Today's Slides: June 6

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Last Class of the Jabbamatheez.

Today was the end of of our classes together. It's been quite fun learning about everything from circular functions to Geometric sequences, especially because of all the skits Mr. K put in.

Well, we started the class by solving a geometric sequence, given the 2nd and 3rd term. We had to find the 8th term given this. Since it was a geometric sequence, we know that there's a common ratio. To find this common ratio, we found the product of the 2 terms. Which was 12/24 = 1/2. We then found where y = 0, if we were to graph this. We found the 1st term by going backwards, which means you would have to multiply the 2nd term (24) by 2. 24 x 2 = 48. Then to find the term of 0, we would double that, which was 96. We then multiplied 96 by (1/2)⁸ because we need to find the 8th term, and it had a common ratio of 1/2. The answer was 0.375

Last class we found out where the infinite geometric sequence came from, but today we learned that when the common ratio is greater than 1, then it is a diverging series, because it will keep going on and on, thus meaning it is infinity. So when |r| > 1 then r^∞ = ∞
When the common ratio is less than 1, it's called a converging series, because it hones in on one value, which is 0. So when |r|<>∞ = 0.

This was pretty much all we learned, so we did some questions after. There was a question about a super ball. At this point Paul pulled out a super ball for Mr. K to use, and it gave the class a little interaction. The ball would start at 200 cm and drop to the ground, it would rebound 3/4 of the distance it fell. We had to then find the total distance travelled by the 4th bounce. This question was easily accomplished using a diagram.
The initial distance being 200 cm, we can multiply by 3/4 to find the distance it bounces back up, then multiply that distance by 2, because it travels that distance twice, bouncing up and falling down, then we multiply the distance from the second bounce by 3/4 (150 x (3/4) = 112.5) to get the distance going up, then multiply by 2 for it going up and down,(150 x (3/4) = 112.5 x 2 = 225 cm) then multiply the distance from the 3rd bounce by 3/4 again (112.5 x (3/4) = 84.375) then multiply that by 2 because it goes up that distance and down hat distance, so it would be (84.375 x 2 = 168.75) add all the values together to get 893.75
You can also use the geometric sequence equation to find this also.
After this we had to find the distance travelled until the ball stopped, so we used the infinite geometric series which was 1200, given all the variables, and you would add the initial value which was 200, and that would be 1400 cm. This was the last question.

That was all we did for our last class. I know, we need something special, because I'm the last scribe and this whole course has been oh so special, because of this reason, I wrote a poem. It's a Kyrielle poem. With 8 syllables each line, and rhyming scheme being aabB ccbB ddbB. Enjoy.

One More Step

Unluckily today's the last,
Of this most enjoyable class.
Though it's not the end for us yet,
As we move on just one more step.

Mr. K, we give you homage,
For expanding on our knowledge.
Never dull, you were fun instead.
As we move on just one more step.

Leaving with a fantastic smile,
As every thing's been so worthwhile.
Now to continue, and to prep.
As we move on just one more step.

-Francis Bowers

Today's Slides: June 5

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Infinite Geometric Series

OVERVIEW:

• Finding the Sum of Numbers in a Sequence
• Sigma Notation
• Infinite Geometric Series Formula
  

Finding the Sum of Numbers in a Sequence

(a) What is the sum of the integers from 1 to 5000?

Credit to Paul for being our Gauss today.

We imagine the sequence of all the numbers 1 to 5000 in the top row while the bottom row has all the numbers from 5000 to 1.

Example:
1, 2, 3, ... 5000
5000, 4999, 4998, ... 1

We find the sum of each column in our table and see that they're all 5001. Refer to slide 4 of June 3's slide for a better understanding of what I'm talking about.

We know that there are 5000 of these terms. We also know that if we want to find the sum of the integers between 1 to 5000, then we would have to multiply the number of terms by the total we have then divide by 2. We divided by 2 because there were two sequences we added.

Example: 5000 * 5001 / 2 = 12 502 500

Another way of solving this problem is to pull out the equation (found on June 3's slide #7) and plugging in the appropriate numbers, which Mr. K has a distaste for using this method because we're not really understanding the concept but just plugging numbers into an equation.

(b) What is the sum of all multiples of 7 between 1 & 5000?

The lowest multiple of 7 between 1 and 5000 is 7.
The highest multiple of 7 between 1 and 5000 is 4998.
There are 714 terms that are multiples of 7 between 1 and 5000.

