| Transcript
for the Math
Jam "AoPS classes Math Jam"
on Feb 26. |
| Math Jam hosted by nsato
(Naoki Sato ). |
nsato (19:29:42)
Welcome to tonight's math jam.
nsato (19:30:13)
We will be discussing the Special AIME Problem Seminar and the Intermediate Algebra class.
nsato (19:30:35)
Special AIME Problem Seminar
nsato (19:30:41)
The Special AIME Problem Seminar is on Saturday, March 3, and Sunday, March 4 from 3:30 - 6:30 PM ET (12:30 - 3:30 PM PT) each day. During this class we will discuss both general test-taking strategies and specific mathematical tactics for the AIME. We will discuss several problems both from past AIMEs and from other challenging competitions that exemplify some of the most common problem-solving tactics on the AIME.
nsato (19:30:52)
Richard Rusczyk will be teaching the Special AIME Problem Seminar. He earned the only perfect score on the AIME in 1989 (yeah, that long ago), and he'll be sharing some of the general approaches he used to do so.
nsato (19:31:04)
The following problems are examples of problems that we'll discuss in the class.
nsato (19:31:10)
nsato (19:31:18)
http://www.artofproblemsolving.com/Classes/AIME/Images/1983aime6k14.gif
nsato (19:31:30)
robinhe (19:31:49)
6=7-1 and 8=7+1
nsato (19:32:14)
Good observation.
Ubemaya (19:32:25)
and binomial expand, half the terms cancel
Ubemaya (19:32:38)
and then you only need to look at the last few because most are congruent to 0 mod 49
nsato (19:32:50)
nsato (19:33:10)
nsato (19:33:24)
nsato (19:33:39)
When we add these two quantities, which part of the sum is not divisible by 7^2?
nsato (19:34:45)
nsato (19:35:04)
What is our final answer?
Ubemaya (19:35:12)
35?
nsato (19:35:36)
Our final answer is the remainder when 1162 is divided by 49, which is 35.
nsato (19:35:50)
nsato (19:35:58)
http://www.artofproblemsolving.com/Classes/AIME/Images/1990aime9NoHH.gif
nsato (19:36:20)
There are 2^10 = 1024 possible outcomes for the series of ten tosses. How can we count the number of tosses that include no consecutive heads?
Ubemaya (19:36:31)
casework: there are 0, 1, 2, 3, 4, 5 heads
nsato (19:36:59)
Is there anything we could try before delving into case work?
tjhance (19:37:04)
recursion?
nsato (19:37:29)
That's one idea, but how could we track down a recursion?
nsato (19:37:42)
We could begin by examining smaller sequences of tosses.
nsato (19:37:49)
One toss outcomes without HH:
nsato (19:37:54)
H
T
nsato (19:38:05)
Two toss outcomes without HH:
nsato (19:38:07)
HT
TH
TT
nsato (19:38:22)
In what ways could we toss a coin three times without ending up with HH in the sequence?
robinhe (19:38:50)
HTH
HTT
THT
TTH
TTT
suvisa (19:38:53)
hth, ttt, htt, tht, tth?
Ubemaya (19:39:09)
H T T, T H T, T T H, T T T
Ubemaya (19:39:12)
H T H
nsato (19:39:27)
Three toss outcomes without HH:
nsato (19:39:30)
HTH
HTT
THT
TTH
TTT
nsato (19:39:37)
In what ways could we toss a coin four times without ending up with HH in the sequence?
panjia123 (19:40:34)
TTTT, TTTH, TTHT, THTT, HTTT, HTHT, HTTH, THTH
rnwang2 (19:40:43)
would there be 8 total?
nsato (19:41:43)
Four toss outcomes without HH:
nsato (19:41:56)
HTHT
HTTH
HTTT
THTH
THTT
TTHT
TTTH
TTTT
Super Naterul (19:41:42)
TTTT
HTHT
THTH
HTTT
THTT
TTHT
TTTH
HTTH
ruth-bruce-ted@sbcglobal. (19:41:44)
htth htht thth ttth ttht thtt httt tttt
nsato (19:42:04)
Do we notice anything from examining these smaller cases?
nsato (19:42:24)
How are we constructing each sequence as we add an additional flip?
nsato (19:42:47)
Each flip of n coins either ends in an H or a T. How can we construct new sequences from these possibilities?
nsato (19:43:16)
First we should note that any sequence of n + 1 flips without HH occurring is a sequence of n flips without HH occurring plus one more flip (such that the new sequence does not end in HH).
