As they are learning all about Quarks and String theory and the big bang.. maybe we could teach our kid a little about applications of science to solving world problems.. Here is a MIT prof using her students to do just that, and the results are intriguing.. If you watch nothing else, listen to the last five minutes when she talks about her vision... see the world in a whole new way...

## Sunday, 31 January 2010

### A Course on Street Science

Labels:
Amy Smith,
Sustainable technology,
TED

## Saturday, 30 January 2010

### Don't Miss These

Dan MacKinnon over at Math Recreations has a nice blog about Appoloinian Gaskets formed by Ford Circles (

Also Robert Talbet over at Casting Out Nines has a nice video about Sierpinski's gasket as part of a Dorito Ad for the Super Bowl..... Fractals make the big time...

Addendum... Cory Poole, the teacher of the students who created the fractal commented on this post, Nice job Cory, you have ever reason to be justly proud of your students.

After getting Cory's message, I tracked down one of his web sites where he explains the process of construction (and some notes on motivation)... see it here

*what you get when the steering wheel comes off a Model A*) and the relationship of course to Farey Sequences..(If you ever added fractions the way they told you was wrong... this is your revenge).Also Robert Talbet over at Casting Out Nines has a nice video about Sierpinski's gasket as part of a Dorito Ad for the Super Bowl..... Fractals make the big time...

Addendum... Cory Poole, the teacher of the students who created the fractal commented on this post, Nice job Cory, you have ever reason to be justly proud of your students.

After getting Cory's message, I tracked down one of his web sites where he explains the process of construction (and some notes on motivation)... see it here

Labels:
Farey Sequence,
Ford Circles,
Sierpinski's gasket

### Fair Division......A Piece of Cake...

Many really complex problems of mathematics are founded in simple social practices, such as the fair division of assets. A company dissolves and the partners need to divide up the assets. The last surviving parent dies without a will and the children have to decide how to distribute the home, boat, car, golf-club membership...etc. Such issues keep lawyers in business and lead to terrible fighting between the parties because of the different ways they value the things to be divided.

One of the most commonly talked about is the idea of dividing a cake fairly between two people. Most people think they know how to do that. Here is a quote from Alan D. Taylor's introduction to The Geometry of Efficient Fair Division, by Julius B. Barbanel

--------------------------------------------

Discussions of cake cutting almost always begin with the procedure known as divide-and-choose. Historically, this two-person scheme traces its origins back 5000 years to the Bible’s account of land division between Abram (later to be called Abraham) and Lot, and it resurfaces more explicitly two-and-a-half millennia ago as Hesiod, in his Theogony, describes the division of meat into two piles by Prometheus, with Zeus then choosing the pile that he preferred.

Mathematical investigations of fair division date from the early 1940s. The constructive vein was first opened by the Polish mathematician Hugo Steinhaus (see [40]) and his colleagues Stefan Banach and Bronislaw Knaster. Steinhaus appears to have been the first to ask if there is an obvious extension of divide-and-choose to the case wherein there are three participants instead of two, and he derived the scheme referred to in a number of mathematical texts for non-majors (see [18] and [42]) as “the lone-divider method.” But extending this procedure to four or more participants is somewhat complicated, and was not actually achieved until Harold Kuhn [30] did so in 1967. Banach and Knaster, however, took an entirely different tack and devised a fair-division scheme for any number of participants that is known today as the “last-diminisher method.

----------------------------------------------------------------------

The idea that the "cut and choose" solution is "fair" assumes that the object to be divided is uniform and that both people approach the division with the same value system. Consider, for instance, a cake baked in a square pan with frosting on the top and sides. Perhaps the cake has one half that is chocolate and one half is vanilla. Now what if one of the people really likes frosting, or really LOVES chocolate, and other is indifferent to the frosting amount or flavoring(you can play with the implications of levels of these factors). Is there an advantage to being the one who does the cutting or does the choosing? To further complicate the process, consider the different ways you might cut the cake if you were one of these people and you knew the other person's preferences. What if the information access was not equal? How would it effect the choice of who cuts if you knew they had more information about your preferences than you had about theirs? What if you had the information advantage? And what if there were three people.. how would you extend the cut-and-choose method (it can be done)?

And what if the cake is not a cake, but the remains of the family estate, A car, a house, a boat, membership in the golf club...? Now what would be "Fair"?..

Ok, maybe "Piece of Cake" is not such a piece-of-cake after all. Maybe I'll take time to talk soon about the more complicate situation when an estate contains non-divisible items, a family home, the boat in which your dad took you out fishing when you were a kid... which half of your mom's antique Ming Dynasty vase would be more valuable...perhaps not simple when it is your family.

Labels:
Discrete math,
fair division

## Thursday, 28 January 2010

### Baptize Me Brother, I'm a Believer

It started with Dan MacKinnon's blog at mathrecreation one of those blogs I read regularly (he writes, I read, that's regularly).

In this one he mentioned a speech in Ottawa... OK, I'm in England, and not likely to pop over to Ottawa for a quick speech, but he had this quote about what Alfie Kohn was all about..."Our knowledge of how children learn – and how schools can help -- has come a long way in the last few decades. Unfortunately, most schools have not: They’re still more about memorizing facts and practicing isolated skills than understanding ideas from the inside out; they still exclude students from any meaningful decision-making role; and they still rely on grades, tests, homework, lectures, worksheets, competition, punishments, and rewards. Alfie Kohn explores the alternatives to each of these conventional practices, explaining why progressive education isn’t just a realistic alternative but one that’s far more likely to help kids become critical thinkers and lifelong learners."

Wait... that sounds like me talking... no grades (not like we have to give now anyway)... Self directed students making their own educational decisions... HEY... I think I found my Edu-church...

Sooooo, I went to Alfie's web page...wow, this guy is a radical from way back..(you go guy)... and I liked what he was saying... so I read some articles he had published.. and I found one that may save me the price of an international flight from here to there... It may be the same talk he is giving..more or less... If this is not your bag, that's OK too, but here is a clip that I think will help you understand whether you may like this guy or not...

"A friend of mine, who is a teacher-educator, had a daughter in fifth grade at the time of this story. She came home, he wrote me, with a worksheet on simple machines—ball bearings, inclined planes, pulleys, that sort of thing. As he came home from work, she said, “Dad, test me, test me!”

“Well,” my friend said, “Why don’t you just tell me what you’ve been learning about?”

“OK, but first ask me what these things are!”

“OK, if you insist. What is a ball bearing?”

“OK, easy dad! A ball bearing is blah blah blah.” A verbatim repetition of the definition she had learned from the teacher.

My friend said, “But Rachel, what is a ball bearing?”

“I told you, Dad. A ball bearing is a blah blah blah blah.”

To make a long story short, he continues, “I turned over the ottoman, which is on wheels, and showed her the ball bearings, and her eyes got wide.

“‘Cooool! That’s what a ball bearing is? How does it work? Can we take apart the ottoman? Oh, I get it. Why didn’t Mrs. Lambert just tell us this is what it was? Can you buy these at the store? Where do they sell these things anyway? Hey, wanna help me make something that rotates? Hey, cool, watch what happens if I hold one of these things and try to spin this thing. What would happen to this thing if the balls were really big? Would the wheels go faster?’”

A progressive school is not about memorizing the definition of ball bearings, or the date at which an event happened in history, or the difference between a simile and a metaphor. That’s not to say that these topics aren’t covered. It’s to say that questions that kids have drive the education."

That's the kind of education I want to be part of.... Thanks Dan...

In this one he mentioned a speech in Ottawa... OK, I'm in England, and not likely to pop over to Ottawa for a quick speech, but he had this quote about what Alfie Kohn was all about..."Our knowledge of how children learn – and how schools can help -- has come a long way in the last few decades. Unfortunately, most schools have not: They’re still more about memorizing facts and practicing isolated skills than understanding ideas from the inside out; they still exclude students from any meaningful decision-making role; and they still rely on grades, tests, homework, lectures, worksheets, competition, punishments, and rewards. Alfie Kohn explores the alternatives to each of these conventional practices, explaining why progressive education isn’t just a realistic alternative but one that’s far more likely to help kids become critical thinkers and lifelong learners."

Wait... that sounds like me talking... no grades (not like we have to give now anyway)... Self directed students making their own educational decisions... HEY... I think I found my Edu-church...

Sooooo, I went to Alfie's web page...wow, this guy is a radical from way back..(you go guy)... and I liked what he was saying... so I read some articles he had published.. and I found one that may save me the price of an international flight from here to there... It may be the same talk he is giving..more or less... If this is not your bag, that's OK too, but here is a clip that I think will help you understand whether you may like this guy or not...

"A friend of mine, who is a teacher-educator, had a daughter in fifth grade at the time of this story. She came home, he wrote me, with a worksheet on simple machines—ball bearings, inclined planes, pulleys, that sort of thing. As he came home from work, she said, “Dad, test me, test me!”

“Well,” my friend said, “Why don’t you just tell me what you’ve been learning about?”

“OK, but first ask me what these things are!”

“OK, if you insist. What is a ball bearing?”

“OK, easy dad! A ball bearing is blah blah blah.” A verbatim repetition of the definition she had learned from the teacher.

My friend said, “But Rachel, what is a ball bearing?”

“I told you, Dad. A ball bearing is a blah blah blah blah.”

To make a long story short, he continues, “I turned over the ottoman, which is on wheels, and showed her the ball bearings, and her eyes got wide.

