Combinatorics

We were covering some elementary combinatorics in the Pre-Calculus course this past week and were discussing some problems from VK Balakrishnan’s Discrete Mathematics textbook.  I took a Discrete Math course at University of Maine in 2000 from Balakrishnan and really enjoyed the class.

I needed to make a quiz question for the Pre-Calculus class and tried to remember some of the more interesting ways Balakrishnan had constructed his problems and came up with this one:

Seven people (A, B, C, D, E, F, G) want to ride into Portland but the car will only hold four.  Person “A” owns the car and will always be driving.  People “F” and “G” don’t get along and won’t ride together.  If it makes no difference where in the car people sit, how many ways are there for four people to ride into Portland together?

Confounding Calculus

Another fun Calculus problem we ran into the other day involved using the quotient rule versus using the product rule.  This problem also came from the Briggs and Cochran book in the section on Implicit Differentiation (pg. 162 #15).  So, the problem this time started out as

Use implicit differentiation to find .

So, if we start out by differentiating the expressions on each side, we get

If we expand and simplify a little -

So then

and

This is the answer that appears in the text – but, if we assume that , then we can clear the denominator in the original problem and use implicit differentiation on the result.

Then, differentiating

or

Then, to isolate :

and

I spent some time playing around with these, and believe me, they are not the same.  I graphed them and here is what the first answer looks like:

And here is the second version,

I tried plugging values for and into these different expressions for the derivative of the same function to see if I got the same answer.  This led me to realize the solution to the conundrum!

I started by plugging in some simple values for and like but quickly realized that this wasn’t valid because .  So then I tried and for these values, the two expressions were the same!

and

These are the same – because the point (1,0) satisfies the original function: .

So I got out the Excel spreadsheet and found a series of points that satisfied the original function and tried a few of these points in the two expressions for the derivative and for each one the value of the two expressions is the same!

So next I was determined to show that assuming the original function was true would show the equality of the two different expressions for the derivative.

So, assuming we need to turn into .

First I substituted in for , so:

or

Then I cleared the fraction in the numerator and denominator:

OK, so this is

or

I saw that where I was headed with this had a and a in it, so I moved a few things around and ended up with:

Going back to the original function I saw that I could substitute in for the :

Factor out from the everything except the and this is starting to look good:

And there it is:

- sigh of relief -

Inequalities – Graphical and Algebraic

I’m teaching a Calculus course this year and came across an interesting problem the other day.  We were working from the Briggs & Cochran Calculus book and discussing the quotient rule.  Problem #40 on page 127 says given the function:

a) Find the values of for which the slope of the curve is zero.

b) Does the graph of have a slope of at any point?

OK – so we need to apply the quotient rule and find

OR

OK – so part (a) is easy enough.  The derivative is equal to zero when .  But what about part (b) ?  We can graph the derivative and see that it’s always less than , but what about algebraically?

The problem can be posed in a number of different ways.  I chose to set it up as:

so

and

Again, we can graph this and SEE clearly that the graph of is definitely above the graph of , but I like to know what’s happening algebraically.  In fact, I used to give my College Algebra students algebraic inequalities as an introductory project, so I really wanted to figure this out.

Of course, this came up in the middle of class, so we looked at the graph and I began to consider the algebraic reasoning, but didn’t get far, so I told the class that I would think about it while they had their quiz – but still to no avail.  I worked on it during my free time yesterday and made some progress, then when I came in this morning it all fell together:

OK, done there.

Again – ok, done.  What was the sticking point for me yesterday were the positive values of . So:

So,

Also,

So,

So now we know that and can work our way back to

For some reason, I find this beautiful!

The Pirate’s Treasure

I was remembering today a contest problem I encountered back in 1999-2000 about pirates and buried treasure.  At first, I couldn’t remember the problem, then when I found multiple on-line versions of it (physicsforums.com, mathpages.com, Bradley University, geometer.org, the mathfactor podcast, and University of Georgia), I couldn’t remember how I had solved it!  It apparently appears originally in George Gamow’s One, Two Three,…Infinity.

The essential features of the problem are that two pirates arrive on a desert island on which there are three prominent trees.  Choosing the most prominent tree as their starting place, the pirates walk first to one tree, counting their steps.  When they arrive at the tree, they turn 90 degrees to the right and walk the same number of steps and mark the spot.

