Growth/Decay Series

1. Sep 2, 2006

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?

2. Sep 3, 2006

Random333

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 3. Sep 3, 2006 HallsofIvy Staff Emeritus 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 nth 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 nth[/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. 4. Sep 3, 2006 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?

5. Sep 3, 2006

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

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