“‘Cooool! That’s what a ball bearing is? How does it work? Can we take apart the ottoman? Oh, I get it. Why didn’t Mrs. Lambert just tell us this is what it was? Can you buy these at the store? Where do they sell these things anyway? Hey, wanna help me make something that rotates? Hey, cool, watch what happens if I hold one of these things and try to spin this thing. What would happen to this thing if the balls were really big? Would the wheels go faster?’”

A progressive school is not about memorizing the definition of ball bearings, or the date at which an event happened in history, or the difference between a simile and a metaphor. That’s not to say that these topics aren’t covered. It’s to say that questions that kids have drive the education."

That's the kind of education I want to be part of.... Thanks Dan...

Labels:
Alfie Kohn,
Dan McKinnon,
progressive education

## Wednesday, 27 January 2010

### Geometric Explorations in Three Stages

Here are some geometry explorations I think that you could present to kids in three levels.. For example:

If you had a triangle and marked one of the points of concurrency, perhaps the intersection of the medians or the intersection of the perpendicular bisectors.

First you challenge the student to think about what would happen if??? What if we held two of the vertices fixed, and moved the third along a line parallel to the side joining the first two; what would be the path traced out by the int

It is good for them to ponder the "what might happen" without a definite goal or a right or wrong answer. Maybe doodle with a pencil and paper... Push for a little detail.

Now (this would be stage two) let them actually construct the locus with interactive software and see what happens...this is not trivial, even with the magic of modern software like Geogebra or Cabri or Geometer's Sketchpad... and there are often surprises that lead them to wonder their own "what if" questions.

Now the final stage... prove it analytically. This will be the hardest part. For example, If you picked the centroid, or intersection of the perpendicular bisectors of the sides, then what they will see is that the locus it follows is a straight line parallel to the fixed side; but you can be a little more definite.. Let's call the two fixed points A and B, and put them on the x-axis. Now the third point will have a changing value of x, but the y-coordinate will be constant at some value, call it C. If they have trouble starting, we can ask them to think what is the coordinates of the point M that is the median of AC, and N that is the median of BC..... again the x values will change, but the y-values will be C/2 for both. So now we can challenge them to describe the geometric figure that is AMNB, whose diagonals are the medians, and whose intersection we seek.

So what can we see in the trapezoid??... what do we know??... what about the size of the two parallel bases??... Do you see any triangle relations that would help? Eventually we hope they will figure out that triangles AFB and EFD are similar.. and since the two bases are in a ratio of one to two, the perpendicular distance to the centroid will be in that same ratio... or 2/3 of the way from the lower base to the upper base.... SO??? aha, but the upper base of the trapezoid has a y-coordinate of C/2, so 2/3 of that means the centroid will be at C/3...

For an advanced challenge, what happens if we follow the intersection of the angle bisectors, or the intersection of the altitudes, the locus will not be a straight line. Can you visualize it? Construct it.. what does it look like... Prove it?

If you had a triangle and marked one of the points of concurrency, perhaps the intersection of the medians or the intersection of the perpendicular bisectors.

First you challenge the student to think about what would happen if??? What if we held two of the vertices fixed, and moved the third along a line parallel to the side joining the first two; what would be the path traced out by the int

It is good for them to ponder the "what might happen" without a definite goal or a right or wrong answer. Maybe doodle with a pencil and paper... Push for a little detail.

Now (this would be stage two) let them actually construct the locus with interactive software and see what happens...this is not trivial, even with the magic of modern software like Geogebra or Cabri or Geometer's Sketchpad... and there are often surprises that lead them to wonder their own "what if" questions.

Now the final stage... prove it analytically. This will be the hardest part. For example, If you picked the centroid, or intersection of the perpendicular bisectors of the sides, then what they will see is that the locus it follows is a straight line parallel to the fixed side; but you can be a little more definite.. Let's call the two fixed points A and B, and put them on the x-axis. Now the third point will have a changing value of x, but the y-coordinate will be constant at some value, call it C. If they have trouble starting, we can ask them to think what is the coordinates of the point M that is the median of AC, and N that is the median of BC..... again the x values will change, but the y-values will be C/2 for both. So now we can challenge them to describe the geometric figure that is AMNB, whose diagonals are the medians, and whose intersection we seek.

So what can we see in the trapezoid??... what do we know??... what about the size of the two parallel bases??... Do you see any triangle relations that would help? Eventually we hope they will figure out that triangles AFB and EFD are similar.. and since the two bases are in a ratio of one to two, the perpendicular distance to the centroid will be in that same ratio... or 2/3 of the way from the lower base to the upper base.... SO??? aha, but the upper base of the trapezoid has a y-coordinate of C/2, so 2/3 of that means the centroid will be at C/3...

For an advanced challenge, what happens if we follow the intersection of the angle bisectors, or the intersection of the altitudes, the locus will not be a straight line. Can you visualize it? Construct it.. what does it look like... Prove it?

Labels:
triangle concurrency

## Monday, 25 January 2010

###
Different Question, Same Idea, *Sort of*

A couple of days ago I wrote about the probability of a 28 Year Snow and 100 year Rain. It reminded me of another problem that, seemingly completely different, is somewhat related.

Here is a version of the question... Suppose you were looking at a group of n (a large number of numbers) numbers presented one at a time. Your goal is to guess which is the largest number in the list. You get to look at one number, and say "That one." or pass, if you pass, you can never go back and the next number is presented.. How do you decide which number to pick. You could just guess one of the numbers, and when it comes along pick it.. the probability you would win is 1/n. Or you could guess that there will be approximately ln(n)+.58 numbers.. this is the limit for the harmonic series, $\sum_{k=1}^{n}\frac{1}{k}$= [1+

But what if you decided to wait until a certain number, call it r, of events had passed, then pick the next number higher than any of the first r... But what r would you pick.

If there are three numbers, 1, 2, and 3, they can occur in six orders. If you decide to always pick the first, or second or third you would win 1/3 of the time.. Or you could wait and after the first, pick the next one that is higher... Well if 3 is first, you lose, but if the order is 1-3-2; 2-3-1; or 2-1-3 you would win, hey, that's a fifty percent chance of winning..

If there are four numbers, they can be in 24 different orders. Would it be best to let one number go, or two. If we pick the first number, we know the winning probability is only 1/4, and if we wait until the last number, it is again 1/4. But what if we let one go, then pick the next that exceeds that value... We would pick the right number in 11/24 of the orders.. If we let two go, then picked the next that exceeds those two, we would only win in 5/12 of the trials.

So how do you decide? How many numbers would you let go before you picked the next higher number. If there were ten numbers in all, for example, and you let r of them go by, the probability that the next one is the biggest (and hence you win) is 1/n. But you would also win if the next one was smaller than the max of the first r numbers [prob r/(r+1)] and the one after that WAS the largest in the set (prob 1/n again)... Or ...you could win if the next two after you stopped counting were not greater than the max of the first r numbers [probability r/(r+2)] and the following one (yep, prob 1/n) was the biggest value. Ok, you see a pattern here. So the probability of winning after waiting for any r-values is P(r) =

It turns out that the max happens when r=n/e... so for n= 100, we would wait for the first 1/e = 36.7 (ok, 37) numbers and then pick the next number that is bigger than any of those...and the probability you win.... 1/e or about 36.8%

And how is that related to the 100 year snow? Well, 1-.368 = .632... So the probability of picking the max value from a string of 100 numbers by this strategy is the same as the probability that you do NOT hava 100-year snow in the next hundred years... . Bundle up... the snow is more likely.... rule of the day for multiple choice.... Pick e.... ;-}

Here is a version of the question... Suppose you were looking at a group of n (a large number of numbers) numbers presented one at a time. Your goal is to guess which is the largest number in the list. You get to look at one number, and say "That one." or pass, if you pass, you can never go back and the next number is presented.. How do you decide which number to pick. You could just guess one of the numbers, and when it comes along pick it.. the probability you would win is 1/n. Or you could guess that there will be approximately ln(n)+.58 numbers.. this is the limit for the harmonic series, $\sum_{k=1}^{n}\frac{1}{k}$= [1+

^{1}/_{2}+^{1}/_{3}+^{1}/_{4}+...+^{1}/_{n}(for big n...and that doesn't mean huge...at n=20, the actual value is 3.597 and the approximation is 3.572) approaches ln(n) + .0.57721, which will estimate the number of record breaking events in N events, for example, record breaking floods in a 100 year period (look for about 5... ln(100)+.577 = 5.182).. and when that many record results occur, you could pick it... (this would have been my first choice)... unfortunately, the standard deviation of the number of records that occur is pretty wide, so you are not likely to win more than about 20% of the time... Still, in most cases that is better than 1/n.But what if you decided to wait until a certain number, call it r, of events had passed, then pick the next number higher than any of the first r... But what r would you pick.

If there are three numbers, 1, 2, and 3, they can occur in six orders. If you decide to always pick the first, or second or third you would win 1/3 of the time.. Or you could wait and after the first, pick the next one that is higher... Well if 3 is first, you lose, but if the order is 1-3-2; 2-3-1; or 2-1-3 you would win, hey, that's a fifty percent chance of winning..

If there are four numbers, they can be in 24 different orders. Would it be best to let one number go, or two. If we pick the first number, we know the winning probability is only 1/4, and if we wait until the last number, it is again 1/4. But what if we let one go, then pick the next that exceeds that value... We would pick the right number in 11/24 of the orders.. If we let two go, then picked the next that exceeds those two, we would only win in 5/12 of the trials.

