Numerical Analysis - Method of Undetermined Co-efficients

**MickQ** · March 30th 2010, 03:39 PM

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?

**HallsofIvy** · March 31st 2010, 02:40 AM

Originally Posted by MickQ

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.

**simplependulum** · March 31st 2010, 03:20 AM

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 $x_1 , x_2$ .

**simplependulum** · March 31st 2010, 07:57 PM

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 !

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