Deriving the Fourth-Order Adams-Moulton Formula Using Undetermined Coefficients

[muso07]

Homework Statement

Use the method of undetermined coefficients to derive the fourth-order Adams-Moulton formula
x_k+1=x_k+ (h/24)(9f_k+1+19f_k-5f_k-1+f_k-2

Homework Equations

Adams-Moulton: $$x_{k+1}=x_{k}+h\Sigma ^{n}_{i=0}\beta_{i}f_{k+1-i}$$

The Attempt at a Solution

It has to be exact for x(t)=1, x(t)=t, x(t)=t², x(t)=t³

So I have to solve for the coefficients of $$x_{k+1}=x_{k}+h(\beta_{0}f_{k+1} + \beta_{1}f_{k} + \beta_{2}f_{k-1} + \beta_{3}f_{k-2})$$

I know I need to set up a system of equations, but I'm confused as to what f_k is.. Is it just the derivative of x_k?

edit: Nevermind, I figured it out. :)

 

[mighty2000]

Hello there,

Thank you for your post and for using the method of undetermined coefficients to derive the fourth-order Adams-Moulton formula. You are correct in setting up a system of equations to solve for the coefficients. In this case, fk represents the derivative of xk, which can be written as x'(t).

To begin, we can substitute the values given in the problem into the Adams-Moulton formula:
x_{k+1}=x_{k}+h\Sigma ^{n}_{i=0}\beta_{i}f_{k+1-i}
x_{k+1}=x_{k}+h(\beta_{0}f_{k+1} + \beta_{1}f_{k} + \beta_{2}f_{k-1} + \beta_{3}f_{k-2})

We can then plug in the values for xk, xk+1, and fk+1:
x_{k+1}=x_{k}+h(\beta_{0}f_{k+1} + \beta_{1}f_{k} + \beta_{2}f_{k-1} + \beta_{3}f_{k-2})
1 = 1 + h(\beta_{0}f_{k+1} + \beta_{1}f_{k} + \beta_{2}f_{k-1} + \beta_{3}f_{k-2})
t = t + h(\beta_{0}f_{k+1} + \beta_{1}f_{k} + \beta_{2}f_{k-1} + \beta_{3}f_{k-2})
t^2 = t^2 + h(\beta_{0}f_{k+1} + \beta_{1}f_{k} + \beta_{2}f_{k-1} + \beta_{3}f_{k-2})
t^3 = t^3 + h(\beta_{0}f_{k+1} + \beta_{1}f_{k} + \beta_{2}f_{k-1} + \beta_{3}f_{k-2})

We can then solve for the coefficients by equating the coefficients of the same power of t on both sides of the equations. This will give us a system of equations to solve for