Results 1 to 3 of 3

Math Help - Help with 2 Other Questions ??

  1. #1
    Newbie
    Joined
    Oct 2006
    Posts
    17

    Help with 2 Other Questions ??

    i If someone sends an answer
    please write that in mathematical form

    i have problems with my english
    Attached Thumbnails Attached Thumbnails Help with 2  Other  Questions ??-untitled-2.gif  
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    For any n>0

    Consider the vector space \mathbb{R}^n over \mathbb{R}.

    Now, let
    \bold{u}=\left<1,\frac{1}{2},\frac{1}{3},...,\frac  {1}{n}\right>

    \bold{v}=\left<1,\frac{1}{2^3},\frac{1}{3^3},...,\  frac{1}{n^3}\right>

    Then by Cauchy-Schwartz Inequality we have,
    \bold{ab}\leq ||\bold{a}||||\bold{b}||

    Now, (the dot product)
    \bold{ab}=1+\frac{1}{2^4}+...\frac{1}{n^4}

    And the norms,
    ||\bold{a}||=\sqrt{1+\frac{1}{2^2}+...+\frac{1}{n^  2}}
    ||\bold{b}||=\sqrt{1+\frac{1}{2^6}+...+\frac{1}{n^  6}}
    Thus,

    1+\frac{1}{2^4}+...+\frac{1}{n^4}\leq \sqrt{1+\frac{1}{2^2}+...+\frac{1}{n^6}}\cdot \sqrt{1+\frac{1}{2^6}+...+\frac{1}{n^6}}

    Square both sides (non-negative terms),
    \left(1+\frac{1}{2^4}+...+\frac{1}{n^4}\right)^2\l  eq  \left(1+\frac{1}{2^2}+...+\frac{1}{n^2}\right)\cdo  t \left(1+\frac{1}{2^6}+...+\frac{1}{n^6}\right)
    Last edited by ThePerfectHacker; November 2nd 2006 at 07:28 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Let,
    x\geq 1
    Then,
    \frac{1}{x^n}-\frac{1}{x^{n+1}}=\frac{x^{n+1}-x^n}{x^{2n+1}}\leq x^{n+1}-x^n
    Thus,
    x^{n+1}-x^n\geq \frac{1}{x^n}-\frac{1}{x^{n+1}}
    Thus,
    x^{n+1}+\frac{1}{x^{n+1}}\geq x^n+\frac{1}{x^n}

    Now, if 0<x<1
    Then,
    \frac{1}{x}>1
    Thus, (by the previous discussion),
    (1/x)^{n+1}+\frac{1}{(1/x)^{n+1}}\geq (1/x)^n+\frac{1}{(1/x)^n}
    Thus,
    x^{n+1}+\frac{1}{x^{n+1}}\geq x^n+\frac{1}{x^n}
    Last edited by ThePerfectHacker; November 2nd 2006 at 07:30 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. More log questions
    Posted in the Algebra Forum
    Replies: 1
    Last Post: March 31st 2010, 04:58 PM
  2. Please someone help me with just 2 questions?
    Posted in the Algebra Forum
    Replies: 3
    Last Post: May 4th 2009, 04:55 AM
  3. Some Questions !
    Posted in the Geometry Forum
    Replies: 1
    Last Post: May 3rd 2009, 03:09 AM
  4. Replies: 4
    Last Post: July 19th 2008, 07:18 PM
  5. Replies: 3
    Last Post: August 1st 2005, 01:53 AM

Search Tags


/mathhelpforum @mathhelpforum