Results 1 to 5 of 5

Math Help - existence of root

  1. #1
    Junior Member
    Joined
    Nov 2007
    Posts
    32

    existence of root

    Prove that if an/(n+1) + an-1/n + ... + a1/2 + a0 = 0 (ak in R), then the equation anx^n + an-1x^(n-1) + ... + a1x + a0 = 0 has at least one real root between 0 and 1.
    This is my work.
    Let p(x) = (an/n+1)x^(n+1) + ..... a0x.
    Then p'(x) = anx^n + ......... + a0 (precisely what we need to show has a root). This part I'm not sure about. If p(x) has two distinct real roots (call them a and b and WLOG b < d), then p(b) = p(d) = 0. Rolle's Theorem on the interval [b,d] tells us that there exists a number r such that p'(r) = 0, which is the root I'm looking for. Does this make sense? Then, all I'm unsure about is how to show p(x) has two distinct real roots and how the interval [0,1] plays in the problem.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by agentZERO View Post
    Prove that if an/(n+1) + an-1/n + ... + a1/2 + a0 = 0 (ak in R), then the equation anx^n + an-1x^(n-1) + ... + a1x + a0 = 0 has at least one real root between 0 and 1.
    Consider f(x) = a_nx^n+...+a_1x+a_0 on [0,1]. But the Mean-Value theorem for integrals we have \int_0^1 f(x) dx = (1-0)f(c) for some c\in (0,1).
    Thus,
    \frac{a_n}{n+1}+\frac{a_{n-1}}{n}+...+\frac{a_1}{2}+a_0 = 0 = f(c).
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Nov 2007
    Posts
    32
    Thanks for helping, but I can't use that theorem (we haven't done anything with integrals yet). Is my idea hopeless?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by agentZERO View Post
    Thanks for helping, but I can't use that theorem (we haven't done anything with integrals yet). Is my idea hopeless?
    Consider f(x) = \frac{a_n}{n+1}x^n+\frac{a_{n-1}}{n}x^{n-1}+...+\frac{a_1}{2}x+a_0x on the interval [0,1]. This function is continous on the closed interval and differenciable on the open interval. Thus, by mean value theorem: f'(c)(1-0) = f(1)-f(0) but f(1)=f(0)=0. Thus, f'(c)=a_nc^n+...+a_0 = 0 for some c\in (0,1).
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Nov 2007
    Posts
    32
    I realized after I wrote the second post that my p(0)=p(1)=0. Thanks for the help.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Prove the existence of the square root of 2
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: October 17th 2011, 02:50 AM
  2. The existence of root 2
    Posted in the Differential Geometry Forum
    Replies: 9
    Last Post: October 1st 2009, 07:35 AM
  3. Existence of a root
    Posted in the Math Challenge Problems Forum
    Replies: 1
    Last Post: September 9th 2009, 12:08 AM
  4. Replies: 2
    Last Post: April 6th 2009, 04:51 AM
  5. Replies: 1
    Last Post: March 29th 2008, 12:11 PM

Search Tags


/mathhelpforum @mathhelpforum