overidon.com Central Database for Overidon Omnimedia

February 3, 2017

Math Formulas

Filed under: Tips and Tricks — Tyler @ 10:43 pm
Math Formulas

Math Formulas

Sinusoudal Functions:


Cosine Functions

y == a*cos(bx+c)+d

The Period of this function would be:

( 2π / b )


FEATURES OF A Trapezoid:

Formulas of a Trapezoid:

Area of a Trapezoid:

A = (1/2) * (b + c) * (h)

A = Area
b = Base 1 of Trapezoid
c = Base 2 of Trapezoid
h = height of Trapezoid

Features of a Circle: 

Standard form of a circle:


(x-a)2 + (y-a)2 = r2

If the circle is in expanded form and there are a lot of x-sqared and also x- try to factor. Remember the standard form is equal to r-squared so if you find the equation equal to zero then you might need to backtrack and do some investigating.


Euler’s Formula: 

eix = cos*x + i*sin*x


To determine the determinant in a 2 x 2 matrix, we must multiply the diagonals of the matrix and then subtract those products.

[2  4]

[1 -5]     … would be: -10 – (4)  or -14.


To determine the Adjugate of a 2 x 2 matrix, we must first flip the first item in the first column with the 2nd item in the 2nd column. Then we have to take the negative of the first item in the 2nd column and the negative of the 2nd item in the 1st column.

[ 4  15]

[ -3  11]    … the adjugate of this would be….


[11  -15]

[3      4]  ;


The find the inverse matrix of a 2 x 2 matrix, we must multiply the reciprocal of the determinant by the Adjugate.

(1 / Determinant) * (Adjugate);


Vectors are primarily subjects of a direction and a magnitude.


A unit vector is a vector with a magnitude of 1.

Therefore even though it can have any direction, the “length” or magnitude of that vector can only be one.

In order to find the unit vector, first we need to determine the magnitude. To do that we must first find out what the “length” of the vector is.

Then, divide the vector components by the magnitude.

[ (x / ||m|| )  , (y / ||m||) ]  == â;


Formula for an ellipse:

[(x2) / (a2)] + [(y2) / (b2)] = 1

NOTE: x is the center x coordinate for the center of the ellipse. y is the center y coordinate of the ellipse.
If there are modifiers such as (x-1)2 that would mean to move the center point accordingly which would bring the “x” value to zero.

As for the “a” and “b” aspects of the formula:
The “a” is the x consideration for the ellipse. To determine the size of the x “semi-axis” one must take the square-root of the “a” value etc.

Determine the Foci length of an Ellipse

If a > b:

f = sqrt( a2 – b2 )


Formula for a Hyperbola:

[(x2) / (a2)] – [(y2) / (b2)] = 1;

The asymptotes are defined by the formula:

y ± (b2/a2) * x;

This appears to be the asymptotes for a “left” / “right” opening hyperbola. But I will verify this.

NOTE: This is similar to the formula for an Ellipse, but there is a subtraction in the middle.
One thing to note is that in the case of the “x” system being subtracted by the “y” system, the Hyperbola will “open” to the left and to the right.

This is important because a system such as this:

[(y2) / (b2)] – [(x2) / (a2)] = 1

… is a hyperbola as well. In this case the hyperbola will open “up” and “down”


Factoring Polynomials: 


(x + a) (x + b) == x^2 + (ab)*x + ab;

Dividing Numbers of the Same Base with Exponents: 

(xa) / (xb) == xa-b
Raising Numbers that already have exponents to an Exponent: 

xaj == xa * j


(+) sine
(+) cosine
(+) tangent
(+) cosecant
(+) secant
(+) cotangent


(+) sine
(-) cosine
(-) tangent
(+) cosecant
(-) secant
(-) cotangent


(-) sine
(-) cosine
(+) tangent
(-) cosecant
(-) secant
(+) cotangent


(-) sine
(+) cosine
(-) tangent
(-) cosecant
(+) secant
(-) cotangent



Law of Cosines

c2 = a2 + b2 – 2*a*b*cos(theta);

