Results 1 to 8 of 8

Math Help - bisection method .. solving equation ?

  1. #1
    Newbie
    Joined
    Apr 2011
    Posts
    13

    bisection method .. solving equation ?

    Bisection method .. solving equation:

    x^3-9x+1
    Last edited by mr fantastic; April 6th 2011 at 01:29 AM. Reason: Copied title into main body of post.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,832
    Thanks
    1602
    First of all, what you have is not an equation, but an expression.

    Are you trying to solve for \displaystyle x in the equation \displaystyle x^3 - 9x + 1 = 0? Or is it equal to some other number?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Mar 2011
    From
    Awetuouncsygg
    Posts
    182
    Thanks
    12
    What you wrote is not an equation, it's an expression.

    "bijection method" - well, you have to transform *stuff*=*stuff, stuff* in something like f(x)=f(y) and then prove that f is bijective (evidently, you chose a convenient function), so x=y. Then solve the new equation

    Edit: o.o I was absolutly sure there is write "bijection". Sorry.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,832
    Thanks
    1602
    Quote Originally Posted by veileen View Post
    What you wrote is not an equation, it's an expression.

    "bijection method" - well, you have to transform *stuff*=*stuff, stuff* in something like f(x)=f(y) and then prove that f is bijective (evidently, you chose a convenient function), so x=y. Then solve the new equation
    The OP is asking for help with the BiSection method... In other words, continually cutting the region that the root falls in in half so that you close in on the root.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Apr 2011
    Posts
    13
    does it belong to integration or differentiation ?

    and what is a differential equation ... this is tough
    Last edited by mr fantastic; April 6th 2011 at 01:26 AM.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,832
    Thanks
    1602
    Quote Originally Posted by thomaztriz View Post
    does it belong to integration or differentiation ?

    and what is a differential equation ... yehova my lord .. this is tough
    It belongs to neither. And a differential equation is an equation that involves derivatives. Your equation does not have any.

    I suggest you read this.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,453
    Thanks
    1868
    The first thing you need to do is find an interval in which you know there is a solution.

    If we let f(x)= x^3- 9x+ 1, it's easy to see that f(0)= 0- 9(0)+ 1= 1> 0 and that f(1)= 1- 9+ 1= -7< 1. Since the value of this continous equation is postive at 0 and negative at 1, there must be a solution between them.

    Now "bisection" means "cutting in half. Half way between 0 and 1 is, of course, 1/2. What is f(1/2)?

    (Actually this is a surprisingly simple problem!)
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Apr 2011
    Posts
    13
    thanks a lot .. nice

    for a moment i was thinking about that equation ..

    and the derivatives ... and the integrals .. and the fundamental theorem of calculus .. ( do i make any sense?)

    then the differential equations ?

    and after that the numerical methods ..

    i wish all this stuff was much clearer ..
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Bisection method help
    Posted in the Calculus Forum
    Replies: 2
    Last Post: March 27th 2010, 11:38 AM
  2. Bisection method
    Posted in the Calculus Forum
    Replies: 3
    Last Post: February 24th 2010, 05:31 AM
  3. Replies: 4
    Last Post: October 30th 2009, 01:29 PM
  4. More Bisection MEthod
    Posted in the Calculus Forum
    Replies: 0
    Last Post: September 28th 2009, 12:15 AM
  5. Bisection Method Help
    Posted in the Math Topics Forum
    Replies: 4
    Last Post: November 20th 2008, 03:41 PM

Search Tags


/mathhelpforum @mathhelpforum