So how do you decide? How many numbers would you let go before you picked the next higher number. If there were ten numbers in all, for example, and you let r of them go by, the probability that the next one is the biggest (and hence you win) is 1/n. But you would also win if the next one was smaller than the max of the first r numbers [prob r/(r+1)] and the one after that WAS the largest in the set (prob 1/n again)... Or ...you could win if the next two after you stopped counting were not greater than the max of the first r numbers [probability r/(r+2)] and the following one (yep, prob 1/n) was the biggest value. Ok, you see a pattern here. So the probability of winning after waiting for any r-values is P(r) =

^{1}/_{n}[1 +^{r}/_{r+1}+^{r}/_{r+2}+ ...+^{r}/_{n-1}] but if we factor out the r on top.. this is just^{r}/_{n}[^{1}/_{r}+^{1}/_{r+1}...+^{1}/_{n-1}] and that last part is the harmonic series between^{1}/_{r}and^{1}/_{n-1}... If we make the assumption that r and n are pretty big then we can approximate the sum of the harmonic series out to 1/(n-1) as Ln(n-1)+ .577 and out to 1/r it would be Ln(r)+.577. Subtracting this last from the first, we see that it will be about Ln(n-1) - Ln(r) or, using the log properties you learned in Alg II, we can write that as the Ln(n-1/r)... so for any number of numbers n, if we let the first r go by and pick the next number greater than any of those, our probability of having that be a winning number is (^{r}/_{n}) Ln(^{n-1}/_{r}). All we have to do is find the maximum value of Ln(^{n-1}/_{r}) and we know how many r to let pass. For calculus students, n is a constant here, r is the variable, so we want find the maximum of (r/n)*Ln(n/r).... and for algebra students, graph it and find the max probability.It turns out that the max happens when r=n/e... so for n= 100, we would wait for the first 1/e = 36.7 (ok, 37) numbers and then pick the next number that is bigger than any of those...and the probability you win.... 1/e or about 36.8%

And how is that related to the 100 year snow? Well, 1-.368 = .632... So the probability of picking the max value from a string of 100 numbers by this strategy is the same as the probability that you do NOT hava 100-year snow in the next hundred years... . Bundle up... the snow is more likely.... rule of the day for multiple choice.... Pick e.... ;-}

Labels:
1/e,
harmonic series,
records

## Sunday, 24 January 2010

### Dear Fox News

Joshua Zucker sent this to the AP Stats EDG, and I thought it worth sharing....

It is from the PhD comic site.

It reminded me of this post I made in December.

They did have another nice comic about the recent financial debacle... http://www.phdcomics.com/comics/archive.php?comicid=1077

It is from the PhD comic site.

It reminded me of this post I made in December.

They did have another nice comic about the recent financial debacle... http://www.phdcomics.com/comics/archive.php?comicid=1077

Labels:
Fox News,
statistics

## Saturday, 23 January 2010

### The 28 Year Snow and The 100 Year Flood

Europe just went through one of its worst winters in years, and according to the press, the largest snowfall in England for 28 years (and apparently similar effects in the US.. that darn El Nino). The last this big in Europe was 1982. That doesn't actually make it a 28 year Snow, in the sense of a 100 year flood. Most people don't realize that the probability of having a "100 year flood" in any 100 years is only about 63% (go on, say "Huh?").. Officially, a 100 year flood is a flood that has a 1/100 chance of happening in any given year (I think, technically, they say the probability of a flood that great or greater). You can use a little easy probability to see why it would only have a 63% probability of occurring in the next 100 years (even if it hasn't occurred in 200 years, or if it happened yesterday.. the assumption is that the great floods are essentially independent events). So the probability of having (at least one) "100 year flood" in the next 100 years is just 1- (the probability that we don't have one)..which is easier to compute. The probability of NOT having a 100 year flood this year is .99 or 1- the probability that we do.. The probability we don't have one in the next 100 years is just (.99)

^{100}which my trusty calculator tells me is about .366. So it there is a .366 chance of NO great flood, then there is a .634 chance of having at least one.

"You mean there might be MORE than one?" I hear you ask. Well yes, now we just apply the binomial distribution to find, for example, the probability of great floods in two of the next 100 years (and not in 98 others) and accounting for all the possible orders, we get $\dbinom{100}{2}(.01)^2(.99)^9^8$ or about 18.5%; and if you extend that to exactly three great floods, the probability comes out to about 6%... so there is something like a 36% chance of NOT having a great flood, and about a 40% chance of just one. (It takes about 230 years to be 90% sure of having a 100 year flood; and almost 300 years to be 95% sure).

Now if we said, Wait, a "100 year flood" should have a probability of almost one of happening (would you settle for the proverbial 95% standard from statistics?) then we could work backwards and compute the probability of such a flood this year (or any random year). Since the probability of not having one in 100 years is 1-.95, all we have to do is solve the problem (1-p)

^{100}= .05 and we get the annual probability of such a flood. Surprisingly, that puts the probability up to about 3% (.0295..)

So if we assume that this winter was the traditional "28 year snow" by the normal definition (ie, one that has a 1/28 chance of occurring) then the probability that it happens again next winter is 3.57% ( a little greater than our "certain 100 year flood". So how likely is THAT to happen in the next 28 years...... (you are soooo gonna love this) .... about 63%, to be slightly more precise, about 63.87%...

Hmmm (wheels turning in your mind)... that's almost the same answer..... "COINCIDENCE? " I ask my students; and they know the proper reply..." I THINK NOT!". So what is happening here. If the probability of an n-year flood is 1/n, then the probability of one occurring in the next n years is given by Prob(flood)=1-$[\frac{n-1}{n}]^n$.. and if we plot this we see that it very quickly becomes asymptotic to about .63... or 1-(

^{1}/

_{e}); which is perfectly understandable since the limit as n goes to infinity of $[\frac{n-1}{n}]^n$ is e

^{-1}; or about .36.

Which means, for anything more than a ten year event or so, the probability that it happens in the next N years is always about the same...about 63%... Many questions, one answer.... ahhhh ... gotta love it.

Labels:
100 year flood,
e,
geometric probability

## Friday, 22 January 2010

### Math Humor

One of my ex-students, Ali B, sent me a link to the comic web site Spiked Math. Some of the comics are a little ... um...er... "ripe" for my HS students (Ali, I am telling your mom)... but here are a couple I thought were pretty nice..

I like this first one because I have a buzz-lightyear doll on my file cabinet that I can press the button and hear lines from the movie, like the one in this comic...not sure if it has a flaw or if they all work that way, but after you press the button and it delivers a line, it will wait three or four minutes and then do another... really freaks kids out when it happens...

This second, I thought, had the wrong title, it should have been something like non-transitive..(look it up children)...

I like this first one because I have a buzz-lightyear doll on my file cabinet that I can press the button and hear lines from the movie, like the one in this comic...not sure if it has a flaw or if they all work that way, but after you press the button and it delivers a line, it will wait three or four minutes and then do another... really freaks kids out when it happens...

This second, I thought, had the wrong title, it should have been something like non-transitive..(look it up children)...

*and thanks, Jeffo*
Labels:
buzz lightyear,
math comics,
non-commutative,
Spiked Math

## Thursday, 21 January 2010

### You Can Count on It... A Brief History of Tally Sticks

The term "tally" comes from the name of a stick on which counts were made to keep a count or a score. The Latin root is

Another candidate for the oldest tally record in history is a wolf bone found in Czechoslovakia with 57 deep notches cut into it, some of which appear to be grouped into sets of five.

In

discontinued they were left with a huge residue of wooden tally sticks, so in 1834 they decided to have a bonfire to get rid of them. The bonfire was such a success that it burned the parliament buildings to the ground. What Guy Fawkes could not do with dynamite the Exchequer did with tally sticks.... The power of math.

The story, as improbable as it seems, is verified by a speech by Charles Dickens 1855. [Charles Dickens, Speech to the Administrative Reform Association, June 27, 1855, in Speeches of Charles Dickens, ed. K.F. Fielding, Oxford: The Clarendon Press, 1960, p. 206, ] The somewhat clipped version below is taken from

Around 1960 an ancient mathematical record on bone was uncovered in the African area of Ishango, near Lake Edward. While it was at first considered an ancient (9000 BC) tally stick, many now think it represents the oldest table of prime numbers. Here is a link with a picture where you can see and read more about the "Ishango bone"

*talea*and is closely related to the origin of tailor, "one who cuts". Many math words have origins that reflect back to the earliest and most primitive uses of number. Compare the origins of compute, digit, and score. The first record existing of tally marks is on a leg bone of a baboon dating prior to 30,000 BC. The bone has 29 clear notches in a row. It was discovered in a cave in Southern Africa. It is sometimes called the Lebombo Bone after the Lebombo mountains in which it was found. The exact age of such artifacts is a subject of debate, and their mathematical usage is somewhat speculative. Some sources have stated that the bone is a lunar phase counter, and by implication that African women were the first mathematicians since keeping track of menstrual cycles requires a lunar calendar.Another candidate for the oldest tally record in history is a wolf bone found in Czechoslovakia with 57 deep notches cut into it, some of which appear to be grouped into sets of five.

In

__Mathematics Galore__by Budd and Sangwin, there is a story of much more recent tally sticks. It seems that until around 1828 the British kept tax and other records on wooden tally sticks. When the system wasdiscontinued they were left with a huge residue of wooden tally sticks, so in 1834 they decided to have a bonfire to get rid of them. The bonfire was such a success that it burned the parliament buildings to the ground. What Guy Fawkes could not do with dynamite the Exchequer did with tally sticks.... The power of math.

