Monday, March 31, 2014

A Squirrel’s Tale

A Squirrel’s Tale

By Bobby Neal Winters
I received the following report from the Bureau of Unlikely Seeming Happenings, Bulsh for short.
Cletus T. Gnasher was out hunting squirrels one day in order to put a little meat on the table when he came upon a strange sight. He saw a squirrel go into one end of a ten foot long hollow limb.  He took out his trusty 22 rifle and aimed it toward the squirrel.  One minute later the squirrel came out the other end.  
Cletus was about to squeeze off a round when the squirrel ducked back into the hollow limb.  Not about to give up this easily, Cletus kept his eyes on the tree.  At the end of 30 seconds, the squirrels head was out of the other end.  Again, before Cletus could squeeze off a shot, the squirrel had ducked back inside the hollow limb.
The squirrel went again to the other end of the hollow limb and this time in only took 15 second before the little head poked out the other side.
It is at the point some began to doubt the veracity of the report because Cletus said that something funny began to happen. The squirrel continued to go back and forth  between the ends of the hollow limb until at one point it appeared he was looking out of both ends at once.  How long did this take?
This was presented to the Bulsh board and reports were requested.  The first report came from a mathematician.  It was very succinct.  This is an infinite series problem.  To determine the time expended, sum one minute plus one-half minute plus one-fourth minute and so on, i.e. 1+ 1/2 + 1/4+ ... .  This is a geometric series whose sum is 2.
The second response was from another mathematician who didn’t offer a solution, but said that this one couldn’t be correct because it assumed the squirrel would be transversing the tree limb an infinite number of times.  As the length of the limb was fixed, being 10 feet, transversing it at infinite number of times would require the squirrel to go an infinite distance.  That is to say that while the infinite series for time-passed converged, the series for distance traveled diverged.
At this point, a physicist chimed in, saying that the problem was reached long before infinity.  The first time the squirrel went through the limb, he was going at a speed of 0.1136 miles per hour.  However, by halving the length of time through the limb each time, he was doubling his speed. Taking the liberty of converting this to 0.05784 meters per second to make sure no one forgot he was a physicist, he said that after 33 more times through the limb, the squirrel would be going 4.36 times 10 to the 8 meters per second which is greater than the 3.0 times 10 to the 8 meters per second that light travels.  At this point, one of the mathematicians corrected him to 2.998 times 10 to the 8 meters per second.  The physicist wrote back something that best not be repeated here and a flame war ensued.  
After the verbal abuse subsided, the physicist calculated that, because of relativity, a one pound squirrel would weigh 1.45 pounds after 32 trips through the limb.  One of the mathematicians asked him whether he meant a one kilogram squirrel would have the mass of 1.45 kilograms, but fortunately the irony was lost on him, and then he was distracted by the following question from an engineer.
The engineer noted that Cletus only said that it looked as if the squirrel were looking out of both ends at the same time.  An image persists on the retina for one-sixteenth of a second.  According to his calculations, it would only take 10 trips through the limb to be making it in that amount of time.  This is a speed of 58 meters per second.
The first mathematician then calculated that this would take 1.998 minutes which was only 2 one thousands off his original estimate. The second mathematician said, “Yeah, your original WRONG estimate.”
This was beginning to degenerate into something nasty then the physicist said, “You know this would require an acceleration of almost 12 times the force of gravity to accomplish.  The squirrel would be  mush.”
At that point, one of the animal-rights folks on the committee said this was something you couldn’t even talk about doing not even to a theoretical squirrel. As this was a supposedly real squirrel, an investigation would have to be launched.
One of the mathematicians said, “By academic freedom..."
And the reply was that he could be free to seek another academic job if he said one more word.

