Pilot Exam!

We took the pilot exam yesterday. After taking it, I learned that there are certain areas that I badly need some practice in:

Conics
Logarithms
Probably
Graphing Trig Functions
Pascal's Triangle

I don't want to get into the details of it, but I found that the probability questions were really hard. I think that there was a sufficient amount of time to do the exam, and I don't think the other questions were that hard. The only really hard ones were the ones where my math knowledge is deficient.

Ho boy, this is going to be tight. I also need to get accelerated math handed in. In fact, that's what I'm going to try and practice with. Even after the exam, I'll do accelerated math.

Partial sums of infinite geometric series.

Find the sum of a geometric series (implies the whole infinite series).

{ 2, 2/5, 2/25, . . . }

becomes 2 + 2/5 + 2/25 + 2/125

The formula of a partial sum is:

Sn = t1 / 1 - r; r = 1/5 and t1 = 2
2 / 1 - 1/5
= 2 / (4/5)
= 2 * 5/4
= 5 / 2
= 2.5

e.x. An infinite convergent geometric sequence has as its infinite sum, 16 with a common ratio 1/2. What is the first term in the sequence?

S = 16; r= 1/2

S = t1 / 1 - r
16 = t1 / 1 - 1/2
16 = t1 / (1/2)
8 = t1

This reminds me of Zeno's paradox: if an archer shoots an arrow at a target, it'll travel 1/2 way to its target. So, for the first round, it's 1/2 from the target, 1/4 from the target, 1/8, 1 /16, 1 / 36, 1 / 72 . . .

So, it never reaches its target. Yet real world experimentation disproves this. All this proves is that if we break a movement into smaller steps, the movement becomes impossible. That's kinda the same with these infinite geometric series. It keeps adding up and up and up infinity. Yet it reaches a certain number. Interesting. . . .

Sum Formulas and Sigma Notation.

There was more stuff to learn about geometric sequences. The New and Innovative thing was called sigma notation:


What does this all mean?

Well . . . how about the sum a sigma? It'll ask us to expand it:


t1 + t2 + t3 + t4 . . .

Evaluate n = 5, k = 1, 2^k

= 2 + 4 + 8 + 16 + 32

in summation notation:

How do we get the sum of a sequence without adding all the numbers? We have a formula for that.

Sn = t1 (1 - r^n) / 1 - r

We can use this formula to measure the sum of a sequence without having to go through the tedium of adding every number in there.

Geometric Seqences.

We learned about Geometric Sequences today. We did arithmetic sequences back in grade 11, and the difference between arithmetic and geometric sequences is that the former is linear and the latter is exponential.

To determine whether a sequence is geometric or not, we look for a common ratio.

Ex:

r = common ratio

1, 2, 4, 8, 16 . . . Is there a common ratio? YES! r? 16/8, 8/4, 4/2, 2 / 1 = 2

4, 2, 1, 1 /2, 1 /4, 1 / 8, Is there a common ratio? YES! r? 1/8 / 1/4 . . . 2/4 = 1/2

5, 15, 20, 25 Is there a common ration? NO! Arithmetic sequence.

How do we find n term in a geometric sequence?

t = { 3, 6, 12, 24 . . . }

First, we find the common ratio:

24 / 12, 12 / 6, 6 / 3 = 2

How do we solve for n when n is 10?

tn = t1*r^(n -1)
t10 = 3 * 2 ^(10 - 1)
t10 = 3 * 2 ^ (9)
t10 = 3 * 512
t10 = 1536

How about something like this?

t = { 4, 2, 1, 1 /2, 1 /4, 1 / 8 . . . }

Common ratio? (1/8) / (1/4), (1/4) / (1/2), (1/2) / 1 = 1/2

tn = t1*r^(n -1)
t10 = 4 * 1 / 2 ^(10 - 1)
t10 = 4 * 1 / 512
t10 = 4 / 512
t10 = 1 / 128


We can also use neat sequences like:

t = { 3, -6, 12, -24 . . . }

How do we figure out 10 in this one?

