Today's Scribe By: daniel g and Jorge V.

04/06/2007 Scribe Post

As usual the class started with a POD

Today’s was:
If the angle is in the 4th quadrant, and its cosine is 4/5, find the values of cosecant and tangent of the angle.
First we had to find y:
We used the Pythagorean Theorem (x^2+y^2=y^2).
4^2+y^2=5^2
16+y^2=25
16+y^2-16=25-16
y^2=9
We then got the square root of each side which ultimately crossed out in one side and in the other side it gave 3, so y equals 3.
We then needed to get the cosecant of the angle. Since we know that cosecant is the inverse of sin, which is y/r, it meant that cosecant must be r/y.
That made the cosecant = -5/3
Then we got tangent. Tangent equals x/r, so the answer was -3/4.
As you can see both answers were negative. This happens because since in the 4th quadrant all y’s are negative and all x’s are positive the division of both will result a negative answer.


On Friday we left on this problem:

Find the sum of this infinite series: 3+2.4+1.92+1.536+…
Today we wrote it more efficiently like this:
Σ 3(.8) ^ (k-1)
The expanded form for this formula is what we already know as:
3(.8) ^(1-1) + 3(.8) ^(2-1) + 3(.8) ^(3-1) + …
The Σ stands for Sigma Notation.

Sigma notation: a fancy but efficient way to write complicated addition problems.
Vocabulary for Sigma Notation:
Sigma: Greek letter- means add the following.
Summand: the expression that is being added. n the last example: (3(.8)^(k-1))
Index: the variable that is changing. Example: k.
Limits of summation: upper and lower bound of the index (1,∞).

Example:

For the following:
Σ-2k its limit bounds are (1, 6).

Write the expanded form.
Evaluate the expression.
Identify the summand.

Solution:
-2(1) + -2(2) + -2(3) + … + -2(6)
-2 + -4 + -6 + -8 + -10 + -12 = -42
the summand is -2k

Shortcut
Since the series changes arithmetically because every time there is being added -2 more we can use the arithmetic series formula as a shortcut.
S6 = 6/2 (-2-12)
S6 = -42

We then started studying the properties of the Sums

This are:
When the limit is (1,n) Σcai it is the same as cΣai.
What is being done here is that a constant of multiplication is being removed where ai is some expressions dependent on i.

Example: find the sum of the 1st 30 multiples of 7.
If the limit of Σ7k is (1,30).
It’s the same as: 7Σk To solve this we can use Gauss’ technique: [n/2(n+1)]
7 [30/2 (30+1)]
7*15*31= 3255

Rule # 2 states that:
ΣAk+Bk = ΣAk + ΣBk and ΣAk-Bk = ΣAk - ΣBk
This means that the sum of a sum is the sum of the sums.
And the sum of a difference is the difference of the sums.

Class was over and we finished there.
There is a new homework which is:
Read 506-507 and do in page 508 # 1-15 odd.

Tomorrow’s scribe since we get to choose would be ………………………………………………………..........................
Alejandro Covo