Monday, January 20, 2014

On One Hand But Then on the Other

On One Hand But Then on the Other

By Bobby Neal Winters
I was approached yesterday by a friend of mine at church in Opolis in the following way.
“You’re a math professor, so I got a question for you,” he said.  “It’s from my grandson.”
I remained calm outside, but on the inside I had the reaction a gunfighter in the Old West had whenever someone said, “They say you’re pretty fast.”  You never know when the person asking the question might be faster.
I decided to take my chances and listen.  He held out his right hand and began to count extending his fingers one by one:
“One, two, three, four, five,” he said, extending his pinky last. “And five is ten,” he said extending all of the fingers of his left hand at once.
I nodded because I knew we weren’t at the hard part.
He then held out the fingers of his left hand and started extending them thumb first:
“Ten, nine, eight, seven, six,” he said, giving emphasis to the six. “And five is eleven, so you got eleven fingers.  How can I explain to my grandson why this is wrong.”
And my friend knows it’s wrong.  But knowing it’s wrong and being able to explain why are two different things.  And I could go on to a digression about politics here, but my point is math, or at least the explaining of math.
My friend knows this is wrong because we have ten fingers and ten does not equal eleven, no matter how fast you talk.
Let’s first analyze how this is presented because that is very important to how the confusion comes in.  We start off with something that is true: one, two, three, four, five, and five is ten.   What has happened there?  We’ve listed off the names of the first five numbers that we learned when we learned how to count.  We’ve set up a correspondence between those names and our fingers.  
The next part is where something subtle is done which sets us up for the confusion. When he says, “And five makes ten,” he’s made a very subtle shift.  He’s using the name five to refer to the quantity of fingers on his left hand.”  The five he said when he held out his pinkie was referring to a place in order; the five he said when he held out all his fingers at once was referring to the quantity of fingers.  Mathematics call the first one ordinality and the second cardinality; think of these as order and quantity.
In listing the numbers in standard order, “one, two, three, etc,” the name of the last number listed is also the name of the quantity of the items in the list.
This does not work when you count backwards, and that is one of the things that the example my friend brought me illustrates. In counting backwards from ten, there is no point at which the number counted with also be the quantity of things counted. (Ironically, if he’d started counting at eleven it could be made to work, but you can’t do that because you have ten fingers. It is the center of this trick that you start counting backwards at ten because you know there are ten fingers.)   
When you hold out the pinkie on your left hand as you count backwards from ten until you get to six, you are giving its order as if all the fingers were counted beginning with the other hand.  It can refer to a quantity, but that quantity would be the pinkie itself and the fingers on the left  hand.
Now, you must understand that I didn’t tell my friend all this.  I said, “It’s a confusion between cardinality and ordinality.  Don’t let your grandson play poker with Bill.”
Bill is a man who’s given me poker lessons. After $15 worth, I learned: don’t play cards will Bill. But that is a different story.

Monday, December 30, 2013

Casting Out Mathematical Demons

Casting Out Mathematical Demons

By Bobby Neal Winters
The other day on his Word of the Day blog, Anthony Esolen made reference to the divisibility test for 9 and 3.  The basic idea is this: take the base 10 expression of a number and add up its digits.  That is called the digit sum/ mathematicians don’t waste their imagination on names for things.  If the digit sum is divisible by 9 (or by 3) then the original number was.  
For example, 5643 is divisible by 9 because 5+6+4+3 is 18 and 18 is divisible by 9.  If you aren’t good enough at arithmetic to know that 18 is divisible by 9, note that 1+8=9.  If you don’t know that 9 is divisible by 9, then I can’t help you.  
There is a name for this beyond the Division Test for Divisibility by 9.  It’s called Casting Out Nines.  The reason for this name might become apparent after you’ve done a lot of testing.  Say that you have a number 9954968 and you want to determine whether 9 divides it.  You only need to calculate 5+4+6+8 because those 9s will not add any information and will only add complication to the addition,  so you cast them out.  Indeed, as that 5+4 is a pat 9, you can leave them out too and only worry about the 6+8=17 which is not divisible by 9.
This test is so good we sometimes there are other division tests.  
  • The number 5 divides a number only if the last digit is 0 or 5;
  • the number 2 divides a number only if the last digit is even;
  • the number 4 divides a number only if the the number that is last two digits are divisible by 4;
  • the number 8 divides a number only if the number that is the last three digits are divisible by 8;
  • the number 6 divides a number that number is divisible by 3 and 2; see the tests above
This takes care of all of the numbers from 2 to 9 with the exception of 7; this demon 7 became a mystery to me until I become math major and was given the keys to the priesthood.  I am now going to share those with you because I don’t think you will believe me, and if you do believe me, you will just forget about it anyway.  
I dare you to believe me. I double dog dare you.
Okay, take the number you want to divide by and call it d and take the number you want to test and call it T. I’ve lost a bunch of people right there because I’ve used variables. Tough.  When you divide T by d you get a remainder r.  The number T is divisible by d exactly when r is 0. I’ve not said anything now that ought to surprize you.  This process is called reducing modulo d. Forget that bit I said about not wasting imagination on names.
The surprize comes next.  You can reduce modulo d in an amazingly simple way. It is called modular arithmetic.
Let’s go back to our first example.  Is 5543 divisible by 9?  Well, 5643= 5x1000+6x100+4x10+3.  Take every term and factor of this an replace it with its remainder when divided by 9.  It is easy in this case because the 5, the 6, the 4, and the 3 don’t change and the 1000, the 100, the 10, and the 1 are all 1.  So 5x1000+6x100+4x10+3=5x1+6x1+4x1+3x1 which is exactly what we had done before.
Now consider the test for divisibility by 4 which only uses the last two digits of a number.  Why?  Well consider that the third digit will be multiplied by 100, the fourth by 1000, the fifth by 10000, and so on.  Each of these number is evenly divisible by 4 and so leaves a remainder of 0. So 5643= 5x1000+6x100+4x10+3 reduces to 5x0+6x0+4x10+3=43. Here, I am not using the full power of the technique for the sake of demonstrating the classical test.  The 4x10+ 3 would reduce to 4x2+3=11 which would reduce to 3.
You can see how the test for 5 would work out neatly as well because all powers of 10 leave a remainder of 0 when divided by 5.
Let me now go though a more complicated example to show you the full power.  Is 5643 divisible by 7?  I don’t know of any classical test for this. If you do, let me know.  Note that 10 divided by 7 leaves a remainder of 3;  then 100=10x10 will leave the same remainder as 3x3=9 which is 2; so 1000=100x10 will leave the same remainder as 2x3=6, i.e. 6. So 5643=5x1000+6x100+4x10+3 reduces to 5x6+6x2+4x3+3=30+12+12+3 which reduces to 2+5+5+3 which reduces to 5+3 which reduces to 1.  
Therefore 5643 is not divisible by 7, and, indeed, leaves a remainder of 1.   A very short calculation on a calculator will confirm this.
If you have a number like 75867 you can forget about the 7s and reduce it to 5860 which will reduce to 5160 which will reduce to 5x1000+ 100+6x10 which reduces to 5x6+2+ 6x3 which reduces to 2+2+4 which again reduces to 1.  
While I find this to be a very entertaining activity, there does come a point where simply dividing by 7 by hand becomes easier and I suspect that is why this is not taught to mere mortals.

