Results 1 to 5 of 5

Math Help - Arithmetic/Geometric Means

  1. #1
    Newbie
    Joined
    Jun 2006
    Posts
    4

    Unhappy Arithmetic/Geometric Means

    I'm supposed to prove that:

    If 0<y(1)<x(1) and x(n+1)=(x(n)+y(n))/2 and y(n+1)=sqrt(x(n)y(n)
    then 0<y(n)<x(n).

    I can't figure out how to get started on this proof. I can see that it should be an induction proof...but I can't figure out how to prove it. Any help would be great.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by dancingshoes63 View Post
    I'm supposed to prove that:

    If 0<y(1)<x(1) and x(n+1)=(x(n)+y(n))/2 and y(n+1)=sqrt(x(n)y(n)
    then 0<y(n)<x(n).

    I can't figure out how to get started on this proof. I can see that it should be an induction proof...but I can't figure out how to prove it. Any help would be great.
    First the result to be proven is true for n=1, the base case as its give.

    You do know the arithmetic geometric mean inequality?

    For any positive numbers a, b:

    (a+b)/2 >= sqrt(a.b)

    with equality if and only if a=b.

    (in this two variable version this is a corrolary of Pythagoras's theorem)

    Now applying this you should see that if for some k:

    0<y(k)<x(k),

    as x(k+1) is the arithmetic mean of y(k) and x(k), and y(k+1) is
    the geometric mean of x(k) and y(k) we have by the arithmetic
    geometric mean inequality:

    y(k+1)<x(k+1)

    (we have strict inequality because y(k)!=x(k)), and y(k+1)>0
    as it is the square root of the product of two positive numbers.

    That is sufficient, together with the base case, to conclude that

    0<y(n)<x(n)

    by mathematical induction.

    RonL
    Last edited by CaptainBlack; September 16th 2006 at 02:47 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jun 2006
    Posts
    4
    So to prove the arithmetic/geometric mean inequality, I should look at the pythagorean theorem and go from there?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by dancingshoes63 View Post
    So to prove the arithmetic/geometric mean inequality, I should look at the pythagorean theorem and go from there?
    Just consider a right triangle with sides (a+b)/2, (a-b)/2, sqrt(a.b), and
    the fact that the length of the hypotenuse is >= any other side.

    RonL
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Or you can be more elegant.
    You know that,
    (sqrt(a)-sqrt(b))^2=>0
    Thus,
    a-2\sqrt(ab)+b>=0
    Thus,
    a+b>=2\sqrt{ab}
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 0
    Last Post: December 14th 2011, 06:02 AM
  2. Replies: 3
    Last Post: April 30th 2011, 02:14 AM
  3. Replies: 4
    Last Post: November 9th 2010, 02:45 PM
  4. Graph Theory: Arithmetic and Geometric Means
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: September 22nd 2010, 06:23 AM
  5. inequality relating to arithmetic and geometric means
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: December 9th 2009, 09:51 PM

Search Tags


/mathhelpforum @mathhelpforum