Results 1 to 4 of 4

Thread: Numerical Analysis - Method of Undetermined Co-efficients

  1. #1
    Newbie
    Joined
    Dec 2009
    Posts
    22

    Numerical Analysis - Method of Undetermined Co-efficients

    Having a lot of trouble with this one, not even sure where to start:

    Q3: Use the method of undetermined coefficients to prove that, for two given points $\displaystyle x0, x1$ (different) and any gives values $\displaystyle y0, y1, z0$ and $\displaystyle z1$ there exists a unique polynomial p(x) of degree 3 such that:

    $\displaystyle p(x0) = y0; p(x1) = y1; p'(x0) = z0; p'(x1) = z1 .$ [25 marks]

    My lecturer has not specified whether or not MATLAB may be used. Is this question possible without it? If so, how do I go about it?

    P.S - This is the correct forum for this, isn't it?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,731
    Thanks
    3011
    Quote Originally Posted by MickQ View Post
    Having a lot of trouble with this one, not even sure where to start:

    Q3: Use the method of undetermined coefficients to prove that, for two given points $\displaystyle x0, x1$ (different) and any gives values $\displaystyle y0, y1, z0$ and $\displaystyle z1$ there exists a unique polynomial p(x) of degree 3 such that:

    $\displaystyle p(x0) = y0; p(x1) = y1; p'(x0) = z0; p'(x1) = z1 .$ [25 marks]

    My lecturer has not specified whether or not MATLAB may be used. Is this question possible without it? If so, how do I go about it?

    P.S - This is the correct forum for this, isn't it?
    The problem says "polynomial of degree 3" and "use the method of undetermined coefficients"- have you tried that?

    Any polynomial of degree 3 is of the form $\displaystyle p(x)= ax^3+ bx^2+ cx+ d$ which has four "undetermined coefficients" and you are given four conditions. That will give you four linear equations to solve for a, b, c, and d. Since the problem only says "prove there exist", all you need to do is show that the four equations have a unique solution.

    For example, "$\displaystyle p(x_0)= y_0$" means that $\displaystyle y_0= ax_0^3+ bx_0^2+ cx_0+ d$, "$\displaystyle p(x_1)= y_1$" means that $\displaystyle y_1= ax_1^3+ bx_1^2+ cx_1+ d$, "$\displaystyle p'(x_0)= z_0$" means that $\displaystyle z_0= 3ax_0^2+ 2bx_0+ c$, and "$\displaystyle p'(x_1)= z_1$" means that $\displaystyle z_1= 3ax_1+ 2bx_1+ c$.

    Those are your four linear equations for a, b, c, and d. There are many different ways to show that such a set of equations has a unique solution and I don't know which you have learned. One might be to just go ahead and find the solution, in terms of $\displaystyle x_0, x_1, y_0, y_1, z_0$, and $\displaystyle z_1$, of course. Another, more sophisticated and probably simpler, would be to show that the coefficient matrix could be row reduce to the identity matrix. Still another would be to show that the determinant of the coefficient matrix is not 0.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Jan 2009
    Posts
    715
    Just my opinion , i am not going to answer your question


    Given that $\displaystyle p(x) $ is a polynomial of degree $\displaystyle 3 $ so its derivative is a polynomial of degree $\displaystyle 2$


    By Lagrange's Interpolation formula ,

    but wait ! Don't we need to find three points to construct a quadratic function ? We only have two points $\displaystyle p'(x_0) = z_0 $ and $\displaystyle p'(x_1) = z_1 $ . Where is the third point ?

    Now , i let $\displaystyle x_3 =0 , p'(x_3) = p'(0) = t $ (undetermined)

    and hope that neither $\displaystyle x_1 $ nor $\displaystyle x_2 $ is also zero . haha


    Now , we have

    $\displaystyle p'(x ) = \frac{x(x-x_2)}{x_1 (x_1 - x_2)} z_1 + \frac{x(x-x_1)}{x_2 (x_2 - x_1)} z_2 + \frac{(x-x_1)(x-x_2)}{x_1 x_2} t $

    followed by integration and substitution $\displaystyle p(x_1) = y_1 ~,~ p(x_2) = y_2 $ , we obtain two linear equations with two unknowns .

    Therefore , we can confirm the two unknowns $\displaystyle t , C $ from them .

    I guess even though one of $\displaystyle (x_1,x_2) $ is zero , it is still ok because i believe we could finally eliminate the denominators $\displaystyle x_1 , x_2 $ .
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Jan 2009
    Posts
    715
    I thought of another method in my dream last night , i'm still using Lagrange's formula .