Thursday, November 21, 2013

Long Division

Long Division
By Bobby Neal Winters
Long division is hard.  I want to state that up front. I remember when I first encountered it in grade school that it hurt my head.  There were a lot of rules and you had to be able to know your multiplication tables and guess and try and guess again. I learned how to act pitiful to get my mother to do it for me.
More than forty years later, I will admit that it’s a useful technique.  Let’s talk about long division long enough to get into trouble. Consider the easiest case: dividing a number by a smaller number.  For example, 7 divided by 2.
Two time three is six is the largest multiple of two that is less than seven, so we put down a 3 on the top line and subtract 6 from this leaves 1.   When you are a little kid who doesn’t know about decimals, you stop there and say there is a remainder of 1, but when you know about decimals, you put a decimal point after the 3 on the top line and you put a 0 down after the one to make it ten.  Two times five is ten exactly, so you put down a 5 after the decimal and stop.  This yields that 7/2=3.5, which, of course, is the correct answer.  You can always check your answer--but our students never do--by multiplying 3.5 by 2 to get 7.  
This is the easy case not because we are dividing a bigger number by a smaller number, but because the process terminates after a finite number of steps.  Consider a case that doesn’t like two divided by seven.  This is illustrated below:

Note that since seven is larger than two, we have to put down a 0 followed by a decimal on the top line. We then put zero times seven on the line below 2 getting--surprise, surprise--0.  We then take zero from two and put down 2 on the next line.  We put a 0 by it to make it twenty.  We then note that seven times two is fourteen so we put 14 below the 20.  Taking fourteen from twenty leave six, so we put down 6 on the next line and follow it by a 0 to make sixty.  Now seven times eight is fifty-six, so we put 56 below the 60.
You know the drill.  This will literally go on forever.  But before too long this will begin to repeat so that we get 2/7 is equal to 0.285714285714285714285714... .
This works nicely because we have a place-value system to represent numbers.
Mathematicians are never satisfied with just numbers, however.  We like to use letters too.  We work with entities (I almost wrote things there.  Things is sloppy writing, so I used entities there instead.  It means things, by the way.) called polynomials.  If you’ve had an algebra class you’ve seen something like 2+x or 2-x-x2.  You can divide on of these by the other two.  For example, let’s divide 2-x-x2 by 2+x:

I was tempted to explain this division. Indeed I wrote a paragraph of it, but I lost consciousness in the process.  If you’ve had the course, you can do it, but if not, I am not going to abuse your good nature by teaching it here.  Let it be sufficient to say that we go through the same motions as we do when we do the long division of numbers.
Just as in the case of long division of numbers, we can have non terminating cases here too.  Consider 2+x divided 2-x-x2:

