Intermediate Value Theorem

 Intermediate Value Theorem

If f is continuous on the closed interval [ a , b] and K is any number between f(a) and f(b), then there is at least one number c in the interval such that f(c) = K.

You can use this theorem to determine if a polynomial has at least one real root.

To begin using this we first need to know if

f(a) < 0 < f(b)    or  f(b) < 0 <  f(a)

If either of those cases are true. We know that there is a c in the closed interval such that f(c) = 0.

As an example:

we set f(x) = x^2 - 2. We will now test the function for some closed interval. Lets say [1 , 2]. Then lets test these numbers 1 and 2.

f(1) = -1 < 0 and f(2) = 2 > 0.

Therefore we know by the theorem that there exists a number c in the interval [1 , 2] such that f(c) = 0.

Since f increases on [1 , 2], there is only one such number. This number we can see is the square root of 2.

April Fools Day and my 100th Follower

First of all i would like to mention it is April Fools Day.So to start off this morning for my house. I decided to pull the fire alarm, run into my significant others rooms. Scream Fire at the top of my lungs. Get them running outside half nude. To realize that it's ridiculously cold outside and that i just trolled them. So have fun trolls, play safe.

Also  on a side note Donal is my 100th Follower. His blog is http://songsyoucandanceto.blogspot.com/
At the moment he only has two posts. But damn are they ever songs you can dance to. 

Mathematical Induction

Mathematical induction is a method of mathematical proof typically used to establish that a given statement is true of all natural numbers (non-negative integers). It is done by proving that the first statement in the infinite sequence of statements is true, and then proving that if any one statement in the infinite sequence of statements is true, then so is the next one.
- Stolen from http://en.wikipedia.org/wiki/Mathematical_induction

In my own words. Mathematical Induction is the concept of having a ladder that goes on to infinite. If you know that you can reach the first rung. Then odds are you are able to reach the next rung. If you can reach the "n-th rung", then you can touch the "n+1-th rung". 

I will show how this can prove various equations to show that they are true for all values of your variable.

Ex.

We will prove by Mathematical Induction that the equation 

1+2+...+n = [n(n+1)]/2   is true for all n > 0.

So to start. Let p(n) be the propostion => 1+2+...+n=[n(n+1)]/2

Our objective is to show that p(1) is true(the first value that n can be), and that the conditional statement p(k) implies p(k+1) also denoted as p(k) -> p(k+1), is true for all k > 0.

So we first need to show that the proposition p(n) is true for the very first value of n. In this case 1.
This step is called the Basis step. 

Basis Step: verifying that p(1) is true. As seen by inspection p(1) is true for

1 = [1(1+1)]/2 = 1(2)/2 = 2/2 = 1

Since the basis step is true we move onto the next step called the Inductive step. This is where all the nitty gritty algebra comes into play.

Inductive step: In this step, we assume that p(k) holds for an arbitrary positive integer k. That is we assume that 

1+2+...+k = k(k+1)/2

under this assumption, it must be shown that p(k+1) is true.

To show that p(k+1) is true we first must find what p(k+1) is. 

p(k+1) => 1+2+...+(k+1) = ((k+1)[(k+1)+1]) / 2 = [(k+1)(k+2) ]/2

(k+1)(k+2) / 2 is the equation that we will need to show for our next step . The next step we call the Inductive Hypothesis.

Inductive Hypothesis: This was the hypothesis that if any value of n is true then n+1 must also be true. To show this we write the proposition p(k) as the following.

1+2+...+k+(k+1) = k(k+1)/2 + (k+1)             (we added (k+1) to both sides of this equation)
                                                                     
The idea in this step is to show that the left side can equal the same answer we determined in the Inductive step. 

So as the following working with the left side of the equation.

=   k(k+1)/2  + (k+1) 
=  [  k(k+1) + 2(k+1) ] / 2   (By adding the fractions, the 2 came from the base difference of 1 and 2)


=  [k^2 + k + 2k + 2 ] / 2  (Textbook feels it's unnecessary to skip this step, i find it confusing without)


= [ k^2 + 3k +2 ] / 2 ( By factoring we get)

= [ (k+1)(k+2) ] / 2  (which hopefully looks familiar)

The answer is the same as the Inductive step. Thus proving the hypothesis correct.

So by Mathematical Induction we know that p(n) is true for all positive integers n. That is we have proven that

1+2+...+n = [ n(n+1) ] / 2   for all n>0.   

My Halo 3 Dueltage

This a montage that features myself(Annihilate)spelled(AnnihiIate) and my former MLG team mate Beware.
I hope you enjoy.

Limits and Continuity

Limits are fundamental for both differential and integral calculus. The formal definition of a derivative involves a limit as does the definition of a definite integral.

The limit of a function(if it exists) for some x-value, a is the height  the function gets closer and closer to as x gets closer and closer to a from the left and the right. Lets take a look at what that even means.

The limit of the function f(x) = 3x + 1 as x approaches 2. When writing a limit we denote this as the following equation.

lim f(x) = lim 3x+1.
x->2         x->2

Now for such an easy limit we can plug in the number 2 and solve this limit. Plugging in the 2 we see.


lim f(x) = lim 3(2)+1  = 7
x->2         x->2


 That the limit is 7. So as x approaches 2 from the left and right side. The limit of this function is 7.