    $\displaystyle p(x_1) = y_1$
    $\displaystyle p(x_2) = y_2 $


    $\displaystyle p'(x_1) = z_1$
    $\displaystyle p'(x_2) = z_2 $

    Consider $\displaystyle p'(x) = \lim_{a\to 0 } \frac{ p(x+a) - p(x) }{a} $

    so $\displaystyle p'(x_1) = z_1 = \lim_{a\to 0 } \frac{ p(x_1+a) - p(x_1) }{a}$

    $\displaystyle y_1 + a z_1 = p(x_1 + a) $ and

    $\displaystyle y_2 + a z_2 = p(x_2 + a) $ and the above

    $\displaystyle y_1 = p(x_1) $

    $\displaystyle y_2 = p(x_2 )$

    Now , we have four points , so that a cubic polynomial could be easily obtained .


    Consider

    $\displaystyle \frac{ (x-x_2)(x-x_1-a) (x-x_2 -b)}{(x_1-x_2) (x_1-x_1-a) (x_1-x_2-b) } y_1 + $ $\displaystyle \frac{ (x-x_1)(x-x_2) (x-x_2 - b)}{(x_1 + a -x_1) (x_1 + a-x_2) (x_1 + a -x_2 - b ) } (y_1 + a z_1 ) $ $\displaystyle a,b \to 0$


    $\displaystyle = z_1 \frac{ (x-x_1)(x-x_2)^2 }{(x_1 - x_2)^2} $ $\displaystyle + (x-x_2-b)(x-x_2)(x-x_1)(x-x_1-a) \frac{y_1}{a} $ $\displaystyle ( \frac{1}{ [ (x_1 + a ) -x_2 ] [ (x_1 + a) - x_2 - b ] [ x - (x_1 + a) ] } $ $\displaystyle - \frac{1}{ [ (x_1 ) -x_2 ] [ (x_1 ) - x_2 - b ] [ x - (x_1 ) ] } )$



    $\displaystyle = z_1 \frac{ (x-x_1)(x-x_2)^2 }{(x_1 - x_2)^2} $ $\displaystyle + y_1 (x-x_2-b)(x-x_2)(x-x_1)(x-x_1-a) \frac{ \partial }{ \partial t } \left [ \frac{1}{ (t-x_2)(t - x_2 - b)(x-t) } \right ] |_{t=x_1}$


    $\displaystyle = z_1 \frac{ (x-x_1)(x-x_2)^2 }{(x_1 - x_2)^2} $ $\displaystyle + y_1 \frac{(x-x_2-b)(x-x_2)(x-x_1)(x-x_1-a) }{ (t-x_2)(t - x_2 - b)(x-t) } \left( \frac{1}{x-t} - \frac{1}{t - x_2} - \frac{1}{t - x_2 - b} \right ) |_{t=x_1} $


    $\displaystyle = z_1 \frac{ (x-x_1)(x-x_2)^2 }{(x_1 - x_2)^2} $ $\displaystyle + y_1 \frac{ ( x-x_2)^2 (x-x_1)}{(x_1 - x_2)^2} \left( \frac{1}{ x - x_1} - \frac{2}{ x_1 - x_2 } \right ) $


    $\displaystyle = z_1 \frac{ (x-x_1)(x-x_2)^2 }{(x_1 - x_2)^2}$ $\displaystyle + y_1 \frac{ ( x-x_2)^2 }{(x_1 - x_2)^2} - 2 y_1 \frac{ ( x-x_2)^2 (x-x_1)}{(x_1 - x_2)^3} $


    This is a part of $\displaystyle p(x) $ , the other part is similar to it but we have to exchange between $\displaystyle 1 $ and $\displaystyle 2$.


    $\displaystyle p(x) = z_1 \frac{ (x-x_1)(x-x_2)^2 }{(x_1 - x_2)^2}$ $\displaystyle + y_1 \frac{ ( x-x_2)^2 }{(x_1 - x_2)^2} - 2 y_1 \frac{ ( x-x_2)^2 (x-x_1)}{(x_1 - x_2)^3} $ $\displaystyle + z_2 \frac{ (x-x_2)(x-x_1)^2 }{(x_2 - x_1)^2}$ $\displaystyle + y_2 \frac{ ( x-x_1)^2 }{(x_2 - x_1)^2} - 2 y_2 \frac{ ( x-x_1)^2 (x-x_2)}{(x_2 - x_1)^3} $


    This method looks so complicated but it is not , the dizzy thing is just the typing ,but the idea is simple indeed !
    Last edited by simplependulum; Mar 31st 2010 at 08:10 PM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Numerical Analysis: Newton's Method Problem
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Jan 31st 2009, 03:39 PM
  2. Numerical Analysis Bisection Method Problem
    Posted in the Calculus Forum
    Replies: 5
    Last Post: Jan 19th 2009, 11:46 AM
  3. Secant Method (Numerical Analysis)
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Oct 27th 2008, 04:47 AM
  4. Newton's Method: Numerical Analysis
    Posted in the Advanced Applied Math Forum
    Replies: 3
    Last Post: Jun 12th 2008, 01:33 PM
  5. Replies: 3
    Last Post: Nov 3rd 2007, 01:43 PM

Search Tags


/mathhelpforum @mathhelpforum