Exponential growth

eop=((b^x)/(x^n))

Log[eop]//PowerExpand

Assuming[
{b>1&&n≥1},
Reduce[Log[eop]>0//PowerExpand,Reals]//Simplify
]

How does x log b-n log x behave as x->∞?

Fibonacci sequence and the golden ratio

FunctionExpand[Fibonacci[n]]

FunctionExpand[GoldenRatio]

Assuming[
n∈Integers&&n≥1,
Simplify[Cos[n π]]
]

fib[
n_,
ϕ_:GoldenRatio
]:=
(1/(Sqrt[5]))((ϕ^n)-(((-ϕ))^(-n)))

fib[n]

Assuming[
n∈Integers&&n≥1,
Simplify[FunctionExpand@fib[n]]
]

{
Table[FullSimplify[fib[n]],{n,1,20}],
Table[FullSimplify[Fibonacci[n]],{n,1,20}]
}

fib[n,ϕ]+fib[n+1,ϕ]

FullSimplify[(1/(GoldenRatio-1))==GoldenRatio]

-(((-ϕ))^(-n))-(((-ϕ))^(-n))(((-ϕ))^(-1))==-(((-ϕ))^(-n))+(-(((-ϕ))^(-1-n)))//Simplify

-(((-ϕ))^(-n))(1+(((-ϕ))^(-1)))==-(((-ϕ))^(-n))+(-(((-ϕ))^(-1-n)))//Simplify

-(((-ϕ))^(-n))(1-(ϕ^(-1)))==-(((-ϕ))^(-n))+(-(((-ϕ))^(-1-n)))//Simplify

N[1-(ϕ^(-1))==(ϕ^(-2))/.ϕ->GoldenRatio]

FullSimplify[-(((-ϕ))^(-n))(ϕ^(-2))==-(((-ϕ))^(-n))+(-(((-ϕ))^(-1-n)))/.ϕ->GoldenRatio]

FullSimplify[-(((-ϕ))^(-n-2))==-(((-ϕ))^(-n))+(-(((-ϕ))^(-1-n)))/.ϕ->GoldenRatio]

So, -(((-ϕ))^(-n))+(-(((-ϕ))^(-n-1))) simplifies to -(((-ϕ))^(-n-2)), which is clearly what I need. What about (ϕ^n)+(ϕ^(n+1))? That is

FullSimplify[(ϕ^2)==1+ϕ/.ϕ->GoldenRatio]

Solve[(ϕ^2)==1+ϕ,ϕ]

N[Solve[(ϕ^2)==1+ϕ,ϕ]]

{fib[0],Fibonacci[0]}

If I know that ϕ+1=(ϕ^2), do I automatically know that 1-(ϕ^(-1))=(ϕ^(-2))? Dividing ϕ+1=(ϕ^2) by (ϕ^2) gives (ϕ^(-1))+(ϕ^(-2))=1, which can be rearranged to the desired form.

fib[1,ϕ]

ϕ+(ϕ^(-1))=(ϕ^2)-(ϕ^(-2))

FullSimplify[ϕ+(ϕ^(-1))==(Sqrt[5])/.ϕ->GoldenRatio]

(ϕ^2)-ϕ-1=0
(SubPlus[ϕ])+(SubMinus[ϕ])=1
(SubPlus[ϕ])(SubMinus[ϕ])=-1
(SubPlus[ϕ])-(1/(SubPlus[ϕ]))=1

FunctionExpand[GoldenRatio]

FunctionExpand[1/GoldenRatio==((2((Sqrt[5])-1))/4)]

FunctionExpand[1/GoldenRatio==(((Sqrt[5])-1)/2)]

FunctionExpand[1/GoldenRatio==(((Sqrt[5])+1)/2)-1]

FunctionExpand[1/GoldenRatio==GoldenRatio-1]

ϕ+(1/ϕ)/.ϕ->GoldenRatio//FullSimplify

Numbers

N[E,20]

N[(Sqrt[2]),20]

Plots

Block[{is=200},
{
Plot[(((1/2))^x),{x,-2,2}
,PlotLegends->{"(0.5^x)"}
,ImageSize->is
],
Plot[(1^x),{x,-2,2}
,PlotLegends->{"(1^x)"}
,ImageSize->is
],
Plot[(2^x),{x,-2,2}
,PlotLegends->{"(2^x)"}
,ImageSize->is
]
}
]//Row

Block[{is=200, b=0.5},
Plot[(((1/2))^x),{x,-2,2}
,PlotLegends->{"(0.5^x)"}
,ImageSize->is
]
]

Block[{is=200},
Plot[(1^x),{x,-2,2}
,PlotLegends->{"(1^x)"}
,ImageSize->is
]
]

Block[{is=200},
Plot[(2^x),{x,-2,2}
,PlotLegends->{"(2^x)"}
,ImageSize->is
]
]

