Access Your own Mind…

Blocks to creativity as variables…

The path to creativity as a formula…

What other people or experience doesn’t need to be fair in comparison to your situation…

Making it Fair probably won’t benefit your art…

Samurai swords get a lot of T.L.C…

## February 12, 2017

### ASMR – How to Crush Writer’s Block

## February 3, 2017

### 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} = r^{2}

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: **

e^{ix} = cos*x + i*sin*x

**MATRIX DATA:**

**Determinant**:

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.

**Adjugate: **

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] ;

**INVERSE MATRIX: **

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:**

Vectors are primarily subjects of a direction and a magnitude.

UNIT VECTORS:

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:**

[(x^{2}) / (a^{2})] + [(y^{2}) / (b^{2})] = 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( a^{2} – b^{2} )

**Formula for a Hyperbola:**

[(x^{2}) / (a^{2})] – [(y^{2}) / (b^{2})] = 1;

The asymptotes are defined by the formula:

y ± (b^{2}/a^{2}) * 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:

[(y^{2}) / (b^{2})] – [(x^{2}) / (a^{2})] = 1

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

ALGEBRA FORMULAS:

**Factoring Polynomials: **

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

**Dividing Numbers of the Same Base with Exponents: **

(x^{a}) / (x^{b}) == x^{a-b}

**Raising Numbers that already have exponents to an Exponent: **

x^{aj} == x^{a * j }

**TRIG QUADRANT DATA: **

**QUADRANT I (ONE):**

(+) sine

(+) cosine

(+) tangent

(+) cosecant

(+) secant

(+) cotangent

**QUADRANT II (TWO):**

(+) sine

(-) cosine

(-) tangent

(+) cosecant

(-) secant

(-) cotangent

**QUADRANT III (THREE):**

(-) sine

(-) cosine

(+) tangent

(-) cosecant

(-) secant

(+) cotangent

**QUADRANT IV (FOUR):**

(-) sine

(+) cosine

(-) tangent

(-) cosecant

(+) secant

(-) cotangent

**BASIC TRIGONOMETRY FORMULAS:**

**Law of Cosines**

c^{2} = a^{2} + b^{2} – 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)

**PYTHAGOREAN IDENTITY:**

(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)^{2 })

**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;

**QUADRANT I** (ONE) – [TOP RIGHT]

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;

**QUADRANT II** (TWO) – [TOP LEFT]

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;

**QUADRANT III** (THREE) – [BOTTOM LEFT]

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;

**QUADRANT IV **(FOUR) – [BOTTOM RIGHT]

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*x^{2} + 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 = x^{2}

**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*x^{2} + bx + c

**Finite Series ( Arithmetic):**

S_{n} = [(S_{0} + S_{n}) / 2] * n

**Finite Series (Geometric and Exponential Patterns)**

Finite Geometric Series – S_{n}

S_{n} = (a(1-r^n)) / 1 – r

S_{n == 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

**SIGMA NOTATION**

(n_{final})

∑ a*r^{k}

(k=z)

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

log_{a}^{b} = x

is equivalent to:

a^x == b;

Evaluate Properties –> adding log terms to each other

( log_{c}^{a} ) + ( log_{c}^{b} ) == ( log_{c}^{a * b} )

Evaluate Properties –> subtracting log terms to each other

( log_{c}^{a} ) – ( log_{c}^{b} ) == ( log_{c}^{a / b} )

Multiplying Log -> adding additional exponent to log expression

A * log_{b}^{c} == log_{b}^{(c^A)}

Special Property – Evaluating non-base 10 Log using division

log_{b}^{a} = ( log_{c}^{a} ) / ( log_{c}^{b} ) ;

**Complex Number Theory:**

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

If z^{n} = i ……

Modulus of z^{n} relates to r^{n} in the form of:

z^{n} == r^{n}[cos(n * theta) + i * sin( n * theta) ];

This next part is interesting:

If z^{n} == 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.

## October 15, 2016

### Preparation for Squirrel

Perhaps my existence has become synonymous.

The land and the king may be one. But my subservience to the many has drained me, only to be renewed on a daily basis…just enough.

It was such hubris and perhaps slothlike greed which made me want to withold technologies such as Rocketless Semi-orbital Satellite Insertion.

Monetize, what a joke.

How could I ever monetize a system which will require several lifetimes to complete.

…especially at my rate of development.

Although I must say confining options has proven extremely effect for creative construction. This of course hasn’t come by any stretch of will-power or spiritual fortitude of myself.

On the contrary my most recent leaps in progress have been due to the destruction quite by accident and attrition of my primary computer.

Perhaps a simpler maze is all that is required for one to find a complex solution. Or maybe it isn’t about the maze being simpler, maybe it’s about segmenting the maze.

Quantized units of struggle that test rather than antagonize. And when the test taker quits…antagonize rather than test.

And even with all the encouraging progress that people have had with urban farming, vertical agriculture and open source technologies…one would think I’d be both hopeful and happy.

And of course I am.

But there is a fruitlessness to it all.

Why push the envelope when the letter is never going to be sent in the first place?

I guess emotions are challenging because the same ones that allow for rapid development are the same ones that are painful to activate.

Packaging.

That’s been my Achilles heel for years. It doesn’t matter how good an idea or item is, bad packaging is bad packaging.

My raw, unpolished packaging and perhaps attitude is probably going to hold me back forever.

It’s been two months and I haven’t even put all my paintings up.

The last time I was in this position was…11 years ago.

I think I can manage much better now, but it’s different. Loneliness isn’t there, I think the internet quite sickly defeated loneliness a long time ago.

There is still frustration and apathy. But compared to loneliness, those emotions are like bugs, squashed after a good night’s sleep and a bowl of cereal.

I can remember what loneliness felt like then, it was so strange. There is this urge to seek outward rather than dig in and create.

Feeling cold, that’s a terrible feeling. If MECE did anything for me, removing that cold feeling definitely has value. A cold that no flame can warm.

No hearth can melt. A cold that aches and tears apart until there are no structures left to resist the abyss.

Yep, glad I’m done with that one.

Yet what is left?

If I age too quickly and don’t achieve anything, will my life be a waste?

What would my mitochondria say about this?

Edits, communication, sacrifice, modification, growth, consumption, explanation, work, contraction, rest…these are all valid methods.

What do these all have in common? They all seem to be active words. Active little mitochondria…it’s no wonder.

Rest, even rest…maybe that’s the link in the chain.

## October 9, 2016

### Why are Spiritual Rules Tough

Spiritual Rules are tough. They tend to require discipline and also can create cognitive dissonance. This video discusses some of these issues…