### Can you say..... Patterns are IMPORTANT??

Hi guys its me again!! I hope everyone is enjoying their evening... so

this is what we did today in class.....

We started class off by going over some of the things we did last

Thursday, so Mr. K wrote these on the board.

(a+b)^0 = 1

(a+b)^1 = 1a+1b

(a+b)^2 = 1a^2+ 2ab+1b^2

(a+b)^3 = 1a^3+ 3a^2b+ 3ab^2+ 1b^3

(a+b)^4 = 1a^4+ 4a^3b+6a^2b^2+4ab^3+1b^4

1

11 1

11211

1 13 131 1

11 4 16 14 11

1115110 10 15 11

oCo

Co 1C1

2Co 2C1 2C2

3Co 3C1 3C2 3C3

4Co 4C1 4C2 4C3 4C4

5Co 5C1 5C2 5C3 5C4 5C5

Do you remember those... if you don't you might want to

check Ashley's lovely scribe post from yesterday for reference

After he finished putting all of these back up on the board, we went

over patterns that we fount yesterday like:

Powers of 2

1

1 1

1 2 1

1 5 10 10 5 1

2^3 = 8 111111111111+3+3+1= 8

2^4 = 16111111111+4+6+4+1= 16

1 1

1 2 1

1 3 3 1

1 4 6 4 1 1 5 10 10 5 1

2^3 = 8 111111111111+3+3+1= 8

2^4 = 16111111111+4+6+4+1= 16

And powers of 11

1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

1 5 10 10 5 1

11^2 = 121

11^3 = 1331

11^5 = 161051

1 1

1 2 1

1 3 3 1

1 4 6 4 1

1 5 10 10 5 1

11^2 = 121

11^3 = 1331

11^5 = 161051

Whoa that doesn't look

the same... well it still works, you just

have to make sure to do the carrying.

The 5 turns into 6 and the tens turn

into 0 and 1 because 10 carry the 1

leave the 0, and 11 carry the 1 and

leave 1 behind and that 5+ the carried

1 = 6

Ok now here's the new pattern that Mr. K showed us... it shows the

**Fibonicci**numbers in Pascal's triangle (even though it wasn't pascal

who made it :P )

Pay attention cause its hard to see at first!!

The fibonicci numbers can be found on the diagonal edges... the numbers that are cut by this line are added together to show these fibonicci numbers.

Can you see the diagonal edges... they are the pink lines.... the first line crosses through only 1... the next crosses only 1... the next passes through 1 and 1... add these together to get 2... the next row passes through 1 and 2.. add these together to get the 4th fibonicci number 3... and on and on

Can you see it?? if not feel free to come and ask me in class, i'll do

my best to show you and make sure you understand.

After everyone found this pattern Mr. K wanted to point out other interesting

facts... so he told us to take out our calculators...

if you take one fibonicci number, and divide it by the fibonicci number before

it, you will find that as you travel down the triangle, the values get closer and

closer to the "GOLDEN RATIO" or PHI !!! ∅(phi) =

**1.61803399**

1/1 = 1

2/1= 2 Can you see how the numbers seem to be

3/2= 1.5 converging (narrowing down) to a single value?

5/3= 1.6667 All getting closer and closer to∅!!

1/1 = 1

2/1= 2 Can you see how the numbers seem to be

3/2= 1.5 converging (narrowing down) to a single value?

5/3= 1.6667 All getting closer and closer to

**8/5= 1.6**

13/8=1.625

13/8=1.625

As I've already mentioned, ∅ is the GOLDEN RATIO... but why is

this so you ask... well because if an item has a ratio that is equal to, or

close to ∅ than it just looks better... unexplainably but it really does..

For example have you ever wondered why 4 by 6 pictures look better

than 5 by 7s... they're smaller, yet they look better... this is because

6/4 = 1.5 * because this has a ratio closer to ∅ it is more visually appealing*

7/5 = 1.4

Another example of this is the Parthenon. This is a temple in Greece dedicated

to the goddess Athena (who is the goddess of wisdom). This temple is also

known as 1 of the 7 wonders of the world as it is one of the most visually

appealing structures in the world. This is because if you look at the Parthenon

it is constructed of rectangles. If you take the larger length of each rectangle and

divide it by the shorter length you will find that the ratio is 1.618... or really

close to the GOLDEN RATIO !

