or a failure. Random variables will be measurement that are related to these successes or failures, and we must choose an appropriate measurement that will allow us to answer the questions of interest.

On our quiz last week, I gave the following scenario: "Apples are packaged in 3-pound bags. Suppose that 4% of the time, the bag weighs less than 3 pounds. You select bags randomly and weigh them in order." In principle, there is an unlimited supply of bags of apples, and we indefinitely select a random bag and weigh it. If we were to weigh enough bags over a long enough time, we would find that the fraction that are underweight appears 4% of the time. However, for any particular number of weighings, we could have found any number of underweight bags. The random experiment is described by an infinite sequence of F's and U's, with an F if the bag was full and a U if the bag was underweight.

So now consider the first question: "What is the probability that you must weigh at least 20 bags before you find an underweight bag?" One method that we could use to do this is to directly understand the event in question. That is, we could create a tree diagram that would eventually give enough information to answer the question. Unfortunately, this tree diagram would be much to cumbersome to answer a question about the first 20 bags weighed: 2^20 different outcomes after 20 weighings. So we try to see if there is a random variable for this random experiment that has enough information to answer the question instead.

There are actually multiple ways to choose a random variable to answer the question. The best random variable is a geometric random variable, which counts the number of Bernoulli trials until we see the first success. Since our question considers when we find the first underweight bag, we define success for our purposes as weighing a bag as underweight. Thus, X counts the number of weighings until we find the first underweight bag. If the first bag is underweight, X=1. But if the first 5 bags are full and the 6th bag is underweight, X=6. Every path on our hypothetical tree diagram is associated with a particular value of the random variable. Since an underweight bag will be chosen with probability p=0.04, we use X~Geometric(p=0.04). Now, having chosen a random variable, we must determine how we answer the question in terms of that random variable. The event of interest, "weigh at least 20 bags before you find an underweight bag," corresponds to an event, "X is greater than or equal to 20". Thus, we wish to compute P[X≥20].

Another way that we could answer the question is to consider that finding an underweight bag for the first time on or after the 20th weighing means that the first 19 bags must have all been full. Consequently, we could answer our question using a Binomial random variable with n=19. So let X count the number of underweight bags out of the first 19 that are weighed. Again, a "success" is finding an underweight bag, so that X~Binomial(n=19,p=0.04). Our event can now be restated as saying that "X is equal to 0". But remember that the X in this paragraph is different from the random variable in the previous paragraph. For this random variable, our question will be answered by computing P[X=0].

The third question on the quiz asked, "What is the probability that you find the 5th underweight bag before you weigh 80 bags?" The best choice for a random variable that will answer our question is to count the number trials until you do find the 5th underweight bag. If we say that a trial is a success if the bag is underweight, then we are counting trials until the 5th success. That is, our random variable X is a negative binomial random variable with r=5 (the number of successes needed to stop counting) and p=0.04 (the success probability). We write this: X~Neg.Binom.(r=5,p=0.04). To answer the question, we must realize that the event of interest is to say that X<80. (We stop counting before we reach 80.) Thus, the answer would be P[X<80].

And of course, as I mentioned earlier, we could actually compute the probability using another type of random variable. For this problem, we must find another way to describe the event of interest. One way to do this is to realize that we find the fifth bag prior to the 80th weighing if there are at least 5 underweight bags by the time we weigh the 79th bag. This can be represented using a binomial random variable. Let our random variable X count the number of underweight bags in the first 79 weighed. Thus, X~Binomial(n=79, p=0.04). The probability we wish to compute is P[X>5].

In summary, to do well in probability, you will need to think probabilistically. Before you can actually compute a quantity (which usually involves fairly straightforward mathematics), you must identify a random variable and understand how to answer the question in terms of that random variable. Then you can use one of the appropriate functions to finally answer the question, whether it is a probability or an expectation.

Random Variables

2008-02-04T09:54:00.000-08:00

There seems to be a lot of confusion in relation to random variables. Part of this has to do with students being pressed onto new subjects without necessarily understanding the previous material adequately. Part of this has to do with the plethora of new functions that are introduced in relation to these random variables (e.g., a random variable X is itself a function in the probability space, it has a p.m.f., a m.g.f., and a c.d.f.). And part of this confusion has to do with a variety of ways to compute things: probabilities, expectations, and moments. Before you allow yourself to be engulfed by the onslaught of waves of probability theory crashing down on you, take a breath (of air) and consider the following.

