The Economic Vector Space and the Economic Cycle | OMICS International
ISSN: 2375-4389
Journal of Global Economics

# The Economic Vector Space and the Economic Cycle

Paul T E Cusack*

*Corresponding Author:
Cusack PTE
Independent Researcher
BSc E, DULE, 1641 Sandy Point Rd
Tel: (506) 214-3313
E-mail: [email protected]

Received Date: January 17, 2017; Accepted Date: March 22, 2017; Published Date: March 29, 2017

Citation: Cusack PTE (2017) The Economic Vector Space and the Economic Cycle. J Glob Econ 5: 243. doi: 10.4172/2375-4389.1000243

Copyright: © 2017 Cusack PTE. This is an open-access article distributed under the terms of the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original author and source are credited.

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#### Abstract

A vector space is a series of vectors of different magnitudes and direction which are connected head to tail. There is a vector space that applies to the US economy, for example. My previous paper on the Cusack US Economy Equation contains the 11 variables with the 12th one being t

#### Keywords

Vector space; Macroeconomics; Physical economics

#### Introduction

I illustrate here the vector space for the economy as opposed to the physical universe. I've written yet another paper on Cusack's Physical Economics (Figure 1) [1,2].

Figure 1: Illustrate 1 on Cusack's Physical Economics.

1/81=0.012345679 81=c4=speed of light4

Each successive digit is adding 1 less power of 10. So, from logic, a binomial tree yields:

(1+t)11=e0+e1+e1...

=(1/7+7et)

=1/81

0.01234567-1/7=7et

t=398~396

The plot above is 396 × 396

396+396+396+396=4(386)=0.1584=1-sin 1=1-cos 1

1-1584=0.8416

This is the ideal balance for maximum sustainable output Y.

Sin and cos are a two pole problem since Y=1 at full output and Y=0 at no output (Figure 2).

Figure 2: Balance for maximum sustainable.

Now,

The cross product is:

||E||||t||cos 60={G,ec}

E=t

E2×1/2=G =0.

So there is a gravitational equivalent in the economy (Figure 3).

Why is it?

Figure 3: Gravitational equivalent in the economy.

From Newton, we know:

F=GM1M2/R2

2.667=8/2=2|D|/c={Gec}(Infinity)/R2

Gec=0.251~0.253=Period T (seconds)

R=0.3068

1/R=3.25=13/4

The economy is like a pendulum ranging between high and low outputs.

x=1/[x-1]

x2-x-1=0

x=1.618 =Golden mean

Eigen value=c=3

Eigenvector=√3=1.73

Eigen vector2=eigenvalue

Continuing,

y=y' is telescopic

y=ex=y'=y''....

y=1/+7et

y'=7et=1/c2=0.111

et=0.111/7=0.1586

sin1=cos1

This is where sin meets cos or the settling point between Y=0 and Y=1 (NO output, Full output)

So,

F=Gec × M1M2/R2

0.253 × (Infinity)(0)/13/4)2

0.86=8/3=2|D|/c=E × |D|/eigenvalue=0.778x

x=0.1111=1/c2=1/eigenvalue2

from above

y'=1/7 +7et

0.86=Y Max sustainable

QED

If you didn’t understand that, try this:

Assumption: Money is stored energy. Energy cannot be created nor destroyed. Money cannot be created nor destroyed. Governments cannot create money. The simply devalue their currency by printing money.

From the vector space, we know the economy has a resistance r which is the costs of producing output.

when:

R=dR/dt NPV=0 Y=y’ R=r’ integrate R2/2=R

R=0, 1/2dR/dt=π

E/tπ/2π=1/2=R

When E=π

A=πR2; A’=2πr

Circle=2πr A’=Circle R2=A/π R=√A/√π

R=1/2=√A/1.7725

A=π/4=45 degrees which is the minimum energy level.

The economy seeks the minimum energy to produce maximum sustainable possible output.

So y=y’

E=1/t

Work W=F × d; d=s

=J/s=E/s

W=E/t=E=1/t

E/t=E2

E=0; 1/t=Wt=1/t

Y=W Y’={F × d}’

Y’=F × (ds/dt Y’=Fv=(Ma)(P/v)

P=Mv

Y’=MP; Y’=M(Mv)

M2=1M=√1=-1 M=1

Now, Wt2=1 W=1

W=J/s; W=y/1

W=y=1; W=1; y=1; M=1

t=1

y=F; y’=-F

y=W and y’=W

Recall Money is stored work. So Economics follows the same laws as the universe.

Y’=F(ds/dt) Y’=(1)(ds/dt)

Y’=ds/dt

Y=Integral ds/dt Y2/2=s+C1 Y=√2+ C1

Y=sin1

Y’=–cos1

Y=–y’=sin1=cos1

Vector spaces components:

Period: T=1/t=E=W × t; 1/t=Wt

T2W=1; T(F×d)=1

T2 × y × s=1

(1)(y(1))=1 Y=1

Speed of light: C2-c-1=2c-1 C=1

Force

Y=F

W=F × d

1=(sin1)(d)

=1/sin1=d=s

csc1=s

s=1.11884

1-sin 1=0

–sin=–F; 0.84=F; F=y

Y=sin1

(π-e)=m=E/cuz

Y=mx+b; Y=0.4233x+0

Y=cuz × x

Y=R × x; A=π × R2; A’=2π × R; C=2πR; A’=C

2πR=2πu; R=rx; X=2π; T=2π;

The economic cycle:

Boom=2.09 years stagnation=4.18 years decline=2.09 years=1 economic cycle=8.36 years (Figures 4, 5 and 6).

Figure 4: The economic cycle.

Figure 5: Graph for economic cycle.

Figure 6: Cycles from depression to depression.

So, we had a decline starting in 2007.9. Add a cycle we get 2014.17=March 2014.

Profits go toward zero. So e-x=0 x=Ln 0 x=t=1 Y=e1cos[2π(1)] Y=2.71828

y=y’

And,

9.44 cycles from depression to depression Y=e-9.44 × cos(2π × 9.44)

Y=7.5

1/Y=0.1334=s

Circle=s2

Circle=0.13342 Circle=0.0178

C=2πr Circle=2.8323

#### Conclusion

The mathematics models of physics can be used to solve outstanding economic problems.

#### References

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