 Now your probably think well that's silly. Any function that's continuous you just plug in and solve for x. Now limits are important for functions with holes.

Definition of a Limit:

lim f(x) exists if and only if

x -> a

1. lim f(x) exists
     x->a- (a from the left side)

2. lim f(x) exists, and
    x->a+(a from the right side)

3. lim f(x) = lim f(x)
    x -> a-      x->a+

Number three is the most important.


A function with an infinite limit and vertical asymptotes.

f(x)  = (x+2)(x-5) / (x-3)(x+1) This function has a VA at x = 3, x = -1

So the lim f(x) = lim (3+2)(3-5) / (3-3)(3+1) =  infinity as x will never reach 0 thus it heads closer to 0
           x-> 3-      x->3-

for infinity.

From the right side the limit will be negative infinity. Thus we know by the limit definition that the limit does not exist for x approaches 3.


Definition of Continuity:

A function is continuous at a point x  = a if the following three conditions are satsified.

1.f(a) is defined
2.limf(x) exists and,
    x-> a
3. f(a) = lim f(x)
              x->a

Limits to Memorize:

lim c = c
x->a

lim 1/x = infinity
x-> 0+

lim 1/x = negative infinity
x-> 0-

lim 1/x = 0
x -> infinity(same for both - and + infinity)

lim sin x/x = 1
x->0

lim cos x - 1/x =0
x->0

lim (1+1/x)^x = e
x -> infinity


Plugging and Chugging Limits:

Any limit where it's continuous and straight forward.

lim (x^2 - 10) = -1
x-> 3

lim 10/x-5= 10/0  if you get an answer that has any numerator thats not zero divided by zero. The limit
x - > 5

does not exist.

Now in the below case if you get 0/0 you have a real limit problem.

lim (x^2 - 25) / (x - 5) = 0/0 which is undefined
x-> 5

undefined is not the limit. This is not the answer we are looking for. Four things we can do to find a limit that is undefined. Calculator, algebra, limit sandwhich, L'Hopitals rule.

I will only show the algebra and L'Hopitals rule. I will not explain L'Hopitals Rule though.

Using algebra to solve for 
lim (x^2-25)/(x-5) we can rewrite this by factoring (x^2 - 5) = (x-5)(x+5) knowing this we can
x-> 5

divide out the (x-5) which leaves us with the equation lim x+5 = 10
                                                                                       x->5

Using L'Hoptials Rule

lim (x^2-25)/(x-5) = lim d/dx(x^2 - 25) / d/dx(x-5) = lim 2x = 10.
x-> 5                        x->5                                             x->5




      





             

Boolean Operations

In computer science a Boolean is a primitive data type. This data type consists of whether something is True or False. In this blog 1 or 0.
//section (11.1) personal note

Definitions:
                                                                                                        
The Boolean compliment: 
                                                                 __   __
The negation of  a Boolean. Denoted as  1 or 0.  Spoken as 1(bar) and 0(bar).

 0(bar) = 1
1(bar) = 0

Boolean Addition:

This works exactly like the logical operator OR.

0+0 = 0
0+1= 1
1+0= 1
1+1 =1

Boolean Multiplication:

This works exactly like the logical operator AND.

0*0 = 0
0*1 = 0
1*0 = 0
1*1 = 1

As you can see zeros dominate in any multiplication

Example:
__
 1  *(0+1) = 0 * 1 = 0
(1+1)+(1+0) = 1 + 1 = 1

Silly itunes

Ever have one of those itune moments where after closing your itune's. The silly program loads back up. So what do you do. You decide to close it because your done with it. But now it becomes even  more persistent and reloads more rapidly. If anyone has experienced this problem with itunes. You know how ridiculously annoying this is. For those who have not had any experience with this problem. You may read, or choose not to. But for those who have this problem. This is a must read blog.

My hypothesis to this problem. Is that the process may have been initially closed, but a linked process related to the itunes process is still running. Then it realizes hey the itunes process was terminated. But i need it to keep running. So i think i'm going to reload it, to make both myself and the itunes process happy.

From my experience with itunes i noticed one of two things. The process itunes.exe, and ituneshelper.exe are closely related to my hypothesis. It seems that these two processes keep wanting to start up if at least one of them is running. Now you can open up task manager and close the processes right from there. Now i have done this with some success, but not always. It seems to be a hit and miss for myself. So i decided well there has to be a solution or atleast a quick fix. So i don't have to keep closing it every 30 seconds it decides to pop up again.

The solution i call a batch file i named SillyItunes. This batch file forcefully ends the processes from the command line and from my experience with this batch that i just wrote. It stops the random reload of itunes to keep popping up with no hope of ever ending it's cycle.

so copy the below source code into a text file. Then save the text file as a .bat instead of a .txt.

Source Code:

-----------------------------------------------------------------------------------------



@ echo off
echo "SillyItunes" a program made by DisposedCheese
echo.
echo.
taskkill /F /T /IM  itunes.exe
taskkill /F /IM ituneshelper.exe
echo.
echo.
echo I hope that solved the problem
echo.
set /p M="Press any key to continue..."


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Suggestions can be left as a comment. If you think this program is useless. Well so do i. But i still find it to be an easy solution to an annoying problem. A very handy file to keep on the desktop for your itunes mishaps.


Also i am not responsible if your computer takes a random dump on you after using said program.