Function Transformations

Hey guys ! got something here for the class..Hope this helps.. :p

FUNCTION TRANSFORMATION

Let's start this by knowing what function is....

      *A Function is a relation between a set of inputs and a set of permissible outputs with the property that each input is related to exactly one output.

means that...

     " In each value of x, there is only one value of y"

sample??..

   IN  f(x) = x² 

     ^{*note: f(x) = y}

 ^{IF x = 2, THEN y = 4}

 IF x = 3, THEN y = 9

 IF x = 4, THEN y = 16 ...and so on...

If we graph it, it should look like this...

    

f(x) = x²

^{NOW THAT'S A FUNCTION...}

   ^{*SO IS A CIRCLE A FUNCTION??..}

      

 

...well its

HOW TO KNOW IF A GRAPH IS A FUNCTION??.. :D

  Well there's a TEST to know if a GRAPH is a FUNCTION....

ITS V. L. T OR VERTICAL LINE TEST.. :D

If we draw a vertical line in a circle, the line INTERSECTS TWICE. So it means, its NOT a function.

NOTE: If the vertical line intersects the graph only once, the graph is a FUNCTION.

NOW THAT WE KNOW WHAT FUNCTION IS.

*Can we move or resize the graph of a function?..YES!!..

Let us start with a function, in this case it is f(x) = x², but it could be anything:

f(x) = x²

Here are some simple things you can do to move or scale it on the graph:

You can move it up or down by adding a constant to the y-value:

g(x) = x² + C

Note: if you want to move the line down, just use a negative value for C.

• C > 0 moves it up
• C < 0 moves it down

 

You can move it left or right by adding a constant to the x-value:

g(x) = (x+C)²

Adding C moves the function to the left (the negative direction).

Why? Well imagine you are going to inherit a fortune when your age=25. If you change that to(age+4) = 25 then you would get it when you are 21. Adding 4 made it happen earlier.

• C > 0 moves it left
• C < 0 moves it right

An easy way to remember what happens to the graph when you add a constant:

add to y: go high
add to x: go left

BUT you must add C wherever x appears in the function (you are substituting x+C for x).

Example: the function v(x) = x³ - x² + 4x

Move C spaces to the left: w(x) = (x+C)³ - (x+C)² + 4(x+C)

You can stretch or compress it in the y-direction by multiplying the whole function by a constant.

g(x) = 0.35(x²)

• C > 1 stretches it
• 0 < C < 1 compresses it

 

You can stretch or compress it in the x-direction by multiplying x (wherever it appears) by a constant.

g(x) = (2x)²

• C > 1 compresses it
• 0 < C < 1 stretches it

Note that (unlike for the y-direction), bigger values cause more compression.

 

You can flip it upside down by multiplying the whole function by -1:

g(x) = -(x²)

This is also called reflection about the x-axis (the axis where y=0)

You can combine a negative value with a scaling.

Example: multiplying by -2 will flip it upside down AND stretch it in the y-direction.

 

You can flip it left-right by multiplying the x-value by -1:

g(x) = (-x)²

It really does flip it left and right! But you can't see it, because x² is symmetrical about the y-axis. So here is another example using √(x):

g(x) = √(-x)

This is also called reflection about the y-axis (the axis where x=0)

Summary

y = f(x) + C	C > 0 moves it up C < 0 moves it down
y = f(x + C)	C > 0 moves it left C < 0 moves it right
y = C·f(x)	C > 1 stretches it in the y-direction 0 < C < 1 compresses it
y = f(Cx)	C > 1 compresses it in the x-direction 0 < C < 1 stretches it
y = -f(x)	Reflects it about x-axis
y = f(-x)	Reflects it about y-axis

All In One ... !

You can do all transformation in one go using this:

 a is vertical stretch/compression 

• |a| > 1 stretches 
• |a| < 1 compresses
• a < 0 flips the graph upside down

 b is horizontal stretch/compression

• |b| > 1 compresses 
• |b| < 1 stretches
• b < 0 flips the graph left-right

 c is horizontal shift

• c < 0 shifts to the right 
• c > 0 shifts to the left

 d is vertical shift 

• d > 0 shifts upward 
• d < 0 shifts downward

Thats it...

Sources:

GOOGLE :D

http://www.mathsisfun.com/sets/function-transformations.html

         

Binomial Theorem

Write a Blog?

 

Binomial Theorem

What happens when you multiply a binomial by itself ... many times?
BINOMIAL-CEPTION

 

 

What is a Binomial?