Everything is connected!!! The Fibonicci ratio is not only encoded Pascal's

Triangle, but it is also in various aspects of life, including the human body.

This is the Vitruvian Man done by Leonardo Davinci. (Found on Flickr,

posted by absolutesteven) It is known as the most accurate drawing of

a human being ever done. Why is this... i bet you can guess... that's right

Davinci made sure to measure every and make sure that every

part came out close or equal to the GOLDEN RATIO ....

ex.

Man's height / distance from ground to navel

Man's arm / distance from elbow to tip of middle finger

Don't you just love it when everything fits together like that?? :D

After his speech on the GOLDEN RATIO, Mr. K brought us back to

the (a+b)^x expansions....

What are the patterns???

1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

1 5 10 10 5 1

oCo

1Co 1C1

2Co 2C1 2C2

3Co 3C1 3C2 3C3

4Co 4C1 4C2 4C3 4C4

5Co 5C1 5C2 5C3 5C4 5C5

1 1

1 2 1

1 3 3 1

1 4 6 4 1

1 5 10 10 5 1

oCo

1Co 1C1

2Co 2C1 2C2

3Co 3C1 3C2 3C3

4Co 4C1 4C2 4C3 4C4

5Co 5C1 5C2 5C3 5C4 5C5

Firstly the co-efficents for the expansion can be found within

the triangle (look at above triangle with matching colour)

(a+b)^4 = 1a^4+ 4a^3b+6a^2b^2+4ab^3+1b^4

Can you see the co-efficents??

1, 4, 6, 4, and 1

according to the 5th row

and 4Co row...

tells us the exponents for the expansion

a^(e-x) b^x

e = what it is being chosen out of.. in this case 4

4Co 4C1 4C2 4C3 4C4

so this expansion is

a^(4-0) + a^(4-1) b^1 + a^(4-2) b^ 2 + a^(4-3) b^3 + a^(4-4) b^4

a^4 + a^3b + a^2 b^2 + a b^3 + b^4

If you add these two patterns together you will get...

1a^4 + 4 a^3b + 6a^2 b^2 +4 a b^3 + 1b^4

the expansion of (a+b)^4

Now, if asked to find the 3rd term of (a+b)^4 do you have to

figure, and write out the whole expansion. Sure why not,

(a+b)^4 is easy... but what if i asked for the 7th term of

(a+b)^11?? do you really want to write that out??

I can guess that your answer, like mine, is NO WAY!!!

Ok now its time to show you how to do this much QUICKER

and much EASIER....

First of all i must make sure that it is clear to you that as we go along

the triangle from the left to right the terms are numbered from 1

increasing... 1, 2, 3, 4, 5 etc.

oCo

1Co 1C1

2Co 2C1 2C2

3Co 3C1 3C2 3C3

4Co 4C1 4C2 4C3 4C4

5Co 5C1 5C2 5C3 5C4 5C5

11112222333444455556666

So,1Co 1C1

2Co 2C1 2C2

3Co 3C1 3C2 3C3

4Co 4C1 4C2 4C3 4C4

5Co 5C1 5C2 5C3 5C4 5C5

11112222333444455556666

5Co is term 1

5C1 is term 2

5C2 is term 3

5C5 is term 6

Do you see the pattern, the term number is always 1 larger

than the amount being chosen!!

Therefore, if the number being chosen is always 1 larger,

when we ask for the Xth term, we must subtract 1 so it will

look like this nC(x-1) where n equals the amount being

chosen from.

Ok given this question, 7th term of (a+b)^11

lets use what we know, and the new equation

11 is the number being chosen out of and takes the place of

the n. looking like this 11C(x-1)

7 is the term we want, and takes the place of x.

looking like this 11C (7-1)

Do the math... 11C6 is what you are left with. Now use

rules from before that state that give exponents to a and b

a^(e-x) b^x

e = what it is being chosen out of.. in this case 11

b = the number chosen, in this case 6

a^(11-6) b^6

a^5 b^6

Now put both portions together!!

11C6 a^5 b^6

This will be the 7th term in the expansion of (a+b)^11

Lets try this again to make sure you've got

it....

find the 5th term of (a+b)^13

13 is amount being chosen from

5-1 = 4 this becomes the number being chosen!!

