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Notes on Exponential Grow

 

Motivated by prof. Albert Bartlet Lecture on Aritmetic, Population and Energy.

Lecture available on youtube in 8 parts. The first part inserted here:

Part two in youtube. Click on the logo below:



On how to derive the formulas professor Bartlet explains
.

Whenever something grows at a given rate, the new quantity is equal to the original quantity plus what it grew at the given rate:

N1= P0 + P0 r       extracting the common factor P0 yields:

N1= P0 (1+r) and since we assume that the original quantity grows continuously, we get some other N2 which also grew at the same rate; then,

N2=N1+N1r   or N2=N1(1+r) —again extracting the common factor— and substituting N1 by P0(1+r) we have:

N2= P0 (1+r)(1+r) = P0(1+r)2   since this process never stops we are allow to represent the Nn quantity as:

Nn= P0 (1+r)n   

which is the well known formula to calculate compound interest, an example of a constantly growing quantity. Rates are expressed as annual rates, so the given rate is divided into the number of times it is applied, namely monthly, daily, or even at every instant, this is, continuously compounded. That’s why the rate will be divided by “n” and the whole expression raised to the “tth” power, where t represent the number of years.
The formula then becomes:

Nn= P0(1 +r/n)nt    

The quantity  (1 + 1/n)n    has a peculiar  property. When n increases to a very large number, the whole expression converges to a specific number. This is what is called “the limit” of the quantity  (1 + 1/n)n when n increases without bound, or, as mathematicians says, approaches infinity. The number, to be denoted as e, stands for exponential — known as Euler’s number— is the base of the exponential growth and is defined as follows:

e= lim (1 + 1/n)n  


It’s value is  2.718281828459045… up to infinity.  It also can be demonstrated (see notes below) that

ex  = lim (1 + x/n)n          
       

Now we are allowed to substitute the limit of  (1 + r/n)n    by er when n approaches infinity.

Nn=P0ert    which is the formula to calculate the N quantity knowing the initial population, the rate of growth, and the time elapsed.
A very specific problem is the one that calculate the time needed to duplicate the initial population or value under exponential growth.
In this case Nn= 2P0 which leads to:

2P0=P0ert   dividing both sides of the equation by P0 and taking natural log, it yields: 

ln2=rt (see properties of logs to understand the two missing steps)

0.693/r= t

When rate is expressed as a percentage then %/100= r and the value of t equals  69.3/r

Note 1:

In finance, instead of 69 or 69.3 the numbers 70 or 72 are used since they both have many small divisors. The error introduced
in the calculation for small percentages can be ignored.

Note 2:

Demonstration of 

ex  = lim (1 + x/n)n          
       

We start from de definition of e:

e= lim (1 + 1/n)n  

Therefore,

 ex= [lim (1 + 1/n)n]x          
                 

Then, since the nth power of the limit is the limit of the nth power (the power function is continuous):

 ex= [lim (1 + 1/n)nx]          
               

now multiply 1/n by x/x:

 ex= [lim (1 + x/nx)nx]          
               

Since x is a fixed number, nx is a growing quantity that only depends on n and that quantity can be identified by any other variable, say, m.
The expression of the limit becomes:

ex= [lim (1 + x/m)m]          
      

Which is another expresion of

ex  = lim (1 + x/n)n          
       

Q.E.D

[Quod Erat Demonstrandum].

...............................................................................................

A more descriptive approach to derive a formula for exponential grow:

Firstly, let’s describe the system we are modeling: we have a quantity that grows continuously without environmental restrictions (population, energy consumption, etc). Denote the quantity by  x  and assert that it grows at a given rate k.  The simpler linear equation that describe this behavior is y=kx ( a given value of x is increased by a factor k, yields a new quantity y); we also know that our system changes at every instant, the quantity x increases in a limiting quantity that can be denoted by dx on every limiting fraction of time, dt. Now we are ready to express the information on this system as:

dx/dt = kx

[An equation that combines functions and its derivatives to model the behavior of a system is called a differential equation].

How can we move forward?  We have in the left side of the equation the rate of change of x. In order to know what happens when time pass by, we need to add all those instantaneous changes in the quantity x. This is, we need to integrate the equation. Recall: integration put together smalls parts, yield the limiting quantity if an infinite sum of terms. We now rewrite the equation so that both variables ( x and dx) are on the same side of the equation, then integrate:

dx/x = kdt   

A very basic knowledge of rules of integrations leads to:

ln x = kt + c

In order to determine the value of the constant c, let’s evaluate the equation at the initial time, t = 0 then c = lnx0  where is x0 is the initial quantity. Now we may get rid of the subscript by introducing a = x0 just to facilitate the notation. Also knowing that elnx = x (see definition of log) and that any base b raised to a+d powers is equal to the product of the powers: b(a+c) = ba bc
  
We eliminate the ln x by raising both sides of the equation to the e power:

elnx = e(kt+lna)

x = aekt

Which is equivalent to the formula previously obtained:

Nn= P0ert