Page 1 of 2 12 LastLast
Results 1 to 15 of 17

Math Help - Prove x^n > x for all x > 1.

  1. #1
    Member
    Joined
    Oct 2009
    From
    Canada
    Posts
    128

    Prove x^n > x for all x > 1.

    Suppose n > 1 is a positive integer. Prove x^n > x for all x > 1.
    (hint: suppose we know f and g are differentiable on the interval (-c,c) and f(0) = g(0), if f'(x) > g'(x) for all x \in(0,c), then f(x) > g(x) for all x \in (0,c).
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Mar 2009
    Posts
    378
    Do you know how to prove something by induction? If not I am unsure how to do this for general n.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Oct 2009
    From
    Canada
    Posts
    128
    Quote Originally Posted by lvleph View Post
    Do you know how to prove something by induction? If not I am unsure how to do this for general n.
    we have done a couple of induction prove, not so familiar with that, can we apply the thereom in the hints that is given?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member
    Joined
    Mar 2009
    Posts
    378
    What you want to do is use the hint on a base case (n=2). Then you will make the assumption that the problem hypothesis holds for n=k. Use this assumption to prove that it is true for n=k+1.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Oct 2009
    From
    Canada
    Posts
    128
    Quote Originally Posted by lvleph View Post
    What you want to do is use the hint on a base case (n=2). Then you will make the assumption that the problem hypothesis holds for n=k. Use this assumption to prove that it is true for n=k+1.
    why do we set n = 2?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Senior Member
    Joined
    Mar 2009
    Posts
    378
    Because it is easy and the problem statement says n > 1.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Oct 2009
    From
    Canada
    Posts
    128
    Quote Originally Posted by lvleph View Post
    Because it is easy and the problem statement says n > 1.
    then how are we gonna apply the hint to the proof, I can't see the relation between them
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by 450081592 View Post
    Suppose n > 1 is a positive integer. Prove x^n > x for all x > 1.
    (hint: suppose we know f and g are differentiable on the interval (-c,c) and f(0) = g(0), if f'(x) > g'(x) for all x \in(0,c), then f(x) > g(x) for all x \in (0,c).
    Note that if f(x)=x^n\quad n>1 and g(x)=x. That f(1)=1=g(1) and f'(x)=nx^{n-1}\ge n>1=g'(1) and the concluso follows.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Senior Member
    Joined
    Mar 2009
    Posts
    378
    Okay, I will show you the base case.

    Assume n=2.
     f(x) = x^2, \; g(x) = x
    Now let's show that these are equal at x=0
    f(1)=1^2 = 1 = g(1)
    Now we differentiate the two
     f'(x) = 2x
    <br />
g'(x) = 1
    For x>1
    <br />
2x > 2
    Therefore we see that f'(x)=2x>1 =g'(x) for x>1.
    Last edited by lvleph; November 20th 2009 at 09:46 PM.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Senior Member
    Joined
    Mar 2009
    Posts
    378
    Quote Originally Posted by Drexel28 View Post
    Note that if f(x)=x^n\quad n>1 and g(x)=x. That f(1)=1=g(1) and f'(x)=nx^{n-1}>n>1=g'(1) and the concluso follows.
    In fact, I guess this is a much quicker and simpler proof.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Member
    Joined
    Oct 2009
    From
    Canada
    Posts
    128
    Quote Originally Posted by lvleph View Post
    In fact, I guess this is a much quicker and simpler proof.

    so with this method, we do not need to prove by induction, right?
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Member
    Joined
    Oct 2009
    From
    Canada
    Posts
    128
    Quote Originally Posted by Drexel28 View Post
    Note that if f(x)=x^n\quad n>1 and g(x)=x. That f(1)=1=g(1) and f'(x)=nx^{n-1}\ge n>1=g'(1) and the concluso follows.
    how do we know g(x) is x?

    That f(1)=1=g(1) and f'(x)=nx^{n-1}\ge n>1=g'(1)

    and I dont get this line
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Senior Member
    Joined
    Mar 2009
    Posts
    378
    We know g(x)=x because you are trying to prove x^n > x and the hint was suggesting you f(x) > g(x)
    Wouldn't you agree  nx^{n-1} \ge n for  x>1 and  n>1? But we know that f'(x) = nx^{n-1} and  g'(x) = 1.
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Member
    Joined
    Oct 2009
    From
    Canada
    Posts
    128
    Quote Originally Posted by lvleph View Post
    We know g(x)=x because you are trying to prove x^n > x and the hint was suggesting you f(x) > g(x)
    Wouldn't you agree  nx^{n-1} \ge n for  x>1 and  n>1? But we know that f'(x) = nx^{n-1} and  g'(x) = 1.
    so we are just assuming g(x) = x?
    Follow Math Help Forum on Facebook and Google+

  15. #15
    Senior Member
    Joined
    Mar 2009
    Posts
    378
    Yes, because the hint was to prove f(x) > g(x) and we wanted to prove x^n > x so it makes sense to assume g(x)=x.
    Follow Math Help Forum on Facebook and Google+

Page 1 of 2 12 LastLast

Similar Math Help Forum Discussions

  1. Prove a/b and a/c then a/ (3b-7c)
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: March 23rd 2010, 06:20 PM
  2. prove,,,
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: March 1st 2010, 10:02 AM
  3. Prove |w + z| <= |w| +|z|
    Posted in the Algebra Forum
    Replies: 3
    Last Post: February 28th 2010, 06:44 AM
  4. Replies: 2
    Last Post: August 28th 2009, 03:59 AM
  5. How to prove that n^2 + n + 2 is even??
    Posted in the Algebra Forum
    Replies: 3
    Last Post: November 30th 2008, 02:24 PM

Search Tags


/mathhelpforum @mathhelpforum