Probabilities most fundamentally describe random experiments. The probability space (or sample space) describes the possible outcomes of the experiments in as much detail as necessary to completely characterize that experiment. A random variable is a single numeric summary of the nature of the experimental outcome. There are typically many different outcomes of the experiment that can lead to the same value for the random variable. Furthermore, a single experiment can provide the information for many different random variables, each of which summarizes a different aspect of the experiment.

For example, suppose our experiment consists of rolling two standard 6-sided dice, one of which is red and the other is green. There are 36 distinct outcomes. We might be interested in the value of the red die, which value we could assign to a random variable R. We might instead be interested in the value of the green die, which value we could assign to a random variable G. Other values of interest might be the largest of the two values, the smallest of the two values, the sum of the values, the difference between the dice, the greatest common factor between the values, etc. The list could go on forever, with each summary value corresponding to a different random variable.

The random variables R and G are independent because the roll of one does not influence the other. They are also identically distributed. The probability mass function (p.m.f.) for the random variable describes the probabilities of individual values for R. We either need a formula for an arbitrary value or a table showing all of the values to describe this function. For this random variable, we have R(x)=1/6 for each of the values in the support S={1,2,3,4,5,6}. The expected value is defined as a sum over all possible values in the support
For a single die roll (R or G), we have E[R] = 1/6(1)+1/6(2)+...+1/6(6) = 3.5.

Let us now consider another random variable, S, which represents the sum of the two values. That is, S=R+G. The support for S is the set {2,3,...,12}. We could compute the p.m.f. for this random variable: f(7)=6/36, f(6)=f(8)=5/36, f(5)=f(9)=4/36, f(4)=f(10)=3/36, f(3)=f(11)=2/36, and f(2)=f(12)=1/36. And we could compute the expected value of the random variable using the definition of mathematical expectation:

E[S]=2(1/36)+3(2/36)+...+10(3/36)+11(2/36)+12(1/36).

However, since S=R+G, we have a lovely little theorem that allows us to use a sum rule:

E[S] = E[R+G] = E[R]+E[G] = 3.5+3.5 = 7.

This is much easier than calculating using the definition.

So, we learn a lesson: if we can express a random variable as a sum of easier random variables, expected value may be more effectively calculated using these individual terms.

To continue, let us consider the distance between the two rolls. That is, we introduce a random variable X = |R-G| to represent the distance between the rolls. The smallest possible value for X is 0, which occurs when the dice are the same. The largest possible value for X is 5, which occurs when one die is 1 and the other is 6. So the support for X is the set S={0,1,2,3,4,5}. In this example, a table for the p.m.f. is much easier, computed based on the number of ways to obtain each distance:
f(0)=6/36, f(1)=10/36, f(2)=8/36, f(3)=6/36, f(4)=4/36, f(5)=2/36

For this problem, the random variable X is not easily expressed as a sum. So we must compute expected value using the original definition:

E[X]=0(6/36)+1(10/36)+2(8/36)+3(6/36)+4(4/36)+5(2/36)=70/36.

In summary, a random variable is a single number that summarizes some aspect of a random experiment. The p.m.f. of the random variable gives probabilities of individual outcomes. The expected value (or mathematical expectation) computes the average of a random quantity weighted by appropriate probabilities of those values.

Statistical Independence

2008-01-23T07:46:00.000-08:00

I was talking with a student about independence the other day and realized that the student was thinking of independence as being unrelated. In usual speech, we probably think of two outcomes as independent as being that they are not connected. For example, we say that the American states won their independence from England when they broke the governing ties with England.

However, statistical independence should be viewed in a different way. It refers to events A and B so that A provides no information on whether the event B occurs or not. If A and B (remember, these are sets) have no overlap (mutually exclusive or disjoint), then if you knew that the outcome was in A, then it would be impossible that the outcome is in B. Thus, the event A is providing information regarding event B and these are not statistically independent. Similarly, if I knew that the outcome was not in A (i.e., the outcome is in A'), then we know that B is a larger portion of the remaining possible outcomes. Again, that the outcome is in A' is giving information whether the outcome is in event B.

