1. modular exponentiation/recurrence relation

Assume the pop. of world in 2002 is 6.2B and growing at a rate of 1.3% each year.

a. Set up a recurrence relation for pop. n years after 2002
b. find an explicit formula for the pop. of the world after n years
c. what will be the pop. on 2022

The recurrence relation I came up with is
A(n+1)= 1.3*A(n) , where A(n) is 6.3B
But how do I incorporate modular exponentiation?

2. Re: modular exponentiation/recurrence relation

Hello, delgeezee!

Assume the pop. of world in 2002 is 6.2B and growing at a rate of 1.3% each year.

a. Set up a recurrence relation for pop. n years after 2002.

b. find an explicit formula for the pop. of the world after n years.

c. What will be the pop. on 2022?

The recurrence relation I came up with is
A(n+1)= 1.3*A(n) , where A(n) is 6.2B

You don't understand percents, do you?

The population increases each year by a factor of 1.3% or 0.013.

$(a)\;A(n+1) \:=\:1.013A(n),\;\;A(0) \,=\,6.2B$

$(b)\;A(n) \:=\:(6.2B)(1.013)^n\;\text{ where }n\text{ = number of years after 2002.}$

$(c)\;A(20) \:=\:(6.2B)(1.013)^{20} \;\approx\;8.03B$

3. Re: modular exponentiation/recurrence relation

Originally Posted by Soroban
Hello, delgeezee!

The population increases each year by a factor of 1.3% or 0.013.

$(a)\;A(n+1) \:=\:1.013A(n),\;\;A(0) \,=\,6.2B$

$(b)\;A(n) \:=\6.2B)(1.013)^n\;\text{ where }n\text{ = number of years after 2002.}" alt="(b)\;A(n) \:=\6.2B)(1.013)^n\;\text{ where }n\text{ = number of years after 2002.}" />

$(c)\;A(20) \:=\6.2B)(1.013)^{20} \;\approx\;8.03B" alt="(c)\;A(20) \:=\6.2B)(1.013)^{20} \;\approx\;8.03B" />

Sorry about that. It increase by 1.3% each year
so A(n+1)=A(n)*1.013
But I am having trouble grasping modular expression. I need to make this into a program and have the computer compute the answer. The only problem I am having is using the large interger 6.2Billion.

4. Re: modular exponentiation/recurrence relation

My apologies again. I see now that i don't need to use modular expression.