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February 3, 2017

Math Formulas

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

Math Formulas



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Derivatives are related to slopes.

Raw Derivative Base Formulas regarding a function “g” where x = t:

Variation 1:

{lim{x–>t}}  (g(x) – g(t)) / (x – t)

Variation 2:

{lim{h–>0}} (g(t+h) – g(t)) / h


Differentiation Rules:

Exponent and Differentiation Rule: 

If you have a function (f(x)) that is raised to an exponent such as (n).

To find the derivative (d/dx) of f(x) = ( x^n) is equivalent to (  n *x^(n-1))

Sum and Difference Rules:

When taking derivatives of a sum of terms, or a difference of terms, separately perform the derivative on each term.

Constant Rule: 

Any Derivative of a constant has a zero slope.


(d/dx) *(constant) = 0

Sine Rules:

(d/dx)[cos(x)] = -sin(x)

(d/dx)[sin(x)] =   cos(x)

Product Rule: 

(d/dx)[f(x)*(g(x)] == f ‘(x)*g(x) + f(x)*g ‘(x)

LOG Rules:

(d/dx) [ln(x)] = (1/x)

(d/dx) [ex] = (ex)

Derivative Trigonometric Rules: 

(d/dx) tan(x) = sec^2(x)

Chain Rule: 

if ( (f ‘ (x)) ==  v ( u (x) ) )

Outer Function related to inner function….

(f ‘ (x)) = (v ‘ (u(x))) * ( u’ (x) )



Basic formula for probability:


P = (Favorable Outcomes / Total Possible Outcomes)



A combination is a process, where order of items in a set does not matter like in a permutation.

The goal is to find out the total number of different ways to express the set.

Even though the order does not matter, and this makes it seem like it would be less convoluted of a process to determine the number possible sets. A combination actually has a more convoluted formula and process.

Also, there is a specific form of notation for combinations.

N choose K;

[NOTE: These items are supposed to directly on top and below each other.)

(n k) (8 4)  N choose K of “8 choose 4”:

This means:

(8!) / (8-4)! * (4!)

So what does this mean?

We have a set of choices of items which is 8 deep. This means we have 8 different types of gems to choose from.

We also have four boxes that one gem can go into.

In the case of a combination, we have to take the factorial of the number of choices for things. So in the case of  our example, since we have 8 different types of gems, we take 8 factorial.

But the factorial is not enough to go on, since we need to only go down our factorial list by the number of “boxes” in which our items will be entered into.

So in the case of our gem example, since there are only 4 boxes, we go down our factorial to 5.

That is represented in the formula by (8-4)!

But we also need to multiply that factorial expression by the factorial of the number of boxes themselves.

So as you can see there is a lot more multiplication and factorial-ing that needs to be done to complete a combination.




A factorial is a process by which a number all of its integers proceeding down to 1 are multiplied by each other.

This process is denoted by the “!” notation.

For example: 5! or “Five Factorial” is equivalent to: 5 * 4 * 3 * 2 * 1 – which is: 120.



In a permutation, the order of the items in the set matter.

This means a set which has three different items can be rearranged. But even if the same items are in a subset and they are in a new order, it will be considered a unique sub-set.

For example, if there are four boxes where seven different colored gems are available. Each box can contain one gem each. And there can be only one gem color represented per four box set. Also, no sets can be repeated. How many unique sets of gems can be generated?

The answer is: (7 * 6 * 5 * 4) this is because the total number of potential sets decrease as each set is created.


Simplified Expanded Form vs Factored Form:

Simplified Expanded Form means whatever is not removed from an expression due to canceling out of factors etc…is expressed without needing parenthesis etc.


ODD functions vs Even Functions 

Even Functions are symmetrical in respect to the y-Axis.

Odd Functions are symmetrical in respect to the origin.

Exponential Function Graphing: 

If an exponential function had a form of:

a * (b^x) + d

“d’ is the horizontal Asymptotic line.

IF THE FUNCTION “x” IS POSITIVE:  The “sign” of “a” will determine if the function has an ending behavior of negative for a negative sign.

A positive sign for “a” give the function a ending behavior of going “up” or positive on the y Axis.


The Sign of “x” will determine which “direction” the function opens . If it is positive the function will open to the right. If “x” is negative the function will open to the left.


Properties of Exponents:

Multiply two of the same base numbers with Exponents:

(b^a) * (b^c) == (b^(a+c))

Dividing Numbers of the Same Base with Exponents: 

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

xaj == xa * j


Volume of a PYRAMID:


V = (1/3)*(AH)

A = Area of the base

H = Height of the Pyramid from the Base



A = (V)/ ((1/3)H)



Sinusoudal Functions:

Cosine Functions

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

The Period of this function would be:

( 2π / b )


y == f(x)

a == Amplitude of function
(cos or Sin are interchangeable depending on rework of formula (it is not directly interchangeable but there can be offsets using either))

To find the period of a function that is in radians you use (2 * pi ) / k

To find “k” one must first the length of the period. So if the period or distance between the beginning and end of a cycle is “8” then the b would be in Radians would be (2 pi) / (8) or 1/4 pi.

c is a horizontal offset variable.

