zcheridan-math

[zcheridan-math]

PWT - Particle Wave Theory

Wave–particle duality is the concept that every particle or quantic entity may be partly described in s not only ofparticles, but also of waves. It expresses the inability of the classical concepts "particle" or "wave" to fully describe the behavior of quantum-scale objects. 

Imagine a piano key~ 1/2kx^2

Set the particle at 1/2 of a calculated space, a spring.

XY, X is the worth of an oscillation, and Y is the displacement 

Each spring is the width of an oscillation x~

in the y direction, a measurement of the displacement of x^2 can be made to reflect a calculable difference of the potential energy as well as its presence can be made.

The X-AXIS being a displacement 

F*D = 1/2KX^2

Permutations soon

[zcheridan-math]

TRAVELING PERSON’S PROBLEM

LOG[2PI*R](2PI*R) = 1 & 0

Imagine concentric circles, but we only draw the diameter.

All of these fit in the palm of your hand. yay.

Like a photon glowing, whatever that is

each ray extends only 1,2,3,4,5... infinite away.

LOG[Bigger Circle] (smaller circle) = 1

something something,

These can only fit a specific way from the zero that all rays know as a term.

One is 1 base away from Two, and 2 is Two bases away from zero, Three is 2 bases away from One, etc.  This can only fit one way in the palm of one’s hand if each ray were labeled and termed LOG[T] (2*PI*R) = 0

or is it T*LOG[1](2*PI*R)

Can someone check this? My head really hurts.

[zcheridan-math]

5 MILLENIUM PROBLEMS

Everything becomes solveable with this introduction to a new understanding of a forgotten (or overlooked) math.  One that had no real-life example up until computer science as discrete mathematics, linearly, resonantly and equivocally can be interpreted by all preceding relationships to different mathematics such as TRIG, GEO, ALG, CALC, PEMDAS, BIN, ORD, SET & now I present to you, LOG.

The missing piece for a new generation of .... problems. lol~

LOG(A) = 1 or 0 (or) 1 and 0

BASE 2PI*R for any explanations.

or

BASE 1 for any explanations.  And A is equal to 2PI*R

    so that tangents are recognized up to 2 decimal points, 1 more than LOG Base 10.

Reminds one of [-B +- ([B^2 - 4AC]^1/2)/]2A To be re-formatted later to show you it’s resemblance to solving A^2 + B^2 = C^2

Will be done with this by 7/24/2017 I think~

Ah, Prime numbers are around LOG[2PI*R + 1] (1) = 1

If you didn’t scroll further.  There is a school of thought... my thought.... that also thinks the solution can exist as LOG[2PI*R + 1] (2PI*R) = 1 & 0.  However, this assumes you believe that a BASE can exist only as an integer.... which is obviously not true.  We shove 1/0′s all over the place in math all the time and overlook it during Calculus.

Without further ado~

BSD

T* LOG[2PI*R] (2PI*R) = 1 * T

LOG[1](2PI*R) = 1

I can’t explain anything unless if you look at the RHIEMANN Solution.

Elliptic Curve

Y^2 = X^3 + AX + B

LOG[2PI*R] (Y^2) = LOG[2PI*R] (X^3 + AX + B)

2 * LOG [2PI*R] (Y) = LOG[2PI*R](X^3 + AX + B)

LOG[2PI*R]( Y ) =1/2 (LOG[2PI*R](X^3 + AX + B)

Y = 1/2  (LOG[2PI*R](X^3 + AX + B)

           * LOG[2PI*R] (Y) ^ -1

The conjecture is not held~

Reason why.  If you look closely.

(1/0)^S makes a big difference when calculating RHIEMANN.

S as 1, is a significant value now.

S as anything else is a significant value now.

Easy right?

RHIEMANN

S*[LOG(1) - LOG(N)]

S*[0 - LOG[BASE](N)] per integer of a base, [2PI*R]

LOG[BASE EQUATION = 2PI*R) (2PI*R) = 1 == PRIMES

Easy right?

P=NP

1 - 0 = LOG(2*PI*R) (2*PI*R + 1)

I forget why I do this +1 here.  Solving 5 of these at a time keeps throwing me off.

1 - 0 = LOG[2PI*R[(2*PI*R)  = LOG[2PI*R](2PI*R)

Keeps checking until the highest and longest sequence of terms is accounter for. If I write anymore, you won’t pay attention~

(other millenium problems have  or (1 or 0) or (1&0) Pretty cool right?

LET 1, a distance from 0 upon every single axis, be the value of completeness.

It is possible to calculate completeness.

Polynomial Function in any base, binary to canary (aha, a bird) * time.

HODGE

T*LOG[2PI*R](2PI*R) = 1 or LOG[2PI*R](2PI*R)^T = 1

This is the same as

1 - 0 = 1/T*LOG[2PI*R](2PI*R)

The funny thing is that, R is already a base away from everything. T is a period away from everything. Or was it 1/2? Depends on your understanding, I won’t tell you which one is right. You know that this is the correct way of doing this depending on your geometric series shape for cycles. 0 - 1 = 1/T * (LOG[2PI*R[(2PI*R))

Algebraic cycles if you like to see it as frequency.

But if you like to see it as a multiplier of the completely reduced original algebraic expression.

T on both sides is correct.

PARTICLE-WAVE

1/(2Y)^2

The prize will require “ Progress in establishing the existence of the Yang-Mills theory and a mass gap will require the introduction of fundamental new ideas both in physics and in mathematics. “ clay mathematics YANG-MILLS MASS GAP.

Mass Gap is not a fundamental, therefore I will start where they started.  Gaussian and Particle Wave Theory.

