Results 1 to 5 of 5

Math Help - Inequality puzzle

  1. #1
    Member
    Joined
    Sep 2009
    Posts
    177
    Thanks
    1

    Inequality puzzle

    Prove that \forall n \epsilon N, \prod_{i=1}^{n}(1+x^{2^i})<\frac{1}{1-x}
    0<x<1
    I know of 2 proofs. One, which is mine, and anouther that is better. Try to find both!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Quote Originally Posted by Chris11 View Post
    Prove that \forall n \epsilon N, \prod_{i=1}^{n}(1+x^{2^i})<\frac{1}{1-x}
    0<x<1
    I know of 2 proofs. One, which is mine, and another that is better. Try to find both!
    I think the problem looks better if the product start at i=0 rather than i=1. That makes the product bigger, but it's still less than \frac{1}{1-x}.

    Use the difference of two squares: (1-x^{2^i})(1+x^{2^i}) = 1-x^{2^{i+1}}. Therefore

    \prod_{i=0}^{n}(1-x^{2^i})\prod_{i=0}^{n}(1+x^{2^i}) = \prod_{i=1}^{n+1}(1-x^{2^i}). Divide both sides by \prod_{i=1}^{n}(1-x^{2^i}) to get

    (1-x)\prod_{i=0}^{n}(1+x^{2^i}) = 1-x^{2^{n+1}} < 1.

    Is that your proof, or is it better?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Sep 2009
    Posts
    177
    Thanks
    1
    That's actualy neither! I like it though. My proof expaned the darn polynomial via induction, showing that the coefficents were all one. Then, I subtracted that expansion from the geometric series for \frac{1}{1-x}. The slicker proof is just a multiplication by (1-x) ! You didn't have to multiply by that product. Of course, you can do a lot to the exponent and maintain that inequality!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Jun 2010
    From
    Israel
    Posts
    148
    Quote Originally Posted by Chris11 View Post
    Prove that \forall n \epsilon N, \prod_{i=1}^{n}(1+x^{2^i})<\frac{1}{1-x}
    0<x<1
    I know of 2 proofs. One, which is mine, and anouther that is better. Try to find both!
    Write S=\prod_{i=0}^{\infty}(1+x^{2^i}). (Only later did I notice that it's actually n and not \infty, but I wanted to post this anyway.)

    If we expand the infinite product S , a typical term is x^N, where N is a sum of powers of two.

    The following is the Basis Representation Theorem for base two.
    Every positive integer N can be represented uniquely as the sum of powers of two. Example: 5=2^2+2^0.

    Since the exponents of the terms in the infinite series S consist of all possible combinations of powers of two, it follows from the theorem that the exponents range over all the positive integers.

    Therefore, S=\prod_{i=0}^{\infty}(1+x^{2^i})=1+x^1+x^2+x^3+x^  4+x^5+\cdots=1/(1-x).

    We have \prod_{i=1}^{\infty}(1+x^{2^i})<(1+x)\prod_{i=1}^{  \infty}(1+x^{2^i})=S, i.e. \prod_{i=1}^{\infty}(1+x^{2^i})<1/(1-x).

    Here, \prod_{i=1}^{n}(1+x^{2^i})<\prod_{i=1}^{\infty}(1+  x^{2^i})<1/(1-x).

    So, a third one?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Sep 2009
    Posts
    177
    Thanks
    1
    Quote Originally Posted by melese View Post
    Write S=\prod_{i=0}^{\infty}(1+x^{2^i}). (Only later did I notice that it's actually n and not \infty, but I wanted to post this anyway.)

    If we expand the infinite product S , a typical term is x^N, where N is a sum of powers of two.

    The following is the Basis Representation Theorem for base two.
    Every positive integer N can be represented uniquely as the sum of powers of two. Example: 5=2^2+2^0.

    Since the exponents of the terms in the infinite series S consist of all possible combinations of powers of two, it follows from the theorem that the exponents range over all the positive integers.

    Therefore, S=\prod_{i=0}^{\infty}(1+x^{2^i})=1+x^1+x^2+x^3+x^  4+x^5+\cdots=1/(1-x).

    We have \prod_{i=1}^{\infty}(1+x^{2^i})<(1+x)\prod_{i=1}^{  \infty}(1+x^{2^i})=S, i.e. \prod_{i=1}^{\infty}(1+x^{2^i})<1/(1-x).

    Here, \prod_{i=1}^{n}(1+x^{2^i})<\prod_{i=1}^{\infty}(1+  x^{2^i})<1/(1-x).

    So, a third one?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: January 11th 2011, 09:20 PM
  2. Replies: 3
    Last Post: December 12th 2010, 02:16 PM
  3. A new puzzle?
    Posted in the Math Puzzles Forum
    Replies: 1
    Last Post: October 23rd 2010, 01:29 PM
  4. Need help with puzzle please
    Posted in the Math Puzzles Forum
    Replies: 1
    Last Post: January 15th 2010, 08:19 PM
  5. A new Puzzle :)
    Posted in the Math Challenge Problems Forum
    Replies: 3
    Last Post: October 13th 2009, 07:37 AM

Search Tags


/mathhelpforum @mathhelpforum