We then imagine, again, that table, listing all the numbers in the sequence from 1 to 5000 (that are multiples of 7) in the top row and all the numbers in the sequence from 5000 to 1 (that are multiples of 7) in the bottom row, and find the sum of each column.

Example:
7, 14, 21, ..., 4998
4998, 4991, 4984, ..., 7

Add the top row and the bottom row, and you find that under each column in our imaginary table is 5005.

We multiplied the number of multiples of 7 between 1 and 5000 by the sum of the first and last terms of the sequence and divided that by 2 because we've counted the sequence twice and we get 1 786 785.

(c) What is the sum of all integers from 1 to 5000 inclusive that are not multiples of 7?

Well, we have the sum of all the integers from 1 to 5000. We have the sum of all the multiples of 7 between 1 and 5000. So if we find the difference of those two numbers, we find the sum of all the integers from 1 to 5000 inclusive that are not multiples of 7.

12 502 500 - 1 786 785 = 10 715 715

Sigma Notation

Left Question:

• The k = 1 means that you start with 1 and evaluate 3k.
• The number above the sigma symbol, 4, means to keep evaluating 3k until k = 4, then add all the terms.

Right Question:

• The k = 0 means that you start with 0 and evaluate 2^k.
• The number above the sigma symbol, 3, means to keep evaluating 2^k until k = 3, then add all the terms.

Then Mr. K says to find another sigma notation that is equivalent to the two questions on the slide. Of course, "mathematics is the science of patterns" is the common phrase that Mr. K always says, and yes, there're patterns here.

Also, a quick way of figuring out the answer to the sigma notation in question is to add the first and last term, then multiply by 2. Example: In the left question, [3(1) + 3(4)] * 2 = 30

Infinite Geometric Series Formula

The formula given in the above slide is something that we already derived from the previous class.

To answer the question "why is that the formula?" let's first use an example to help better our understanding of infinite geometric series. We make a square, cut it in half, and shade the half. We take the unshaded region of the square, cut that in half, and shade the half. Then we take another unshaded region of the square, cut that in half, and shade that half, and so on. So, firstly, we got 1/2 of the square shaded, then 1/2 + 1/4 shaded, then 1/2 + 1/4 + 1/8 shaded, then 1/2 + 1/4 + 1/8 + 1/16 shaded, and so on. If we continue this pattern and keep cutting and shading the square in half an infinite number of times, then we would've a shaded square.

1/2 + 1/4 + 1/8 + 1/16 + 1/32 ... = 1

The formula that we derived from the previous class was

Sn = t1*(1-r^infinity) / (1-r)

If we take r and raise it to the power of infinity, that would equal zero.
Example: r = 1/2

1/2 * 1/2 * 1/2 * 1/2 ... = 0

So, in the part of the formula where it says (1-r^infinity), that part of the formula is going to equal (1-0) = 1 anyways, and multiplying a number by 1 won't really have a visible effect. It's more efficient to just not include the (1-r^infinity) into the formula.

Plug in the numbers into the equation, BING! BANG! BOOM! you have the result. Have a nice day.


END NOTES

• Start on Exercise 47: Infinite Geometric Series. You should have finished all the questions in the exercise book up to 47.
  
• Tomorrow, we will finish off the lesson the infinite geometric series.
  
• Hopefully, we will be on task and finish the course tomorrow, and do a review on Friday.
  
• Next scribe is Francis.

Today's Slides: June 4

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Introducing Series

The first thing we did in class today is answer some simple questions. Pretty straight forward, should be easy to do. Even if you weren't in class yesterday. Noo excuses, 'cuz I wasn't in class yesterday.

We also graphed the function 'y = [square root] x', which looks like half a sideways parabola.

Next, we learned about 'young Gauss', who was a pretty clever seven-year-old. His teacher told the class one day to add up all the numbers from 1 to 100. Instead of writing it all out and adding it, he found a pattern.

When the numbers are laid out like above, their sums equal 101. There should be a hundred 101's, hence the reason we multiply 100 and 101. We divide it because you added one hundred twice. That's what Karl Gauss figured out. If you want to know the details, you can go ahead and visit this link: http://www.sigmaxi.org/amscionline/gauss-snippets.html


Using the same idea as above, we found the formula for an arithmetic series.
Sn = (n/2)[2a+(n-1)d]

A series is the sum of numbers in a sequence to a particular term in a sequence (definition can be found on blog). You can't have a series if you don't have a sequence first, or else, what are you adding?