Here the final quotient is the infinite series 1+x+x2+x3+x4+... .  This might ring a bell for some of you because not only is it 2+x divided by 2-x-x2 it is also 1/(1-x).  As PeeWee Herman used to say, “I meant that to happen.”
The equation 1/(1-x)=1+x+x2+x3+x4+... is quite famous among math geeks.  It is called the Geometric Series.  I can literally talk for hours about this.  My students will testify to this, as will the piles of legs in the classroom which my students have gnawed off like coyotes in futile attempts at escape.
You can have a lot of fun with this equation.  Let x=0.5.  The left hand side, i.e. 1/(1-x), becomes 2 while the right hand side becomes 1+0.5+ (0.5)2+ (0.5)3+... .  If you add enough terms of that right hand side, you get as close to the sum of 2 as you desire.  
If you let x=1, then the left hand side becomes 1/0 by zero while the right hand side becomes 1+1+1+.... which is infinity.  If you make yourself believe that 1/0 is equal to infinity, then this is not a problem.  Let’s push it further, though.  Let x=-1.  Then the left hand side is equal to 1/2.  The right hand side, however, becomes 1-1+1-1+1-1+.... . Let’s refer to this right hand side as S.  It is possible to argue that S is equal to zero; it is possible to argue that S is equal to one.  All of the arguments make the assumption that you can treat arithmetic operations done an infinite number of times the same way you can those done a finite number of times.
What happened to resolve this is a good example of how mathematicians operate.  We have a formula here that gives good answers in certain circumstances and nonsense in others; we determine exactly what those conditions are.  In this case, we get good answers exactly when the absolute value of x is less than one and nonsense at all other times.  There is a whole language created to talk about it using words like convergent and divergent and the Greek letters epsilon and delta.
So we began with a method for dividing one number by another.  We then extend that method to polynomials.  This led to an infinite series where work had to be done to delineate what could be said.  It would appear to be the end of the matter.  Yet.
Given the theory developed to talk about convergence,  the idea that the series S=1-1+1-1+... might be equal to anything appears to be nonsense, but consider the calculation below.  It is like a twisted coda at the end of a horror movie, Jason lives:

What was it that Scotty said, “If one man calls you an ass, pay him no mind. If two men do, buy a saddle.”
One can talk about Cesaro sums. These are like running averages of series.  The Cesaro sum of S is 1/2.
Let’s go back to the equation 1/(1-x)=1+x+x2+x3+x4+... .  Those who know a little calculus will recognize that taking the derivative of each side will yield 1/(1-x)2=1+2x+3x2+4x3+... .
Putting in x=0.5 makes the left hand side equal to 4 and the right hand side 1+2(0.5)+3(0.5)2+4(0.5)3+... can be made as close to 4 as we desire. This is all good.  It becomes more interested when we let x=-1.  Then the left hand side becomes 1/4 and the right hand side becomes 1-2+3-4+5-... .
But this can be made meaningful as well.  There is a theory of divergent series that has been worked out.  It is not my point here to expound on it--though if I learn any more I might--but to bring it up as an example of what mathematics sometimes do.
We take a technique in one area where it is clearly defined and makes sense and then apply it--sometimes playfully--in another area, where sometimes we get a mixture of good results and nonsense.  We then perform investigations to separate the good results from the nonsense.

It’s not a bad way of doing business.

Tuesday, July 23, 2013

M&M Epilog: Sex-Sex-Sex

I've commented from the time to time that the standard deviation is the problem child of statistics because, relative to what introductory statistic students have done before, it is calculationally intense.

Its calculation can be simplified if we make a study of the quantity that I called "Fred" in one of the earlier entries.  It is also sometimes called the second moment of x and referred to by S subscript xx.  When I say this aloud, it sounds like "sex, sex, sex."  That might make it more amusing to you, but it might create strange associations with math for you. It might be best not to think on this too much.

Below I've worked out the calculations that simplify Sxx.  If you are geeky enough to want to see how it's done, you are probably geeky enough to follow them.  If not, let me know.


M&M Activity: Part 4

For this, we will again return to the data from the first table:
This time we are interested in the variation of the total number of M&Ms in each bag.  This is given in the row along the bottom. In order to find the average, we could simply take the total 177 and divide by 10 to get 17.7, but we will need the standard deviation too, and we will be doing something more complicated later. Therefore, we need to be a little more sophisticated.

