No shortage of marriagable women with polygyny due to exponential population growth rate

Additional Note added September 26, 2015.  This work says nothing about whether or not the practice is ethical or unethical it is simply addressing a mathematical fallacy commonly made.

Copyright

2010

Carl Janssen

First created September 3, 2010

Modified September 6, 2010

Are there enough women?  Looking at the relationship between population growth rate and polygyny through basic differential equations.

I just posted this quickly.   Let me know if I made math errors or other errors.

A common statement I have seen is similar to the following, “if men start having polygamous marriages, there would be some men who would have to be unmarried, because for every one man who is born approximately one woman is born.”

I wish to show that this is not true if the average age of marriage for men is higher than the average age of marriage for women as has been practiced commonly in many cultures in which polygamy is practiced.   I will use differential equations to show that the number of women of marriageable age and older is greater than the number of men of marriageable age and older, if men typically marry younger women.

The most simplified population growth model I will start with assumes that the number of births-deaths per time is directly proportional to the population at that time

Pretend an exactly equal number of males and females are born and die

R is proportionality constant

K is a constant for the differential equations solution

T is time

Let P = the number of males = the number of females

P is to be treated as dimensionless.  That is the population divided by 1 person.

dp/dt=R*p

(dp)/p=R*dt

(take the anti-derivative that is integrate and put in a constant)

Ln(p) = Rt + K

e^(Ln(p))=e^(Rt+K)

P=e^(Rt+K)

If all men marry after age M and all women marry after age F

Then the number of marriageable men equals

e^(K+ [R*(t-m)])

And the number of marriageable females equals

e^(K+ [R*(t-f)])

So if each man is to marry exactly N women and marry all N women at the same time.  Then

Marriageable Women>= Marriageable Men*N

So

N*e^(K+[R*(t-m)]) <= e^(K+[R*(t-f)])

N<=e^(R*[m-f])

As m-f increases the highest N possible increases.  M-F equals how much older the men are than the women.

Ln(n)<=R*[m-f]

m-f=>[ln(n)]/R

notice that when n=1 the ln(n)=0

Now let us assume

That the number of births-deaths per time is directly proportional to N (the number of wives per man) times the population P.

dp/dt= N*R*P

(dp)/p=N*R*dt

Ln(p)=[T*N*R]+K

P=e^([T*N*R)+K]

Number of marriageable men

e^(K+[(T-m)*N*R])

Number of marriageable women

e^(K+[(T-f)*N*R])

At least N marriageable women for every marriageable man

N* e^(K+[(T-m)*N*R])<= e^(K+[(T-f)*N*R])

N<=e^([m-f]*N*R)

Ln(N)=[m-f]*N*R

[m-f]=[Ln(N)]/[N*R]

Note that the limit as N approaches infinity of [ln(N)]/N = 0 from above

Note that when N =1 the ln(N)= 0

For N => 2 as N increases the lowest m-f required for each N decreases. 

That is the greater the number of wives each man marries the closer in age the men can be to their wives, but the lower number of wives each man marries the greater the age gap required for each man to be able to have at least N wives.  Note also that I am talking about the minimum m-f required there is nothing to prevent a greater m-f from being used.

m-f is how many years older every man would have to be than his wives, in order for every man to be able to marry at least N wives, in the type of mathematical model described above.