13-6

13-6 is all about sigmas. A sigma is simply a series written in condensed form. Since I'm not fancy with my computer I'm going to use an E to represent a sigma and I will try to express the rest as best as I can.

On the top of the sigma is the limit of summation (#). Underneath the sigma is where the index is placed (n). And to the right is the summand (f(x)).
Example
7
E 2k
k=1

Here the limits are 7 and 1
The summand is 2k
The index is k

In order to evaluate the sigma replace k with the numbers 1-7 like so
2*1+2*2+2*3+2*4+2*5+2*6+2*7
simplify
2+4+6+8+10+12+14
Then evaluate and you get 56


Now, expand
8
E 2i(^2)-1
i=5

2*5(^2)-1+2*6(^2)-1+2*7(^2)-1+2*8(^2)-1
In expanded form this ^ is 49+71+97+127

Simple enough, right?

Week 6 Prompt

How do you determine if a sequence is arithmetic or geometric? What are the rules for finding limits? Give examples of each.

Braindead

The title says it all, I am braindead. The only reason I even remember what Chapter we are on is because I read other people posts. So what I gathered from that is we are on Chapter 13. Chapter thirteen is about sequence and series and all this other stuff. I think we have exams coming up soon, oh joy. I need to do really good on these exams so if anybody is thinking about having study groups or something let me know. I hope everyone had a Happy Thanksgiving. I ate a lot of food. Then we lost to John Curtis officially ending the 2010 football season. I'm bored now. Goodnight.

chapter 13- section 3

Happy late Turkey Day everyone! Lucky for us, we get to go back to school tomorrow! We were taught all Sections of Chapter 13 before the holidays, and its the last chapter before our exam review for the next few weeks. This wasn't the easiest chapter for me, and ill need some review when we get back, because i have a lot of unclear problems and a lot of confusion. Section 13-3 was the most simple of the whole chapter because it was mostly just finding the answers from the guildlines in your notes.

Guidlines for this Chapter:

For Fractions:
1. If the top degree = bottom degree the answer is a coefficient.
2. If the top degree > bottom degree the answer is +/- infinity.
3. If the top degree < bottom degree the answer is 0.

If the rules above do not apply to the given, then you simply use the table function in your calculator.

For example:
1. Lim/n->infinity n^7 +6/ 9n^2 – 7n
The degrees are 7 and 5.
The top one (7) is larger than the bottom one (2).
When you look at your rules, they state that the answer will be +/- infinity.

see you tomorrow (:

Before the holidays we did chapter 13. 13-2 was about recursive- defines the terms of what came before.

FORMULAS:
to find:
previous term---tn-1
two terms back---tn-2

1)find the certain number of terms.
2)give a recursive definition.

EXAMPLES:
find the 3rd, 4th, and 5th terms.
tn=4tn-8
t1=10

just multiply the number in front of t and the first terms and subtract 8.

t2=4(10)-8 = 32
t3=4(32)-8 = 120
t4=4(120)-8 = 472
t5=4(472)-8 = 1,880

It's very simple to find terms like this especially when you are given the 1st term already.

The other way is to find a recursive definition.

Give a recursive definition for 2,4,6,8....
**you can obviously see that 2 is being added each time.
*ARITHMETIC PATTERN!

RECURSIVE DEFINITION:
tn=tn-2+2

SEE ALL OF YOU TOMORROW :)

13-4

I hope everyone had a great Thanksgiving holidays. But now that we have school tomorrow it is time to get your mind focus to school again. Section 13-4 is basically the rules for fractions and how to apply it.
Rule for fractions are:
-If top degree = bottom degree; answer is coeff
-If top degree > bottom degree; answer is +/- infinity
-If top degree < bottom degree; answer is 0

If rules don’t apply you use a table and figure out what it is approaching.


Lim
n->infinity r^n=0 iff absolute value r<1


Examples:

Lim n->infinity(.99)^n=0
absolute value .99^n<1
.366,..,.0004,.000….2=0 is your answer


lim n->infinity n^2+1/2n^2-3n=1/2

lim n->infinity sim(1/n)=0

lim n->infinity 5n^2+square root n/3n^3+7=0

lim n->infinity 7n^3/4n^2-5=infinity


Basically section 13-4 is really easy and common sense. If you know your rules and know how to apply them and use them you will not get any trouble at all in section 13-4 and this section should be a piece of cake for you.

13-2

Recursive Definitions( Recursive means define interms of what came before)

tn-1/previous terms/ tn=tn-1 - 3

tn-2/two terms back

tn-3/three less than preceding term

Ex:1 tn=4tn-1 +1 and t1=7 find the 3,4,5 terms.

t2=4(7) + 1=29

t3=4(29)+1=117

t4=4(117)+1=469

t5=4(469)+1=1877

Ex:2 Give a recursive definiton for 7,10,13,16..

• since the pattern is +3, the definition would be tn=tn-1 + 3

Arithmetic and Geometric Series and their Sums

I hope everyone had a wonderful break :) I know the past week off was appreciated, and for me, needed. But anyways back to business...

Section thirteen three is basically on the sums of the geometric and arithmetic series. This section, like the rest of the chapter is relatively simple. Some important terms to remember are Finite, which is a certain number, and infinite, which is an unlimited number of terms. There are two formulas; one for each sequence and you simply replace, or substitute the numbers given. Once you have learned to solve these types of problems, it will be easier to solve the more difficult ones.

Sum of arithmetic series :
Sn=n(t1+tn) /2
Sum or geometric series:
Sn=t1(1-r^n)/1-r
Where r is the common ratio and is not equal to one.

