Results 1 to 13 of 13

Thread: roll theorem question..

  1. #1
    MHF Contributor
    Joined
    Nov 2008
    Posts
    1,401

    roll theorem question..

    i am given that
    $\displaystyle
    \frac{{a_0 }}{{n + 1}} + \frac{{a_1 }}{n} + \frac{{a_2 }}{{n - 1}} + .. + a_n = 0 \\
    $
    prove that there is a solution for 0<x<1
    using roll theorem
    $\displaystyle
    a_0 x^n + a_1 x^{n - 1} + .. + a_n = 0 \\
    $
    Last edited by transgalactic; Feb 5th 2009 at 09:45 PM.
    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 transgalactic View Post
    i am given that
    $\displaystyle
    \frac{{a_0 }}{{n + 1}} + \frac{{a_1 }}{n} + \frac{{a_2 }}{{n - 1}} + .. + a_n = 0 \\
    $
    prove that there is a solution for 0<x<1
    using roll theorem
    $\displaystyle
    a_0 x^n + a_1 x^{n - 1} + .. + a_n = 0 \\
    $
    Consider the function $\displaystyle f(x) = \tfrac{a_0}{n+1}x^{n+1} + \tfrac{a_1}{n}x^n + ... + a_n x $.
    Notice that $\displaystyle f(0) = f(1) = 0$.
    Now apply Rolle's theorem and get $\displaystyle f'(x_0) = 0$ for some $\displaystyle 0 < x_0 < 1$.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Nov 2008
    Posts
    1,401
    if i put x=1 on your function
    f(1) doesnt equal zero

    how you get f(1)=0
    ??
    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 transgalactic View Post
    how you get f(1)=0
    ??
    That is the hypothesis you were given.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Joined
    Nov 2008
    Posts
    1,401
    i cant see the equation that if i will put x=1
    i will get 0

    ?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Nov 2008
    Posts
    1,401
    there is no such hypothesis in the posted question
    ??
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by transgalactic View Post
    there is no such hypothesis in the posted question
    ??
    $\displaystyle f(1) = \frac{{a_0 }}{{n + 1}} + \frac{{a_1 }}{n} + \frac{{a_2 }}{{n - 1}} + .. + a_n = 0$
    That is the hypothesis.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Joined
    Nov 2008
    Posts
    1,401
    Quote Originally Posted by ThePerfectHacker View Post
    $\displaystyle f(1) = \frac{{a_0 }}{{n + 1}} + \frac{{a_1 }}{n} + \frac{{a_2 }}{{n - 1}} + .. + a_n = 0$
    That is the hypothesis.
    so its only an assumption
    its not a fact??
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by transgalactic View Post
    so its only an assumption
    its not a fact??
    You are given this assumption part of the problem.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor
    Joined
    Nov 2008
    Posts
    1,401
    ok if f(0)=f(1) then there is a point c for which
    f'(c)=0
    0<c<1
    but its not a solution for f(x)

    its an extreme point
    ??
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by transgalactic View Post
    ok if f(0)=f(1) then there is a point c for which
    f'(c)=0
    0<c<1
    but its not a solution for f(x)

    its an extreme point
    It is a solution to $\displaystyle f'(x) = 0$ (with $\displaystyle x=c$).
    But $\displaystyle f'(x) = a_0x^n + ... + a_n$.
    Thus, you found a solution to the polynomial equation $\displaystyle a_0x^n + ... + a_n = 0$.
    Follow Math Help Forum on Facebook and Google+

  12. #12
    MHF Contributor
    Joined
    Nov 2008
    Posts
    1,401
    so you take a function
    $\displaystyle
    f(x) = \tfrac{a_0}{n+1}x^{n+1} + \tfrac{a_1}{n}x^n + ... + a_n x\\
    $
    which is an anti derivative of my given equation that i need find solution to.
    $\displaystyle
    f'(x)=a_0 x^n + a_1 x^{n - 1} + .. + a_n
    $

    so the extreme point of f(x) is the solution of f'(x)

    thanks
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Yes, finally you understand.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Counterfeit dice roll question
    Posted in the Statistics Forum
    Replies: 6
    Last Post: Apr 15th 2010, 01:04 PM
  2. roll theorem prove question..
    Posted in the Calculus Forum
    Replies: 12
    Last Post: Mar 10th 2009, 10:45 PM
  3. probability die roll
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: Jan 29th 2009, 09:05 PM
  4. roll a die
    Posted in the Advanced Statistics Forum
    Replies: 3
    Last Post: Apr 9th 2008, 08:35 AM
  5. roll of the die
    Posted in the Statistics Forum
    Replies: 8
    Last Post: Jul 18th 2006, 08:48 PM

Search Tags


/mathhelpforum @mathhelpforum