Today's Slides: June 8

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Pre-Cal 40S June 8, 2009

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Scribe...i accidentally put BOB earlier-__-

*note* Mr. K. made a correction to the formula we learned in previous sessions:
the correct formulas are now and
If you look the june 5 slides, pg 6 and 7 shows how we get to these formulas (sorry, my print srn doesn't really work)

On friday, the main thing we learned was solving infinite geometric series!...here's the deal...basically, any fraction that is below 1 and over 0, to the power of infinity will be equal to zero...of course it's not REALLY zero, but our calculators don't have the potential to hold a number less than google, and there is no exact value for it since infinity is an idea and not a number...since it's really ultimately infinitely close to zero, we make it equal zero, making our lives easier. Look at the June 5th slides, page 13 for an example.

...to be honest i don't really know what else to add...i'm a little preoccupied thinking finishing up my final project and worrying/studying for the exam...if anyone needs more clarification, just leave a comment within the next week and i'll fix this up some more...

-Jonno out

Today's Slides: June 5

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Pre-Cal 40S June 5, 2009

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june/4/09

Arithmetic and Geometric series.

We started the class with a quick probability question, one that was brought up by a student.

When solving questions like these, always try to go back to the basics. Probability is the number of favorable outcomes over the possible outcomes. From that, we have something to work with.
Finding the possible outcomes, or Sample Space.

• The question states that "4 men and 4 women" would be chosen. This means that out of the 7 men, 4 will be chosen (7C4), and out of 10 women, 4 will be choose (10C4).

Finding the favorable outcomes.

• Since the question said that, "Allen and Bridget will be among these 8 chosen people", we already know that they are part of the possible outcomes. They are represented by the green and blue "1"'s.
• Out of the 7 men, only one has been chosen, (Allen). This leaves us with 6 more men, and 3 more spots for men, 6C3.
• Out of the 10 women, only Bridget has been chosen. This leaves us with 9 more women, and 3 more spots for women, 9C3.

As a result, you would get what is shown on the image above. (:

---------------------------------------------

Today's class was focused on:

1. Arithmetic and geometric sequences
2. Series
3. Arithmetic and geometric series.
4. Sigma Notation

Arithmetic sequences
An arithmetic sequence is a sequence (ordered list of numbers), where a fixed number(common difference) is found between two consecutive terms. (negative numbers are added too! )
This means each term is going up or down by the same number.
When wanting to find the nth term in an arithmetic sequence, refer to the equation below. (use Carl Friedrich Gauss's 7year old story to help you remember the equation.)

Geometric sequences
Geometric sequences are like arithmetic sequence, but instead of adding its multiplying. This means instead of a common difference, there's a common ratio.
When finding the nth term in a geometric sequence, refer to...
Series
Series is defined as, the sum of terms in a sequence. (Sn, where S reads as "sum of" and n would be the rank of the nth term. ex. S4 = sum of the first 4 terms.)

If we were to have a sequence of 1,2,3,4,5,6,7 etc, the series would consist of 1,3,6,10,15. Why? Well the first term is a given, 1. The second term would be 3, because that was the sum of the first and second term from the sequence, (1+2). The third term is the sum of the first, second, and third terms from the sequence. (1+2+3). The orange circled numbers in the above picture are the ranks.

Arithmetic series
Arithmetic series is the sum of numbers in an arithmetic sequence. This would be helpful if you are asked to find the "sum of integers from 1 to 5000" for example. Where n would be 5000, because there are 5000 terms, a would be 1, because that is the value of the first term, and d would be 1, because that's the common difference. (numbers are going up by one)

Or you might be asked "What is the sum of all multiples of 7 between 1 & 5000".
What you know:

• From the sequence of multiples of 7 between 1-5000, first term is 1.
• Last term is 4998
  
• common difference is 7.

What we need to know:

• number of terms within that sequence.
  

By using the arithmetic equation, tn = a + (n-1)d, plug in known values, do some grunt work, and you'll be able to find n. (or you can just take the last number, 4998, and divide by 7, to see how many times 7 can go into it.4998/7 = 714) Once you've found the number of terms, plug it into the arithmetic series equation. ta-da~
Now don't get too carried away with questions a and b. These kind of questions wont be asked on the exam, but you'll need to know the methods of a&b in order to solve c. (c = a question likely asked on the exam).

Logic:( The sum of all integers 1-5000) - (sum of all multiples of 7) = sum of all integers not multiples of 7. This "build up" to a question, is called scaffolding.


Geometric series.
Sum of numbers in a geometric sequence.