We also found the formula for a geometric series.
Sn = [a(1-r^n)]/(1-r)



Lastly, we touched a little bit on 'Sigma', which is located on the second-last slide (8).



Mmm, yeah, that's it. I'm tired now, I'm sorry. Hahaha, next scribe is Joseph, since he asked.. again.

Today's Slides: June 3

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Geometric Sequences

WARNING: The geometric sequences unit does not have any new material; rather the unit gives a new perspective to what we already know from previous learning.

OVERVIEW:

• What is a sequence?
  
• Recursive definition vs. implicit definition
• Common difference vs. common ratio
  
• Determining any term (the nth term) in an arithmetic sequence vs. determining any term (the nth term) in a geometric sequence
  

SLIDE 2
An example of an arithmetic sequence is the first sequence of numbers found on SLIDE 2. Examples of geometric sequences is the second and the third sequence of numbers. Note that arithmetic sequences and geometric sequences aren't the only sequences that exists, such as the Fibonacci sequence, but the main scope of the unit focuses on geometric sequences.

We find the next three terms by determining a rule for each sequence of numbers.

1. For the first sequence of numbers (4, 7, 10, 13...), we see that 3 is added to any term to get the next term.
2. For the second sequence of numbers (3, 6, 12, 24...), we see that 2 is multiplied to any term to get the next term.
3. For the third sequence of numbers (32, 16, 8, 4...), we see that 1/2 is multiplied to any term to get the next term. (Remember, in grade 12, we multiply by 1/2 instead of dividing by 2; it's to make our calculations easier to do.)
   
4. For the fourth sequence of numbers (1, 1, 2, 3...), we see that we have to add the previous two terms to get the next term.

SLIDE 3
Let's take a closer look at the first sequence of numbers (4, 7, 10, 13...).

How did we determine that the rule is to add 3 to any term to get the next term? We find that out by determining that the common difference is 3 (green font). We then find out that if we're given any term, n, then 3n+1 is the equation we use to get what n equals. For example,

3n+1 = 3(1)+1 = 4.

So if the rank (n) is 1, then its term is 4. If the rank is 2, then its term is 7. If the rank is 7, then its term is 22. This is expressed with the equation 3n+1.

The rank represents the term. For example, in this sequence, the rank of the first term is 1, the rank of the second term is 2, the rank of the third term is 3, the rank of the nth term is 3n+1, and so on.

Graph 3n+1, and we find that it is the equation of a line. Also, we can determine the 0th term by plugging 0 into the equation.

3n+1 = 3(0)+1 = 1.

For the zeroth term, the output 1. It is also the y-intercept of the graph.

Term 1 (t1), or in some textbooks it's called a, equals 4. The common difference, d, is 3.

The recursive definition, in this sequence of numbers, is to take a term and add 3 continuously.

The implicit definition, in this sequence of numbers, is the equation of the sequence, tn = 3n+1.


SLIDE 4

Let's take a closer look at the second sequence of numbers (3, 6, 12, 24...).

How did we determine that the rule is to multiply 2 to any term to get the next term? Firstly, there isn't a common difference in this case but a common ratio. The common ratio is 2. We then find out that if we're given any term, n, then (3/2)(2^n) is the equation we use to get what n equals.

tn = (3/2)(2^n) = [3 * (1/2) * 2^n] = [3 * 2^(n-1)]

(Refer to the bottom-right corner of SLIDE 4.)

So if the rank (n) is 1, then its term is 3. Similarly, if the rank is 2, then its term is 6. If the rank is 7, then its term is 192. Etcetera.

Graph [3 * 2^(n-1)], and we find that it is the equation of an exponential function. Also, we can determine the 0th term by plugging 0 into the equation.

tn = [3 * 2^(n-1)]
t(0) = [3 * 2^(0-1] = 3/2

For the zeroth term, the output 3/2. It is also the y-intercept of the graph.

The recursive definition, in this sequence of numbers, is to take a term and multiply it continuously by 2.

The implicit definition, in this sequence of numbers, is the equation of the sequence, tn = 3*2^(n-1).


SLIDE 5
Let's take a closer look at the third sequence of numbers (32, 16, 8, 4...).

The recursive definition, in this sequence of numbers, is to take a term and multiply it continuously by 1/2.

The implicit definition, in this sequence of numbers, is the equation of the sequence.


VOCABULARY/SUMMARY is found on SLIDES 6 to 9.


HOMEWORK

• Exercise 45: Geometric Sequences
  

Next Scribe is Nelsa.

Today's Slides: June 2

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