In this scenario “c^2” is referring to the length of a triangle “side” which is opposite to the angle “theta”;

Law of Sines

[(sin(A)) / a] == [(sin(B)) / b] == [(sin(C)) / c]


Trig Identities

Pi Subtraction Identities (Supplements):

sin(Θ) == sin(π-Θ);

-cos(Θ) == cos(π-Θ);

-tan(Θ) == tan(π-Θ);

Pi Addition Identities (Working Periods):

sin(Θ) = sin(Θ + 2π)

cos(Θ) = cos(Θ + 2π)

tan(Θ) = sin(Θ + π)


(sin Θ)2 + (cos Θ)2 = 1

Other Sum and Difference Identities (Ptolemy):

sin(x + y) == sin(x) * cos(y)+cos(x) * sin(y)

sin(x – y) == sin(x) * cos(y) – cos(x) * sin(y)

cos(x + y) == cos(x) * cos(y) – sin(x) * sin(y)

cos(x – y) == cos(x) * cos(y) + sin(x) * sin(y)

Other Identities Relating to tan(Θ) :

tan(α – β) = ( tan(α) – tan(β) ) / (1 +  tan(α) * tan(β) )

tan(α + β) = ( tan(α) + tan(β) ) / (1 –  tan(α) * tan(β) )

tan(2 * Θ) = ( 2 * tan(Θ) ) / ( 1 – tan(Θ))

Double Up Combo Identity:

sin(2*Θ) == 2*sin(Θ)*cos(Θ)

cos(2*Θ) == 2*cos(Θ)2-1


Special Angles and their Unit Circle Data

note = For the degree measurements, your graphing calculator or cal needs to be set to degree or regular depending on the calculator.

For the radians or (pi) based measurements, you need to have your graphing calculator set to radians.

If you use radians when in degree mode, you can get different results after using cosine etc. on angles.

Zeroth Quadrant – QUADRANT 0 (ZERO) – [ALONG x – AXIS]

0° or 0π —> cos(0° or 0π) == 0 , sin(0° or 0π) == 1;


30° or (π/6) —> cos(30° or (π/6)) == (sqrt(3)/2) , sin(30° or (π/6)) == 1/2 , tan(30° or (π/6)) == (sqrt(3) / 3 );

45° or (π/4) —> cos(45° or (π/4)) == (sqrt(2)/2) , sin(45° or (π/4)) == (sqrt(2)/2) , tan(30° or (π/4)) == (1);

60° or (π/3) —> cos(60° or (π/3)) == (1/2) , sin(60° or (π/3)) == (sqrt(3)/2), tan(60° or (π/3)) == (sqrt(3)) ;

Zeroth Quadrant – QUADRANT 0 (ZERO) – [ALONG y – AXIS]

90° or (π/2) —> cos(90° or (π/2)) == 0 , sin(90° or (π/2)) == 1;


120° or (2π/3) —> cos(120° or (2π/3)) == (-1/2) , sin(120° or (2π/3)) == (sqrt(3)/2);

135° or (3π/4) —> cos(135° or (3π/4)) == (-sqrt(2)/2) , sin(135° or (3π/4)) == (sqrt(2)/2);

150° or (5π/6) —> cos(150° or (5π/6)) == (-sqrt(3)/2) , sin(150° or (5π/6)) == 1/2;

Zeroth Quadrant – QUADRANT 0 (ZERO) – [ALONG x – AXIS]

180° or (π) —> cos(180° or (π)) == -1 , sin(180° or (π)) == 0;


210° or (7π/6) —> cos(210° or (7π/6)) == (-sqrt(3)/2) , sin(210° or (7π/6)) == -1/2;

225° or (5π/4) —> cos(225° or (5π/4)) == (-sqrt(2)/2) , sin(225° or (5π/4)) == (-sqrt(2)/2);

240° or (4π/3) —> cos(240° or (4π/3)) ==  (-1/2), sin(240° or (4π/3)) == (-sqrt(3)/2);