The story, as improbable as it seems, is verified by a speech by Charles Dickens 1855. [Charles Dickens, Speech to the Administrative Reform Association, June 27, 1855, in Speeches of Charles Dickens, ed. K.F. Fielding, Oxford: The Clarendon Press, 1960, p. 206, ] The somewhat clipped version below is taken from

__Number, The Language of Science__by Tobias Dantzig (pgs 23&24)Ages ago a savage mode of keeping accounts on notched sticks was introduced into the Court of Exchequer and the accounts were kept much as Robinson Crusoe kept his calendar on the desert island. A multitude of accountants, bookkeepers, and actuaries were born and died... Still official routine inclined to those notched sticks as if they were pillars of the Constitution, and still the Exchequer accounts continued to be kept on certain splints of elm-wood calledSeveral images of the fire was painted by J.M.W. Turner who watched the fire from a boat on the Thames. I have a clip that I can not credit that says, "The fire of 1834 burned down most of the Palace of Westminster. The only part still remaining from 1097 is Westminster Hall. The buildings replacing the destroyed elements include Big Ben's tower (tallies. In the reign of George III an inquiry was made by some revolutionary spirit whether, pens, ink and paper, slates and pencils being in existence, this obstinate adherence to an obsolute custom ought to be continued, ..... All the red tape in the country grew redder at the bare mention of this bold and original conception, and it took until 1826 to get these sticks abolished. In 1834 it was found that there was a considerable accumulation of them; and the question then arose, what was to be done with such worn-out, worm-eaten, rotten old bits of wood? The sticks were housed in Westminster, and it would naturally occur ot any intelligent person that nothing could be easier than to allow them to be carried away for firewood by the miserable people who lived in that neighborhood. However, they never had been useful, and official routine required that they should never be, and so the order went out that they were to be privately and confidentially burned. It came to pass that they were burned in a stove in the House of Lords. The stove, over-gorged with these preposterous sticks, set fire to the paneling; the paneling set fire to the House of Commons; the two houses were reduced to ashes; architects were called in to build others; and we are now in the second million of the cost therof.

*oooh, side bar... Big Ben is not the name of the tower at Westminster, it is the name of the great Bell in the Chimes there.. admit it, you did NOT know that*), with it's four 23 feet clock faces, built in a rich late gothic style that now form the Houses of Commons and the House of Lords. These magnificent buildings are still the subject of many paintings, including my own Parliament, with the grand Westminster Abbey on their north." The one below hangs in the Tate Gallery; while another, I believe, is in a gallery in Cleveland, Ohio.*WIkipedia |

Around 1960 an ancient mathematical record on bone was uncovered in the African area of Ishango, near Lake Edward. While it was at first considered an ancient (9000 BC) tally stick, many now think it represents the oldest table of prime numbers. Here is a link with a picture where you can see and read more about the "Ishango bone"

Labels:
big ben,
ishango bone,
london parliament fire

## Wednesday, 20 January 2010

### Typing Monkeys

More observations stimulated by John Barrows new book (see my recent blog)

Everybody has heard the suggestion that a million (or some other number) of monkeys typing continuously for many millenia would eventually produce a) Shakespeare, b) all of known science, c) the bible, d) all of the above).

It began with Jonathon Swift and Gulliver's Travels, 1872, according to Professor Barrow. In the tale "a mythical professor of the Grand Academy of Lagado who aims to generate a catalogue of all scientific knowledge by having his students continuously generate random strings of letters..." (I think, see emphasis in the excerpt below, that it was random strings of words).. Anyway, according to the good Professor Barrow, the story was embellished in different forms until French Mathematician Emile Borel{

*there is a street and a square named for him in the 17th District in Paris*} suggested that random typing monkeys could duplicate the French national library. A few years later(1929), Arthur Eddington Anglicised that to "books in the British Museum."

By 1972, Arthur Koestler writing in The Case of the Midwife Toad, New York, 1972, page 30, refers to Monkeys

typing Shakespeare as "proverbial":"Neo-Darwinism does indeed carry the nineteenth-century brand of materialism to its extreme limits--to the proverbial monkey at the typewriter, hitting by pure chance on the proper keys to produce a Shakespeare sonnet."

Ok, so eventually someone had to put this to a more scientific test, and they did. "A website entitled The Monkey Shakespeare Simulator, launched on July 1, 2003, contained a Java applet that simulates a large population of monkeys typing randomly, with the stated intention of seeing how long it takes the virtual monkeys to produce a complete Shakespearean play from beginning to end. For example, it produced this partial line from Henry IV, Part 2, reporting that it took "2,737,850 million billion billion billion monkey-years" to reach 24 matching characters:"RUMOUR. Open your ears; 9r5j5&?OWTY Z0d... "

Even more impressive, to me, is the fact that 'in another part of that book, Swift tells of how the astronomers on the flying island of Laputia had: "discovered two lesser stars, or satellites, which revolve around Mars, whereof the innermost is distant from the center of the primary exactly three of his diameters, and the outermost five: the former revolves in the space of ten hours, and the latter in twenty-one and a half".

Swift wrote this in 1726, but it was not until 1877 that Asaph Hall discovered the two moons of Mars.'... I just did a little checking on the orbit and periods he "predicted?" and the actual periods are about 7 hours for Phobos, and about 30 for Deimos... and their distance from the planet were about 9 x 10

^{3}km and 23.5 x 10

^{3}km. Mars has a diameter of 6.794 x 10

^{3}km so they are closer to 1.5 and 3 radii away it seems, but wow, for 100 years before the actual discovery??? Don't you wonder what made him use Mars instead of Venus or ??? Wait, maybe authors typing randomly can describe the true nature of the universe (with some limits of error)...

------------------------------------------------------------------------------------------

If you have your copy, here is what I found in Chapter Five of Gulliver's Travels. The whole thing is available at the Guttenburg Project.

"The first professor I saw, was in a very large room, with forty pupils about him. After salutation, observing me to look earnestly upon a frame, which took up the greatest part of both the length and breadth of the room, he said, "Perhaps I might wonder to see him employed in a project for improving speculative knowledge, by practical and mechanical operations. But the world would soon be sensible of its usefulness; and he flattered himself, that a more noble, exalted thought never sprang in any other man's head. Every one knew how laborious the usual method is of attaining to arts and sciences; whereas, by his contrivance, the most ignorant person, at a reasonable charge, and with a little bodily labour, might write books in philosophy, poetry, politics, laws, mathematics, and theology, without the least assistance from genius or study." He then led me to the frame, about the sides, whereof all his pupils stood in ranks. It was twenty feet square, placed in the middle of the room. The superfices was composed of several bits of wood, about the bigness of a die, but some larger than others. They were all linked together by slender wires. These bits of wood were covered, on every square, with paper pasted on them; and on these papers were

**written all the words of their language**, in their several moods, tenses, and declensions; but without any order. The professor then desired me "to observe; for he was going to set his engine at work." The pupils, at his command, took each of them hold of an iron handle, whereof there were forty fixed round the edges of the frame; and giving them a sudden turn, the whole disposition of the words was entirely changed. He then commanded six-and-thirty of the lads, to read the several lines softly, as they appeared upon the frame; and where they found three or four words together that might make part of a sentence, they dictated to the four remaining boys, who were scribes. This work was repeated three or four times, and at every turn, the engine was so contrived, that the words shifted into new places, as the square bits of wood moved upside down.

Six hours a day the young students were employed in this labour; and the professor showed me several volumes in large folio, already collected, of broken sentences, which he intended to piece together, and out of those rich materials, to give the world a complete body of all arts and sciences; which, however, might be still improved, and much expedited, if the public would raise a fund for making and employing five hundred such frames in Lagado, and oblige the managers to contribute in common their several collections.

He assured me "that this invention had employed all his thoughts from his youth; that he had emptied the whole vocabulary into his frame, and made the strictest computation of the general proportion there is in books between the numbers of particles, nouns, and verbs, and other parts of speech."

## Tuesday, 19 January 2010

### Lewis Carroll, Happy Birthday, and RIP

January was the month in which Lewis Carroll was both Born (27th) and Died (the 14th)

It seems a fitting time to remember some of my favorite (not all true) stories about him.

Carroll was of course the pseudonym by which he wrote, but in his day to day life he was an instructor of mathematics at Oxford by the name of Charles Lutwidge Dodgeson (Lewis Carroll is an alteration of the Latinization of Charles into Carroll, and the replacement of Lutwidge with Lewis). He was a good (but not great) mathematician, but in the way of the world, he is remembered most for his children's stories... and had he not written them, he would probably be remembered for his photography... And if he had avoided that also... Maybe people would know he was a mathematician, but probably not; how many people on the street have heard of Euler?.

An interesting, but seemingly false, story circulated about a gift of a book on determinants to the Queen of England by Lewis Carroll. Here is the version as it is told on the Mathworld page.

=========================

Several accounts state that Lewis Carroll (Charles Dodgson ) sent Queen Victoria a copy of one of his mathematical works, in one account, An Elementary Treatise on Determinants. Heath (1974) states, "A well-known story tells how Queen Victoria, charmed by Alice in Wonderland, expressed a desire to receive the author's next work, and was presented, in due course, with a loyally inscribed copy of An Elementary Treatise on Determinants," while Gattegno (1974) asserts "Queen Victoria, having enjoyed Alice so much, made known her wish to receive the author's other books, and was sent one of Dodgson's mathematical works." However, in Symbolic Logic (1896), Carroll stated, "I take this opportunity of giving what publicity I can to my contradiction of a silly story, which has been going the round of the papers, about my having presented certain books to Her Majesty the Queen. It is so constantly repeated, and is such absolute fiction, that I think it worth while to state, once for all, that it is utterly false in every particular: nothing even resembling it has occurred" (Mikkelson and Mikkelson).

