Sunday, October 30, 2016

Alternative math

I've claimed that there were other things you could do in high-school math besides learn how to divide polynomials in algebra.

One possibility: learn about the zoo of sequences.

If you have a rule for generating numbers: a first, second, third, fourth, and so son--what does the list look like? The order matters, so you can number the terms in the sequence if you need to: term 1, term 2, etc: T(1), T(2), T(3), etc. You usually need to number them.

The answer to that question has applications in measurement, chaos theory, control systems, and on and on.

Zeno knew about this one 2500 years ago: (1, 1/2, 1/4, 1/8, 1/16, 1/32 ...). The next term in the sequence is half the previous term. The numbers get small pretty quickly--closer and closer to 0 but never reaching it. This kind of sequence "converges:" it has a limit.

A boring example is (1, 1, 1, 1, ...).

You've probably heard of Fibonacci's sequence: (1, 1, 2, 3, 5, 8, 13, 21, 34, ...). The next term in the sequence is the sum of the previous two, and it gets larger without any limit ("diverging"). You can show that if you go farther and farther along, that the ratio of a term to the one before it is (1+√5)/2.

Some do neither: (1/2, 1/2, 1/4, 3/4, 1/8, 7/8, 1/16, 15/16, ...). The first, third, fifth, etc (odd) terms converge to 0, while the second, fourth, sixth,etc (even) terms converge to 1.

It isn't hard to predict what this will do: (1, -1/3, 1/9, -1/27, 1/81, ...)

Pick a rule. How about a continued fraction? Start with something something simple: T(n) = (1+1/(1+1/T(n-1)))

Let the first term be 1. The second is 1+1/(1+1/1) = 3/2. The third is 1+1/(1+1/(3/2)) = 8/5. Look at the Fibonacci sequence and see if you can guess that the fourth term of this sequence will be.

Suppose we use T(n) = 1 + 1/(2 + 1/T(n-1)). Start at 1. (1, 4/3, 15/11, 48/37, 173/133, ...). That converges to (1+√3)/2 or about 1.36602...

You can start at other numbers besides 1. How does the sequence change?

Things can get downright weird. Suppose you define a sequence T(1) =1, T(2) = 1, T(n) = T(n -T(n-1)) + T(n -T(n-2)). This is "Hofstader's Q Sequence." When a sequence gets a name, you know that there's either something very important or something very weird about it. The first numbers are pretty easy: (1, 1, 2, 3, 3, 4, 5, 5, 6, 6, 6, 8, 8, 8, 10, 9, 10, ...). As you keep on going, though, there can be wild swings. Mostly the numbers stay close to T(n) ≈ n/2, but they jump around a lot. These aren't random numbers--they're all well-defined. However, it isn't even known if the sequence dies somewhere along the line, by trying to refer to an undefined term (like the -1'th term).

The heaviest algebra I've referred to so far is the quadratic equation.


I'm not the first to think of different math directions; I got the idea from learning about some people who were trying to develop materials for exactly that purpose.

The New Mathematical Library series of books was "written by professional mathematicians in order to make some important mathematical ideas interesting and understandable to a large audience of high school students and laymen." The book from that series on continued fractions is available as a PDF.

1 comment:

The Mad Soprano said...

You can't take three from two,
Two is less than three,
So you look at the four in the tens place.
Now that's really four tens,
So you make it three tens,
Regroup, and you change a ten to ten ones,
And you add them to the two and get twelve,
And you take away three, that's nine.
Is that clear?

Now instead of four in the tens place
You've got three,
'cause you added one,
That is to say, ten, to the two,
But you can't take seven from three,
So you look in the hundreds place.

From the three you then use one
To make ten ones...
(and you know why four plus minus one
Plus ten is fourteen minus one?
'cause addition is commutative, right.)
And so you have thirteen tens,
And you take away seven,
And that leaves five...

Well, six actually.
the idea is the important thing.

Now go back to the hundreds place,
And you're left with two.
And you take away one from two,
And that leaves...?

Everybody get one?
Not bad for the first day!

Hooray for new math,
New-hoo-hoo-math,
It won't do you a bit of good to review math.
It's so simple,
So very simple,
That only a child can do it!
Now that actually is not the answer that I had in mind, because the book that I got this problem out of wants you to do it in base eight. but don't panic. base eight is just like base ten really
if you're missing two fingers. shall we have a go at it? hang on.

You can't take three from two,
Two is less than three,
So you look at the four in the eights place.
Now that's really four eights,
So you make it three eights,
Regroup, and you change an eight to eight ones,
And you add them to the two,
And you get one-two base eight,
Which is ten base ten,
And you take away three, that's seven.

Now instead of four in the eights place
You've got three,
'cause you added one,
That is to say, eight, to the two,
But you can't take seven from three,
So you look at the sixty-fours.

"sixty-four? how did sixty-four get into it? " I hear you cry.
Well, sixty-four is eight squared, don't you see?
(well, you ask a silly question, and you get a silly answer.)

From the three you then use one
To make eight ones,
And you add those ones to the three,
And you get one-three base eight,
Or, in other words,
In base ten you have eleven,
And you take away seven,
And seven from eleven is four.
Now go back to the sixty-fours,
And you're left with two,
And you take away one from two,
And that leaves...?

Now, let's not always see the same hands.
One, that's right!
Whoever got one can stay after the show and clean the erasers.

Hooray for new math,
New-hoo-hoo-math,
It won't do you a bit of good to review math.
It's so simple,
So very simple,
That only a child can do it!