Ex. Find the sum of the arithmetic series .
1. S10: t1=3,t10=39
S10=10(3+39)/2
S10=10(42)/2
S10=420/2
S10=210

Ex. Find the sum of the geometric series.
1. Find the sum if the first 50 terms.
2,4,8..
• Because the t50, or the fiftieth term is not given you must go back to section one to find it.
T50=2+(50-1)2
T50=2+(49)2
T50=100
S50=2(1-2^50)/1-50
Finally you just solve to get your answer.
:)

13-2

Recursive Definitions

Section 13-2 was the second easiest section in chapter 13. The term recursive means to define in the terms of what came before.

Previous term – tn-1
Two terms back – tn-2
….. – tn-3

The two main directions you will follow for this section is to give a recursive definition for a series of numbers and finding a certain number of terms. Let’s try some examples:

Example 1:
Find the 3rd, 4th, 5th, and 6th terms

Tn = 2tn-1 + 7 and t1 = 3

T2 = 2(3) + 7 = 13
T3 = 2(13) + 7 = 33
T4 = 2(33) + 7 = 73
T5 = 2(73) +7 = 153
T6 =2(153) + 7 = 313

Example 2:
Give a recursive definition for

2, 6, 10, 14

Tn = tn-1 + 4

10, 20, 40, 80

Tn = tn-1(2)

Hope everyone had a great Thanksgiving holiday! Now only 2 more weeks and 1 week of exams (unless your exempt) until Christmas holidays! 

13-4

I hope everyone had a great Thanksgiving holidays! But unfortunately, we have to go back to a few more weeks of school before Christmas. Before the holidays, we learned chapter 13, which is the last chapter before our exam. This chapter was pretty difficult for me, so I will need a good review when we go back to school tomorrow. Section 13-4 was sort of simple because MOST of the time there were answers given by looking at your notes. The notes we were given are:

For Fractions:

1. If top degree = bottom degree; answer is coeff
2. If top degree > bottom degree; answer is +/- infinity
3. If top degree < bottom degree; answer is 0

If none of these rules apply, the you use the table in your calculator to find out which number it is approaching.
For example:

1. Lim/n->infinity n^5 +4/ 6n^2 – 7n

The degrees are 2 and 5. The top one (2) is smaller than the bottom one (5).
When you look at your rules, they state that the answer will be 0.
That is your final answer, so these types of problems are pretty simple.

13-2

13-2
Section 13-2 was about recursive definitions. Recursive means define in terms of what came before.
tn-1: previous term
tn-2: two terms back
tn-3: …
EXAMPLE 1: tn=3tn-1 + 1 and tn=6. Find the 3, 4, and 5 terms.
A). t2=3(6) + 1 = 19
t3=3(19) + 1 = 58
t4=3(58) + 1 = 175
t5=3(175) + 1 = 526
B). 58, 175, and 526 are your 3, 4, and 5 terms. All you do is replace tn-1 with tn to begin. Then you take your answer and place is where tn-1 is each time.

EXAMPLE 2: Give a recursive definition for 6, 10, 14, 16..
A). Looking at the numbers above, you can easily see that you’re adding 4 each time to make this arithmetic pattern.
B). Your recursive definition is tn=tn-1 +4

Hope everyone had a great, relaxing Thanksgiving break :)

Week 6 Prompt

Due to the fact that the internet was not dependable on my laptop I was unable to post the blog prompt for the holidays. Therefore it is a freebie for everyone. Just make sure to post your 2 regular blogs. They can come from any review topic.

13-6

This week in advanced math we learned chapter 13. In 13-1 we learned geometric and arithmetic sequences. In Chapter 13-2 we learned Recursive Definitions. In chapter 13-3 We learned Arithmetic and Geometric Series and their sums. In chapter 13-4 we learned Limits of Infinite Sequences. In chapter 13-5 we learned Sums of Infinite series. And finally in 13-6 we learned Sigma Notation. The greek letter sigma is often used in mathematics to express a series or its sum in abbreviated form. The Sigma contains three parts. The summand, the limits of summation, and the index. The number to the right of the Sigma is the summand. The number on top of the Sigma is the limits of summation. The number on the bottom of the Sigma is the index. You can either be asked to expand the Sigma or evaluate the sigma. To expand you only plug the numbers were they need to go. To evaluate you solve the whole thing.

Ex.Evaluate k=1; summand = (k+4); ;limit of summation = 25

1+4=5+4=9+4=13+4=17+4=21+4=25

Ex. Expand k=1; summand (k^3); limits of summation = 15

1^2 + 2^2 + 3^2... 15^2

Chapter 13.

What's up people? I hope you all get to enjoy your thanksgiving vacation. I will be spending mine practicing and I couldn't be happier. We mad it to the third round baby. I know none of ya'll believe we can win but poo on ya'll. We got a shot baby and I'm pumped. This week we started Chapter 13. This Chapter should be easy but of course the Math Gods had to screw it up. These sequences should just be a simple matter of using logic but instead it has been turnedd into an arduous process with multiple formulas to memorize. It is possible to figure out anything that has to do with these sequences by using the formulas. The easiest part of 13 was 13-6. It has to do with something called enigmas. These are a simple matter of pretty much reading a graph. Happy Thanksgiving everybody.

13-4! Uhhh.....

I am super pumped for this week off!! My brain needs a rest. Haha.
Now it's math time....

