*BScE, DULE, 23 Park Ave. Saint John, NB E2J 1R2, Canada*

***Corresponding Author:**Paul TE Cusack, BScE, DULE, 23 Park Ave. Saint John, NB E2J 1R2, Canada.

**Received:**October 27, 2020;

**Published:**November 20, 2020

**Keywords:**AT Math; Golden Mean Parabola; Sugar Intake; E Coli; Cholera; Parkinson’s; Schizophrenia; ALS; Alzheimer’s; Cancer.**Golden Mean Parabola**

Ln t=e

^{-t}

Derivative:

1/t= (-e

^{-t-1})/ (-1)

1/t=e

^{-t-1}

Integrate both sides:

Ln (1/t) =Ln e

^{(-1-t)}

Ln (1/t) =-1-t

Aside:

Ln t=e

^{-t}

Derivative:

1/t=-e

^{-t}/-1=e

^{-t}

t=1/e

^{-t}

t=e

^{t}

1/e

^{t}=e

^{exp(-1-t)}

Ln (e

^{-t})= e

^{(-1-t)}

Ln (-t) =Ln e

^{(-1-t)}

Ln -t=-1-t

t=-1

∫dt/dt=∫1+? =t+?

t+?=1

t=?-1

Let ?=0

t=-1

-t=e

^{exp(1-t)}

Take ln of both sides twice:

-Ln Ln t=(-1+t)

Ln Ln t-t=-1

1/Ln t+1=t

Integrate both sides:

∫1/Ln t=∫t-1

Ln t=t²-t

From above:

e

^{-t}=t²+t

Take Ln of both sides:

-t=Ln t²+Ln t

Take derivative:

?=1/t²+1/t

?=(1+t)/t²

Let ?=1

Golden Mean Parabola

t²-t-1=0

t=1.618; -0.618

2π/3=212

M=Ln t

=Ln 212

=1.550

TE=M [0.15915]

= (0.1550)(0.15915)

=246.68

E=1/t=405.38

E=Mc²

405.4=Mc²

M=451

M=Ln t

t=166.99~1.57=π/2

PE=KE=t

Mc²=1/2Mv²

1/2=c/v

1/2=9/v

v=18

t=KE=1/2Mv²

=1/2(Ln π/2) (18)²

=73.156

E=1/t=0.01367

E²+E-2=t

(0.01367)²+0.01367-2

=-1.986~-2

t²-t-1=E

t²-t-1=--1.986

t²-t+1=0

t=1.314; 3.1429=Pi

C

_{8H11NO2 + C7H16NO2+ C13H16 N2O2 + C8H ON4O3 + C5H5N5 Fe3+ +NaCl +H2O2 → Dopa. + Ach. + Melatonin + Caffeine and Adenine+ Iron and Chlorine + Hydrogen Peroxide → C41H48N13O12 + 3 Fe+2+2NaCl+5O2 ==> 41CH + 12NO-+Fe2O3 +2NaOH + FeCl2-+ H2NOH +2H2O2-2 }

a CH+ C

_{22}H

_{30}O

_{5}+ 10 NO

^{-}+ 22 H

_{2}O

**→**C

_{22}H

_{44}O

_{22}+10 H

_{2}NOH + 2O

_{2}+OH

^{-}

Carbene +Cortisol (Stress) Sugar Hydroxylamine

The excess sugar makes the bodily system acidic. Bacteria need iron Fe +3 to grow.

amu= (14.016+362.46+30.006+396.33=1072.866

1072.866 x 6.026=6461.87 g

6461.87g x/ 0.994 g/mole of blood=65.008 moles

Normal pH=7.35-7.45 Say 7.4

pH=Ln [H+]

e

^{-7.4}=6.1125

[H+]=moles of H+/moles of solute

6.1125=Mol 0f H+/6500

Mol of H+=3973

H+ =1.008 g/mol x 4 moles=4.00 =|D|

4.00 x 6.023=24120.8

e

^{0.241208}=127.3=4/π=ρ

Moles of Sugar:

C22H44O22=660.33 g/mol x 6.023=397.716 g~4g of Sugar

=Moles of H+

H+1.008 g/mol x 397.716=4.008977~4.00g of H+

We examine E Coli in more detail.

E Coli double in growth every 20 minutes (1/3 hour)

Since it follows exponential growth, we have:

2=a

_{0}e

^{t}

2=a

_{0}e

^{(1/3)}

a

_{0}=1.4335=1/698=1/7 Economic Multiplier.

A full cycle is 2π rads

2π/7=0.8976=c²

GMP: t=1/3

(1/3)²-(1/3)-1=1.22222

E Coli has 20% of its protein in its envelope

M=Ln t

=Ln (1/3)

=1.096

(20%/100%) M=21972/χ

χ=1.0986

M

_{T}=21972

TE=M[0.15915]

=1.0986[0.15915]

=174.842~1 rad

t=E²-E-2

t=30.743

E=1/t=3.2527

100% M=1.0986/30 genes=0.03663 gm/gene

3.2527/0.03662=0.88823

GMP: = E=1.099~11 base pairs

4 mol of sugar=amu x mol x Avg.

=660.33 x 4 x 6.023=1590.86~1591

The chemical mol balance equation is:

C

_{5}H

_{5}N

_{5}+C

_{22}H

_{44}O

_{22}+22NaCl +3H

_{2}O

_{2}+ 8H

_{2}==>

(Adenine+ Sugar +Salt + Hydrogen Peroxide + Proton ==>

27CH+22HCl+22NaOH +5NO+ (1/2) O

_{2}

Carbene + Hydrochloric Acid + Sodium Hydroxide+ Nitrate +Oxygen

pH = Moles of [H

^{+}] /Mol of solute.

pH =7.4

Ln 7.4=2

2 = (8x2)/Moles of Solute

Moles of solute=8

8 x 6.023 =481.8

GMP: E=-1.249~-1.25=E

_{min}

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**Paul TE Cusack. (2020). E Coli & Electrolyte Depletion: The Cause of Parkinson’s; Schizophrenia. ALS; Alzheimer’s; & Cancer.**

*Citation:**Journal of Brain and Neurological Disorders*2(1).

**Copyright:**© 2020 Paul TE Cusack. 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.