==============================

Ok, but still a neat story...

I love this letter, written by Dodgeson to a young man named Wilton Cox.

"Honoured Sir,

Understanding you to be a distinguished algebraist (that is, distinguished from other algebraists by different face, different height, etc.), I beg to submit to you a difficulty which distresses me much.

If x and y are each equal to 1, it is plain that

2 × (x2 - y2) = 0, and also that 5 × (x - y) = 0.

Hence 2 × (x2 - y2) = 5 × (x - y).

Now divide each side of this equation by (x - y).

Then 2 × (x + y) = 5.

But (x + y) = (1 + 1), i.e. = 2. So that 2 × 2 = 5.

Ever since this painful fact has been forced upon me, I have not slept more than 8 hours a night, and have not been able to eat more than 3 meals a day.

I trust you will pity me and will kindly explain the difficulty to Your obliged, Lewis Carroll."

Another interest of Dodgson's was the analysis of tennis tournaments:

"At a lawn tennis tournament where I chanced to be a spectator, the present method of assigning prizes was brought to my notice by the lamentations of one player who had been beaten early in the contest, and who had the mortification of seeing the second prize carried off by a player whom he knew to be quite inferior to himself." Carroll set out the guidelines for a seeding system well before the good folk at Wimbledon ever thought of it. "Good on ya" as they say over here.

And from a letter he wrote in 1868 with suggestions for essentials of math instruction

Here are links to several other blogs I have written that involve Lewis Carroll in some way..

A Brief History of Logic Diagrams

Searching for Snarks

and Two times Two is Five Enjoy...

## Monday, 18 January 2010

### The Tale of the Creole Pig

This is not a math blog, except that it has to do with logic, or the lack thereof, but I teach kids, and this is a story my bright kids need to read.....and thanks to JD2718 in New York for passing this along.

Just before the recent storm hit Haiti, Kendra Pierre-Louis wrote this blog about the Creole Pigs that were once literally everywhere in Haiti, and how they came NOT to be there. It is a story of the worst indifference to sustainable development, and needs to be shared. ... This is me sharing. Since Blogs come and go, I am copying the whole thing below, but I do encourage you to read the original:

===================================================================================

Growing up in the United States, I grew up listening to my Haitian father speak longingly of two things that he said we couldn't get quite right in the US. The first were mangoes. Most of the mangoes that are found in the US are vaguely round like a Nerf football, and have a mostly deep reddish hue when ripe. They are beautiful, but to hear my father speak, are to the mango what the Red Delicious is to the apple: overproduced and vaguely generic.

The mango of his childhood, the Madame Francis mango, is flatter and green - like an overgrown lima bean. Even at its ripest it only hints at a dusky yellow color. It is also unique to Haiti. I've had it, and he's right it is delicious a queen among mangoes.

My father's other long lost food craving, pork from the Creole Pig, was also unique to Haiti. Unlike the pink pig encapsulated in the image of Wilbur, the pig from Charlotte's web, the Creole Pig was not pink. It, like the population of Haiti, was black and thus unlike American pigs did not sunburn. Raised by eighty to 85% of rural households, the relatively small but dense Creole pig subsisted not on grain, but on the detritus of the island's human population. It could thrive on the husk of rice, the cob of corn. In a nation without consolidated trash pickup the Creole pig acted as the nation's garbage men playing a key role in maintaining the fertility of the soil. And, because it was not dependent on feed for its survival, it functioned for the peasant population as a sort of mobile, literal piggy bank - the animals were sold or slaughtered to pay for school, for marriages, for unexpected medical expenses.

All of this is spoken in the past tense because between the 1970s and the 1980s the Creole pigs were systematically eradicated under pressure of the US government.

Like most of development history some of the facts are in contention, but this much is certain. In the 1970's the African Swine River Virus had spread from Spain to the Dominican Republic and then to Haiti by virtue of the Artibonite River which straddles the two countries.

Now comes the contentious part.

By 1982, says the United States government almost 1/3rd of Haiti's pig population was infected. A lot of Haitians (and many independent organizations) argue otherwise. What is not in contention is that the US in fear of the virus spreading to its own pig population pressured Haiti's government to seize all of the pigs and kill them.

Everyone who had pigs seized were supposed to be compensated in the form of replacement pigs - fat, pink pigs from the American Midwest, deemed 'better' by the USDA. These pigs needed clean drinking water (which 80% of Haitians did not have access to), $90 dollars a year in feed (in a nation where per capita income was $130 dollars a year), vaccination, and special roofed pens to serve as protection from the harsh Caribbean sun.

Does anyone see a problem with this?

Never mind the fact that many Haitians who had their pigs seized were never actually compensated (more on that in a second) - they couldn't have afforded the compensation anyway. In fact, many of those who received pigs found that their new pigs rapidly died.

So much for 'better'.

The eradication of the Creole pig only served to further impoverish Haitians. It forced many children to quit school, forced small farmers to mortgage and eventually lose their land, and forced many Haitians to cut down trees, rapidly increasing the Island's rate of deforestation, to create cash income from charcoal. All simply to save an already rich country from the small risk (and by most independent accounts the number of pigs infected in Haiti was much smaller than the 33% cited by the US) posed to it by a poor, tiny isla

nd nation.

It was, however, a boon to US pig farmers who generated millions in revenue according to grassroots international offloading these ill suited pigs on poor Haitian peasants. How?

In order to get a replacement pig, Haitians were required to pay a princely sum of $50 dollars per pig.

Labels:
Creole Pigs,
Haiti,
JD

## Sunday, 17 January 2010

### Why the three-halves Power

One of the Christmas presents from my sweetheart was John D. Barrow's "One Hundred Essential Things You Didn't Know You Didn't Know." Perhaps it is a little "Math-Lite", but it is still a nice read.

Anyway, reading a section on "A sense of Proportion" in which he points out that if you plot world-record weight lifting records against the weights of the lifters, they fall into a line along the 3/2 power rule, that is (weight of athlete)

The obvious first choice, is Kepler's third law... The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit, or as I think of it to make it easy, year

From there I wondered about other examples I had heard of..For instance the Beaufort scale gives a relationship between wind speed and wave height on the oceans as v = 0.836 B

There is another that shows up in forestry... I couldn't remember it offhand, so I found a note that says, "The 3/2 power law states that the relative rate of self-thinning with respect to biomass growth per unit area is a universal constant" so maybe there is more of a geometric relationship here that would make sense.. Maybe you can explain it to me.

ADDENDUM: I got a tweet from tom@liverbubble that told me about a similar result with the 3/4 power called Kleiber's Law. The principal is named after a agricultural chemist named Max Kleiber. In 1932 he came to the conclusion that the ¾ power of body weight was the most reliable basis for predicting the basal metabolic rate (BMR) of animals and for comparing nutrient requirements among animals of different size. (Send more examples please)

Anyway, reading a section on "A sense of Proportion" in which he points out that if you plot world-record weight lifting records against the weights of the lifters, they fall into a line along the 3/2 power rule, that is (weight of athlete)

^{2}=(weight lifted)^{3}. OK, that isn't THAT surprising, but it got me thinking about other things that fall into a square to cube relation, that don't seem to be as easy to understand.The obvious first choice, is Kepler's third law... The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit, or as I think of it to make it easy, year

^{2}=(distance from sun)^{3}. That has always seemed unexpected to me. One is dependent on the circumference (sort of) and one on the radius, so they might be expected to be essentially equal.. but that inverse square effect of gravity, somehow makes it work.From there I wondered about other examples I had heard of..For instance the Beaufort scale gives a relationship between wind speed and wave height on the oceans as v = 0.836 B

^{3/2}m/s .. ;There is another that shows up in forestry... I couldn't remember it offhand, so I found a note that says, "The 3/2 power law states that the relative rate of self-thinning with respect to biomass growth per unit area is a universal constant" so maybe there is more of a geometric relationship here that would make sense.. Maybe you can explain it to me.

ADDENDUM: I got a tweet from tom@liverbubble that told me about a similar result with the 3/4 power called Kleiber's Law. The principal is named after a agricultural chemist named Max Kleiber. In 1932 he came to the conclusion that the ¾ power of body weight was the most reliable basis for predicting the basal metabolic rate (BMR) of animals and for comparing nutrient requirements among animals of different size. (Send more examples please)

Labels:
Beaufort scale,
tree growth,
wave height,
wind speed

## Thursday, 14 January 2010

### Tic Tac Toe.... Looks Like the Number 15...

Playing around with the mathematics of Tic-Tac-Toe with some kids to talk about symmetry and counting... got me thinking..... and searching....

I found the following (slightly edited) interesting history notes at the ADIT sight..

"It may be that the ancient Roman game of Terni Lapilli was an identical game although the evidence is somewhat mixed. It is certainly true that identical grids to the noughts and crosses grid have been found scratched and etched into surfaces all over the ancient Roman empire. However not a single Nought or Cross has been found to confirm the link. It seems probable that Terni Lapilli was played with simple pieces and may have been played with the same rules but in my mind it’s sheer popularity casts doubt upon the connection.

The first software program designed to play Noughts and Crosses (Which is how the English describe T-T-T,... and I recently was told that many Irish folk call the game Boxin' Oxen) was written by A.S. Douglas as part of his PhD dissertation on Human-Computer interaction. The computer was the EDSAC machine built at Cambridge University in 1949. The EDSAC machine was the first true programmable computer as we would understand it today." Here is what the output looked like on the old Edsac..

