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Math Help - Numerical Analysis - Method of Undetermined Co-efficients

  1. #1
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    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 x0, x1 (different) and any gives values y0, y1, z0 and z1 there exists a unique polynomial p(x) of degree 3 such that:

    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?
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  2. #2
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    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 x0, x1 (different) and any gives values y0, y1, z0 and z1 there exists a unique polynomial p(x) of degree 3 such that:

    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 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, " p(x_0)= y_0" means that y_0= ax_0^3+ bx_0^2+ cx_0+ d, " p(x_1)= y_1" means that y_1= ax_1^3+ bx_1^2+ cx_1+ d, " p'(x_0)= z_0" means that z_0= 3ax_0^2+ 2bx_0+ c, and " p'(x_1)= z_1" means that 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 x_0, x_1, y_0, y_1, z_0, and 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.
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  3. #3
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    Just my opinion , i am not going to answer your question


    Given that  p(x) is a polynomial of degree  3 so its derivative is a polynomial of degree 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  p'(x_0) = z_0 and  p'(x_1) = z_1 . Where is the third point ?

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

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


    Now , we have

     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  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  t , C from them .

    I guess even though one of  (x_1,x_2) is zero , it is still ok because i believe we could finally eliminate the denominators  x_1 , x_2  .
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  4. #4
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    I thought of another method in my dream last night , i'm still using Lagrange's formula .


     p(x_1) = y_1
     p(x_2) = y_2


     p'(x_1) = z_1
     p'(x_2) = z_2

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

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

     y_1 + a z_1 = p(x_1 + a) and

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

     y_1 = p(x_1)

     y_2 = p(x_2 )

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


    Consider

     \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 + \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 )  a,b \to 0


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



     = z_1 \frac{ (x-x_1)(x-x_2)^2 }{(x_1 - x_2)^2}  + 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}


     = z_1 \frac{ (x-x_1)(x-x_2)^2 }{(x_1 - x_2)^2}  + 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}


     = z_1 \frac{ (x-x_1)(x-x_2)^2 }{(x_1 - x_2)^2}  + 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 )


     = z_1 \frac{ (x-x_1)(x-x_2)^2 }{(x_1 - x_2)^2}  + 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  p(x) , the other part is similar to it but we have to exchange between  1 and 2.


     p(x) = z_1 \frac{ (x-x_1)(x-x_2)^2 }{(x_1 - x_2)^2}  + 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}  +  z_2 \frac{ (x-x_2)(x-x_1)^2 }{(x_2 - x_1)^2}  + 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; March 31st 2010 at 08:10 PM.
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