So, here are some rules for fractions
**If the top degrees equals the bottom degrees, the answer is the coefficient
**If the top degrees is greater than the bottom degrees, the answer is plus or minus infinity
**If the top degrees is less than the bottom degrees, the answer is zero

****If none of these rules apply use a table and figure out what it is approaching

-limit rⁿ=0 and n=infinity if |r|<1 and r is a real number

I am not too sure how to do this but here is an example of how it looks like it would work.
Limit
n=infinity=(-.99)ⁿ=0
|-.99| < 1

Limit
n=infinity n²+ 1/2n²-3n = 1/2

Answers are .50766, .50075, .50008

13-6

In advanced math this week we started chapter 13. I found that chapter 13 was very hard. However, the easiest section was section 13-6. This section is about a symbol called sigma. A sigma is a series written in condensed form. Around the sigma are its index, summand and limits of summation. The two main things you do with sigma is evaluate and expand it. Let’s try a few examples.

Example 1:

Identify

6
∑4k
K = 2

Find the Summand? 4k
What is the index? k
What are the limits of summation? 5 and 2

Evaluate the Sigma

4(2) + 4(3) + 4(4) + 4(5) + 4(6)
8 + 12 + 16+ 20 + 24 = 80

Example 2:

Expand the Sigma

10
∑ 5k4
K = 3

5(3)4 + 5(4)4 + 5(5)4 + 5(6)4 + 5(7)4 + 5(8)4 + 5(9)4 + 5(10)4
405 + 1280 + 3125 + 6480 + 12005 + 20480 + 32805 + 50000

Last week we were pretty much taught all of chapter 13. It's all very easy as long as you remember your formulas.

The easiest for me was 13-4 so ill do a few problems from that.

FORMULAS:
if the degrees at the top is = to the degrees at the bottom then the answer is coefficient.
if the degrees at the top is > than the degrees at the bottom then the answer is +/- infinity.
if the degrees at the top is < than the degrees at the bottom then the answer is 0.
--if none of these rules apply then you would just plug it into your calculator using a table.

EXAMPLES:
2n^4/2n^3
**2nd rule applies!
since 4 is greater than 3 your answer is infinity.
* but is is positive or negative?!
POSITIVE INFINITY.

4n+1/4n
4n is = to 4n so you are left with 1.
your answer is 1!

HAVE A GOOD WEEK OFF EVERYONE :)

13-5

Last week we were taught sections from Chapter 13. In 13-5 we learned the sum of an infinite geometric problems. 13-5 is dealing with alot of things such as the following stuff: infinite geometric and repeating decimal and infinite series converage and sequence converage.

The formula for a infinite geometric is sn=ti/1-r

When you want to write a repeating decimal as a faction you do #repeating/lastplace^-1

Examples:

-Find the sum of the infinite geometric series of 9-6+4-
-6/9=-2/3
-4/6=-2/3

Sn=9/1-(-2/3)=27/5 is the answer.


-Write .infinite5 as a faction

5/10-1=5/9 is the answer.


-For what values of x does the following infinite series converage
1+(x-2)+(x-2)^2+(x-2)^3+

x-2/1=x-2

(x-2)^2/x-2=x-2

(x-2)^3/(x-2)^2=x-2

-1+2 +2 +2=

1

Section 13-5 is really easy and not hard at all. If you know the rules and the formulas and know how to do the repeating and whatever else you have to do in section 13-5. This section will be a breeze and easy for you.

Chapter 13- Section 6

This past week we learned Chapter 13, and completed 13-6 on friday to finish up the chapter before the holidays. We had to learn this chapter quick because we needed to finish so we could review before our midterm. For the unfortunate students who are not exempt, we will be taking a chapter test and preparing for the midterm exam that alot of us are worried about passing. The most recent section we learned is 13 – 6. Overall this chapter has been prettty simple. Please excuse my examples for this section, it is kind of hard to do without an image like we've drawn in our notebooks.

• For the problems we are solving in this sectin a “sigma” is drawn for every problem.

• A sigma may look like an E to you, but it is not.

• A sigma is a series written in a condensed (not as long) form.

• At the top of the sigma, there is a number called a limit of summation.At the bottom, there is a letter that will always equal a number.

Ex. K=9i The k would be the index and the number is another limits of summation.And in the middle there is an equation called the summand.

• Just follow directions to complete the problems you are given of this type. If it asks for a particular piece of the sigma, you just find it. If it asks you to evaluate, you expand the sigma and solve it. If it tells you to expand, you just expand and stop there.

It's pretty difficult to show you exactly what to do on the computer, and without a picture, but i tried my best to sum it up.

13-6

This week we finished up learning chapter 13 before the holidays. This is the last chapter we are learning before our midterm. For the next weeks, we will be taking a chapter test and preparing for the midterm exam that we are all hoping to be exempt from. The most recent section we learned is 13 – 6. It was pretty simple, but pretty hard to do on the computer, so I will do my best.

For this section a “sigma” is drawn for every problem. A sigma looks sort of like an E. A sigma is a series written in a condensed form.

At the top of the sigma, there is a number called the limits of summation.
At the bottom, there is a letter that equals a number. Ex. K=4 The k would be the index and the number is another limits of summation.
And in the middle there is an equation called the summand.

To complete problems in this chapter all you have to do is follow directions. If it asks for a particular piece of the sigma, you just find it. If it asks you to evaluate, you expand the sigma and solve it. If it tells you to expand, you just expand and stop there. It is pretty simple, but without me being able to draw a sigma on the computer, it would be difficult for me to work any problems.

13-4

13-4
Last week we were taught sections from Chapter 13. In 13-4, we were taught how to figure out what an equation’s limit by three simple rules.
1. If degrees top = degrees bottom, answer is coefficient.
2. If degrees top > degrees bottom, answer is +/- infinity.
3. If degrees top < degrees bottom, answer is 0.
*If none of these rules apply, use your calculator and figure out what it is approaching.

EXAMPLE 1: limit: n – infinity.. n^2-1/n^2
A). According to the rules, n^2 is = to n^2.
B). Therefore, take the number in front of n, which is 1 – 1/1
C). Your answer is 1.

