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The Second Iteration

To begin the second iteration, we must define the new basepoint:

    x0:=4/5: y0:=7/10:
For this basepoint we have:

\begin{eqnarray*}linf(x,y)&=&-\frac{213}{100} + \frac{8}{5}x + \frac{7}{5}y=0\\
ling(x,y)&=&49 + x - \frac{14}{5}y=0.
\end{eqnarray*}


Once again we can solve the system obtained by setting these functions equal to zero. The solution is (82/105,1849/2940).

We can add the zero level sets for linf and ling for the basepoint (4/5,7/10) to our previous plot:

(Make sure A, B, and C have been defined above before executing these commands.)

    B2:=implicitplot({linf(x,y)=0,ling(x,y)=0},x=0..2,y=0..2,color=green):
    display([A,B,C,B2]);

Notice that the point of intersection of the new lines is very close to the true point of intersection. Indeed, it is hard to distinguish them on this plot. It appears that the points that we are producing are converging to the solution. We verify this analytically by taking the norm of the differences of the coordinates of the points for each of the three approximations. Here we use the norm function in the package linalg:

    with(linalg,norm):
    evalf(norm([(-1 + sqrt(17))/4,sqrt((-1 + sqrt(17))/8)]-[1,1]));
    evalf(norm([(-1 + s*rt(17))/4,sqrt((-1 + sqrt(17))/8)]-[4/5,7/10]));
    evalf(norm([(-1 + sqrt(17))/4,sqrt((-1 + sqrt(17))/8)]-[82/105,1849/2940]));
These approximations are converging to the root and we might anticipate that we can obtain even better approximations by carrying out further iterations. Before we do this, let us formalize the process.


next up previous
Next: The Sequence of Approximations Up: Differentiation of Real-Valued Functions Previous: The First Iteration

2000-08-31