Saturday, 17 March 2012

Derivatives


Derivatives are all about change ...
... they show how fast something is changing (called the rate of change) at any point.
Let us get straight into an example: the function x2:
f(x) = x2
Here are some values:
 
xf(x)
00
0.50.25
11
1.52.25
24
2.56.25
As x increases, so does f(x) ... but how fast?
Can we find the slope (or rate of change) at any point?
With Derivatives we can!

Differences

First, let's calculate some differences.
The symbol for difference is the greek letter "delta": "Δ"
Example: if t goes from 3 to 3.1, then Δt=0.1
Here are some differences for x and f(x)
xΔxf(x)Δf(x)
0 0 
0.50.50.250.25
10.510.75
1.50.52.251.25
20.541.75
2.50.56.252.25
The table shows that, for example, between 1.5 and 2 f(x) grew by 1.75
I kept Δx at 0.5, but Δf(x) is getting larger and larger, meaning the rate of change is increasing.
But now we introduce the first of our important steps:

1. Make a Formula

Instead of long tables of numbers, we want a formula for calculating Δf(x).
What happens between x and x+Δx?
At x: At x+Δx:
f (x) = x2right arrowf (x+Δx) = (x + Δx)2
  
So, the difference is:  
Δf(x) = f (x + Δx) - f (x)
= (x + Δx)2 - x2

2. Simplify Formula

   
We can expand "(x + Δx)2": Δf(x) = x2 + 2x Δx + Δx2 - x2
   
And then simplify the formula: Δf(x) = 
Now we can calculate Δf(x) in one go:
Example: What is Δf(x) for x=1.5 and Δx=0.5 ?
Δf(x) = 2 x Δx + (Δx)2
= 2 × 1.5 × 0.5 + 0.5 × 0.5 = 1.5 + 0.25 = 1.75
That tells us how much f(x) changes, but not how fast. We still need to compare it to the change in x.
Example: if my weight increased by 1kg, was that fast or slow?
You can only answer if you know how long it took to increase. If it happened in 1 day, then it is fast, but if it took 1 year, it is slow.

3. Rate of Change

To work out how fast (called the rate of change) we divide by Δx:
Rate of change = Δf(x) / Δx
That tells us "how much f(x) changes for every change in x"
The Rate of Change can be seen as the slope of a line:
In our x2 example, this becomes:
Δf(x) / Δx = 
Example: What is the rate of change for Δf(x) when x=1.5 and Δx=0.5 ?
Rate of change = Δf(x) / Δx
= ( 2 x Δx + (Δx)² ) / Δx
= (2 × 1.5 × 0.5 + 0.52)/0.5
= 1.75/0.5
= 3.5
This tells us the slope of the line between x and (x+Δx).

But Not At A Point

 
But it doesn't yet tell us the slope at any specific point ...
... for example what is the slope at 1.5?
But we can find the slope at a point if we:
Reduce Δx towards 0. 
That will show us the slope just where 1.5 is.

However, Δx can't be 0, because dividing by 0 is wrong! ...
... but we can try getting closer and closer:
Δx = 0.5: Rate of change (we already calculated) = 3.5
Δx = 0.05: Rate of change = (2 × 1.5 × 0.05 + 0.052)/0.05 = 0.1525/0.05 = 3.05
Δx = 0.005: Rate of change = (2 × 1.5 × 0.005 + 0.0052)/0.005 = 3.005
We seem to be heading for Rate of change = 3

4. Use Limits to get Δx close to 0

In fact we just used the idea of limits to find the Rate of Change!
Because we weren't allowed to have Δx = 0, we tried approaching it closer and closer.
So let's try putting our Rate of Change formula into a limit, and see what we get:
The first thing we can do is simplify it, because Δx is at both the top and bottom:
Now, as Δx goes towards zero we have:
And that is it ... the rate of change (the slope of the line) is simply 2x
We have solved it! We can find the slope of the line at any point.
The slope at 1.5 is 2 × 1.5 = 3, ... the slope at 10 is 2 × 10 = 20, etc ...
And even better, we have done our first derivative!

Those Steps Again

So how did we do this again?
  1. We wrote a formula for Δf(x) = f(x+Δx) - f(x)
  2. We simplified that formula (to "2x Δx + Δx²")
  3. We divided that by Δx to get the rate of change
  4. We used limits to get Δx close to 0.