EXAMPLE 2: limit: n – infinity.. 2n^3/ 2n^2
A). According to the rules, n^3 is > n^2
B). Therefore your answer is + infinity.

EXAMPLE 3: limit n – infinity.. log[sin(1/n)]
A). Since none of the rules apply to this problem, plug it into your calculator in the table.
B). Your answer is 0.

When plugging into your calculator, look at the table and see what the number is approaching. For instance, if it’s .01, … , .001, …, .0001 then your answer would be 0.

thirteen six.

chapter thirteen six.
this week we began chapter thirteen. this is the last chapter before our accumulated midterms. chapter thirteen explains sequences and series. there are many type of sequences but the most common are arithmetic (the adddition sequences) and geometric (the multiplication sequences).
section 13-6 is on the sigma. The sigma is apart of the Greek alphabet, but in math it is used to describe a condensed series. the sigma has three parts.. the top, bottom, and he right side. at the top of the sigma is the limit of summation. at the bottom is k= # where the "k" is the index and the numbers located at both the top and bottom of the sigma are limits of summation. to the side of the sigma is the summand,which is a function.

ex. expanding a sigma..
if k=1 ,the ending number ( above the sigma) is 3, and the summand is 2(k) then the expansion of the sigma would be :
2(1) + 2(2) + 2(3)+ 2(4) * simplify

2+4+6+8

Week 5 Blog Prompt

What is a famous sequences and series? What is it used for and who discovered it? Everyone should find a different type.

13 - 1

This past week we started chapter thirteen. It is about sequences. You have to determine whether or not it is geometric, or arithmetic. There are formulas that you use to determine this. They are:

Arithmetic – tn = t1 + (n – 1)d
Geometric – tn = t1 x r^n-1

Example 1:
Determine whether the sequence is geometric, arithmetic, or neither and find a formula.
17, 21, 25, 30. . .
It is neither because you do not subtract or add the same number every time. You cannot find a formula, because it is neither geometric or arithmetic.

Example 2:
Determine whether the sequence is geometric, arithmetic, or neither and find a formula.
4, 6, 8, 10, 12. . .

This sequence is arithmetic because you add 2 every time. To find the formula, you plug into the arithmetic formula. d = 2, because that is the number you add everytime.

Tn = 4 + (n- 1)2
Tn= 4 + 2n -2
Tn =2 + 2n <--the formula you find.

Example 3:
Determine whether the sequence is geometric, arithmetic, or neither and find a formula.
16, 8, 4, 2. . .

This sequence is geometric because if you put 2 over 4, 4 over 8, and 8 over 16, you get ½. r = ½, because that is the number you multiply by to get the sequence.
Tn = 16 x (1/2) ^n-1 <--the formula you find.

13-1

This week in advanced math we took a test on chapter 11 which involved lots of formulas. For the most part the test were easy as long as you knew the formulas. But at the end of the week we learned 13-1 which involves number sequences. There's arithmetic sequence, a sequence where the same number is added each time. And geometric, a sequence where the same number is multiplied each time.

Formulas:

arithmetic- tn=t1 + (n-1)d

geometric- tn= t1 * rn^-1

I don't really understand this yet so I don't know if this is right.

Ex.

How many multiples of 5 are there between 25 and 75

30, 35 , ... 75

d=9

75= 25 + (n-1)(9)

50=(n-1)(9)

49=(n)(9)

n=7

Playoffs Week 2

 Well everyone, we are coming off a big win over Mangham. Now we play away at Pope John Paul in Slidell. Rebels all the way baby. Now, I don't remember much about this week. We did take a few tests. Not sure how good I did on those. Up now I remember. This week we learned about sequence. In the past sequences were always figured out with your brain and this is still the method I use, but there are formulsa which I was shocked to learn. This formulas completely simplify the process. I hope all of you had a good weekend. Peace.

13-1

This week we learned a section 1 in chapter 13. 13-1 was all about a list of numbers better known as sequence.

There are two sequences in 13-1:

Arithmetic-a sequence where the same number is added each time

Geometric-a sequence where the same number is multiplied each time

Arithmetic formula
tn=t1+(n-1)d

Examples:
5 7 9 11

is it arithmetic, geometric, or neither?

It is arithmetic because you add 2.
D=2


T1=4 t2=10 t75=?
T75=4(75-1)(6)
T75=448


Geometric formula
tn=t1=r^n-1

Examples:
6 12 24 48
is it arithmetic, geometric, or neither?

Geometric because you multiply 2

R=2

Find t10 if t1=4 t2=12 and the sequence is geometric.
Tn=4*(3) ^10-1
4*3^9=78,732


This section is easy if you know your formulas and stuff and now the sequence stuff section 13-1 shouldn’t be hard at all. Remember some problems in 13-1 would not be geometric and arithmetic because it can be neither. Know all the steps of how to identify and solve both methods. The methods are the arithmetic and geometric sequences.

Chapter 13- Section 1

Chapter 13, Section 1 was taught on Thursday. I was not in class on Thursday due to the volleyball game so I am way behind and kind of lost, unfortunately. I knew if I missed the class I would be lost, but i really wanted to go and support the volleyball girls. I have done the homework, and followed as you did examples on Friday but I still dont feel comfortable with what I've learned. I am not confident enough to make my own examples for this chapter so I will do a few that I feel are simpler and easier for me to do. I will be sure to ask many questions on Monday, to make sure i get clarification on the things i feel lost about. Chapter 13-1 applies finding arithmetic and geometric sequences.

When doing these problems, you will first need to identify the sequence and its type.