Notice that in his table only four different values appear: 16,17,18, and 19.  I've made e frequency table with columns for x, f, x-squared, f times x, and f times x-squared:
It is comforting that the sum of the f times x column comes to 177 because that is what is was when I added it in the usual way. In my calculations at the bottom of the page, I get a value of x-bar=17.7 as before and I get a value of s=1.1 for the sample standard deviation.  The calculations all follow the same principals as for the single color.

Your answers for x-bar and s should be similar to what got.  If you've followed my directions, you should all have a sample size of exactly 10.  This is a small sample.  When we learn more about taking samples, it will be shown that a sample size of 30 or more is to be desired.

However, there are several groups and each of you took a sample of 10 that means if you pooled your samples you would have a much larger sample. But the candy is all eaten now and it would be a lot of work to do this all again. If only there were some easy way.....

Well, in fact, there is an easy way.  The groups will now share their data with each other.  Let me show you how this works by showing you an example with some faked data.

I used faked data rather than run the experience 5 more times.  When I faked the data, I fiddled around with the frequencies in a table similar to the one above and obtained values of the sum of the f column, the sum of the f times x column, and the sum of the f times x-squared column form each.  I put the results in the table on the sheet below:
I am being careful to tell you this is faked data because (1) we must practice intellectual honesty and (2) I don't want the M&Ms folks going all commando on me.

Once the data is gathered, the calculations are simple.  The sum of the Sigma f column is 50.  There are 5 groups and each took a sample of 10, so the pooled sample is 50.

The sum of the Sigma f times x column is the sum of all of the f times x entries in all of the groups.  For this fake data, it is 875.  Similarly, the sum of the Sigma f times x-square column is equal to the sum of all of the entries in all of the tables.  With this fake data, it has a value of 15,412.

We calculate x-bar by dividing 875 by 50.  The sum of the pooled data by the total number of items in the pooled data.

Notice when we calculate "Fred," we give him a different name.  We use the variable Sxx.  When I say this out in class, it sounds like I'm saying "sex, sex, sex." 

Note that the n value is 50, so the n-1 value is 49.

Your assignment is two-fold:
  1. Do the calculations as described to calculate the sample mean and the sample standard deviation for the number of M&Ms in each bag.
  2. Share your data with the other groups; pool it; calculate the sample mean and the sample standard deviation from the pooled data.
Every group's answer on number 2 should be the same if every group does it correctly.  Each groups should make its own submission.

Monday, July 22, 2013

M&M Activity: Part 3

Let us now revisit the original table of data:
Notice that I've added a Totals column to the far right and a Total row to the bottom.   Other than those additions, the data is exactly the same as I've been using.

Look along each row labeled by color.  Notice that there is variation.  In the Red row, for example, the values vary from 1 on the low end up to 4 on the high end.  I've take this separated this data in the table below.


While there is variation, there is not all that much. We calculated in the last activity that the sample standard deviation is only 0.9 which is relatively small.  The range of the data (high minus low) is 3.  While there are rigorous techniques to justify this, it is not unreasonable to assume by just looking at it that each bag contains approximately the same proportion of red as any other. 

We will now examine this assumption in more depth.

Notice that at the bottom of the Reds table, I've calculated a quantity that I've called Expected Proportion and have labeled it with the variable p-hat.  Note that p-hat is 17/177 which rounds off to 0.096.  This means that 9.6 percent of the M&Ms we counted are red.  The 17 is obtained as the total of the red row and the 177 is the total number of M&Ms.

The question we ask now, are 9.6 percent of the M&Ms in each bag red?  How do we figure that because, as you may have noticed from our first table, the number of M&Ms in each bag varies from 16 to 19?  Well, if a bag contains 16 M&Ms and 9.6 percent of them are red, then 16 times 0.096 of them (1.7) are red.  I've done this calculation in the table below:
  Compare each entry in the column labeled "Red" to the corresponding entry in the column labeled "Expected Red."  They don't differ too much, do they.

I've carried out this same activity for all of the colors. The results are below.  In carrying out the calculations for Red to three decimals, I noticed they weren't all that much different than using two decimals, so in the subsequent examples, I've only gone to two decimal places.
Take the data that your group gathered, and process it as I have above. 
  • Make a table of you original data, adding a total column and a total row as I have. 
  • You will need a separate table for each color of M&M.  
  • You will need to calculate a value of p-hat corresponding to each color.  It is suggested that at least two people, working independently of each other, perform the calculations involved.
  • You will need to submit your data in electronic form.