A binomial is a polynomial with two terms.

Multiplying

The Binomial Theorem shows what happens when you multiply a binomial by itself (as many times as you want).
It works because there is a pattern ... let us see if we can discover it.

Exponents of (a+b)

Now on to the binomial.

We will use the simple binomial a+b, but it could be any binomial.

Let us start with an exponent of 0 and build upwards.

Exponent of 0

When an exponent is 0, you get 1:

(a+b)⁰ = 1

Exponent of 1

When the exponent is 1, you get the original value, unchanged:

(a+b)¹ = a+b

Exponent of 2

An exponent of 2 means to multiply by itself:

(a+b)² = (a+b)(a+b) = a² + 2ab + b²

Exponent of 3

For an exponent of 3 just multiply again:

(a+b)³ = (a+b)(a² + 2ab + b²) = a³ + 3a²b + 3ab² + b³

 

^{GETTING SMARTER}

 ^{We have enough now to start talking about the pattern.}

The Pattern

In the last result we got:

a³ + 3a²b + 3ab² + b³

Now, notice the exponents of a. They start at 3 and go down: 3, 2, 1, 0:

Likewise the exponents of b go upwards: 0, 1, 2, 3:

If we number the terms 0 to n, we get this:

k=0	k=1	k=2	k=3
a3	a2	a	1
1	b	b2	b3

Which can be brought together into this:

a^n-kb^k

  

Coefficients

So far we have:	a3 + a2b + ab2 + b3
But we really need:	a3 + 3a2b + 3ab2 + b3

We are missing the numbers (which are called coefficients).
Let's look at all the results we got before, from (a+b)⁰ up to (a+b)³:

And now look at just the coefficients (with a "1" where a coefficient wasn't shown):

THIS MAKES PASCAL'S TRIANGLE

The Original Triangle

As a Formula

Our last step is to write it all as a formula.
But hang on, how do we write a formula for "find the coefficient from Pascal's Triangle" ... ?
Well, there is such a formula:

It is commonly called "n choose k" because it is how many ways to choose k elements from a set of n.

Putting It All Together

The last step is to put all the terms together into one formula.
But we are adding lots of terms together ... can that be done using one formula?
Yes! The handy Sigma Notation allows us to sum up as many terms as we want:

Sigma Notation

Now it can all go into one formula:

The Binomial Theorem

 

Sources:

Google: https://www.google.ca/

Math is Fun: http://www.mathsisfun.com/algebra/binomial-theorem.html

Combinations and Permutations

Hello! I found some additional information about Combinations and also some tips for Permutations. 

Combinations and Permutations

What's the Difference?

In English we use the word "combination" loosely, without thinking if the order of things is important. In other words:

	"My fruit salad is a combination of apples, grapes and bananas" We don't care what order the fruits are in, they could also be "bananas, grapes and apples" or "grapes, apples and bananas", its the same fruit salad.
	"The combination to the safe was 472". Now we do care about the order. "724" would not work, nor would "247". It has to be exactly 4-7-2.

So, in Mathematics we use more precise language:

	If the order doesn't matter, it is a Combination.
	If the order does matter it is a Permutation.

So, we should really call this a "Permutation Lock"!

In other words:

A Permutation is an ordered Combination.

Combinations

There are also two types of combinations (remember the order does not matter now):

1. Repetition is Allowed: such as coins in your pocket (5,5,5,10,10)
2. No Repetition: such as lottery numbers (2,14,15,27,30,33)

1. Combinations with Repetition

Actually, these are the hardest to explain, so I will come back to this later.

2. Combinations without Repetition

This is how lotteries work. The numbers are drawn one at a time, and if you have the lucky numbers (no matter what order) you win!

The easiest way to explain it is to:

• assume that the order does matter (ie permutations),
• then alter it so the order does not matter.

Going back to our pool ball example, let us say that you just want to know which 3 pool balls were chosen, not the order.

We already know that 3 out of 16 gave us 3,360 permutations.

But many of those will be the same to us now, because we don't care what order!

For example, let us say balls 1, 2 and 3 were chosen. These are the possibilites:

Order does matter	Order doesn't matter
1 2 3 1 3 2 2 1 3 2 3 1 3 1 2 3 2 1	1 2 3

So, the permutations will have 6 times as many possibilites.

In fact there is an easy way to work out how many ways "1 2 3" could be placed in order, and we have already talked about it. The answer is:

3! = 3 × 2 × 1 = 6

(Another example: 4 things can be placed in 4! = 4 × 3 × 2 × 1 = 24 different ways, try it for yourself!)