Arithmetic- when the sequence of numbers has the same number added to them. Arithmetic Formula: tn= t1 + (n – 1) (d)

Geometric- when the sequence of numbers has the same number multiplied to them.Geometric Formula: tn= t1 x r^n -1

Examples:

1.State whether it is arithmetic or geometric. 1, 2, 4, 8, 16
It is geometric because 2 is being multiplied

2. How many multiples of 5 are there between 30 and 525?
The sequence is: 30, 35, 40, 45,…..525
The numbers are being added, so you use the arithmetic formula
525 = 30 + (n-1) (5)525 = 30 + 5n – 5 525 = 25 + 5n500 = 5nN = 100

13-1 Sequences

This Thursday when everybody was at the volleyball game, Lauren, Kaitlyn, Ty, and I learned about sequences. It was a very small class which actually made it easier to learn.This section isn't hard just has a lot of steps which is hard when you have the attention span of a fly.

So, a sequence is just a list of numbers. There are two types of sequences: arithemetic which is a sequence where the same number is added or subtracted each time and geometric which is a sequence where the same number is multiplied or divided each time.

t_n = whichever term

To find t _n in arithemetic- t_n = t₁+ t(n-1)d and in geometric- t_n = r^n-1

These formulas are helpful when you're trying to figure out whether a problem is arithemetic, geometric, or neither.

For example
9/2, 3, 2, and 4/3 is that sequence arithemetic, geometric, or neither?
First you must divide each number with the second number over the first.
3/9/2 = 6/9 = 2/3
If all of the numbers equal the same thing it's geometric and the r = 2/3.

Chapter 13-1 was all about sequences-a list of numbers. There are 2 types of sequences: arithmetic and geometric.

ARITHMETIC:
a sequence where the same number is added each time.

FORMULA:
tn=t1+(n-1)d

EXAMPLES:
8, 10, 12, 14, 16
is it arithmetic, geometric, or neither?
its ARITHMETIC since 2 is being added each time.
d=2.
NOW FIND THE NTH TERM:
tn=8+(n-1)2
tn=8+2n-2
tn=2n+6.

GEOMETRIC:
a sequence where the same number is multiplied each time.

FORMULA:
tn=t1=r^n-1

EXAMPLES:
4, 8, 16, 32
is it arithmetic, geometric, or neither?
its GEOMETRIC because you can obviously tell that 2 is being multiplied.
so r=2.

I was at school thursday so I was able to learn this with you and it's actually really easy.

To be honest I wish i knew what was going on in class. Therefore, im not really sure what to blog about. I would love to explain something we learned that i know best, but it just doesn't make any sense to me. So, in that case i will just explain 13-1 the best i can.

In Chapter 13 it is all about sequences. A sequence is simply a list of numbers (ex 4,8,12) There are two types of sequences:
Arithmetic: a sequence where the same number is added each time
Geometric: a sequence where the same number is multiplied each time
The formula for each are:
Arithmetic--> tn+t1+(n-1)d
Geometric-->tn=t1-r^n-1
To help you out "d" is the sequence, for example d of 4,6,8,10... is 2 because you add 2 each time. "n" is the last number given, for example 4,10...75 n would be 75. "r" is the ratio, for example ratio of (4,12) would be 3 because you are multiplying by 3.

Examples:
1. Find the indicated term--- t1=4 t2=10 t75=? If it is an arithmetic term
All to do is find d and plug into formula.
t75=4+(75-1)6
t75=448

2.Find t10 if t1=4 t2=12 and the sequence is geometric.
t10=4*3^10-1
t10=4*3^9
t10=78,732

There are many different problems to work with sequences, but that is mostly what you do.

13-1

This week in Advanced Math we are learning chapter 13 section 1. This chapter is extremely easy for me I think because you are just finding the next numbers in a or telling how many are in a sequence. Somewhere in that range. There are many different types of problems that you may run into when doing this section but since we have such a good math teacher she has showed us every kind there is that we need to know, well that is for our homework at least.. no telling what we will see on the test. :) Anyways here are the formulas and following the formulas will be some examples which are really easy to catch on to.

Formulas:
arithemetic: tn=t1+(n-1)d
geometric: tn=t1xr^n-1

Examples:
1. Tell weather each set of sequences is geometric, arithmetic, or neither.
a. 3,6,9,12,15...
aritmetic d=3

b. 4,12,16, 20...
geometric r=4

c. 11, 15, 16, 19...
neither

13-1

This past week in advanced math we took a test on chapter 11 and started chapter 13. On Thursday they learned section 13-1, but I was not there because I attended the volleyball game. On Friday when I came back I was kind of lost. I got the notes I missed and tried to teach myself section 13-1. It was kind of hard to understand so I will have to spend extra time on this chapter. Section 13-1 is about arithmetic and geometric sequences. Like the previous chapters it seems we use formulas. I tried doing the homework we were given over the weekend. I am going to try and work some example problems that I did understand.

Formulas:

Arithmetic – tn = t1 = (n – 1) d

Geometric – tn = t1 = r n-1

Examples:

Tell whether each sequence is arithmetic, geometric, or neither.

4n + 3 is arithmetic
It’s arithmetic because if you plug in the 4 terms you get 7, 11, 15, 19 and d = 4.

8 – 5n is geometric
It’s geometric because if you plug in the 4 terms you get 3, -2, -7, -12 and d = -5.

A sequence is usually neither if you plug in your terms and get a set of random numbers.

13-1

On Thursday, we started learning chapter 13-1. Unfortunately I wasn’t there for 7th hour that day so I had to try and catch up on this section Friday. It didn’t work out too well and I am not very sure what we are doing. I have been trying to teach myself, but I feel like I need a review in class on Monday. Since I am still kind of lost, I’ll show an example of some easy problems that I can do on my own.

