Results 1 to 3 of 3

Math Help - Proof

  1. #1
    Newbie
    Joined
    Feb 2008
    Posts
    13

    Proof

    Any advice?
    Attached Thumbnails Attached Thumbnails Proof-aproof.jpg  
    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 BACONATOR View Post
    Any advice?
    Well you could start by writting out what it is you have to prove for this to be true.

    RonL
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member roy_zhang's Avatar
    Joined
    Mar 2008
    Posts
    64
    First \Theta(g)=\Theta(n) represents the set of functions w(n) such that there exists positive constants c_1,\;c_2 and n_0 such that 0\le c_1 n\le w(n) \le c_2 n,\;\forall n\ge n_0. So in order to show that f\in \Theta(n), we need to find out positive constants c_1,\;c_2 and n_0 such that

    c_1 n \le \frac{n^2+4}{n+4}\le c_2 n,\;\;\forall n\ge n_0

    Dividing n from each parts gives us

    c_1  \le \frac{n^2+4}{n^2+4n}\le c_2

    Consider the function h(n)=\frac{n^2+4}{n^2+4n}, given that n\in Z^{+}, by differentiation we can find the minimum value of function h(n)to be \frac{2\sqrt{5}+10}{6\sqrt{5}+10}\approx 0.618034, also realize that \frac{n^2+4}{n^2+4n}\le 1,\;\forall n\in Z^{+}. Hence we can pick c_1=\frac{1}{2},\;c_2=1 and n_0=1, we have

     \frac{1}{2}\le \frac{n^2+4}{n^2+4n}\le 1,\;\;\forall n\ge 1

    this shows that f\in \Theta(n).

    Roy
    Last edited by roy_zhang; April 5th 2008 at 09:32 PM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 5
    Last Post: October 19th 2010, 10:50 AM
  2. Replies: 0
    Last Post: June 29th 2010, 08:48 AM
  3. [SOLVED] direct proof and proof by contradiction
    Posted in the Number Theory Forum
    Replies: 2
    Last Post: February 27th 2010, 10:07 PM
  4. Proof with algebra, and proof by induction (problems)
    Posted in the Discrete Math Forum
    Replies: 8
    Last Post: June 8th 2008, 01:20 PM
  5. proof that the proof that .999_ = 1 is not a proof (version)
    Posted in the Advanced Applied Math Forum
    Replies: 4
    Last Post: April 14th 2008, 04:07 PM

Search Tags


/mathhelpforum @mathhelpforum