Sigma Notation
Is the shorthand way of writing a series, also known as the weird looking "E". Sigma is really confusing, if you don't know how to read it. The n=1 tells you the value of the first term, which is 1. The 4 on top of the sigma is nth number of term to stop at. The (2n-3) is the "rule" or equation you follow.

bye guys! good luck on the exams and your DEVs!
mary
The next scribe is jonno!

Today's Slides: June 4

Here they are ...

Pre-Cal 40S June 4, 2009

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Today's Slides: June 3

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Pre-Cal 40S June 3, 2009

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June the Third Scribe

First of all we split into groups and Mr.K informed us that we were starting our new unit, sequences. We should be done this unit on Friday or Monday. So be prepared to move really fast!

Okay well apparently today was mostly a review from grade 10 (if you can remember that far back.. good job!)
On the first slide there were four sequences...

4, 7, 10, 13, __, __, __
3, 6, 12, 24, __, __, __
32, 16, 8, 4, __, __, __
1, 1, 2, 3, 5, 8, 13, __, __, __,

Mr. K wanted us to fill in the blanks and explain how we found the missing terms...

4, 7, 10, 13, 16, 19, 22 (Add 3 to the first term to find the second term. This is called an arithmetic sequence. )

3, 6, 12, 24, 48, 96, 192 (multiply the first term by 2 to find the second term. This is called a geometric sequence.)

32, 16, 8, 4, 2, 1, 1/2 (multiply by 1/2.)

1, 1, 2, 3, 5, 8, 13, 21, 34, 55 (this is the Fibonacci sequence! Add the first and second term together to find the third term.)

Now let's take a closer look and try to identify some patterns that can be applied to any other sequence we might encounter in the future.
4, 7, 10, 13, 16, 19, 22

Note: If you are asked to find the 37th term and you plan on adding 3's, you must add 36 threes.How do we get 3n + 1 ?
Well since we found the y-intercept we can graph this.

By looking at this graph we can see that the slope is 3. So for future reference remember that our slope is the constant. So let's make an equation for this sequence.
tn= 3n+1
Where 3 is the slope and 1 is the y-intercept.

Everytime we have an arithmetic sequence it will be a linear function.
This is the graph of our sequence. Here is how you would make that on your calculator! Hit stat, edit. Under L1 enter your rank (1-7) under L2 enter your values (4, 7, 10, 13, 16, 19, 22) Now hit 2nd stat plot (top left corner of the calculator). Hit enter and make sure plot 1 is on, under type make sure the dots are selected. Then hit graph.


Next we covered some definitions:
Recursive: Repeats again and again.
Implicit Definition: This is the teenage way of saying hi, it's an implied hello. Only clear to people who know what they're looking for.
The Common Difference (d): The number that is repeatedly added to successive terms in an arithmetic sequence.
Common Ratio: The number that isrepeatedly multiplied to successive terms in a geometric sequence.

How to find the nth term in an arithmetic sequence.

tn= a + (n-1)d
Where tn is the nth term
a is the first term
n is the rank of the nth term in the sequence
d is the common difference

How to find the nth term in an geometric sequence.
tn= ar^(n-1)
Where tn is the nth term
a is the first term
n is the rank of the nth term in the sequence
r is the common ratio

Here is the next sequence we looked at, 11, 5, -1, -7...
We were asked to find the 51 term.
Look at the difference between the terms.
5-11= -6
-1-5= -6
Therefore we know our constant is -6. Remember we're not subtracting 6 from the first term to get the second term, we are adding -6! :D

Okay so how to find the 51 term. First let's make a formula.

tn= a+ (n-1)-d
Where n is the term. a is the first value and d is the difference. So we have..

t51= 11+ (51-1)(-6)
= -289

Next we looked at the sequence 3, 6, 12, 24, 48, 96, 192
Notice the difference is not constant, so this is not an arithmetic sequence.

To make an equation for this kind of sequence we use the formula, tn= ar^(n-1)
a= the first term r= ratio
The ratio for this sequence is 2 because 6/3=2 and 12/6=2 etc.
So the implicit definition is tn= 3(2)^(n-1)

Next question! 32, 16, 8, 4, 2, 1, 1/2
We multiply by 1/2 to get the next term so 1/2 is our ratio. Now we find the 10 term.
tn= ar^(n-1)
t10= 31(1/2)^10-1)
t10= 1/16

Remember!!
If differences in sequences are different it is NOT arithmetic.
If there are common ratios it is geometric.

And the last slide..

Um next scribe is Mary.

Homework is exercise 44 and 45.