Chapter 13-1 is about finding arithmetic and geometric sequences. The first thing that is usually asked is for you to identify.

Arithmetic- when the sequence of numbers has the same number added to them. The formula is:
tn= t1 + (n – 1) (d)

Geometric- when the sequence of numbers has the same number multiplied to them. The formula is:
tn= t1 x r^n -1

Examples:


1.State whether it is arithmetic or geometric.
1, 2, 4, 8, 16
Answer would be geometric because 2 is being multiplied.

2. How many multiples of 5 are there between 30 and 525?
The sequence is: 30, 35, 40, 45,…..525
The numbers are being added, so you use the arithmetic formula.
525 = 30 + (n-1) (5)
525 = 30 + 5n – 5
525 = 25 + 5n
500 = 5n
N = 100

13-1

We began Chapter 13 this week in Advanced Math. I3-1 is about arithmetic and geometric sequences. This section’s objective was to identify an arithmetic or geometric sequence and find a formula for its nth term.

Arithmetic Sequences - a sequence where the same number is added each time.
(Formula: tn = t1 + (n-1) d) The following sequences are all arithmetic..
Ex: 2, 6, 10, 14, 18.. – difference = 4

Geometric Sequences – a sequence where the same number is multiplied each time.
(Formula: tn = t1 x r^n-1)The following sequences are all geometric..
Ex: 1, 3, 9, 27, 81.. – ratio = 3

EX 1: 2, 5, 7, 10, 12.. Is it arithmetic, geometric, or neither?
A). It would be neither since you are adding 2 and 3 at different times. It isn’t a constant number being added or multiplied.

EX 2: How many multiples of 6 are there between 24and 300?
A). First find the multiples – 24, 36, 42, 48, 56.. 300
B). Since you are adding 6 each time, your difference = 4.
C). Plug the numbers into the arithmetic formula – tn = t1 + (n-1) d
D). 300 = 24 + (n-1)6
E). Solve using order of operations.
F). Your answer is n=47

EX3: Find t9 if t1=3, t2=6 and the sequence is geometric.
A). Since the problem is telling us it’s geometric, use that formula - tn = t1 x r^n-1
B). t9 = 3 x 2^9-1
C). t9 = 768

Chapter 13-1

This week we began learning chapter thirteen, Sequences and Series. Due to certain circumstances, we only got to learn the first section “Finite sequences and series.” I found this section to be a little difficult. And the fact that the book doesn’t give the best examples did not help either. So I’m pretty stuck until Monday when we review. Until then, I guess ill keep doing the problems until I get it.


A sequence is a list of numbers. There are two kinds of sequences : Arithmetic and Geometric. In a arithmetic sequence, the same numbers are being added. On the other hand, in a geometric sequence, the same numbers are being multiplied. Both sequences have formulas that may be used to find nth terms.

Arithmetic- t(n)= t(1) + (n-1) d * here “d” is the common difference.
Geometric- t(n) = t(1)xR^(n-1) * here “r” is the common ratio.

* if a sequence is not arithmetic or geometric, then it is “Neither.”

There are many ways you may be asked to solve sequence problems. The first way you may be asked to solve a sequence problem is to solve for the nth root. Other ways include solving for the t(n)term where the n will be given, and you may even be asked how many terms are in a given sequence.
Ex. Find the formula for t(n)
1, 4, 7, 10..
Each number is being increased by three, therefore this sequence is an Arithmetic one. The common difference is a positive four. ( d=4 ) and it uses the arithmetic formula (t(n)= t(1) + (n-1) d ). To solve, you simply substitute. .
T(n)=1 + ( n-1)3
T(n)=1 + 3n-3
T(n) =3-2n

Ch.11-1

The Chapter that i understood the most so far is Chapter 11 lesson 1. In Ch.11-1, we learned how to plot, give polar coordinates and convert to rectangular. I feel most comfortable converting to rectangular versus the others.

The Formula for converting to rectangular is x=rcos(theta) and y=rsin(theta).
To first convert to rectangular, you must plug in the numbers into the formula. You then solve for cos(ex:2COS90)after you solve cos( already knowing your trig chart) you multiply the number times the answer you get from the second step and you have half of the complete answer. you repeat the exact same steps for solving sin and you will have your complete answer.

EX: Convert(2,30 degrees) to rectangular
1. x=2cos30
2.cos30=square root of 3/2
3. 2 times square root of 3/2= square root of 3
4.(square root of 3,?)
SIN
1.y=2sin30
2.sin30=1/2
3.2 times 1/2=1

Answer:( square root of 3,1)

Week 4 Blog Prompt

What trig concept do you feel most comfortable with from Ch. 9-11? Give an example w/an explanation of how it is worked.

Chapter 11-1

This past week in advance math, we finished learning chapter 11,consisting of formulas and trig chart memory.Two formulas we learned were x=rcostheta y=rsintheta(rectangular) r=sq/root x^2+y^2.tan theta=y/x(polar)

Ex:1: Give the polar coordinates for (2,4)
1. convert to polar using light green formula
2.r=square root of -1^2+12^2
3,r=+/-12
4.Plug x and y into the tan function Tantheta=4/-2
5/Tan i s negative, so now we are finding angles in quad 2 and 4..181 deg/ and 261 deg.
5.answer is ( 12,181 deg.) and (,-12-261)

Ex:2 convert (4,pi/2) to rect.
1.use formula in orange
2.x=4cos pi/2 and y=3 sin pi/2
3.use trig chart..pi/2 is (sin)sq.root of 2 over 2(cos)0
4.x=8sq.root of 2 y=0
5.(8sq.rtof 2,0)

OMG! I UNDERSTAND!

