Results 1 to 11 of 11

Math Help - Function and root

  1. #1
    Junior Member mathfun's Avatar
    Joined
    Mar 2011
    Posts
    35

    Function and root

    Let $$a \in {{\Cal R}^*}$$ and $$b,c \in {\Cal R}$$ such that $$4ac < {\left( {b - 1} \right)^2}$$. Let $$f:{\Cal R} \to {\Cal R}$$ be a function such that $$f\left( {a{x^2} + bx + c} \right) = a{f^2}\left( x \right) + bf\left( x \right) + c,\forall x \in {\Cal R}$$. Prove that $$f\left( {f\left( x \right)} \right) = x$$ has at least one root
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member TheChaz's Avatar
    Joined
    Nov 2010
    From
    Northwest Arkansas
    Posts
    600
    Thanks
    2
    What have you tried so far?
    "Prove that..." is your responsibility, and while many of us (myself included) are willing and able to help, you have to show some effort besides registering for a forum and copying the question down in LaTex markup.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member mathfun's Avatar
    Joined
    Mar 2011
    Posts
    35
    Ok, seeing the hypothesis i said: there are roots $$d,e$$ of the equation $$a{x^2} + bx + c = x$$ because its equivalent to $$a{x^2} + \left( {b - 1} \right)x + c = 0$$. I set d and e in the given relation. and then i get stuck.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member mathfun's Avatar
    Joined
    Mar 2011
    Posts
    35
    Let $${x_0}$$ be such that it is the solution of the equation $$a{x^2} + bx + c = x$$ where $${x_0} = \tfrac{{1 - b + \sqrt {{{\left( {b - 1} \right)}^2} - 4ac} }}{{2a}}$$.
    I set x0 in the given relation $$f\left( {a{x_0}^2 + b{x_0} + c} \right) = a{f^2}\left( {{x_0}} \right) + bf\left( {{x_0}} \right) + c \Rightarrow a{f^2}\left( {{x_0}} \right) + \left( {b - 1} \right)f\left( {{x_0}} \right) + c = 0$$

    Now one of the roots of the last equation are $$f\left( {{x_0}} \right) = \tfrac{{1 - b + \sqrt {{{\left( {b - 1} \right)}^2} - 4ac} }}{{2a}} = {x_0}$$ and we get the wanted.


    Is it correct?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member TheChaz's Avatar
    Joined
    Nov 2010
    From
    Northwest Arkansas
    Posts
    600
    Thanks
    2
    Quote Originally Posted by mathfun View Post
    Ok, seeing the hypothesis i said: there are roots $$d,e$$ of the equation $$a{x^2} + bx + c = x$$ because its equivalent to $$a{x^2} + \left( {b - 1} \right)x + c = 0$$. I set d and e in the given relation. and then i get stuck.
    You're almost there. What is the discriminant of this quadratic? If you subtract 4ac from the given inequality, you will conclude that
    ("the discriminant") >= 0
    and this implies that the quadratic has >= 1 real root.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member mathfun's Avatar
    Joined
    Mar 2011
    Posts
    35
    Actually it has 2 roots, right? Plaese see the solution i gave in post No. 5
    Last edited by mathfun; March 26th 2011 at 04:01 AM.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member TheChaz's Avatar
    Joined
    Nov 2010
    From
    Northwest Arkansas
    Posts
    600
    Thanks
    2
    I don't believe so. If (b - 1)^2 = 4ac, then your "two" solutions are actually the same.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Junior Member mathfun's Avatar
    Joined
    Mar 2011
    Posts
    35
    but $${\left( {b - 1} \right)^2} > 4ac$$
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Super Member TheChaz's Avatar
    Joined
    Nov 2010
    From
    Northwest Arkansas
    Posts
    600
    Thanks
    2
    Hah! All this time I thought it was >=!!
    ok you win
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Junior Member mathfun's Avatar
    Joined
    Mar 2011
    Posts
    35
    Quote Originally Posted by mathfun View Post
    Let $${x_0}$$ be such that it is the solution of the equation $$a{x^2} + bx + c = x$$ where $${x_0} = \tfrac{{1 - b + \sqrt {{{\left( {b - 1} \right)}^2} - 4ac} }}{{2a}}$$.
    I set x0 in the given relation $$f\left( {a{x_0}^2 + b{x_0} + c} \right) = a{f^2}\left( {{x_0}} \right) + bf\left( {{x_0}} \right) + c \Rightarrow a{f^2}\left( {{x_0}} \right) + \left( {b - 1} \right)f\left( {{x_0}} \right) + c = 0$$

    Now one of the roots of the last equation are $$f\left( {{x_0}} \right) = \tfrac{{1 - b + \sqrt {{{\left( {b - 1} \right)}^2} - 4ac} }}{{2a}} = {x_0}$$ and we get the wanted.


    Is it correct?
    Could you please tell me if this solution is correct?
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Member
    Joined
    Dec 2010
    Posts
    107
    Looks fine to me.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Limit of the nth root of a function
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: October 22nd 2011, 08:16 AM
  2. square root function
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: September 16th 2011, 10:32 PM
  3. Cancelling out a root function of y
    Posted in the Pre-Calculus Forum
    Replies: 7
    Last Post: July 3rd 2011, 03:54 AM
  4. Square Root Function
    Posted in the Pre-Calculus Forum
    Replies: 8
    Last Post: May 17th 2008, 11:50 AM
  5. root function analysis
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 6th 2008, 11:19 PM

Search Tags


/mathhelpforum @mathhelpforum