Then they return to the starting place and walk to the other tree, again counting steps.  When they arrive at the second tree, they turn 90 degrees to the left and walk the same number of steps and mark that spot.  They then bury their treasure mid-way between the two marked spots.

Upon returning to the island some time later, they find that the most prominent tree that was their starting point is now gone.  Can they still find the treasure?

Once I was sure I had the right problem, I remembered finally that I had originally used a coordinate proof.

The comments in the Math Factor podcast contain a nice solution using coordinate geometry with a nifty ending that makes finding the treasure fairly simple.

The solution in Gamow’s original apparently uses complex numbers to solve it.

The University of Georgia site asks students to find four separate proofs using 1) complex numbers, 2) Euclidean geometry, 3) coordinate geometry and 4) vector algebra.

Fun Stuff

I was reading Kate Nowak’s blog f(t) yesterday and followed a link to the Phillips Exeter Academy website where they’ve posted a pretty amazing collection of problem sets for all levels of math from Pre-algebra to Calculus (and beyond!).

I’ve only just begun to work on some of these, but I’m having a good time so far.

The first one that caught my eye was this:

Can you find a fraction so that the difference between the fraction and its reciprocal is exactly equal to 1?

That is

The problem gave the example that

D’oh – off by

Then the question asks – Can you find another fraction that gets closer than this?

I approached this from a couple of different ways – first I took the original equation and created a common denominator to get or and .

I didn’t pursue this past that point, but did come back to it later.

Next, I broke out the spreadsheet and set it up to take all the numbers from 1-30 and create fractions and their reciprocals from these and subtract them.

Looking at all that, I noticed a few places where the difference between 1 and the was smaller than .

This happened for and .  Then it was time for class.

The numbers in the fractions that were getting close to 1 had caught my eye yesterday and this morning when I came in, I sat down with it again and saw that they were all consecutive Fibonacci numbers.  So, I made a new spreadsheet with Fibonacci numbers and the fractions and reciprocals and noticed that the difference was approaching 1.

At some point yesterday afternoon I went to Wolfram Alpha and typed in just to see what I would get and it provided a relationship between the two variables that comes from treating one of the variables as a constant so that

In other words a and b cannot both be integers!  And of course, some of you may have picked up on this sooner than I did – the number that produces the exact value of 1 is phi – The Golden Ratio.

.

Pre-Calculus

Over the summer, I was writing posts at the kitchen table math blog on Pre-Calculus curriculum.  I didn’t get as far as I would have liked….but I’m not done yet.

Here’s a link to what I did get done.

More on testing and accountability

My last post described a crisis of testing fraud in many of the big city school systems around the country.

These tests are being used to reward/punish students, teachers, schools and communities.

John Ewing, the former Executive Director of the American Mathematical Society has a great article in the recent Notices of the American Mathematical Society about the use of these tests to make decisions that effect the lives of the students, teachers schools and communities.

But using tests to evaluate teachers, schools, or programs has many problems. (For a readable and comprehensive account, see [Koretz 2008].) Here are four of the most important problems, taken from a much longer list…

4. Inflation. Test scores can be increased without increasing student learning. This assertion has been convincingly demonstrated, but it is widely ignored by many in the education establishment [Koretz 2008, chap. 10]. In fact, the assertion should not be surprising. Every teacher knows that providing strategies for test-taking can improve student performance and that narrowing the curriculum to conform precisely to the test (“teaching to the test”) can have an even greater effect. The evidence shows that these effects can be substantial: One can dramatically increase test scores while at the same time actually decreasing student learning. “Test scores” are not the same as “student achievement”.

This last problem plays a larger role as the stakes increase. This is often referred to as Campbell’s Law: “The more any quantitative social indicator is used for social decision-making, the more subject it will be to corruption pressures and the more apt it will be to distort and corrupt the social processes it is intended to measure” [Campbell 1976]. In its simplest form, this can mean that high-stakes tests are likely to induce some people (students, teachers, or administrators) to cheat…and they do [Gabriel 2010]. But the more common consequence of Campbell’s Law is a distortion of the education experience, ignoring things that are not tested (for example, student engagement and attitude) and concentrating on precisely those things that are.