We thus draw the conclusion that independent events must overlap. In fact, they must overlap in a very significant way. Suppose that A and B are independent and that P(A)=0.25 while P(B)=0.4. By the definition of independence, we must have P(A∩B)=P(A)P(B)=(0.25)(0.4)=0.1. To make this concrete, imagine that the sample space Ω has 100 possible, equally likely outcomes. Then A includes 25 outcomes and B includes 40 outcomes. Our calculation then requires that the intersection A∩B includes 10 outcomes.

Now, notice what this means about conditional probabilities. If we restrict our attention to the set A (i.e., we are given that event A has occurred), then the newly restricted outcome space has 25 outcomes. If we ask the probability that B occurs given this information, we know that 10 of these outcomes belong to B. So P(B|A)=10/25=0.4. Voila! This is exactly the same as P(B). Similarly, if we are given that event B has occured, then 10 of the 40 outcomes available belong to A so that P(A|B) = 10/40=0.25=P(A).

How do we summarize this idea? Well, if events A and B are independent, then each event must have a restricted but proportional representation of the other independent event. One way that we can do this is imagine that A and B are at right angles and overlap. For the above example, we can arrange the 100 items into 5 rows and 20 columns.

Then A might represent the first two columns (5×2=10) while B represents the first two rows (2×20=40). Being given information that A occurs collapses the larger picture into a reduced picture consisting of only two columns. However, the fraction of rows represented by B remains the same and P(B|A)=P(B).

This idea expands to more than two events. To have three independent events A, B, and C, we must imagine a three-dimensional grid corresponding to the outcome space so that A represents a simple division in one directions, B represents a second direction, and C represents the third direction. Conditional probabilities given event A corresponds to collapsing the space in A's direction. But the B and C fractions of A remain exactly in the same proportion as they were originally.

Simplifying with Factorials

2008-01-22T10:36:00.000-08:00

As I was grading the second quiz, I see that I should point out that you should simplify as much as possible whenever you see a fraction involving factorials. For example suppose that you saw a fraction: 10!/(4!6!). You should know that immediately you can cancel the last 6 numbers to get: (10*9*8*7)/(4*3*2*1). But instead of multiplying out, you should cancel the 8 with 4*2 and the 9 with 3 to get: (10*3*7)=210.
If you just multiply things out and do not simplify as you go, you will find that you get some awful numbers that are hard to find factors. In fact, you probably already were looking at the factors.

Discrete Random Variables (Section 2.1)

2008-01-21T08:49:00.000-08:00

My philosophy is that class time should be used to facilitate learning and not simply to reiterate concepts that the book already explains adequately. I readily acknowledge that I am still learning how to accomplish such a feat, especially to help students overcome the tendency to avoid reading their textbook.

As we leave chapter 1 where we learned basic ideas of the probability of events, we begin chapter 2 where we will focus on a family of random variables of the discrete type.
Comparing Definition 2.1-1 with the definition I game in my first week slides, you should notice that there is a distinction between the outcome space of the experiment and the space of the random variable. The outcome space should represent the detailed description of the experiment, while the space of the random variable is the range of the random variable (as a function of the outcome or sample space). Pages 58 and 59 provide an important philosophical guide for what we are trying to accomplish and point out that observations might help us to estimate the probabilities associated with the random variable. However, we can often use basic assumptions to create a mathematical model for these probabilities. This chapter introduces a number of models that describe discrete random behavior.

Definition 2.1-2 is very important, introducing the definition of the probability mass function. Problem 3 in the textbook helps test if you understand the basic ideas. One of the major points you need to remember is that for discrete type random variables, properties (b) and (c) compute probabilities using summations. When we get to continuous type random variables, the corresponding properties will replace summation with integration.

In addition to basic principles (probability mass function (mathematical model' prediction) vs relative frequency (statistical estimate), bar graph vs histogram), we meet the first model for a random variable---the hypergeometric distribution---which describes choosing n objects from a total collection of two types of objects. The model is created by considering exactly the types of calculations used in chapter 1, by counting how many ways to select n objects from a total of N objects (denominator) and then also counting how many ways to choose x of the first type and n-x of the second type (numerator).

Spring 2008 Introduction

2008-01-21T08:33:00.000-08:00

Welcome to my math blog. Last semester, I created a blog that would be specific to one course. But it seems silly to keep creating new blogs for each course. So I'm going to experiment with using one blog for a sequence of courses.
This semester (Spring 2008), I am teaching Math 318, an introduction to probability and statistics. Most of the blog entries will correspond to that course for the next few months.