Ubemaya (19:43:00)
adding a T/H to the end of it?
tjhance (19:42:44)
we add heads to the ones that end in tails and we add tails to all of them
nsato (19:43:47)
If we add one flip to a sequence that ends in H, the new flip must be a T or the sequence will end in HH.
nsato (19:44:01)
If we add one flip to a sequence that ends in T, the new flip can be either a T or an H and there is no new occurrence of HH.
nsato (19:44:06)
What does this tells us about the total number of sequences of length n + 1 without an occurrence of HH?
nsato (19:44:54)
nsato (19:45:13)
What have we learned?
Ubemaya (19:45:47)
b_n = b_n-1 + b_n-2
robinhe (19:42:22)
fibonacci
rnwang2 (19:42:30)
fibonacci
panjia123 (19:42:32)
2,3,5,8...Fibonacci numbers?
Ubemaya (19:44:24)
the sum of the two previous
Ubemaya (19:44:27)
so it's fibonacci
nsato (19:46:09)
Ubemaya (19:46:40)
it's also the fibonacci sequence, except theh index is shifted down
nsato (19:47:14)
nsato (19:47:22)
How can we use these facts to help us solve the problem?
Super Naterul (19:47:42)
look at the 10th
Ubemaya (19:48:06)
the sum of a_n and b_n is f_n+2, so find the 12 fibonacci number
panjia123 (19:48:10)
if they are fibonnaci numbers, we can easily find the 10th term (starting with 2)
nsato (19:48:23)
Ubemaya (19:48:22)
which is 144
nsato (19:49:32)
We can now say that i/j = 144/1024 = 9/64, so i + j = 73.
nsato (19:49:57)
This is one of those problems that shows how important it is to test small cases and then to observe the way that we generate those cases. Making the connection between the way we generated each new sequence and the recursion that we derived gave us a direct calculation to our answer.
nsato (19:50:38)
Now we'll move onto the Intermediate Algebra class.
nsato (19:50:40)
Intermediate Algebra
nsato (19:50:46)
The Intermediate Algebra class will be held on Tuesday evenings (7:30 - 9:00 PM ET, 4:30 - 6:00 PM PT), from March 6 to May 22. It will be taught by Mathlinks website founder Valentin Vornicu.
nsato (19:51:03)
While many students may think that their understanding of Algebra II is enough that they have little need for this course, Intermediate Algebra covers many topics typically avoided in high school curriculum and takes subjects previously covered to a much higher level.
nsato (19:51:16)
Topics covered in the Intermediate Algebra class include algebraic manipulations, advanced factoring techniques, difference equations, techniques for solving hard systems of equations, symmetric polynomial sums, greatest and least integer functions, advanced methods for dealing with logarithms, functional equations, and much more.
nsato (19:51:25)
Let us now take a look at the beginning of the lesson on functional identities.
nsato (19:51:46)
FUNCTIONAL IDENTITIES
nsato (19:51:51)
nsato (19:52:17)
nsato (19:52:23)
http://www.artofproblemsolving.com/Classroom/cbe6/images/lx-39597581.gif
robinhe (19:52:53)
1,3?