There are 255,168 possible games of T-T-T, if you label the positions on the board (from 1-9 for instance), but if you allow for symmetry, there are only 138 possible outcomes. If one (or both) of the two players play the game poorly then it may end in a win on the fifth, sixth, seventh, eighth or ninth move....91 wins by the first player (X) and 44 by the second (o)..... and if you are good at arithmetic, then you now know that out of all the games ever played that ended in a draw.... there are only three different final boards. Somehow I think that is incredible. Can you find them all? It is easiest if you just figure out how to place the four " o " so that X can't win...

You can play against an online computer here...

This is a breakdown of all the endings as presented at Wikipedia...

"Ignoring the sequence of Xs and Os, and after eliminating symmetrical outcomes (ie. rotations and/or reflections of other outcomes), there are only 138 unique outcomes. Assuming once again that X makes the first move every time:

91 unique outcomes are won by (X)

* 21 won by (X) after 5 moves

* 58 won by (X) after 7 moves

* 12 won by (X) after 9 moves

* 44 unique outcomes are won by (O)

* 21 won by (O) after 6 moves

* 23 won by (O) after 8 moves

*

Now for the 15 part... you can play TTT without the board... Players alternate picking numbers from the integers one to nine inclusive. The first who has picked three numbers that add up to fifteen wins...How is it Tic-Tac-Toe? Well label the TTT board like this....Ohhhh, like MAGIC...

I found the following (slightly edited) interesting history notes at the ADIT sight..

"It may be that the ancient Roman game of Terni Lapilli was an identical game although the evidence is somewhat mixed. It is certainly true that identical grids to the noughts and crosses grid have been found scratched and etched into surfaces all over the ancient Roman empire. However not a single Nought or Cross has been found to confirm the link. It seems probable that Terni Lapilli was played with simple pieces and may have been played with the same rules but in my mind it’s sheer popularity casts doubt upon the connection.

The first software program designed to play Noughts and Crosses (Which is how the English describe T-T-T,... and I recently was told that many Irish folk call the game Boxin' Oxen) was written by A.S. Douglas as part of his PhD dissertation on Human-Computer interaction. The computer was the EDSAC machine built at Cambridge University in 1949. The EDSAC machine was the first true programmable computer as we would understand it today." Here is what the output looked like on the old Edsac..

There are 255,168 possible games of T-T-T, if you label the positions on the board (from 1-9 for instance), but if you allow for symmetry, there are only 138 possible outcomes. If one (or both) of the two players play the game poorly then it may end in a win on the fifth, sixth, seventh, eighth or ninth move....91 wins by the first player (X) and 44 by the second (o)..... and if you are good at arithmetic, then you now know that out of all the games ever played that ended in a draw.... there are only three different final boards. Somehow I think that is incredible. Can you find them all? It is easiest if you just figure out how to place the four " o " so that X can't win...

You can play against an online computer here...

This is a breakdown of all the endings as presented at Wikipedia...

"Ignoring the sequence of Xs and Os, and after eliminating symmetrical outcomes (ie. rotations and/or reflections of other outcomes), there are only 138 unique outcomes. Assuming once again that X makes the first move every time:

91 unique outcomes are won by (X)

* 21 won by (X) after 5 moves

* 58 won by (X) after 7 moves

* 12 won by (X) after 9 moves

* 44 unique outcomes are won by (O)

* 21 won by (O) after 6 moves

* 23 won by (O) after 8 moves

*

**3 unique outcomes are drawn**"Now for the 15 part... you can play TTT without the board... Players alternate picking numbers from the integers one to nine inclusive. The first who has picked three numbers that add up to fifteen wins...How is it Tic-Tac-Toe? Well label the TTT board like this....Ohhhh, like MAGIC...

Labels:
15 game,
symmetry,
Tic-tac-toe

## Wednesday, 13 January 2010

### Almost Pythagoras

For some reason I really like little simple geometric relations that remind me of the Pythagorean Theorem. I still remember the first time I saw 3

Just reminded of some other "almost Pythagorean" relations in geometry from an old (1932) Mathematics Teacher article...(Thank you, again, Dave Renfro)

The Two Triangles formed when the median is drawn to any side are

Treat the two sides not cut by the median as if they were the hypotenii (hypotenuses?) of a right triangle. Both are wrong, but in sum, they are right..

That is, while neither of the following are true, AM

AM

Adding the two equations produces a true relation.

AM

The proof is pretty easy using the Law of Cosines.

A second, and about as easy to prove.... Let G be the centroid of Triangle ABC, then AB

A third, which is, I believe, both necessary and sufficient to prove a parallelogram is that the sum of the squares of the diagonals is equal to the sum of the squares of the sides.

And if it is not a parallelogram, but is a trapezoid, then the sum of the squares of the diagonals is equal to the sum of the squares of the non-parallel sides, plus twice the product of the parallel bases (sort of a law-of-cosines look-a-like).

Recently saw the old Wizard of Oz mis-spoken statement of the Theorem by Ray Bolger as the Scarecrow posted at 360, so I thought this was a good time to include it for my students (and anyone else who has never seen it, or just wants to chuckle one more time). [] Play it for your students and see if they can catch all the mistakes...

^{2}+ 4^{2}=5^{2}set next to 3^{3}+4^{3}+5^{3}=6^{3}and wondered.... Could it be????... Is it possible???? Oh go on, you know you are going to check... how could you resist?Just reminded of some other "almost Pythagorean" relations in geometry from an old (1932) Mathematics Teacher article...(Thank you, again, Dave Renfro)

The Two Triangles formed when the median is drawn to any side are

**almost Pythagorean**Treat the two sides not cut by the median as if they were the hypotenii (hypotenuses?) of a right triangle. Both are wrong, but in sum, they are right..

That is, while neither of the following are true, AM

^{2}+MC^{2}= AC^{2}AM

^{2}+MB^{2}=AB^{2}Adding the two equations produces a true relation.

AM

^{2}+MC^{2}+AM^{2}+MB^{2}= AC^{2}+AB^{2}The proof is pretty easy using the Law of Cosines.

A second, and about as easy to prove.... Let G be the centroid of Triangle ABC, then AB

^{2}+BC^{2}+CA^{2}= 3 (GA^{2}+GB^{2}+GC^{2})A third, which is, I believe, both necessary and sufficient to prove a parallelogram is that the sum of the squares of the diagonals is equal to the sum of the squares of the sides.

And if it is not a parallelogram, but is a trapezoid, then the sum of the squares of the diagonals is equal to the sum of the squares of the non-parallel sides, plus twice the product of the parallel bases (sort of a law-of-cosines look-a-like).

Recently saw the old Wizard of Oz mis-spoken statement of the Theorem by Ray Bolger as the Scarecrow posted at 360, so I thought this was a good time to include it for my students (and anyone else who has never seen it, or just wants to chuckle one more time). [] Play it for your students and see if they can catch all the mistakes...

Labels:
Pythagorean-like relations

## Sunday, 10 January 2010

### Zeros and More Zeros,

An Interesting Relationship Between the Zeros of a Polynomial and the Zeros of Its Derivative.

Just reread an article from Dan Kalman at American University a while back which is here if you want to read the original inspiration for what I am blabbing on about.

Now with real valued polynomials, it is easy to show that between any two zeros of the function, the derivative will have a zero as well, this is essentially what Rolle's theorem says... (Strange that he is remembered most for a "calculus" theorem when he spent much of his life speaking out against calculus). I don't think I knew before I first read this, though, that the average value of the zeros of the polynomial is the same as the average of the zeros of the derivative. Dan's paper shows that this property is true, even if we extend to complex polynomials.

Now I don't spend a lot of time in my high school classes on complex polynomials, but the geometry involved was really nice. It seems that if you plot the zeros of the polynomial (real or complex) on a coordinate plane [complex zero a+bi is plotted as the point (a,b)] and then plot the zeros of the derivative, they will always be inside the boundary determined by the derivatives of the original function (This is called Lucas' Theorem... "For an arbitrary not identically constant polynomial, the zeros of its derivatives lie in the smallest convex polygon containing the zeros of the original polynomial."

For a cubic polynomial for instance, if the roots are not all real, then there will be two complex roots so on the complex plane the three zeros will fall in a triangle. What is really clever about the theorem that Dan has called "Marden's Theorem", is that the two zeros of the derivative of the cubic will not just be inside the triangle formed by the three, but it will be the focus of the ellipse inscribed in the triangle touching the midpoints of the three sides. This means they are symmetric about the centroid of the triangle. The image below shows my rough interpretation of the zeros of x

SOOOoooooo, he wonders... what happens if we go to a fourth degree polynomial. Time to play.

Ok, so if the polynomial has real coefficients (and if everyone hasn't already dozed off in boredom by now) I just leaned about something called Jensen's Theorem...

which seems to narrow the location of possible zeros of the derivatives a little more. If the polynomial coefficients are real, then all the complex zeros will occur in conjugate pairs (Alg II, you remember!). It seems that if the derivative has any complex zeros, they will have to occur in a circle whose diameter has endpoints on a pair of these conjugate zeros.

Willing to learn more..HELP!

Just reread an article from Dan Kalman at American University a while back which is here if you want to read the original inspiration for what I am blabbing on about.

Now with real valued polynomials, it is easy to show that between any two zeros of the function, the derivative will have a zero as well, this is essentially what Rolle's theorem says... (Strange that he is remembered most for a "calculus" theorem when he spent much of his life speaking out against calculus). I don't think I knew before I first read this, though, that the average value of the zeros of the polynomial is the same as the average of the zeros of the derivative. Dan's paper shows that this property is true, even if we extend to complex polynomials.