It’s kind of funny because this one looks like all sorts of messed up to me.

L = 1/(4g^s) Integral (T*r F^*F) differential what?

First of all, the first law of Physics is Newton’s First Law of Motion. An Object in Motion stays in Motion.  The object in question is a slant about a wave. Is it not?  Can an object fit it’s sinusoidal curve? No.

Here is proof.  LOG[2PI*R](2PI*R) = 1.... 1 What said another guy in the back.

How can the very object, of unknown dimension, be called to move sinuisoidally in 1/4thG steps let alone 1/4thG^2 steps.  It moves slowly? A wave is meant to move sinusoidally.  If the circle in itself is already inside.  Then doesn’t it have angular momentum?  So we’re talking about a piece of mass within a particle wave, that tells me that it has some spinning particle in it based off it’s gravitation.  But doesn’t the very notion of suspended particles look stupid?  yeah... 

So where is the missing mass? Our integrated overlapping not knowing the constant between mv/mv momentum (angular momentum) waves and a particle’s (property) a natural resonance through a field.

Easy~~!

P.S.

I mean just look at this picture they have

How many axises are there? If you can’t pass this.  You will not want to start from my zero axis theorem. There are 0,1,2,R,Infinite Axises in what seems like 3 different bases. Okay.

All in the same flavor of TRUE ZERO (AXIS) THEOREM, I named that myself :D

If I missed one please check any permutation of

(T)[Zi] LOG [BASE(Zii)](Ziii) = (Ziv)(T)

Where the Z’s can equal [1,0,-1,2PI*R]

Nothing else.

[zcheridan-math]

RHIEMANN 1

1/1^S + 1/2^S + 1/3^S + 1/4^S + 1/5^S + 1/6^S + 1/7^S ..... + (1 / 0^S)

(1 / 0^S)

Also since 1 ^ S is just 1 anyway ~

(1/1)^S + (1/2)^S + (1/3)^S + (1/4)^S + (1/5)^S + (1/6)^S + (1/7)^S.... + (1/0)^S

1/0 is going to be a promblem huh!

PER TERM

S * LOG (1/N)

The nice thing about this is that we have a lot of leeway~

S*[LOG(1) - LOG(N)]

This cycle helps with dealing with 1/0

Okay, so now what do we do...

Oh wait what? LOG(1) is 0!

So that’s how we get rid of those 1/0 in infinite series.  (New Achievement Completed!)

Oh What!!

That is exactly what I needed.

A value for all Integers,

distance from 1, with ANY BASE, for all spots on a line~

This is how I found out the location of every single PRIME Number.

In ideology, the true distribution of prime numbers through infiniti.

LOG [2PI*R + 1] (2PI*R) = 1

I’m writing this up really quickly. It might be.

LOG [2PI*R + 1] (1) = 1 though it should equal 0. ehhehe

It was a great sight~

Another Thought, Eigen Values can be seen as the weighted on front of it’s “ability” to probabilistically land on a value.  With a Base of 1, is the actual % distribution.  From here many extrapolations can be found.

[zcheridan-math]

A

is this.

1/1^S + ½^S + 1/3^S + ¼^S + 1/5^S + 1/6^S + 1/7^S ….. + (1 / 0^S)

(1 / 0^S)

Also since 1 ^ S is just 1 anyway ~

(1/1)^S + (½)^S + (1/3)^S + (¼)^S + (1/5)^S + (1/6)^S + (1/7)^S…. + (1/0)^S

OH NO!

How do I get rid of 1/0 to the S????

([1*0^(-1)] / 1)^S

Special notes while I have your attention~

A CIRCLE contains all rational points by CIRCUMFERENCE because of the following:

It has support fo 2 axises perpendicular to each other like X&Y. Why? It has 2 undefined points and 2 zero points when derived at ¼th 2/4th 3/4th and 4/4th’s of a circle.

All rational points are points are ones that both x and y are able to be created as fractions of rational numbers.  True story. Weird right?

What are the fractions composed of?

Rational Numbers!

Take a Tangent anywhere like Y/X = 1 and this the following *1 holds true~

[zcheridan-math]

Collatz Conjecture

LOG[2](3N+1) = 1 or 0

LOG[2](1/2) = 1 or 0

LOG[2](3N+1) +  LOG[2](1/2) = 2 or 1(0 & 2) or 0 or 1(1)

LOG[2][3N+1] = 1

There will be a number that fits this playbill.

[zcheridan-math]

BSD

LOG[2PI*R](2PI*R) = 1 

RADIUS = Y/X = A natural base from 2PI*R

 if ζ (1) = 0

“there are an infinite number of rational points (solutions)”

  & 

conversely 

“if ζ(1) is not equal to 0, then there is only a finite number of such points.”

[zcheridan-math]

HODGE

T*LOG[2PI*R](2PI*R) = 1 or LOG[2PI*R](2PI*R)^T = 1

[zcheridan-math]

RHIEMANN

0 = 1 - LOG[2PI*R](2*PI*R)

0 = LOG[2PI*R](2PI*R) - 1

[zcheridan-math]

P = NP

LOG(2*PI*R) (2*PI*R + 1) = 1

LET 1, a distance from 0 upon every single axis, be the value of completeness.

Zcheridan: Every axis is orthogonal to a point as a point is only a visually un-acuited circle (is my assumption for this).

[zcheridan-math]

SET OF ALL INTEGERS

LOG [2*PI*R] (2*PI*R) = 1

LOG [2 * PI * R] (1) = 0

An axis drawn left to right from 0.  The circle ends at 1 and only when it intersects with our axis twice by an integer.

2 circles, 1 integer.