Ubemaya (19:52:58)
1 for sure
Ubemaya (19:53:00)
not 2
Ubemaya (19:53:04)
and 3
suvisa (19:53:05)
1 yes 2 no 3 yes?
rnwang2 (19:53:08)
1 and 3
nsato (19:53:28)
The absolute value function gives out the (positive) magnitude of the input.
nsato (19:53:34)
1. The product of the magnitudes is the magnitude of the product.
nsato (19:53:43)
2. When one of x and y is negative and the other is positive, their sum cancels out some of the magnitude of each x and y making the identity f(x+y) = f(x) + f(y) false.
nsato (19:54:06)
3. The magnitude of a magnitude is still just the magnitude because all magnitudes are non-negative.
nsato (19:54:18)
SOLVING FUNCTIONAL IDENTITES
nsato (19:54:23)
So far we have taken a look at a number of different kinds of functions by looking at the identities that satisfy those functions and then solving a variety of problems based on those identities.
nsato (19:54:32)
Now we shall take a look at how to take an identity or a group of identities and determine what kinds of functions satisfy those identities.
nsato (19:54:44)
nsato (19:54:53)
Our goal is to find out what f(x) looks like, but unfortunately we have an identity that relies on two variables instead of one. Is there another way that we could write this identity that might allow us to view the function the way it acts on a single variable?
ra5249 (19:55:07)
f(x)/x = f(y)/y
nsato (19:55:45)
nsato (19:56:47)
Now we have two sides to the identity which each rely on only one variable. No matter what the variable is on either side, the two sides are equal. Does this give us any ideas as to how to solve for f(x)?
nsato (19:56:57)
robinhe (19:57:23)
g(x)=g(y)
rnwang2 (19:57:24)
g(x) = g(y)
suvisa (19:57:11)
g(t) is constant
nsato (19:57:47)
After dividing our original equation by xy we got g(x) = g(y). We know that g(x) = g(y) for all x and y. Thus g(t) is a constant function and we can say g(t) = c, for some constant c.
nsato (19:58:00)
What does that tell us about f(x)?
rnwang2 (19:55:32)
f(x) = cx
Ubemaya (19:56:26)
well if f(x) = cx for c is a constant then it works
suvisa (19:57:28)
so maybe f(x) = kx?
nsato (19:58:47)
nsato (19:59:03)
Thus f(x) = cx is the entire family of solutions that could satisfy the identity
yf(x) = xf(y).
nsato (19:59:11)
Is there anything else that we need to do?
nsato (19:59:42)
We have shown that only functions in the form f(x) = cx satisfy the given identity. However, we have not yet shown that all functions in the form f(x) = cx do satisfy the identity. What do we need to do to test which such functions satisfy yf(x) = xf(y)?
rnwang2 (20:00:30)
the original relation becomes ycx = xcy, which is true
nsato (20:00:48)
We can plug the generic function f(x) = cx into the identity:
y(cx) = x(cy)
Since this statement is always true, we have shown that all functions in the family f(x) = cx do indeed satisfy yf(x) = xf(y). Since we knew only functions in that form could satisfy the identity, we have our complete set of solutions.
nsato (20:01:05)
What exactly did we do to solve this identity? What was the key to the process that brought us to the answer?
nsato (20:01:11)
We began with a two variable equation and we wanted an answer in one variable. The key to solving this identity was isolating each variable x and y to one side of the equation. That allowed us to describe a substitute function, g(t), for each side of the rearranged identity. This is called the method of ISOLATION and it is useful in many situations.
nsato (20:01:33)
In the Intermediate Algebra class we build up several methods for solving functional identities and eventually tackle much harder problems such as
nsato (20:01:57)
nsato (20:02:03)
Find all polynomials p such that p(x+1) = p(x) + 2x + 1.
nsato (20:02:10)
Find all solutions to the functional equation f(1-x) = f(x) + 2 - 4x.
nsato (20:02:15)
The Intermediate Algebra class also involves a variety of other algebraic topics including methods of substitution, functions, polynomials, sequences and series (including the use of difference equations), binomial expansion, logarithms, advanced systems of equations, and greatest/least integer functions. Here are a few harder problems we will tackle in the course:
nsato (20:02:28)
nsato (20:02:34)
nsato (20:02:47)
nsato (20:02:54)
nsato (20:03:12)
nsato (20:03:20)
platinumnerd (20:03:05)
what's the abel summation formula?