Now I don't spend a lot of time in my high school classes on complex polynomials, but the geometry involved was really nice. It seems that if you plot the zeros of the polynomial (real or complex) on a coordinate plane [complex zero a+bi is plotted as the point (a,b)] and then plot the zeros of the derivative, they will always be inside the boundary determined by the derivatives of the original function (This is called Lucas' Theorem... "For an arbitrary not identically constant polynomial, the zeros of its derivatives lie in the smallest convex polygon containing the zeros of the original polynomial."

For a cubic polynomial for instance, if the roots are not all real, then there will be two complex roots so on the complex plane the three zeros will fall in a triangle. What is really clever about the theorem that Dan has called "Marden's Theorem", is that the two zeros of the derivative of the cubic will not just be inside the triangle formed by the three, but it will be the focus of the ellipse inscribed in the triangle touching the midpoints of the three sides. This means they are symmetric about the centroid of the triangle. The image below shows my rough interpretation of the zeros of x

^{3}-7x^{2}+24x-18 which are at 3+3i, 3-3i, and -1. The two zeros of the derivative would then be at 7/3 (+/-) i sqrt(23)/3. And since the second derivative is linear (6x^{2}-14) then the average of the roots will be at x=7/3, which is the centroid of the triangle. Now when the roots are all real, this confirms what most good calculus student's observe pretty quickly, the two zeros of a cubic's derivative are symmetric about the point of inflection (they have the same average).SOOOoooooo, he wonders... what happens if we go to a fourth degree polynomial. Time to play.

Ok, so if the polynomial has real coefficients (and if everyone hasn't already dozed off in boredom by now) I just leaned about something called Jensen's Theorem...

which seems to narrow the location of possible zeros of the derivatives a little more. If the polynomial coefficients are real, then all the complex zeros will occur in conjugate pairs (Alg II, you remember!). It seems that if the derivative has any complex zeros, they will have to occur in a circle whose diameter has endpoints on a pair of these conjugate zeros.

Willing to learn more..HELP!

Labels:
complex zeros,
Dan Kalman,
Jensen's Thm,
Marden's Thm

## Thursday, 7 January 2010

### Rational Roots of a Quadratic Equation

Here is an interesting excursion to try; take a quadratic equation that you know has rational roots, and then permute the values of A, B, and C through all possible variations. What is the probability that they would ALL have rational roots?

For example, you might try 2x

If instead, you had used x

What happened with the choices you made for A, B, and C (you DID do it didn't you???).

Several questions pop up...

Is it possible for All six of the permutations to have rational roots?

Both my examples have an even number of the permutations that have rational roots. Is it possible that the number of permutations with rational roots could be odd?

Is it possible that a choice of A, B, and C would have two permutations with rational roots, two with irrational roots, and two with imaginary roots? (sort of the ultimate root trifecta)

The answer to the first question is yes, in fact, I can prove (and you probably can too) that if we select A, B and C so that A+B+C = 0, then all six permutations will have rational roots. The roots will be rational if b^2-4*A*C is a square of a rational, so we can substitute -(A+B) for C and see that (B,

I like the second question because it can be solved with a simple appeal to symmetry. Since B

For the Third, I resorted to a computer search and the first I found was with the values of 1, -4, and -5. x

For example, you might try 2x

^{2}+7x + 3, which has roots of -1/2 and -3. If you switch to 2x^{2}+3x + 7 there are no real roots at all. In fact, four of the six permutations have imaginary roots, and only 3x^{2}+7x +2 also has rational roots.If instead, you had used x

^{2}+3x - 10 (roots at 2 and -5) you would find three other permutations had rational roots, 3x^{2}+x -10, -10x^{2}+3x + 1, and -10x^{2}+1x + 3 . The other two permutations have real irrational roots.What happened with the choices you made for A, B, and C (you DID do it didn't you???).

Several questions pop up...

Is it possible for All six of the permutations to have rational roots?

Both my examples have an even number of the permutations that have rational roots. Is it possible that the number of permutations with rational roots could be odd?

Is it possible that a choice of A, B, and C would have two permutations with rational roots, two with irrational roots, and two with imaginary roots? (sort of the ultimate root trifecta)

The answer to the first question is yes, in fact, I can prove (and you probably can too) that if we select A, B and C so that A+B+C = 0, then all six permutations will have rational roots. The roots will be rational if b^2-4*A*C is a square of a rational, so we can substitute -(A+B) for C and see that (B,

^{2}-4A(-A+B)=B^{2}-4AB+4A^{2}=(B-2A)^{2}... An unresolved problem at the moment, which I will have to play with some, is whether this works both ways. Is it possible that all six permutations have rational roots when A+B+C is not Zero??? (please advise if you get this before I get back to it..)I like the second question because it can be solved with a simple appeal to symmetry. Since B

^{2}-4AC is equal to B^2-4CA, exchanging A and C will always produce the same value for the discriminant (B^{2}-4AC). So the number of permutations which have rational roots will always be even.For the Third, I resorted to a computer search and the first I found was with the values of 1, -4, and -5. x

^{2}-4x-5 has rational roots, x^{2}-5x-4 has irrational roots, and -4x^{2}+x-5 has imaginary roots.
Labels:
quadratic equation,
rational roots

## Tuesday, 5 January 2010

### Lotteries and Math

Strange how often problems pop up in two or three different ways at once.. recently I was asked by another teacher how often consecutive numbers showed up in the lottery... He had noticed that it seemed to happen pretty often and he thought it might indicate a problem with the way they were being drawn. He was assuming a lottery with 45 numbers, (other places use other numbers)... Fortunately, I had come across the solution only a few days before in two different journal articles dated about 50 years apart.

I have often heard (yes and even said) that the lottery is a tax on the mathematically ignorant. Lately I have been changing my mind a little. I think they are still a VERY bad gamble. The probability of winning, as one pundit put it, is about the same whether you buy a ticket or not. My new appreciation for them is in the way they have contributed to math, and perhaps along the way, to some good social causes also.

My interest was rekindled recently when I came across a letter in the Library of St. Johns College in Cambridge from Brook Taylor, the guy in the calculus theorem, to his friend, the Reverend Newcome in November of 1711 (almost 400 years ago now). What surprised me about the letter was not what he wrote, but what he included... a lottery ticket. I had no idea that the "money prize" kind of lotteries went back so far. Ok, I'm not totally unread. I knew about the soldiers drawing lots from Agamemnon’s helmet During the Persian Wars, but that was not a pay-to-win-a-prize thing... That was a draw-the-lucky-straw and get to fight against the champion kind of lottery. It turns out that except for rare exceptions, most lotteries were of this sort until around the 14-15 th century. The Chinese apparently had one of some form to build their Great Wall, and Augustus Caesar held one to fund public improvements in Rome, but regular lotteries of the modern type began to appear in the middle ages.... the first recorded official lottery was chartered by Queen Elizabeth I, in the year 1566, and was drawn in 1569. Benjamin Franklin organized one to fund the purchase a cannon to defend the state of Philadelphia. THEN?? They got so corrupt and unsavory that they were literally banned by almost every government in the world. Around 1892 the US passed a federal law that lotteries could not use the public mail, and by 1900 they seemed to be totally gone...... until recently. I was simply not aware of their hundreds of years of popularity in the US and England.

Any game that becomes that popular, will find its way into math, and Lotteries did too. In Genoa in the 1500's they used a lottery to replace members of the ninety man Republic Council. Every six months the ninety names would be placed in a bowl and five names drawn would be replaced with new members. Later the process of buying insurance against their loss of position by members,

The Genoese lottery problem came to the attention of Leonhard Euler, most probably, at the interest of Frederick the Great who implemented a lottery to rebuild the state revenues after the Seven Years War. Euler wrote ," On the Probability of Sequences in the Genoese Lottery" and calculated the probability that if five numbers were selected at random from the interval [1, 90], there would be two or more consecutive numbers. With the kind of magical inductive work he is known for, he came up with $\frac{\dbinom{n-m+1}{m}}{\dbinom{n}{m}}$.. Ok, If Euler felt lotteries were worth his mathematical time... I'll consider them worthy of study... but I won't be buying a ticket this week....

I found another article that spelled out a recursive way to create a table to find the number of ways to NOT get two adjacent.

If we define the number of ways of picking m objects from n objects without getting adjacent values as f(n,m) then we can state that f(n,1) = n and f(1,p)=0 for any p>1..

[It is also easy to show that f(n,2)=$\dbinom{n}{2}$-n+1).. all the ways of picking two except the n-1 ways that have two adjacent numbers]

All the members of f(n,m) are of two kinds, then. One kind is those which include item one, but not item two, and there must be f(n-2,m-1) of these.. (if you have a pattern that doesn't include adjacent values when picking 5 things out of 12 for example, then you can make one that works picking 6 things out of 14 by picking the first and not the second and then the remaining pattern from any successful f(12,5) solution).

There are others which do not include ball 1 but do include ball 2; f(n-1,m) of them... SOO the number of solutions for f(n,m)= f(n-2,m-1) + f(n-1,m) Now we can start to make a table,

N ...1...2...3...4...5...6...7...8...9

m

1........1...2...3...4...5....6...7...8...9

2........0...0...1...3...6...10..15..21..28.

3 ......0.....0...0...0...1...4...10..20..35..

4.......0................0....0....1....5...15..

Note that each new entry is the entry to its left, added to the entry two to the left in the previous row...