Therefore this becomes 13C4

next comes the exponents

13-4 = 9 which becomes the exponent on a

4 is the exponent on b

put them together

13C4 a^9 b^4

is the 5th term of (a+b)^13

Mr. K tested the class a few more times, and soon a couple

people could find the Xth term of an expansion in less than

10 seconds, and the rest of the class could do it in under

30 seconds.... can you?? Time yourself... If you work at

it you'll get there... if you have any questions just ask me

i'm always willing to help if i can!!

Ok guys time to turn up the heat!!

How many terms are in each expansion?? How do we know this??

Well the number of terms in an expansion can be found by taking the

power of the expansion...

ex. (a+b) ^11

the power is 11

take this power and add 1

11+1 = 12

12 is the number of terms in the expansion!!

What would happen if we were asked to find the middle term of the

expansion??

for odd numbers this is easy, all you need to do is add 1 to the power,

then divide by 2

13 +1 = 14 14/2 = 7

Therefore the middle term in this expansion is the 7th term!!

* this formula will only work if the power you start off with is odd*

In order to find the middle term for an even power, we take the

power, divide by 2 and add 1

12 / 2 = 6 6+ 1 = 7

Therefore the middle term in this expansion is the 7th term!!

* this formula will only work if the power you start off with is even*

Now that we can find the middle term of the expansions we've practiced,

can we find the middle terms of other binomial expansions??? YES!!!

how about [x-(2/x)]^7 ?? what is the middle term of this??

First of all lets make this into something we can understand shall we??

Let a = x

Let b = - 2/x to make this even simpler we're going to change

this into - 2x^(-1)

so now the binomial looks like this

(a+b)^7 and that is much better....

now find the middle term... 7 is odd therefore we must add 1 to make 8 and divide by 2.. the middle term is 4!!

now show the term 4 of this expansion.

7C3 a^4 b^3 *reference above if you don't remember how

1111111111111to do this step*

now rewrite 7C3 in terms of factorials....

7! / (4!3!) a^4 b^3

4! reduces leaving--> 7x6x5 / 3! a^4 b^3

6 and 3! reduce as 6 = 3! leaving--> 7x5 a^4 b^3

35 a^4 b^3

Now substitute the real values of a and b back in.

35(x)^4 (-2x^-1)^3

35 x^4 (-8x ^-3)

-280x^1

therefore -280x^1 is the middle term in the binomial expansion

of [x-(2/x)]^7

Lets try this again to make sure you've got it!

Find the middle term of (x^2 - 2/x)^8

now make this simpler for yourself

Let a = x^2

Let b = -2/x even simpler -2x^-1

How many terms are in each expansion?? How do we know this??

Well the number of terms in an expansion can be found by taking the

power of the expansion...

ex. (a+b) ^11

the power is 11

take this power and add 1

11+1 = 12

12 is the number of terms in the expansion!!

What would happen if we were asked to find the middle term of the

expansion??

for odd numbers this is easy, all you need to do is add 1 to the power,

then divide by 2

13 +1 = 14 14/2 = 7

Therefore the middle term in this expansion is the 7th term!!

* this formula will only work if the power you start off with is odd*

In order to find the middle term for an even power, we take the

power, divide by 2 and add 1

12 / 2 = 6 6+ 1 = 7

Therefore the middle term in this expansion is the 7th term!!

* this formula will only work if the power you start off with is even*

Now that we can find the middle term of the expansions we've practiced,

can we find the middle terms of other binomial expansions??? YES!!!

how about [x-(2/x)]^7 ?? what is the middle term of this??

First of all lets make this into something we can understand shall we??

Let a = x

Let b = - 2/x to make this even simpler we're going to change

this into - 2x^(-1)

so now the binomial looks like this

(a+b)^7 and that is much better....

now find the middle term... 7 is odd therefore we must add 1 to make 8 and divide by 2.. the middle term is 4!!

now show the term 4 of this expansion.

7C3 a^4 b^3 *reference above if you don't remember how

1111111111111to do this step*

now rewrite 7C3 in terms of factorials....

7! / (4!3!) a^4 b^3

4! reduces leaving--> 7x6x5 / 3! a^4 b^3

6 and 3! reduce as 6 = 3! leaving--> 7x5 a^4 b^3

35 a^4 b^3

Now substitute the real values of a and b back in.