So this weekend, I got some mathematical assistance from my twin, Mary Graci, the math whiz. She helped me better understand chapter 11. One of the things from chapter 11 is converting rectangular to polar and polar to rectangular.

The formula for converting to rectangular is
x = r cos theta
y = r sin theta

The formula for converting to polar is
r = square root of (x^2 + y^2)
theta = tan ^-1 (y/x)

The format for polar is
(r,theta)

The format for rectangular is
(x,y)

This may look confusing, but if you take it one step at a time it really isn't that scary.

For example, give the polar coordinates for (3,3).
First, you must find r.
r= square root of ( 3^2 + 3^2)
r = square root of 18
Now that you have r, you need to find theta.
theta= tan ^-1 (3/3) which is equal to one
And tan of one is 45 according to the trig chart.
So that makes your theta = 45
The answer you get is (square root of 18, 45 degrees) and (-square root of 18, 45 degrees).

Now, here is an example for converting to rectangular
Find (-2,60 degrees) in rectangular
x = -2 cos 60
y = -2 sin 60
x = -1
y = - square root of 3
Answer
(-1, -square root of 3)

11-1 Polars

This week in advanced math we learned chapter 11. We started by learning all of the different Polar graphs. There is Circles in Polar form where the formula is r= a cos(theta) or r= a sin(theta). Limacons where r=a +/- b cos(theta) or sin(theta) where a>0 and b>0. Cardiods where r=a +/- a sin(theta) or cos(theta). Rose Curves where r= a sin n(theta) or a cos n(theta). Lemniscates where r^2= a^2 sin 2(theta) or cos 2(theta). Then there's 11-2 with complex numbers. Which involves the formulas z= x + yi for rectangular and z= rcos(theta) + rsin(theta)i for Polar. The abbreviated version of this formula is z=rcis(theta). 11-3 involves De Moirre's Theorem with the formula (rcis)^n = r^n cisn(theta).

11-1 Polars

Formulas: x=rcos(theta); y=rsin(theta) to convert to rectangular

r=the square root of x^2 + y^2; tan(theta)=y/x to convert to polar

Ex. convert 2, 45 degrees to rectangular

x= 2cos45 degrees = 2(square root of 2)/2= square root of 2

y=2sin45degrees= 2(square root of 2)/2 = square root of 2

[square root of 2, square root of 2]

11-1

This week in advanced math we started and finished chapter 11. Chapter 11 consist of three sections that deal with polar, complex numbers and De Moivre’s Theorem. Section 11-1 was the easiest so let’s review. All you need to know is how to convert to rectangular and polar.

Formulas:

Convert to Rectangular
x = r cos θ
y = r sin θ

Convert to Polar
r = square root x2 + y2
tan θ = y/x

Let’s try some example problems

Example 1:
Convert (4, 45°) to rectangular

x = 4 cos 45° = 4(square root 2/2) = 4
y = 4 sin 45° = 4(square root 2/2) = 4
Your answer (4,4)

Example 2:
Convert (2, π/3) to rectangular

x = 2 cos 60° = 2 (1/2) = 1
y = 2 sin 60° = 2 (square root 2/2) = square root 3
Your answer (1, square root 3)

Example 3:
Give the polar coordinates (6, 8)

1. r = square root 62 + 82
2. square root 100 = +/- 10
3. tan θ = 8/6 reduces to 4/3
4. θ = tan-1(4/3) = 53.130°
5. You want 2 answers so add 180 and get 233.130°

8th seed baby.

Thats right your Riverside Rebels are seeded 8th and that means we have a home playoff game this week. We are playing Mangum. Get ready people its going to be a fun ride. So on to Math. This week we started Chapter 11. This chapter comtains things called polar and rectangular and polar. Rectangular may be easily recognized on account of they look like the regular points on a coordinate plane (x,y). Polar is represented by a radius, which looks like a regular number, and and angle, which is normally represented by theta. We learned this week how to change rectangular to polar and vice versa. Another thing we did this week was learn how to classify graphs. Their can be circles, roses, limacon, if that politically correct, and some other stuff. Limacons are also called snails. I have to say that is pretty legit.

11-1

This week we are learning chapter 11. Chapter 11-1 is converting from polar to rectangular and converting from rectangular to polar. Also you will have to know how to plot the point in polar. I am going to show you how to do this, and give you some example problems and formulas.

FORMULAS:
(r, theta) is polar
(x,y) is rectangular

x=rcos(theta) and y=rsin(theta) is when you are converting to rectangular
r= +/-squareroot(x^2+y^2) and tan(theta) which is y/x is converting into polar

EXAMPLES:
Polar:
Give the polar point for (1,2) and plot the point in polar.
r=squareroot(1^2+2^2)
+/-5 which is going to be your first point, the x point
tan(theta)= 2/1
theta=tan^-1(2)
theta= 63.435
tan is postive in quadrants 1 and 3
Q1(5, 63.435)
Q3(5, 243.435)
To plot the polar point you just draw a number line go to 5(your x) and then do a unit circle of about how much you think 63.435 is and you do the same for the second quadrant.

Rectangular:
Give the rectangular coordinates for the point (4, 45degrees)
x=4cos45
y=4sin45
(2squareroot(2), 2squareroot(2))

11-1

This week we learned section 1 in chapter 11. 11-1 is really easy and quite simple. You will have to know formulas such as polar and rectangular.

To convert to rectangular you do x=rcostheta, y=rsintheta.


To convert to polar you do r=square roots of xsquared+ysquared and after you do tantheta=y/x then you find the inverse of it.

(r,theta)=polar, (x,y)=rectangular


Example Problems:

Give the Polar coordinates for the point (3,4)

R=square root 3squared+4squared=

R square root of 25 which is equal to +-5

Tantheta=4/3=

Theta=tan inverse of 4/3 you get 53.130degrees.