For example, the 4 found for f(6,3) is the one to its left f(5,3) plus the 3 in the location for f(4,2)...

Now we can quickly extend the table for any number of values...

So by either method, the number of ways to NOT get two consecutive numbers (or more) in drawing five balls from a sequence of 45 is given by $\dbinom{45-5+1}{5}$ To find the number of draws that do include adjacent values we subtract this from $\dbinom{45}{5}$ which gives 472,361 ways to get at least one adjacent pair out of the 1,221,759 possible sets of five, a probability of about 38%... so you should expect it pretty often.

I have often heard (yes and even said) that the lottery is a tax on the mathematically ignorant. Lately I have been changing my mind a little. I think they are still a VERY bad gamble. The probability of winning, as one pundit put it, is about the same whether you buy a ticket or not. My new appreciation for them is in the way they have contributed to math, and perhaps along the way, to some good social causes also.

My interest was rekindled recently when I came across a letter in the Library of St. Johns College in Cambridge from Brook Taylor, the guy in the calculus theorem, to his friend, the Reverend Newcome in November of 1711 (almost 400 years ago now). What surprised me about the letter was not what he wrote, but what he included... a lottery ticket. I had no idea that the "money prize" kind of lotteries went back so far. Ok, I'm not totally unread. I knew about the soldiers drawing lots from Agamemnon’s helmet During the Persian Wars, but that was not a pay-to-win-a-prize thing... That was a draw-the-lucky-straw and get to fight against the champion kind of lottery. It turns out that except for rare exceptions, most lotteries were of this sort until around the 14-15 th century. The Chinese apparently had one of some form to build their Great Wall, and Augustus Caesar held one to fund public improvements in Rome, but regular lotteries of the modern type began to appear in the middle ages.... the first recorded official lottery was chartered by Queen Elizabeth I, in the year 1566, and was drawn in 1569. Benjamin Franklin organized one to fund the purchase a cannon to defend the state of Philadelphia. THEN?? They got so corrupt and unsavory that they were literally banned by almost every government in the world. Around 1892 the US passed a federal law that lotteries could not use the public mail, and by 1900 they seemed to be totally gone...... until recently. I was simply not aware of their hundreds of years of popularity in the US and England.

Any game that becomes that popular, will find its way into math, and Lotteries did too. In Genoa in the 1500's they used a lottery to replace members of the ninety man Republic Council. Every six months the ninety names would be placed in a bowl and five names drawn would be replaced with new members. Later the process of buying insurance against their loss of position by members,

*and others who might have depended on their presence on the council*, led to the idea of a lottery with a cash pay-off.The Genoese lottery problem came to the attention of Leonhard Euler, most probably, at the interest of Frederick the Great who implemented a lottery to rebuild the state revenues after the Seven Years War. Euler wrote ," On the Probability of Sequences in the Genoese Lottery" and calculated the probability that if five numbers were selected at random from the interval [1, 90], there would be two or more consecutive numbers. With the kind of magical inductive work he is known for, he came up with $\frac{\dbinom{n-m+1}{m}}{\dbinom{n}{m}}$.. Ok, If Euler felt lotteries were worth his mathematical time... I'll consider them worthy of study... but I won't be buying a ticket this week....

I found another article that spelled out a recursive way to create a table to find the number of ways to NOT get two adjacent.

If we define the number of ways of picking m objects from n objects without getting adjacent values as f(n,m) then we can state that f(n,1) = n and f(1,p)=0 for any p>1..

[It is also easy to show that f(n,2)=$\dbinom{n}{2}$-n+1).. all the ways of picking two except the n-1 ways that have two adjacent numbers]

All the members of f(n,m) are of two kinds, then. One kind is those which include item one, but not item two, and there must be f(n-2,m-1) of these.. (if you have a pattern that doesn't include adjacent values when picking 5 things out of 12 for example, then you can make one that works picking 6 things out of 14 by picking the first and not the second and then the remaining pattern from any successful f(12,5) solution).

There are others which do not include ball 1 but do include ball 2; f(n-1,m) of them... SOO the number of solutions for f(n,m)= f(n-2,m-1) + f(n-1,m) Now we can start to make a table,

N ...1...2...3...4...5...6...7...8...9

m

1........1...2...3...4...5....6...7...8...9

2........0...0...1...3...6...10..15..21..28.

3 ......0.....0...0...0...1...4...10..20..35..

4.......0................0....0....1....5...15..

Note that each new entry is the entry to its left, added to the entry two to the left in the previous row...

For example, the 4 found for f(6,3) is the one to its left f(5,3) plus the 3 in the location for f(4,2)...

Now we can quickly extend the table for any number of values...

So by either method, the number of ways to NOT get two consecutive numbers (or more) in drawing five balls from a sequence of 45 is given by $\dbinom{45-5+1}{5}$ To find the number of draws that do include adjacent values we subtract this from $\dbinom{45}{5}$ which gives 472,361 ways to get at least one adjacent pair out of the 1,221,759 possible sets of five, a probability of about 38%... so you should expect it pretty often.

Labels:
combinatorics,
lotteries,
recursive functions

## Sunday, 3 January 2010

### Weird Al, doing the Palindrome song..

Thanks to Ron Dirkse, A great Stats teacher who still resides in his adopted homeland of Japan..(huge envy here) This is great.. Thanks Ron...

Which leads me to my next question.... Who can come up with the best math palindrome... If you are in my math class, you can get extra credit for this one.... (don't be greedy, take the challenge, then we'll discuss what it is worth)..They don't have to be numbers, but they can... 12(63)= (36)21...

Which leads me to my next question.... Who can come up with the best math palindrome... If you are in my math class, you can get extra credit for this one.... (don't be greedy, take the challenge, then we'll discuss what it is worth)..They don't have to be numbers, but they can... 12(63)= (36)21...

Labels:
palindrome,
Weird Al

## Saturday, 2 January 2010

### A Palindromic Date, and A Problem

01/02/2010.... or 01022010 Ok, technically that is not a Palindrome, or else numbers like 110 would be palindromes. But still,close enough to be special as pointed out by my lovely sweetheart.

How unusual is that??? Well, there are only 19,998 numbers less than 100,000,000 that are palindromes...I just came across this at the Wolfram Mathworld page.. The sum of the reciprocals of the palindromic numbers ( 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 22, 33, 44, 55, 66, 77, 88, 99, 101, 111, 121, 131, 141, 151, 161, 171, 181, 191, 202, 212...) converges to a constant...... approx 3.37018..

Strange, but as numbers get bigger, prime palindromes get very rare it seems...2, 3, 5, 7, 11, 101, 131, 151, 181, 191, 313, 353, 373, 383, 727, 757, 787, 797, 919, 929, 10301, 10501, 10601, 11311, 11411, 12421, 12721, 12821, 13331, 13831, 13931, 14341, and it seems that after eleven, there are no palindromic primes that have an even number of digits (Every palindrome with an even number of digits is divisible by 11).

Next year on Nov 20, we get a real palindromic date, 11/02/2011.... Now that IS unusual.. Ok, and the next one after that is?????

Addendum, Professor Charles Wells of Case Western Reserve suggested a follow up, "Count the number of dates in the American system that are palindromes by something cleverer than brute force. " Well, I didn't actually do that, but I did take a moment to figure out that, if my math doodles are right, there was not such an event from 12/31/1321 (oops, think that should be 09/31/1390... "mia culpa") until 10/02/2001... then after the Jan 2nd this year, and Nov of next year... (and the one you have to figure out that comes after that)... well, there just are not very many of them...

also... I just read somewhere that we who are alive today (those of us over 20) are kind of blessed because we have lived through two palindromic years... 1991 and 2002, and that won't happen to anyone again for about a thousand years unless someone does discover a way not to die in anything like what we consider a normal lifetime now-a-days... lucky me.. lucky you???

How unusual is that??? Well, there are only 19,998 numbers less than 100,000,000 that are palindromes...I just came across this at the Wolfram Mathworld page.. The sum of the reciprocals of the palindromic numbers ( 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 22, 33, 44, 55, 66, 77, 88, 99, 101, 111, 121, 131, 141, 151, 161, 171, 181, 191, 202, 212...) converges to a constant...... approx 3.37018..

Strange, but as numbers get bigger, prime palindromes get very rare it seems...2, 3, 5, 7, 11, 101, 131, 151, 181, 191, 313, 353, 373, 383, 727, 757, 787, 797, 919, 929, 10301, 10501, 10601, 11311, 11411, 12421, 12721, 12821, 13331, 13831, 13931, 14341, and it seems that after eleven, there are no palindromic primes that have an even number of digits (Every palindrome with an even number of digits is divisible by 11).

Next year on Nov 20, we get a real palindromic date, 11/02/2011.... Now that IS unusual.. Ok, and the next one after that is?????

Addendum, Professor Charles Wells of Case Western Reserve suggested a follow up, "Count the number of dates in the American system that are palindromes by something cleverer than brute force. " Well, I didn't actually do that, but I did take a moment to figure out that, if my math doodles are right, there was not such an event from 12/31/1321 (oops, think that should be 09/31/1390... "mia culpa") until 10/02/2001... then after the Jan 2nd this year, and Nov of next year... (and the one you have to figure out that comes after that)... well, there just are not very many of them...

also... I just read somewhere that we who are alive today (those of us over 20) are kind of blessed because we have lived through two palindromic years... 1991 and 2002, and that won't happen to anyone again for about a thousand years unless someone does discover a way not to die in anything like what we consider a normal lifetime now-a-days... lucky me.. lucky you???

Labels:
palindrome

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