Definition of a Derivative

Let us now imagine that f(x) is any function, and follow the steps:

1. A formula for Δf(x)

Δf(x) = f(x+Δx) - f(x)

2. Simplify

(can't simplify in this case)

3. Divide by Δx to get the rate of change

Δf(x)/Δx = (f(x+Δx) - f(x))/Δx

4. Use limits to get Δx close to 0.

And that is the Defintion of a Derivative, for which we often use the little prime mark (') as shown.

Try It On Another Function

Let's try x3

A formula for Δf(x)

Δf(x) = f(x+Δx) - f(x) = (x+Δx)3 - x3

Simplify

(x+Δx)3 - x3 = x3 + 3x2Δx + 3xΔx2 + Δx3 - x3
= 3x2Δx + 3xΔx2 + Δx3

Divide by Δx to get the rate of change

Δf(x)/Δx = (3x2Δx + 3xΔx2 + Δx3)/Δx

Use limits to get Δx close to 0.

We can simplify it, then work out what happens when Δx goes towards 0:
So the derivative of x3 is 3x2

Other Notations

Sometimes the derivative is written like this:
 
We will explore that way of working on Derivatives as dy/dx

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clculus (limit evaluating)


Limits (Evaluating)

You should read Limits (An Introduction) first

Quick Summary of Limits

Sometimes you can't work something out directly ... but you can see what it should be as you get closer and closer!
For example:(x2-1)/(x-1)
  
At x=1:(12-1)/(1-1) = (1-1)/(1-1) = 0/0
But 0/0 is "indeterminate", meaning we can't determine its value. But instead of trying to work it out for x=1 let's try approaching it closer and closer:
x(x2-1)/(x-1)
0.51.50000
0.91.90000
0.991.99000
0.9991.99900
0.99991.99990
0.999991.99999
......
Now we can see that as x gets close to 1, then (x2-1)/(x-1) gets close to 2
We are now faced with an interesting situation:
  • When x=1 we don't know the answer (it is indeterminate)
  • But we can see that it is going to be 2
We want to give the answer "2" but can't, so instead mathematicians say exactly what is going on by using the special word "limit"
The limit of (x2-1)/(x-1) as x approaches 1 is 2
And it is written in symbols as:
So it is a special way of saying, "ignoring what happens when you get there, but as you get closer and closer the answer gets closer and closer to 2"
As a graph it looks like this:
So, in truth, you cannot say what the value at x=1 is.
But you can say that as you approach 1, the limit is 2.
 

Evaluating Limits

"Evaluating" means to find the value of (think e-"value"-ating)
In the example above we said the limit was 2 because it looked like it was going to be. But that is not really good enough!
In fact there are many ways to get an accurate answer. Let's look at some:

1. Just Put The Value In

The first thing to try is just putting the value of the limit in, and see if it works (in other wordssubstitution).
Let's try some examples:
Example Substitute ValueWorks?
(1-1)/(1-1) = 0/0
    
10/2 = 5
It didn't work with the first one (we knew that!), but the second example gave us a quick and easy answer.

2. Factors

You can try factoring.
Example:
  
By factoring (x2-1) into (x-1)(x+1) we get:
 
 
  
Now we can just substitiute x=1 to get the limit:
 

3. Conjugate

If it's a fraction, then multiplying top and bottom by a conjugate might help.
The conjugate is where you change the sign in the middle of 2 terms like this:
Here is an example where it will help you to find a limit:
Evaluating this at x=4 gives 0/0, which is not a good answer!
So, let's try some rearranging:
Multiply top and bottom by the conjugate of the top: 
   
Simplify top using : 
   
Simplify top further: 
   
Eliminate (4-x) from top and bottom: 
So, now we have:
Done!

4. Infinite Limits and Rational Functions

A Rational Function is one that is the ratio of two polynomials: 
   
For example, here P(x)=x3+2x-1, and Q(x)=6x2: 
By finding the overall Degree of the Function we can find out whether the function's limit is 0, Infinity, -Infinity, or easily calculated from the coefficients.
Read more at Limits To Infinity.