So, all we need to do is adjust our permutations formula to reduce it by how many ways the objects could be in order (because we aren't interested in the order any more):

That formula is so important it is often just written in big parentheses like this:

where n is the number of things to choose from, and you choose r of them (No repetition, order doesn't matter)

Here are some tips that can help you in permutations:

Tip #1

0! = 1

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Tip #2

Factorials of negative integers and fractions are not defined.

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Tip #3

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Tip #4

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Tip #5

The number of permutations of n distinct objects taken all at a time is n!.

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Tip #6

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Tip #7

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Tip #8

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Tip #9

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Tip #10

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Tip #11

Number of circular permutations of n different things taken all at a time = (n-1)! (If clockwise and anticlockwise orders are taken as different).

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Tip #12

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Tip #13

In permutation order of objects is important whereas in combination order of objects is not important.

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Tip #14

If m = n, then groups are of equal size. Division of these groups can be given by two types.

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Tip #15

Here are some additional problems that you can practice for combinations:

http://www.youtube.com/watch?v=oQpKtm5TtxU

http://www.youtube.com/watch?v=v9NLtiVt3XY

http://www.youtube.com/watch?v=SbpoyXTpC84

http://www.youtube.com/watch?v=DIjlllgq3dc

Source:

GOOGLE! :D

http://www.mathsisfun.com/combinatorics/combinations-permutations.html

http://www.simplylearnt.com/tips-tricks/Permutation-and-Combination

Permutations: With and Without Repetition

Heyyyy it's me Cedric. I found a blog about the lesson last friday (02/08/13) :)
Here it is:

Permutations

There are basically two types of permutation:

1. Repetition is Allowed: such as the lock above. It could be "333".
2. No Repetition: for example the first three people in a running race. You can't be first and second.

1. Permutations with Repetition

These are the easiest to calculate.

When you have n things to choose from ... you have n choices each time!

When choosing r of them, the permutations are:

n × n × ... (r times)

(In other words, there are n possibilities for the first choice, THEN there are n possibilites for the second choice, and so on, multplying each time.)

Which is easier to write down using an exponent of r:

n × n × ... (r times) = n^r

Example: in the lock above, there are 10 numbers to choose from (0,1,..9) and you choose 3 of them:

10 × 10 × ... (3 times) = 10³ = 1,000 permutations

So, the formula is simply:

nr

where n is the number of things to choose from, and you choose r of them (Repetition allowed, order matters)

2. Permutations without Repetition

In this case, you have to reduce the number of available choices each time.

For example, what order could 16 pool balls be in? After choosing, say, number "14" you can't choose it again.

So, your first choice would have 16 possibilites, and your next choice would then have 15 possibilities, then 14, 13, etc. And the total permutations would be:

16 × 15 × 14 × 13 × ... = 20,922,789,888,000

But maybe you don't want to choose them all, just 3 of them, so that would be only:

16 × 15 × 14 = 3,360

In other words, there are 3,360 different ways that 3 pool balls could be selected out of 16 balls.

But how do we write that mathematically? Answer: we use the "factorial function"

	The factorial function (symbol: !) just means to multiply a series of descending natural numbers. Examples: 4! = 4 × 3 × 2 × 1 = 24 7! = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5,040 1! = 1
Note: it is generally agreed that 0! = 1. It may seem funny that multiplying no numbers together gets you 1, but it helps simplify a lot of equations.

So, if you wanted to select all of the billiard balls the permutations would be:

16! = 20,922,789,888,000

But if you wanted to select just 3, then you have to stop the multiplying after 14. How do you do that? There is a neat trick ... you divide by 13! ...

16 × 15 × 14 × 13 × 12 ...		= 16 × 15 × 14 = 3,360
13 × 12 ...

Do you see? 16! / 13! = 16 × 15 × 14

The formula is written:

where n is the number of things to choose from, and you choose r of them (No repetition, order matters)

Examples:

Our "order of 3 out of 16 pool balls example" would be:

16!	=	16!	=	20,922,789,888,000	= 3,360
(16-3)!		13!		6,227,020,800

(which is just the same as: 16 × 15 × 14 = 3,360)

How many ways can first and second place be awarded to 10 people?

10!	=	10!	=	3,628,800	= 90
(10-2)!		8!		40,320

(which is just the same as: 10 × 9 = 90)

Here's the link to the website:
http://www.mathsisfun.com/combinatorics/combinations-permutations.html