Zeroth Quadrant – QUADRANT 0 (ZERO) – [ALONG y – AXIS]

270° or (3π/2) —> cos(270° or (3π/2)) == 0 , sin(270° or (3π/2)) == -1;


300° or (5π/3) —> cos(300° or (5π/3)) == 1/2 , sin(300° or (5π/3)) == (-sqrt(3)/2);

315° or (7π/4) —> cos(315° or (7π/4)) == (sqrt(2)/2) , sin(315° or (7π/4)) == (-sqrt(2)/2);

330° or (11π/6) —> cos(330° or (11π/6)) ==  (sqrt(3)/2), sin(330° or (11π/6)) == -1/2;


Parabolic Formulas:

NOTE: When in factored form: Remember, the vertex of a parabola is located on the axis of symmetry of the parabola. So if you know the x intercepts for the parabola, then the x coordinate exactly at the midpoint is the x coordinate for the vertex.

Then, to find the y-coordinate for the vertex of the parabola, we must input the “now known” x-coordinate back into the equation and solve for “y”.

FOCUS and Directorix

The Focus coordinates are (a,b) representing the point that is inside the parabola which is equidistant to all the other points on the parabola in the same way that the directorix line is equidistant to the points along the parabola.

The Directorix is the constant – “k” which is a line.


y = (1/2*(b-k)) * (x-a)2 + (1/2) * (b+k)

Base Vertex Form:

y = a(x-h)2 + k

vs: Standard-

y = a*x2 + bx + c

To determine the x coordinate of the parabola vertex, use (-b / 2a) then reinput that result into the formula to find the y coordinate of the vertex.

vertex h=> standard (-b / 2a) => Reinput into vertex formula as necessary


Standard Parabolic Form

This represents the most primative parabola possible. A simple squaring of the x coordinates.

y = x2

Detailed Standard Parabolic (Quadratic) Form

This represents that the details of the parabola can be affected by secondary terms and coefficients.
For example, a negative “a” coefficient can make the parabola “point” in a different direction.

y = a*x2 + bx + c


Finite Series ( Arithmetic):

Sn = [(S0 + Sn) / 2] * n


Finite Series (Geometric and Exponential Patterns)


Finite Geometric Series – Sn

Sn = (a(1-r^n)) / 1 – r

Sn == Evaluated Finite Geometric Series to the “nth” number of  terms or items

a == The first term in the Series

r == the common ratio which everything is acted on by

n == This is the total number of terms – NOTE: This is an exponent in the Geometric Series Formula


∑ a*rk

In sigma notation the top number is the final term.

The bottom number k is assigned to  the first term to be used.

NOTE: If the first term is a negative or a zero, take this into account when plugging “n” into the formula

Honestly, sigma notation can be confusing because “k” and “n” are directly related to each other but, they aren’t the same.

What we really want to do is plug “n” into the first Series Formula. But a lot of times, we have “k” instead. So when the bottom of the sigma says, “k = 0” it is really saying, “if k = 0, then the first term is multiplied by 1 in the iteration. This makes sense because anything to the exponent of 0 is one.

Sometimes there are algebraic instructions in the top and/or bottom. These are to help describe the iteration process.

The information to the right of the sigma is important.

Everything except for the exponent is part of the common ratio or “r”.

The exponent is the true “n” from the previous formula.





Quadratic Equations:


Quadratic Formula:


x =   ( -b ±  [#sqrt] (b^2 – 4ac) ) / 2a


Standard Form for Quadratic Equations:


ax^2 + bx + c = 0


PROPERTIES OF LOG (logarithm properties):

Basic property -> converting log to exponent

logab = x

is equivalent to:

a^x == b;


Evaluate Properties –> adding log terms to each other

( logca ) + ( logcb )  == ( logca * b )


Evaluate Properties –> subtracting log terms to each other

( logca ) – ( logcb )  == ( logca / b )


Multiplying Log -> adding additional exponent to log expression

A * logbc == logb(c^A)

Special Property – Evaluating non-base 10 Log using division

logba = ( logca ) / ( logcb ) ;


Complex Number Theory:

Dealing with Modulus and Arguments of Complex Number based – “Polar Coordinates” etc.

