- You have a quiz Thursday over the Fraudini trick, binary factorization, primitive counting, and the reading. Also everything that we do in class last week! It might pay to review your agendas prior to a quiz, too.
- For the quiz, you may have only hand-written notes (e.g. example problems, summaries of readings, etc.). If you have a calculator, great; if you must use your phone, then you may only use it in calculator mode, and it will be on the table in front of you.
- I've assigned you some homework problems, just to check if you are getting it.
- We did a few extra problems that showed the link between
- The Fraudini trick
- Primitive counting
- The binary factorization (base 2 representation) of a number.
- We then explored Yanghui's triangle, and discovered some of its
secrets. By understanding the mathematical patterns we were able to
learn certain things about the Chinese method of writing their numbers:
- They used both horizontal and vertical "rods" (bamboo counting rods) to represent 1s, 5s, and 10s
- They stacked the rods, both horizontally and vertically.
- They had a symbol for zero (a circle).
- A failure of symmetry suggested that someone had made a mistake in drawing the triangle.
- The triangle contains the counting numbers along diagonals.
- The triangle contains the triangular numbers along diagonals.
- The pattern that we discovered is that each successive row's entries are formed by adding the two values directly above.
- That fact shows us that the sum of the numbers of each row will double each time, and we always find a power of two for the total.
- We looked at some examples of the uses of the triangle:
for example, Pascal used it to answer this question:
What are the chances that a family of X children will have Y girls?
Today's Question of the Day:
- Let's talk a little more about Yanghui's version, in particular
the bamboo counting rods:
Today the triangle is generally called "Pascal's Triangle", after Blaise Pascal, in part because he studied it so thoroughly, and published a paper with many of its properties; he and Pierre de Fermat were working on the foundations of probability and statistics together, and found it singularly useful.
Pascal used "his" triangle to compute probabilities, in part because of his interest in gaming and gambling. It turns out that this "arithmetic triangle" (as he called it) is very useful for counting things.
- Let's take a little field trip up to visit Mathemalchemy....
- Where we've been,
- Where we are, and
- Where we're going!
- Anyone's favorite number 11?
\(11^0\) \(11^1\) \(11^2\) \(11^3\) \(11^4\) 
Notice the powers of 2.... - Primes: if a row's second element is prime, then all the following
elements in the row are divisible by that prime number (not considering
1s), for example 1 7 21 35 35 21 7 1
- Stacking things. Pascal's triangle is full of "figurate numbers":
Let's talk about stacking. In a way, the triangle is about "piling on":
- One-dimensional (in a line):
The counting numbers are what you get when you keep stacking one more onto the line of numbers you've already got.
- Two-dimensional (in the plane):
The triangular numbers are the numbers you get by stacking the counting numbers on top of each other (as line segments).
- Three-dimensional (in space):
The tetrahedral numbers are the numbers you get by stacking the triangular numbers on top of each other, as triangles.
- One-dimensional (in a line):
- Let me make a connection to trees, because Pascal's triangle seems
to resemble a tree. In particular, I want to relate it to
binary trees.
Imagine flipping coins for each ball: heads, go left; tails, go right. Let's draw the "decision tree", which is just a tree which shows the results of some sort of decision at each step.
Pascal's triangle simply doesn't care if you go right then left, or left then right: you end up in the same plinko spot. So Pascal says "2 ways to do that". "1 way to get HH", "1 way to get TT", but two ways to get a head and a tail (HT and TH). Pascal's triangle just does the accounting: \[ 1 \hspace{1in} 2 \hspace{1in} 1 \] The next row is the ways of getting three heads, two heads, one head, and no heads: \[ 1 \hspace{1in} 3 \hspace{1in} 3 \hspace{1in} 1 \]
And, by counting the leaves of the binary tree, we see that it's always a power of 2! So the sum of the rows of Pascal's triangle has to be a power of 2.... Now we understand the "power of 2" rule;
but also the rule about where adding the "two above" to get the value in the next row. That's the plinko rule!:)
- This brings us to Plinko:
One can also think of this triangle as a version "Plinko"
(if you've ever heard of that); and to something called a Galton
Board (e.g. my "Hexstat Probability Generator" --
which shows you that Pascal's triangle is also very closely
tied to probability calculations, too.)
- Now let's look back at the example of how we can use Pascal in the
study of graphs. Consider the concept of simple graphs,
which I'll present in the context of Facebook.
Definition: A simple graph is a graph without loops, and never more than one arc connecting any two vertices (friendship is a binary thing: it either exists or it doesn't; and you're not friends with yourself).
It turns out that Pascal's triangle gives all the ways of having a "Facebook" for any number of individual (distinct) people.- vertices (or points, which we will think of as the people), and
- edges (which are also called arcs, connecting two points -- which we will think of as friendships).
Now, if each vertex represents a unique individual (we name them), then Pascal's triangle tells us all the different Facebooks possible, of each type. If everyone's friends with everyone else, then the graphs we can get will look like these, the complete graphs:
Now how many different Facebook configurations are there for, say
Each choice of a different set of friendships results in a different Facebook. How do we use Pascal's triangle to find the answer for a given number of people? We first find the number of arcs for the given number of people (the appropriate triangular number), and then read off the number of different ways of having a given number of Facebook friendships from the "arc row" of the table.- 1 individual
- 2 individuals
- 3 individuals, with two friendships?
- 4 individuals, with three friendships?
The same process works if we want to decide questions like Pascal's about boys and girls, such as
You have four extra tickets to a concert; in how many ways could you choose a set of four friends from your seven favorite friends?
- Pascal's triangle is useful for counting the ways of
combining terms in the "algebraic expression" (called a "binomial")
\[ (a+b)^n \]
So, for example,
\[ (a+b)^0=1 \] \[ (a+b)^1=1a+1b \] \[ (a+b)^2=1a^2+2ab+1b^2 \] \[ (a+b)^3=1a^3+3a^2b+3ab^2+1b^3 \] \[ (a+b)^4=1a^4+4a^3b+6a^2b^2+4ab^3+1b^4 \]
(see your assigned reading).
I hope that you noticed Pascal's triangle peeking out from the right hand sides above!
Now we can see why the 11 powers work: \[ 11^0=(10+1)^0 = 1 \] \[ 11^1=(10+1)^1= 1*10 + 1*1 \] \[ 11^2=(10+1)^2=1*10^2+2*10*1 + 1^2 \] \[ 11^3=(10+1)^3=1*10^3+3*10^2*1+3*10*1^2 + 1^3 \] etc.!
-
What this allows us to do is calculate the probabilities of
more interesting situations.
For example, suppose that the probability that you have a cold on a given winter day is \(a=1/100\); then the probability that you don't have a cold is \(b=99/100\).
Let's suppose that winter is 90 days long. What is the probability that you have a cold-free winter? We need to compute the 90th row in the table, which has some really big numbers!
\[ \left(\frac{1}{100}+\frac{99}{100}\right)^{90} \]
- Probability of a cold-free winter: \[ 1\left(\frac{99}{100}\right)^{90} \approx 0.4047 \]
- Probability of a one-day cold winter: \[ 90\left(\frac{1}{100}\right)^1 \left(\frac{99}{100}\right)^{89} \approx 0.3679 \]
- Probability of a two-day cold winter (uh oh, we need the rest of the 90th row of the triangle!:): \[ 4005\left(\frac{1}{100}\right)^2 \left(\frac{99}{100}\right)^{88} \approx 0.1654 \]
- Probability of a three-day cold winter: \[ 117480\left(\frac{1}{100}\right)^3 \left(\frac{99}{100}\right)^{87} \approx 0.0490 \]
- etc.