nsato (20:03:54)
That's something that will be covered in the class.
nsato (20:04:39)
Are there any questions about the Special AIME seminar or Intermediate Algebra?
suvisa_2 (20:04:22)
is there a book recommended for the class -- if that info is on the aops website dont bother to reply, i will look it up later
nsato (20:05:19)
Not one in particular. If you want to get some background to prepare, I would recommend AoPS Vol. 2.
notehead (20:04:25)
Do we have to know any of this stuff you're showing us?
nsato (20:05:44)
No, this is just to give a flavor of the material in the course. This will all be covered much more thoroughly.
ra5249 (20:05:26)
What does the ""Special"" in the AIME seminar mean?
nsato (20:06:47)
Nothing much, really. It's just something we hold for the AIME, and only for two days, so it's very different from our other classes.
tjhance (20:05:27)
I would like to sign up for the AIME seminar, but I might have to miss about the first hour of the first session due to a schedule conflict. Would this be a problem?
nsato (20:07:49)
Not at all. And of course, you can always check the transcripts later.
rnwang2 (20:05:56)
would a lot of stuff (not all, of course) in Intermediate Algebra be review if you've taken Algebra 2
nsato (20:09:19)
As I mentioned at the start, we begin with the material in Algebra 2 but then get into it at a much deeper level, and building concepts and techniques to solve problems. So even if you've taken algebra 2, I think there would still be a lot of material you haven't seen.
Ubemaya (20:06:03)
does the class change from year to year?
nsato (20:09:30)
No, it's the same material.
Fairy (20:06:18)
Is it beneficial to do the AIME seminar even if you aren't going to do the exam?
nsato (20:10:37)
It depends. The course is very strongly geared towards the AIME. It might help if you're interested in being able to solve problems at that same level.
suvisa_2 (20:06:33)
is algebra I sufficient for the course or is this kind of above even algebra II ? Also is there a way i can get a feel of the pace of the class -- i just want to be sure that this class doesnt fly over my head
suvisa_2 (20:07:05)
... like for instance is it possible to see the transcripts for the first day of the previous intermed algebra class or something?
nsato (20:12:20)
If you're knowledge of algebra I is very solid, then you are probably ready for Intermediate algebra. You can always take the pre-test as well.
nsato (20:12:57)
We can't show you old transcripts, but the pace is similar to what we have done in today's math jam.
Fairy (20:07:36)
Do you need Intermediate Algebra to do the Intermediate Number Theory course?
nsato (20:13:35)
Nope, they are independent courses.
platinumnerd (20:09:05)
but the transcript is available for viewing only to those that signed up for the course, correct?
nsato (20:13:52)
That's right.
ruth-bruce-ted@sbcglobal. (20:10:56)
wen is the AIME class
nsato (20:14:03)
March 3 and 4.
suvisa_2 (20:13:11)
will the method of instruction in the class be exactly identical to this -- chat based? Once again please ignore this message if this info is avail on the site. I will look over later.
nsato (20:14:50)
Yes, this is the same interface that's used for the classroom.
ruth-bruce-ted@sbcglobal. (20:14:26)
wat time is the class on PT
nsato (20:15:14)
3:30 - 6:30 PM ET, 12:30 - 3:30 PM PT
ruth-bruce-ted@sbcglobal. (20:14:59)
if i want to preapare and AMC 12 and AIME wat classes do i hav to take?
nsato (20:16:57)
We have Problem Series classes that are specifically for training for the AMC 12 and AIME contests, but only the Special AIME Problem Seminar is available in the near future.
nsato (20:17:45)
Are there any other questions?
ruth-bruce-ted@sbcglobal. (20:18:46)
does the intermediate algebra and the number theory a lot of help for AMC 12 and AIME?
nsato (20:19:27)
The level of material is the about the same, so yes, I would say it would help for training for those contests.
nsato (20:18:35)
If there are no other questions, then that concludes tonight's math jam. Thanks for coming!