If zn = i ……

Modulus of zn relates to rn in the form of:

zn == rn[cos(n * theta) + i * sin( n * theta) ];

This next part is interesting:

If zn == i

then the next statement must be true:

n * theta == 90 degrees ( or pi/2 radians) + k * 360 degrees;

This may seem strange but since “i” is not just a variable, it’s also a number. Specifically, “i” is 1i on the imaginary axis.

Also, it is exactly “on” the imaginary axis, so it creates an angle of 90 degrees.

If it was -1i or “negative 1 i” then it would be 270 degrees.

Factors of (i) – complex number theory:


i^1 = i

i^2 = -1

i^3 = -i

i^4 = 1

i ^5 = i

…continue rotation…



Remainder Formula for Polynomials

Let’s say there is a polynomial x^3 + 2x^2 + 17x + 8

and we need to divide that polynomial by x-9

There is a shortcut where the remainder can be found by substituting (x) for the opposite of -9.

So we would input +9 or 9 into the polynomial for x and evaluate.

If the evaluation ends up as 0. Then we know that x-9 is actually a factor for that polynomial. And that there is no remainder.

If the evaluation ends up with a number or another binomial or polynomial etc, then we have that as the remainder.


Standard Form:

Ax + By = C


Point Slope Form:

HERE is an Evaluator program for point Slope Form

(y – y{1}) = m(x – x{1})

KEY: y{1} means the y location of a specific point. (I’m using {} in this instance not to show a set but to denote a base.)

x{1} is talking about the same point of (x{1}, y{1})

y refers to another point which has the x location of x.

To think about this one, we are simply making the rise/run of the line very clear. Point Slope Form makes it really obvious that an average of two points is happening.







August 14, 2012

Air Dusting Floor Fans

Filed under: Tips and Tricks — Tyler @ 10:08 pm

If you air-dust your floor fans using that electronics dusters…it can make them less hot and also improve the air quality of a room

If you live in an environment where you use floor fans, you may have noticed the blades of the fans can get very dusty. Depending on how many windows you have and how dusty your room is…floor fans can get dusty enough to actually hamper their operation. Fans which are extremely dusty can even get hot…which almost completely negates the purpose of using fans to a cool a room. This coupled with less efficiency and noise, makes quickly cleaning your floor fans a priority.

Fortunately, there is an easy solution for this problem. Common air dusters which are sold at electronics store around the country are great for cleaning these fans. I just unplug my fans and then bring them outside by the garage. Then I make sure to air dust the fans to get all the dust out of the electronics and off the grates and blades. This makes the fans less noisy, they generate less heat and they also aren’t just moving dust around.

Where does all this dust come from anyway?

Even though much dust comes from when we track in air from the outside into our work/living spaces…most dust actually comes from our own bodies! That’s right. Human beings and their pets generate a tremendous amount of dust through the shedding of dead skin cells. This dust is actually very fine and it is dangerous to electronics and can make you feel tired if you inhale it all the time.

Another recommendation would be to have your computer devices off the ground. Many people put their desktop computers on the ground but this is a bad idea…regardless of what kind of flooring you have. If you have carpet, it is definitely the worst, but hard-wood floors are bad for a computer as well. Dust loves getting sucked into the fans of desktop computers which are on the ground. So a quick tip would be to get a flat screen monitor and put your computer on a stable desk.

I hope this was a fun Tuesday tip!


August 6, 2012

Making Daily Progress with Your Art

Filed under: Tips and Tricks — Tyler @ 12:06 pm
Daily Progress with your art

Making daily progress with your art can increase your confidence as well as your artistic discipline.

