4th Prestige in Call of Duty: Black ops

 First on a side note. I changed the title of this blog to Programming and Math. This is mainly because of my final exams majority of my content has been about math related content in order to help me study. This content will be mostly about math for a while at least. Until i feel i can post more content like this. Now for the actual post.

So once again i Prestige'd in COD Black Ops. This prestige round this time. I will mostly be using an assault rifle, with extended mag. Any pistol with extended mag. For grenades frag and concussion. Claymore for fun. For perks, lightweight, hardened, hacker. my objective this prestige is to be as big of a meany as possible and shoot people through walls. Shooting people through walls is the most important part of this prestige hence the perk hardened. I have had so much fun today aiming at a wall and just shooting it. Managing to pull of triple kill head shots and such.

Not my favorite prestige emblem. To be honest in game it looks like nothing but a blob. but whatever now I have something to work towards. That is changing it with the 5th prestige emblem.

Rules of Inference and Basic Proofs

This is a small add-on to this post Propositional Logic and De Morgans Law Knowledge of this post will be required for understanding this post. I am posting this as a hope that i will remember it for my Discrete Math Final Exam.

A rule of inference: is a rule that can be used in proving a theorem or some logic directly and polished.

Addition:  is you just add another proposition onto it using the Logical OR.


p  = 'The sky is blue" 
p v q = "The sky is blue or red"
                
Simplification: is you just subtract a proposition

p ^ q = 2+2= 4 and grass is green
p = 2+2 = 4

Conjunction: adding two separate propositions together.

p = Grass is green.
q = Rome is in Italy
p^q = "Grass is green and Rome is in Italy".

Modus Ponens: Basically the end result of an implication.

P - > Q = "If I study, then i get an A".
Q = " I studied so i got an A"

Modus Tollens: Basically the opposite of Modus Ponens.

p - > q "If I go to the store, I buy milk"
~p = " I did not go to the store" Which means
~q = " I did not buy milk"

Hypothetical Syllogism: Say this out-loud to sound cool.Visually you can see how this works.

p -> q = "If i study, then i pass."
q - > r = " If i pass then i grad".
Therefore
p -> r = "If i study then i grad".

Resolution:

p v q = " I will study or eat dinner".
~p v r  "I will not study or go to the party"

Therefore
 q v r = " I will eat or go to the party"

Disjunctive Syllogism:

p v q = "I will study or I will eat"
~p = " I do not study"
q = "I will eat"


Basic Proofs:

Proofs usually involve implications.

So two ways of solving these can be

Contra-positive: p -> q this is equivalent to ~q - > ~p. So instead of proving p - > q. we test ~p - > ~q to see if this is true. If it is true then we know p - > q is true.

Example:

If n + 2 is odd, then n is odd.

Suppose that "if n is even, then n + 2 is even"

let n = 2k for some integer k. n + 2 = 2k + 2 = 2(k+1). Clearly 2(k+1) is divisible by two thus making it even. Therefore since the contra-positive is true then p -> q is true.

Contradiction: This is a little more difficult. We want to prove that. p - > q is true and p -> ~q is true.
If both of them is true we go with the original as being true. For clearly both q and ~q cannot be true.

Example:

If n + 2 is odd, then n is odd.

Assume that n is odd. Then n = 2L + 1 for some integer L. So n + 2 = 2L + 3. Which is odd.

Suppose that  If n + 2 is odd, then n is even.

Let n = 2k for some integer k. Then n + 2 = 2k + 2. which means n is even.

Since both is true, clearly if n + 2 is odd then n is odd is true.

Bam, Poison up in yo body!

If you have not heard this yet, or you have and your confused as to where it came from. It is a line that it said in a Sims commercial and it's actually kinda hilarious.

So i decided hey i enjoy hearing this guy talk in the commercial. The things he says is funny.So i decided I'm gonna make a really mediocre remix. So this is what i did. 

Partial Fractions

This technique is very appropriate for integrals when you have some mess of a fraction with no hopes of being able to integrate it nicely.

A partial fraction is just taking any fraction and breaking it up into smaller fractions that can later be added together to make the same original fraction.

The first thing you must do to determine if you can do te partial fraction technique is see if your equation is a candidate for it. To do this you check if the degree of the numerator is less than the denominator. If it is not you have yourself an improper fraction and will need to do long polynomial division. Else you can begin doing the partial fraction technique.

Example of an improper fraction

2x^3 + x^2 -10
-----------------  <----degrees are the same
 x^3-3x-2

Now to create your partial fraction:

***Important********

These are the forms your broken up fraction will either be. These are in a sense your formula.


         A                                                   Bx + C
   --------------                                ----------------------
     ( x-a)^k                                    (x^2 + bx + y)^k


Where a , b, y are some values. Where k is some exponent.