And “d” is the vertical offset for the mid-line etc.

ALSO: If the wave starts at the minimum point of the function or the maximum, not the midline, then it will require less offset to use cosine rather than sin.


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-b)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




A Matrix has rows and columns.

A matrix that has 4 columns and 2 rows would be called a:

2 x 4 matrix

THUS: = 2 (rows) x  4 (columns)



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

NOTE: Start with the first column spot and the first row spot first. So diagonal top left to bottom right would occur first.

[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);

Also: If Matrix A truly is the Inverse of Matrix B then:

 Matrix A * Matrix B == Matrix B * Matrix A == Identity Matrix

Identity Matrix: 

The Identity 2 x 2 Matrix is simply:

[1    0]

[0   1]


If a matrix requires simply dilation and there aren’t any offsets, modifications to the Identity Matrix can be used.

For example:

To make a square grow to twice its size:

The matrix transformation of:

[2    0]

[0    2]

can be applied.

NOTE: this is simply a modification of the identity Matrix.

ADDITIONAL NOTE: Anything with significant offsets or other combinations of effects will require a more rigorous transformation matrix.

Showing Basic Vectors as a Matrix:

A vector with the location of 4i + 3j could be represented as:



As you can see it is a 1 column Matrix with 2 Rows (2 x 1 matrix).


First we must determine if the number of columns in the first set equals the numbers of rows in the second set.

To Multiply a 2 x 2 Matrix by a 2 x 1 Matrix the technique is as follows.

   (A)    (B)

[1   0]   [2]

[0   1]   [1]

Multiply item in Matrix A at( row 1 column 1 by   B item in the first  row) + (Matrix A Row 1, column 2 multiplied by Matrix B – Row 2 Column 1)


Vectors are primarily subjects of a direction and a magnitude.

Adding or Subtracting Vectors: 

The easy way to add and subtract vectors is to add or subtract their components.


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.


For example: If the formula included:

((x-6)2) / 1 This would mean that the center of the “x-axis” of the ellipse would be located on the point “x = 6”
(note that it is the opposite of its sign similarly to a circle etc.

Also, the “1” at the bottom of the expression would not be a placeholder, it actually determines the “thickness” or minor x-axis or in another words its “x-radius”.

A x-radius of 1 would mean that the ellipse has an x-axis of 2 and a x-radius of 1.

Continuing on:

If the formula included:

((y+3)2) / 16 — This would mean that the center of the “y-axis” of the ellipse would be located on the point “y = -3”

Also, the thickness of the ellipse on its y-axis would be 8 and it would have a y-radius of 4.

NOTE: The radius is the square root of the bottom number. The axis is twice the square root of the bottom number.

Determine the Foci length of an Ellipse

If a > b:

f = sqrt( a2 – b2 )



Other information:

f = focal length

minor radius = q

p= major radius

f^2 = p^2 – q^2


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”


Finding Equation of Hyperbola from Foci and Vertices:

If the Hyperbola opens to the left or right, then (x^2 / a^2) is on the left.

If the Hyperbola opens to the up and down, then (y^2 / b^2) is on the left and then  (x^2 / a^2) is subtracted from that.


If we have the vertices, and there is not offset, and if the hyperbola opens to the top and bottom, then we easily have information to find, “b”.


b^2 is actually the square of the  distance from the center point between  the two vertices and a vertex.

c^2 (which does not exist in the formula but it is important)  is the square of the distance from the center point between one of the two foci.

Using the pythagorean theorem a^2 + b^2 = c^2   — We can find the “a” .

From there it is easy to finish the formula.


Finding Focii of Hyperbola from Equation: 



(x^2)/ 64 –  (y^2)/ 148 = 1


Focal length =


f^2 = a^2 + b^2


a^2 == 64

b^2 == 148

Factoring Polynomials: 

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


Trigonometric Raw Definitions:

Sine: = (Opposite / Hypotenuse)  [sin]

Cosine: = (Adjacent / Hypotenuse) [cos]

Tangent: = (Opposite / Adjacent) [tan]

Cosecant: = (Hypotenuse / Opposite) [sin^-1]

Secant: = ( Hypotenuse / Adjacent) [cos^-1]

Cotangent: = ( Adjacent / Opposite) [tan^-1]



(+) 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);

tan(Θ) == tan(π + Θ);

Useful Cosine Identity:

cos(Θ) = cos(2π – Θ)

Tangent Identity / Definition:

tan(Θ) = ( sin(Θ) / cos(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

(tan Θ)2 = (sec Θ)2 – 1

(cot Θ)2 = (csc 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

cos(2*Θ) == 1-2*sin2(Θ)

cos(2*Θ) == cos(Θ)2 – sin(Θ)2


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(45° 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.