Making daily progress with your art is a huge way to achieve goals and add energy to your process. As the son of a chemist and a fine artist, I find that my ideas tend to pull me around and they want to keep me in the “limbo zone of what-if.” What if’s and wouldn’t it be cool if are great but they should never take away from your actual creative process. But left unchecked, these ideas can become the bulk of the energy absorption for your work. A simple and effective way to combat distracting “genius” ideas is to make daily progress on projects that are at least 10% possible for you to complete in your lifetime.

An important technique is to find out your own personality type. I have the personality where I like to start projects and I also like to complete them. But there is a certain “rush” when a new project is in the beginning stages. It is an almost romantic stage in the process where anything is possible and you can almost taste victory without even typing a single word or drawing a single line. Some of us in the writing world call it “brainstorming” because it definitely feels like a mental synaptic crackling of energy when new ideas get to form.

Since I can identify that I have a personality type that likes the “sizzle” of working on a new project, I create an environment which caters to that predisposition. This environment is created by having several concrete projects that have very real goals and yet require different and interesting disciplines of art in order to complete those projects.

One project for example is to finish the “Javascript” lessons in Codecademy.com. Once I finish this branch of programming tutorials, I’ll be mentally prepared to work on “Actionscript” which is a similar programming language which is used to make animated flash movies and flash games.

Another project is a animated music video which is much longer than my last video, “Frazzle Msnaz” – this video has been taking forever. But working on it definitely is fun. It takes much more energy to work on this video than it takes to go through Codecademy lessons. That’s because it’s always easier to jog through a trail that someone else has carved, rather than blazing one of your own. Don’t be discouraged by how much food and energy it takes to make something new. It will be worth it and it also will increase your creative “gas tank” which will fuel more projects in the future.

And my final project is a pure science project of an hypothesis that it is possible to create a cost-effective method for sending space vehicles into orbit without requiring hydrogen solid fuel rockets. This project requires math and science areas of knowledge which will quite literally take me another 10 to 20 years to master so this is definitely a long-shot. The more distant in the future and the more skills that you may or may not already have…the less “real” the project may become.

Yet I started the Codecademy project a little less than a year ago and now I’m 89% complete with the Javascript tutorials and even a “wall” section on multi-dimensional arrays has finally become palatable enough for me to complete. So there’s really no limit to what daily painstaking progress can do in order to achieve artistic goals.


April 9, 2012

Tips on Codecademy tricky exercises

Filed under: Tips and Tricks — Tyler @ 5:26 pm
codecademy fizzbuzz++ 1.7

codecademy fizzbuzz++ 1.7 - (This exercise seems to run better on Safari for some reason)

Many people including myself have been having a problem with an exercise in Codecademy called, “One more function” in the section of “The Method” within the module of “FizzBuzz++: Return of the Modulus” – There is a simple trick which can make this process go much faster.

The trick is to use multiple browsers. I use Firefox as my primary browser and I was getting this strange error on several of the “Projects.” The error said something like “SandBox error” or “Sandboss is undefined” or something to that sort. I’ve found that switching between Internet Explorer and Safari browser can help with that strange error.

If you need help with 1.7 here’s my answer to the first half, and then I borrowed the code from another dude who made a phenomenal ternary code for the final segment.

Here’s some code which gives a successful “Next Exercise” operation:

//begin code

var comedy = {

schadenfreude: function() {
userResponse = prompt(“How was your day?”);
autoResponse = (“Oh, well it’s a good thing that “) + userResponse + (” happened. Let me make cakes made out of ice cream.”);
return autoResponse;

irony: function() {
userResponse = prompt(“Do you like to be a hero?”);
autoResponse = (“Your path will cause you to eat cheese.”);
return autoResponse;

// comment code – deletable
// begin code from the dude who I borrowed from
// note how excellent his ternary operator/statements are
// I love how he used the || to make the “or” side perfect
// end comment code
slapstick: function(n) {
return(n == “Murdoch” || n == “Gates”) ? “Pie!”:(n == “Hollande”) ? “Flour”: “Make Up!”; }




Older Posts »

Powered by WordPress