Results 1 to 4 of 4

Math Help - modular exponentiation/recurrence relation

  1. #1
    Member
    Joined
    Sep 2011
    Posts
    106
    Thanks
    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?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,683
    Thanks
    615

    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

    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Sep 2011
    Posts
    106
    Thanks
    1

    Re: modular exponentiation/recurrence relation

    Quote Originally Posted by Soroban View Post
    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

    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.}" />

    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.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Sep 2011
    Posts
    106
    Thanks
    1

    Re: modular exponentiation/recurrence relation

    My apologies again. I see now that i don't need to use modular expression.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. modular exponentiation example
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: March 3rd 2012, 12:03 PM
  2. Modular Exponentiation
    Posted in the Number Theory Forum
    Replies: 2
    Last Post: November 13th 2010, 01:54 PM
  3. Modular Exponentiation
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: September 13th 2010, 05:31 PM
  4. Modular exponentiation
    Posted in the Number Theory Forum
    Replies: 0
    Last Post: March 8th 2009, 11:08 PM
  5. Modular Exponentiation
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: August 27th 2008, 04:31 PM

Search Tags


/mathhelpforum @mathhelpforum