35(x)^4 (-2x^-1)^3

35 x^4 (-8x ^-3)

-280x^1

therefore -280x^1 is the middle term in the binomial expansion

of [x-(2/x)]^7

Lets try this again to make sure you've got it!

Find the middle term of (x^2 - 2/x)^8

now make this simpler for yourself

Let a = x^2

Let b = -2/x even simpler -2x^-1

now the binomial looks like

(a+b)^8

middle term for the power of 8 is the 5th term

therefore 8C4 a^4 b^4

write in terms of factorials

8! / (4!4!) a^4 b^4

4! reduces leaving--> 8x7x6x5 / 4! a^4 b^4

Now substitute the real values back in

70 (x^2)^4 (-2x^-1)^4

1111111111111111

A common mistake made is when the exponent

is an even number, the student forgets to make

the coefficient positive. Watch for this!!

Therefore 1120x^4 is the middle term in the binomial

expansion of (x^2 - 2/x)^8

As you have seen through class and scribe posts for this

unit, everything is connected and therefore

(a+b)^8

middle term for the power of 8 is the 5th term

therefore 8C4 a^4 b^4

write in terms of factorials

8! / (4!4!) a^4 b^4

4! reduces leaving--> 8x7x6x5 / 4! a^4 b^4

Now substitute the real values back in

70 (x^2)^4 (-2x^-1)^4

70 x^8 (16x ^-4) 111

1120x^4 11111111111111111111111111

A common mistake made is when the exponent

is an even number, the student forgets to make

the coefficient positive. Watch for this!!

Therefore 1120x^4 is the middle term in the binomial

expansion of (x^2 - 2/x)^8

CONNECTIONS ARE EVERYWHERE!!!!!!

As you have seen through class and scribe posts for this

unit, everything is connected and therefore

Combinations and binomials go hand in hand!!

The last of the class was spent reviewing most of the above one more time, in order to really make sure that we understood!!

I hope everything in this scribe post is useful to you. If you have any questions or concerns don't hesitate to ask me, I'm here to make this scribe post make perfect sense to you!! Enjoy the rest of your evening, and i'll see you all tomorrow!! Good Night!!

p.s please make sure to have at least 5 pictures on your flickr account if you are a new user. If you do not, your account will not be reviewed, and therefore not deemed safe to make your pictures public for the assignment!!

Oh ya... and just to finish this off, jessiccal_198 your scribe tomorrow :)

Ok i believe that the spacing is fixed now... see you later!!

The last of the class was spent reviewing most of the above one more time, in order to really make sure that we understood!!

I hope everything in this scribe post is useful to you. If you have any questions or concerns don't hesitate to ask me, I'm here to make this scribe post make perfect sense to you!! Enjoy the rest of your evening, and i'll see you all tomorrow!! Good Night!!

p.s please make sure to have at least 5 pictures on your flickr account if you are a new user. If you do not, your account will not be reviewed, and therefore not deemed safe to make your pictures public for the assignment!!

Oh ya... and just to finish this off, jessiccal_198 your scribe tomorrow :)

Ok i believe that the spacing is fixed now... see you later!!

Hi this is Mr. Harbeck. If there are some classmates out there this post is Hall of Fame Worthy. There is tremendous detail great use of images and wow so much effort. Step up and nominate this for the Hall.

ReplyDeleteThanks for all your hard work.

Mr. Harbeck

Sargent Park School

is there a day where you don't do scribes so well? j/k YOU DID A GREAT JOB CRAZY! SCRIBE POST HALL OF FAME WORTHY!

ReplyDeleteHi Lisa,

ReplyDeleteI agree that your scribe is definitely Hall of Fame material!!

"Pay attention" "Do you remember" "I can guess that your answer, like mine, is NO WAY!!! ''Ok now its time to show you how to do this much QUICKER and much EASIER...." "ok guys time to turn up the heat"

Those and other comments like those personalized this scribe and keep your readers focused.

Congratulations on such excellent good work!

Best,

Lani

Gheez Lisa how do you do it? Your scribe posts are always awesome and always in depth with what happened in class. You even have that picture of that guy! Sorry about not commenting earlier but definetly Hall of Fame Worthy!!!

ReplyDeletethis post belong to the hall of fame. how nice,it gives me help to understand the lesson. great job. keep it up. thanks.

ReplyDelete