Growth/Decay Series: Deriving C=C1*r^n & C1*(1-r)^n

[Swapnil]

If "C" starts with initial value "C1" and it grows or decays at the rate of "r" every "n" time, then the function that models the growth or the decay of "C" is C = C_1\cdot {(r)}^n and C = C_1\cdot {(1-r)}^n, respectively. I know this makes sense but how do you derive such a forumla?

 

[Random333]

If "C" starts with initial value "C1" and it grows or decays at the rate of "r" every "n" time, then the function that models the growth or the decay of "C" is C = C_1\cdot {(r)}^n and C = C_1\cdot {(1-r)}^n, respectively. I know this makes sense but how do you derive such a forumla?

Well let us do this by using some figures.

Lets say we have $100, and it increases in value at 5% per year.
C_1=100
r=1.05

Using
C = C_1\cdot {(r)}^n

Therefore
C = 100\cdot {(1.05)}^n

After one year it is
C=105

and so on...

Hopefully that helps

 

[HallsofIvy]

The way you have phrased it: " grows or decays at the rate of "r" every "n" times" your equations are not correct. "Growing at the rate of r" means "multiplied by r". "Every n times" means that that happens every n^th step- everytime the variable, t say, is a multiple of n, there is another "whole" multiplication. If n= 5 and t= 15, there have been t/n= 15/5= 3 multiplications. taking t/n for general values of n allows for fractional periods. The formula for process that "increases or decreases by rate r every n<sup>th[/b] time" is
Cr^{t/n} or C(1-r)^{t/n}

If you mean "grows or decays at the rate r for a total of n times, then you are multiplying C by r (or 1- r) repeatedly: C, (C)r= Cr, (Cr)r= Cr2, (Cr2)r= Cr3, etc. The general term is Crn for growth and C(1- r)n for decay.</sup>

 

[Swapnil]

Sorry, its hard to put these things in words for me. Let me explain my question with the aid of an example that Random333 gave.

Say you have $100 dollars in a bank. It increases at the rate of 0.05 every year.

So at the end of the 1st year, the amount is:

A_1 = 100 + 0.05*100 = 105

and at the end of the 2nd, 3rd, and 4th year the amount, respectively, is:

A_2 = 105 + 0.05*105 = 110.25

A_3 = 110.25 + 0.05*110.25 = 115.7625

A_4 = 115.7625 + 0.05*115.7625 = 121.550625

My question is that how can we model this growth by the following formula:

A_n = A_0 (1+r)^n

I mean, how did they derived such a formula?

 

[Moo Of Doom]

Well, you have the relationship A_n = A_{n-1} * (1+r), right? Well then, you can plug in that same formula for A_{n-1} and get

A_n = (A_{n-2}*(1+r))*(1+r)=A_{n-2}*(1+r)^2

and more generally,

A_n = A_{n-m}*(1+r)^m

Plugging in n for m gives

A_n = A_{n-n}*(1+r)^n=A_0 *(1+r)^n