First, we find the common ratio:

-24 / 12, 12 / -6, -6 / 3 = -2

tn = t1*r^(n -1)
t10 = 3 * -2 ^(10 - 1)
t10 = 3 *- 2 ^ (9)
t10 = 3 * -512
t10 = -1536

No real changes here! Just a negative number. But what about an odd number for n? Looking at the sequence, we can predict that the number will be negative. Let's use 19 in this case:

tn = t1*r^(n -1)
t10 = 3 * -2 ^(19 - 1)
t10 = 3 *- 2 ^ (9)
t10 = 3 * -262144
t10 = -786432

A little irrelevant

The United States is home to the greatest education institutions in the world. Nobody is going to argue that MIT, Caltech, Harvard, Princeton, Cornell, Yale,  and Harvard are the top dog institutions on our planet. And, given our increasingly technologically oriented society, the need for students who are technically adept is absolutely necessary. 

Yet, the trends in education is disturbing. Some highschools grad cannot even grasp basic algebra:

http://www.nydailynews.com/ny_local/education/2009/11/12/2009-11-12_cunys_got_math_problem_many_freshmen_from_city_hs_fail_at_basic_algebra.html

In my opinion, math is as important as English for university. If highschool grads are ill-equipped for the rigor of university, what then? Will the university be forced to water down their curriculum in response to the influx of bad students?

Not to imply that our curriculum is bad; it isn't.  But when the United States decides to walk down a new path, Canada isn't very far behind. This article addresses those issues:

http://www.osstf.on.ca/Default.aspx?DN=76e080a5-0ec6-4448-8bba-67494d2add93

More conics!

Assignments: exercises 37, 38, 39. This was assigned previously, but after today's lesson, we can do more.

Today, Mr Max expanded our knowledge on conics. The central idea of today's lesson is that we can change general form conics into standard form conics.


Conversion between general form conic and standard conics.

Ax^2 + Bxy + Cx^2 + Dx + Ey + F = 0


Remember how to complete the square?

Take: x^2 + 4x + 4

We can take a square and
(x+2)(x+2)
(x+2)^2

or:
How do we find x? Well, take b/2, and square the result. sqroot(3/2)







If we try to enter (X^2 + y^2 + 6x - 8y) = 11 into graphimatica directly, it fails to map the equation because it cannot parse the data. So, we use this form after we complete the square

(X^2 + y^2 + 6x - 8y) = 11
(x^2+6x+9)(y^2-8y+16) = 11 + 9 + 16
(x+3)^2 + (y - 4)^2 = 36

Using the general form:

(x-h)^2+(y-k)^2= r^2
(h,k) = center; r = radius
we enter (x+3)^2+(y-4)^2= 36
and get:








c)4x^2 - y^2 - 8x - 4y - 16 = 0
4x^2-8x -y-4y = 16 + 4 -4
4(x^2-2x+1) -1(y^2+4y+4) = 16 + 4 - 4
(4(x-1)^2 / 16) - ((y+2)^2 / 16) = (16 /16)
(x-1)^2/(2^2) - (y+2)^2/(4^2) = 1
(x-h)^2 / a^2 - (y+2)^2/b^2 = 1








The final is only ~27 days away. I'm going to start studying today! Yes, that sounds crazy, but the exam is difficult, and this class is pretty well perpetration for the exam.

Assesment.

Yeah, time for some introspection. Last Friday, we had a test -- I didn't do too well. The problem is that it was half Trigonometry and Logarithms; those are two areas I'm very weak with. I'll confess that I didn't know the trig circle until I noticed the numerous patterns in it. Now, I can derive all the values from it. Also, I'm going through the math 40s website sheets. I'm going to so some of the green book to complement that material with the main exercise sheets. It's obvious that I am not doing enough work. And, to remedy this, I''ll do more work.

The good news? I'm actually doing it. I got through many of the trig sheets and I'm onto the log sheets. I'm well off when it comes to Compressions with Expansions and Perms & Combs. I feel that I need to do more work in all those areas, but the areas that I am well off in do not require much focus, for the material is familiar to me.

Another thing . . . I was in this class as a means to an end -- to get a very important credit. I strongly disliked the material, but for some reason, I really like doing it now. (Perhaps that is the sole reason I started doing the work in this past week?) At any rate, I think that I only have good news to report. As for accelerated math, I'm full steam ahead on that project. It's something I'm behind in, but given this recent turn of events, I'm fully confident that I will catch up and finish all of the exercises.

My reason for this post it really not for anyone else except myself. It's more or less something that forced me to think about my situation.