5. Formal Method

The formal method sets about proving that you can get as close as you want to the answer by making "x" close to "a".
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calculus (limit)


Limits (An Introduction)

Approaching

Sometimes you can't work something out directly ... but you can see what it should be as you get closer and closer!
Let's use this function as an example:
(x2-1)/(x-1)
And let's work it out for x=1:
(12-1)/(1-1) = (1-1)/(1-1) = 0/0
Now 0/0 is a difficulty! We don't really know the value of 0/0, so we need another way of answering this.
So instead of trying to work it out for x=1 let's try approaching it closer and closer:
x(x2-1)/(x-1)
0.51.50000
0.91.90000
0.991.99000
0.9991.99900
0.99991.99990
0.999991.99999
......
Now we can see that as x gets close to 1, then (x2-1)/(x-1) gets close to 2
We are now faced with an interesting situation:
  • When x=1 we don't know the answer (it is indeterminate)
  • But we can see that it is going to be 2
We want to give the answer "2" but can't, so instead mathematicians say exactly what is going on by using the special word "limit"
The limit of (x2-1)/(x-1) as x approaches 1 is 2
And it is written in symbols as:
So it is a special way of saying, "ignoring what happens when you get there, but as you get closer and closer the answer gets closer and closer to 2"
As a graph it looks like this:
So, in truth, you cannot say what the value at x=1 is.
But you can say that as you approach 1, the limit is 2.
 

Test Both Sides!

 
It is like running up a hill and then finding the path is magically "not there"...
... but if you only check one side, who knows what happens?
So you need to test it from both directions to be sure where it "should be"!
So, let's try from the other side:
x(x2-1)/(x-1)
1.52.50000
1.12.10000
1.012.01000
1.0012.00100
1.00012.00010
1.000012.00001
......
Also heading for 2, so that's OK

When it is different from different sides

What if we have a function "f(x)" with a "break" in it like this:
 
This is a function where the limit does not exist at "a" ... !
You can't say what it is, because there are two competing answers:
  • 3.8 from the left, and
  • 1.3 from the right
But you can use the special "-" or "+" signs (as shown) to define one sided limits:
  • the left-hand limit (-) is 3.8
  • the right-hand limit (+) is 1.3
And the ordinary limit "does not exist"

Are limits only for difficult functions?

Limits can be used even if you know the value when you get there! Nobody said they are only for difficult functions.
For example:
We know perfectly well that 10/2 = 5, but limits can still be used (if you want!)

Approaching Infinity

Infinity is a very special idea. We know we can't reach it, but we can still try to work out the value of functions that have infinity in them.

Let's start with an interesting example.

Question: What is the value of 1/∞ ?

Answer: We don't know!

Why don't We know?

The simplest reason is that Infinity is not a number, it is an idea. So 1/∞ is a bit like saying 1/beauty or 1/tall.
Maybe we could say that 1/∞ = 0, ... but that is a problem too, because if we divide 1 into infinite pieces and they end up 0 each, what happened to the 1?
In fact 1/∞ is known to be undefined.

But We Can Approach It!

So instead of trying to work it out for infinity (because we can't get a sensible answer), let's try larger and larger values of x:
x1/x
11.00000
20.50000
40.25000
100.10000
1000.01000
1,0000.00100
10,0000.00010
 
Now we can see that as x gets larger, 1/x tends towards 0
We are now faced with an interesting situation:
  • We can't say what happens when x gets to infinity
  • But we can see that 1/x is going towards 0
We want to give the answer "0" but can't, so instead mathematicians say exactly what is going on by using the special word "limit"
The limit of 1/x as x approaches Infinity is 0
And write it like this:
In other words:
As x approaches infinity, then 1/x approaches 0

When you see "limit", think "approaching"

It is a mathematical way of saying "we are not talking about when x=∞, but we know as x gets bigger, the answer gets closer and closer to 0".
Read more at Limits to Infinity.

Solving!

We have been a little lazy so far, and just said that a limit equals some value because it looked like it was going to.
That is not really good enough!
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Wednesday, 29 February 2012

COMPLEX NUMBER

complex number = real number  + real number X imagine number
the two type of number that real and imaginary number make a complex number
complex number is just a combination of thi two number
so to understand to complex number we have to know what is imaginary number
imaginary number is a number that we cant found our real word but it's exist we have a proof
a^2 -1=0
in this equation there is no real soluation in real world (to solve this equation we must have a negative number for a^2) ,
 if we go out from the real world or real number line we will find a solution and this soluatin is called imagine soluation
so what happend with a^2 in imaginary number or imaginary world
if a is positive then result come positive for a^2   ....(1)
if a is negetive then also result come positive for a^2     .....(2)
there is a twist in imaginary number we get negative result  for (1) and (2)
its actually mean there is another one sign that are not positive and not negative
for more clearly there is two side of a coin head and tail but in imaginary there is one more side .but this side is not exist in our real world, it only exist in imaginary world.
but we did not what is that sign .what it look like .we often call it i. i^2=a negative number or normaly -1


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