Honestly, I think it is better to make your own formula using the distance formula. I’ve tried both custom formulas and previous ones and making a custom formula based on the x and y of the focus and the directorix and setting each side of the equation equal to each other gives more consistent results. This may be due to some mis-usage of previous written down formulas. But at the end of the day I just want to get these things correct.

Base Vertex Form:

y = a(x-h)2 + k

In Base Vertex Form the “-h” is equal to the opposite of the x coordinate of the vertex. So if the formula has (x-2) the x coordinate of the vertex is “+2” or “2” also the “k” is simply the “y” coordinate of the vertex.

Also, the larger the “a” is in the Base Vertex Form, that means that the parabola will be narrower. If the “a” is smaller, it was seem more wide.

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

SIDEWAYS Parabolic Base Vertex Form:

x = a(y – k)2 +k

SIDEWAYS Parabola Directorix and Focus Form:

This form is extremely interesting because it requires a little more thinking than simply plugging in a formula.
Another thing that should be noted about the sideways parabola, is that at first it may seem like a bad function. This is because at first glance the sideways parabola fails the vertical line test.

Normally a vertical line test would tell us if a function in a normal function or not. If you can draw a vertical line and get two points that have the same x-coordinate then you have a problem.

But in this case, the function takes, “y” coordinates as the input. So technically the vertical line test doesn’t apply. In fact, a horizontal line test would apply in this situation.

And in the case of a horizontal parabola, the function does in fact pass the horizontal line test.

So getting back to the sideways formula let’s take a look at the situation.

First we need to realize we are create a “FUNCTION” not just a “formula” at this point in Math it’s important that we are creating little programs that produce desired results.

Simply plugging variable values into formulas aren’t going to help us forever in Math so let’s get cracking!

The Given Information of our function is:

1. We know the “FOCUS” x-coordinate and y-coordinate
2. We also know the “directorix” location.
3. (IMPORTANT) We must follow the “rules” of a parabola.

The desired effect of our function is:

1. To generate “x” coordinates.

That’s it! All we have to do is take input y-coordinates and spit out x-coordinates.

-Let’s look at the “Rules” of the parabola. Maybe these rules are actually clues that can help us.

1. The distance from the Directorix to a specific point on the Parabola must always be equal to the distance from the “FOCUS” to that same point on the Parabola.

So we must set our equation balanced. One side will have the distance formula for the Directorix to the (x,y) coordinate of the parabola.

The other side of the balanced equation will have the distance formula for the Focus and the (x, y) coordinates of the Parabola.


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:


Powers of Complex Numbers: 

To be honest, this is a very convoluted process. It’s no wonder that complex number theory is called, “Complex” because these solutions are not simple. They usually take several steps of alchemy to get them cooking to where you need them to be.

The first step is to determine what the current form is:


If the current information resembles Rectangular Form, then there will be a temptation to simply multiply and distribute and call everything a done day.

Unfortunately this will create erroneous results.


What we need to do is to find the Modulus and Argument.

To find the Modulus we use the formula:


r = [sqrt (a2 + b2 ) ]

Find the argument via: tan-1(b/a)

After that verify and adjust the argument to match the original Quadrants of the original expression.


IMPORTANT: Even if we find that the “multiple of theta is 0, we still have to take into account the “offset” theta for example:

n * theta = theta + k * 360

Even if we find that “k” is actually zero, we still have theta divided by n to account for.

IMPORTANT: Before we solve for the final r[cos(theta) + sin(theta)] we must make sure that we are multiplying by “r” and not by “n”

EXTREMELY IMPORTANT: If dealing with fractions when adding or subtracting these multiples of degrees. It is important to make a single degree measurement by adding or subtracting like fractions. If you are faced with a whole number and a fraction, convert the whole term into a fraction and then simplify before doing any other calculations.

Next raise the modulus and argument to the power in question.

NOTE: You actually multiply the argument by the power —

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


Then finally express the raised argument and Modulus in the desired format.


Forms of Complex Numbers: 

(Rectangular Form) = (a + bi) – [NOTE: This also can be represented graphically on the Argand Grid]

(Polar Form) = r (cos(theta) + i * sin(theta) )


Conjugates in Complex Number Theory:

If (z = 6 + 3i )

z = 6 -3i

NOTE: The purpose of the complex conjugate is to create a non-complex number. Therefore the part of the system which contains “i” will be the one that is made opposite.


Arguments in Complex Number Theory:

An argument in Complex Number theory is basically an angle.

First we need to discuss the Complex plane (imaginary)(argand grid).
The x coordinates are real numbers.
The “y” coordinates are actually in this system, imaginary numbers.

The Quadrant I contains positive Real Numbers and Positive Imaginary Numbers.
The Quadrant II contains negative Real Numbers and Positive Imaginary Numbers.
The Quadrant III contains negative Real Numbers and Negative Imaginary Numbers.
The Quadrant IV contains positive Real Numbers and Negative Imaginary Numbers.


Modulus or “Magnitude” in Complex Number Theory:

The magnitude of a vector in the Complex Plane is the Modulus.

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!”; }




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