1. If the denominator contains linear fractions. Factor them out.

    5                        5
-----------  =   -----------
x^2+x-6        (x-2)(x+3)

2. Use the formulas and break up your equation appropriately. 

  5                        5                             A                                     B
-----------  =   -----------     =      ------------------   +      ---------------------
x^2+x-6        (x-2)(x+3)                   (x-2)                               (x+3)
       
                      

3. Multiple both sides by the left
 
5 = A(x+3) + B(x-2)

4. Takes the roots(the what the x's are) and plug them in to solve for A and B.

x = -3, 2

If x = -3 :    5 = A(0) + B(-5) =  -5B;   B = -1

If x = 2:      5 = A(5) + B(0) = 5A ;   A = 1

5. we have found our A and B. So time to fill in our partial fractions for A and B.

         1                                    -1
------------------   +      ---------------------
     (x-2)                               (x+3)

There you this looks like something you could integrate much easier. Enjoy.

C++ Final Exam Cheat Sheet

So last minute my computer science professor decides to email the class. That we are allowed to bring in a one sided cheat sheet. So seeing as my this is my final exam and is worth 42% of my mark in the class. I decided it might be a good idea to create such a sheet full of info for ease. Last time i heard the exam would only have C++ on it and will deal with mostly classes. So i took a look at all my "class" notes and came up with this cheat sheet/program that goes through everything i learned on classes this semester.

The program i created was rewritten extremely compressed and ugly to save space for future material i decide to put on it. At the moment though here is my cheat sheet.

Source Code
----------------------------------------------------------------------------------------------------------

#include <iostream>
using namespace std;

class study
{
public:
    study(){x = 0,y = 0; n++;}
    study(int a, int b);
    void print(void);
    void displayN(void)const;
    friend study change(study obj){ cin>>obj.x; cin>>obj.y; return obj;}
    friend int area(study);
    study& operator+(study);
    study& setX5(void){this->x = 5; return *this;}
    study& sety10(void){this->y = 10; return *this;}
    friend ostream& operator<<(ostream &output, study &obj){output<<"("<<obj.x<<","<<obj.y<<")"<<endl; return output;}//stream overloading friend
    ~study(){x = 0; y=0; if(n!=0){cout<<"Object deconstruction"<<" Objects remaining = "<<n--<<endl;}}
private://methods in here called utility function
    static int n;
    int x;
    int y;
};
int study::n = 0;

void study::print(void){cout<<x<<endl; cout<<y<<endl;}
void study::displayN(void)const{cout<<"The number of objects created is: "<<n<<endl;}
study::study(int a, int b){ n++; x = a; y = b;}

int area(study obj){ return obj.x*obj.y;}

study& study::operator+(study obj2){this->x = this->x + obj2.x; this->y = this->y + obj2.y; return *this;}

int main()
{
    study obj1;//x = 0 , y = 0
    study obj2(2,4);//x = 2 , y = 4
    const study obj3(5,4);
    study *obj4 = new study(5,5);
    obj1.print();//output: 0  endl 0
    obj2.print();//output: 2  endl 4
    cout<<"The area of obj2: "<<area(obj2)<<endl;//object from area function gets deconstructed
    obj1 = change(obj1);//x = 6 , y = 5
    obj1 = obj1.operator+(obj2);
    obj1.print();// 8, 9
    obj1 = obj1 + obj2;
    obj1.print();//10,13
    obj1.displayN();//3
    //obj3.print(); won't work because const objs can only use const methods
    obj1.setX5().sety10();//cascading member list
    obj1.print();
    cout<<obj1<<endl;//use of operator overloading of <<. Method has to be a friend. ostream for output. istream for input.
    delete obj4;
    return 0;
}

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

As you can see it is pretty ugly, but should really benefit me in terms of knowing the syntax of certain programing styles/commands.

The Area of bounded functions

The Area and Volume of bounded functions: 


This is not the most informative blog, but for those who have a rough idea of whats going on should be able to grasp and understand the examples shown.


Find the area between
y = 2 - x^2 and y = (1/2)x.

from x = 0  to x = 1 .









 Now as seen above we use integration to determine this. Basically you find the integral of the equation shown and solve for the domain of it. Which in this case is  x = 0 to x = 1.

Now the example jumps right to the answer being 17/12. The actual steps behind the seen is

= [ 2x - (x^3 / 3) - (x^2 / 4)] from x = 0 to x = 1

=[ (2(1) - (1^3 / 3) - ( 1^2 / 4) - (2(0) - (0^3)/3 - 0^2 / 4) ]

= [ 2 - 1/3 - 1/4 - 0 - 0 - 0]

= [ 2 - 1/3 - 1/4]

=  [ 6/3 - 1/3 - 1/4]  = [ 5/3 - 1/4] = [20/12 - 3/12] = 17/12 units squared. YAH!

So after these steps we have found the area between these two functions from x = 0 to x  = 1.


Find the area:

y = x^(1/3) , and y = x^3 from x = 0 to x = 2.

First thing we should notice about this is that they intersect at  x= 1. Therefore we need to split this up into two different integrals. One from x = 0 to 1, and the other from x = 1 to 2.







             

= [ 3/4 x ^(4/3) - 1/4 x ^4] from x = 0 to x = 1

= [ 3/4 (1) ^(4/3) - 1/4 (1) ^4 -  3/4 (0) ^(4/3) - 1/4 (0) ^4

= 3/4 - 1/4 - 0 - 0 = 3/4 - 1/4 = 2/4 = 1/2 YAH!



