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