Block[{is=200, b=0.5},
Plot[(((1/2))^x),{x,-2,2}
,PlotLegends->{"(E^(λ x))"}
,ImageSize->is
]
]

Block[{is=200},
Plot[(2^x),{x,-2,2}
,PlotLegends->{"(E^(λ x))"}
,ImageSize->is
]
]

Block[{is=200},
Plot[2 (2^x),{x,-2,2}
,PlotLegends->{"(E^λ)(E^(λ x))"}
,ImageSize->is
]
]

Block[{is=200},
Plot[(2^(x+1)),{x,-2,2}
,PlotLegends->{"(E^(λ(x+1)))"}
,ImageSize->is
]
]

Human population growth

Wikipedia has a table of rate of natural rate of increase (using numbers from the CIA World Factbook), that can be sorted by crude natural increase. Number range from -5.7 (for Bulgaria) to 36.61 for Niger. (These are all per 1000 population.) Looking through the table, it appears that 30 would be a reasonable guess at the maximum rate, but maybe i should take 20 to be conservative. The total land surface area is around 1.5×(10^8)(km^2) (https://hypertextbook.com/facts/2001/DanielChen.shtml). Suppose we started with two humans. How long would it take to reach 10 people per (m^2)?

rhuman=0.02

landSA=Quantity[1.5×(10^8),(("Kilometers")^2)]

landSA/Quantity[0.1,(("Meters")^2)]

Log[landSA/Quantity[0.1,(("Meters")^2)]]

Log[landSA/Quantity[0.1,(("Meters")^2)]]-Log[2]//N

Log[landSA/Quantity[0.1,(("Meters")^2)]]/rhuman

Solve[2(E^(rhuman t))==landSA/Quantity[0.1,(("Meters")^2)],t]

Solve[2(E^(r t))==landSA/Quantity[0.1,(("Meters")^2)],t]

What’s special about e?

Block[{xmin,xmax=1},
xmin = -xmax;
Plot[{
(E^x),
1+x
},{x,xmin,xmax}
,PlotLegends->{"(E^x)","slope = 1"}
,AspectRatio->Automatic
,Epilog->{
PointSize[Large],
Point[{{0,1}}]
}
]
]

Poisson process

From Ladislaus Bortkiewicz (1898). Das Gesetz der kleinen Zahlen. Leipzig: B.G. Teubner. P 24 tabulates for 14 Prussian Army corps in each of 20 years the number of men killed by the kick of a horse. It also includes a summary table that gives the number of corps*years in which 0, 1, ... men were killed:

horseKickDeaths={144,91,32,11,2}

Total[horseKickDeaths]

ndmax=Length[horseKickDeaths]-1

BTW, mules are nowhere mentioned—the word is Pferde.

meanDeaths=((Total[Range[0,ndmax]horseKickDeaths])/(Total[horseKickDeaths]))

{Exp[-meanDeaths],horseKickDeaths[[1]],Total[horseKickDeaths]Exp[-meanDeaths]}//N

So (144/280)=0.514 is nto a bd estimate of the probability of going a year with a man being kicked to death.

From Bortkiewicz’s detailed data, I can glean the following totals for # of years without a death by horse kick in a corps

runs=
{{0,4+3+3+5+4+6+6+2+1+6+4+11+9+2},
{1,5+4+1+2+2+1+3+3+1+3+4+2+4+2},
{2,2+2+3+2+2+2+1+1+1+1+1},
{3,2+2+1+2},
{4,2+1+1+1},
{5,1}
}

norm1=Sum[Exp[-n*meanDeaths],{n,0,∞}]

N[norm1]

runs[[;;,2]]

Show[
ListPlot[runs
,PlotStyle->PointSize[Large]
,PlotLegends->{"# runs"}
],
Plot[((Total[runs[[;;,2]]])/norm1)Exp[-n*meanDeaths],{n,0,5}
,PlotLegends->{"(Subscript[N, 0])(E^(-μ n))"}
,PlotStyle->ColorData[97][2]
]
]

1.0/Exp[-6meanDeaths]

((horseKickDeaths[[1]])/(Total[horseKickDeaths]))//N

meanDeaths//N

Show[
ListPlot[runs
,PlotStyle->PointSize[Large]
,PlotLegends->{"# runs"}
],
Plot[runs[[1,2]](((((horseKickDeaths[[1]])/(Total[horseKickDeaths]))))^n),{n,0,5}
,PlotLegends->{"(Subscript[N, 0])(p^n)"}
,PlotStyle->ColorData[97][2]
]
]

Buffer preparation

trispKa=8.07;
base=Quantity[20,(("Millimoles")/("Liters"))];
acid =Quantity[80,(("Millimoles")/("Liters"))]

pH=trispKa+Log[10,(base/acid)]