= [ 3/4 x ^(4/3) - 1/4 x ^4] from x = 1 to x =2

= [ 3/4 (2) ^(4/3) - 1/4 (2) ^4 -  3/4 (2) ^(4/3) - 1/4 (2)^4]

= 4 - 3/2(2)^1/3 - 1/4 - 3/4

= 4.5 - 1.5(2)^1/3

So add up the values of both these integrals to get

. 5 + 4.5 - 1.5(2)^1/3 = 3.11 approximate units squared.

This is how you find the area between two functions bounded.

Critical Numbers

Critical numbers also know as critical points are numbers that can determine what is happening in the function using derivatives. Now that was poorly explained but i will demonstrate the uses.

Lets  find the critical points of the function f(x) = 3x^5 - 20x^3.

Step 1:

First find the first deriviative of f(x)

d(f(x))/dx = 15x^4 - 60x^2

Step 2:

Now set the derivative to zero

f(x)' = 15x ^4 - 60x^2 = 0

15x^2(x^2-4) = 0

15x^2(x+2)(x-2) = 0

so upon inspection we see that x =  -2 , 0 , 2

These x's are ciritical numbers.

Plugging in these found critical points into f(x) our original function we get

f(-2) = 3(-2)^5 - 20(-2)^3 = 64
f(0) = 3(0)^5 - 20(0)^3 = 0
f(2) = 3(2)^5 - 20(2)^3 = -64

These numbers determine where the function has a local max or local min.

Definition:

Local Maximum: The point where the function has the high mountain or peak in the local area of the function. Being explained the point where y is equal to  a high point of the function or possibly the highest point.


Local Minimum: The point where the function has a low valley or canyon. where y is equal to a lower point on the graph in the local area.

Both definitions are terrible, but if you graph the function using graphing technology you will know what i am talking about.

  A local maximum and local minimum can be determined if the critical points found from the first derivative are plugged into the function. Such that a

local max can be found if f(x) > 0 . By plugging in the critical points of the first derivative.
local min can be found if f(x) < 0 . By plugging in the critical points of the first derivative.

Concavity

concavity is just a fancy way of saying does the function at a point start to turn into a mountain or into a  valley. Also known as concave up or down.

To find concavity can be just as easy as finding the local min and max. Finding concavity is also another way of finding the local min and max. So you can really kill two birds with one stone by finding the second derivative. First find the second derivative of f(x).

d^2(f(x))/dx^2 = d(15x^4 - 60x^2)/dx = 60x^3 - 120x

So now all you have to do is plug in the critical points from the first derivative.

f(x)'' = 60x^3 - 120x

f(-2)'' = 60(-2)^3 - 120(-2) = -240
f(0)'' = 0 (which tells us nothing, you'll see why)
f(2)'' = 240

Just like the same way of determining the local min and max. This determines both that and concavity.


local max can be found if f(x)'' < 0 . By plugging in the critical points of the first derivative.
local min can be found if f(x)'' > 0 . By plugging in the critical points of the first derivative.

concave down can be found if f(x)'' > 0 . By plugging in the critical points of the first derivative.
concave up can be found if f(x)'' < 0 . By plugging in the critical points of the first derivative.

We know that the function is concave up at x = 2 and has a local min.
We know that the function is concave down at x = -2 and has a local max.

As far as x  = 0, this does not work for our rule/theorem. 

Absolute extreme values

Not so extreme. In a function there has to be a maximum value and a minimum value of the function. If the function was horizontal on a graph, there would be a tie for y = constant value.

so far we know that

We know that the function is concave up at x = 2 and has a local min.
We know that the function is concave down at x = -2 and has a local max.

 Now to find the absolute max and min. We just need to know where the function is at it's highest and where it is at it's lowest point.

So by comparison of all our values of critical points. We can see that our local min and local max are also are absolute min and absolute max.

Easy right!


Endpoint values:

 This is pretty much the simplest of all simple. Lets say our function f(x) was continuous between the closed interval of  [ -5 , 5]. For the endpoint values all you do is plug in these numbers into your equation.

If  f(5) > f(-5)  then 5 is an endpoint max else 5 is an endpoint min and -5 is an endpoint max.

simple!

Point of Inflection

A point of inflection is a point such that the function mirror's itself almost. Such a function is x^2 where a point of inflection is (0,0). You can determine a point of inflection easily. To do this you need to find an x such that the second derivative f(x)'' = 0.

So for our second derivative. A point of inflection can be determined by x = 0.

For f(0)'' =60(0)^3 - 120(0) = 0.

A point of inflection is a point such that ( f(x)'' = 0 , f(x)) this is equivalent to ( 0 , f(0))

Where f(0) = 0.

So our point of inflection is (0, 0). Which from observation of graphing technology makes sense. For the function mirrors itself at that point.

Conclusion : 
Knowing these things can help you sketch a graph of your function or understand what your function is doing at certain points. Making them of importance.

Well hopefully this blog helped you understand Critical points. Also i hope this helped me study for my calculus final exam.