Where tan is + is 1st quadrant and 3rd quadrant which u do 180+53.130 =233.130degrees

Your answers are 5, 53.130degrees and -5,233.130degrees

Convert 2, 30 degrees to rectangular
X=2cos30 degrees which you get 2(square root of 3 over 2 because 30 degrees is on the trig chart you get square root of 3 as you answer.

Y=2sin30 degrees
Sin 30degrees is on the trig chart =to ½
2(1/2 you get 1

Your answers are squared root of 3 and 1.

If you know to convert to polar and rectangular and know every formula in 11-1 this section should be a joke for you.

Chapter Eleven

This week our class began chapter eleven and learned chapter complex numbers. Complex numbers have two forms : polar and rectangular.

The rectangular form for complex numbers is z= x + yi

The polar form for complex numbers is z= rcos(x)rsin(x)i or it can be abbreviated as z=rcis(x)

In section one we learned how to convert polar to non polar and vice versa.

Rectangular coordinates are formed by ( x,y)

Polar coordinates are in the form of (r,θ)

In order to convert from rectangular to polar you must solve for “r” and theta.
*r=sqrt of x^2+y^2. Theta is equal to tan y/x
You will always have two answers when converting to polar. One positive and one negative.
EX. A (3,3) CONVERT TO POLAR :
R=sqrt 3^2=3^2 = 9+9
R=sqrt18 = ±2√3
Theta= tan y/x
theta= tan(3/3) = tan 1
theta = 45, 225

*(3,3) is located in the first two quadrants therefore the positive number gets the smaller number. Your answers are (2, 45) (-2,225)
In order to convert to rectangular , you simply use formulas x= rcostheta and y=rsintheta

11-1

This week we started learning chapter 11. A lot of people thought this chapter was sort of easy, but I didn’t. I think this is one of the hardest chapters we have done so far, and doing the pages in the book for a grade really stresses me out because they are a lot harder and will probably bring my grade down. But, I did understand section 11-1 the best out of all of them. For this, you need to be able to convert things into either polar or rectangular.
Convert to rectangular – x = r cos theta and y = r sin theta
Convert to polar- square root x^2 = y^2 and tan theta = y/x
For these problems, you will be given points.
Example 1.
Give polar coordinates for (1, 1)
1. R= square root 1^2 + 1^2
2. R = +/- square root 5
3. Tan theta = 1/1
4. Theta equals tan^-1 (1/1) = 1
5. Since tan is positive, it goes in the first and third quadrants
6. Quadrant 1 = 45 (from the trig chart) quadrant 3 = 225
7. Answer= (square root 2, 45) and ( -square root 2, 225)
Example 2.
Give rectangular coordinates for (1, pi/6)
1. X = 1cos pi/6 and y= 1sin pi/6
2. These are in your trig chart, 1(0) and 1(1)
3. The answer is (0, 1)

Chapter 11- Section 1

This week in Advanced Math, we were introduced to Chapter 11. To be honest i feel this chapter is quite simple. Calculators are still forbidden of use on test but we were allowed to use them for homework. These chapters all feed off of one another, and are accumlative. In 11-1, we were taught two conversion formulas – polar and rectangular.

x= r cos theta and y = r sin theta (formulas used to convert to Rectangular form)
r = square root x^2 + y^2 and tan theta = y/x (formulas used to convert to Polar form)

Points will be given to you like this:

(x , y ) <----Rectangular

(r, theta) <------Polar

EXAMPLE 1: Give the polar coordinates for point (-3, 4).

1. Since we are converting to polar, use the pink formulas.
2. r = square root of -3^2 + 4^2
3. r = +/- 5
4. (Plug x and y into the tan function y/x) tan theta = 4/-3
5. Since tan is negative, find the angles in quadrants 2 and 4 – 126.87 degrees and 306.87 degrees
6. Since the angles are negative, +5 will get the bigger angle.
7. Your answer is (5, 306.87 degrees) and (-5, 126.87 degrees)

EXAMPLE 2: Convert (3, pi/2) to rectangular.
1. Since we are converting to rectangular, use the green formulas.
2. x = 3 cos pi/2 and y = 3 sin pi/2
3. Pi/2 is in your trig chart so continue simplifying.
4. x = 3(0) and y = 3(1)
5. Your answer is (0,3)

11-1

This week in Advanced Math, we covered all of Chapter 11. The whole Chapter consisted of formulas and was pretty simple. In 11-1, we were taught two conversion formulas – polar and rectangular.

x= r cos theta
y = r sin theta
(r, theta)
^ Used to convert to rectangular.

r = square root x^2 + y^2
tan theta = y/x
(x,y)
^Used to convert to polar

EXAMPLE 1: Give the polar coordinates for point (-3, 4).
1. Since we are converting to polar, use the red formulas.
2. r = square root of -3^2 + 4^2
3. r = +/- 5
4. Plug x and y into the tan function – tan theta = 4/-3
5. Since tan is negative, find the angles in quadrants 2 and 4 – 126.87 degrees and 306.87 degrees
6. Since the angles are negative, +5 will get the bigger angle.
7. Your answer is (5, 306.87 degrees) and (-5, 126.87 degrees)

EXAMPLE 2: Convert (3, pi/2) to rectangular.
1. Since we are converting to rectangular, use the blue formulas.
2. x = 3 cos pi/2 and y = 3 sin pi/2
3. Pi/2 is in your trig chart so continue simplifying.
4. x = 3(0) and y = 3(1)
5. Your answer is (0,3)

Week 3 Prompt

What are the different types of polar graphs? Give several examples with pictures or a site to use as a reference. Everyone should have different pictures or a different site.