Hi Everyone I have the following homework assignment problem that I am struggling with:
Consider Newton's method $$ x_{k+1} = \frac{1}{2} \left( x_k + \frac{a}{x_k} \right), \qquad a > 0, $$ for computing $\alpha = \sqrt{a}$ Let $ d_k = x_{k+1} - x_{k}$
Show that $$ x_{k} = \frac{a}{d_k + \sqrt{d_k^2+a}}$$
Use the above to show $$ |d_k| = \frac{d_{k-1}^2}{2 \sqrt{d_{k-1}^2 + a}} $$ Discuss the significance of this result with regard to the overall behavior of Newton's iteration.
I know it's just algebra but I'm having a hard time coming up with that form. I tried getting $x_{k+1}-x_k$ to one side there is always 2$x_{k+1}$ more than $x_k$.
Thank you.
You get also $$ x_k=\sqrt{a+d_k^2}-d_k. $$ Subtracting the expression for $x_{k-1}$ gives $$ d_{k-1}=x_k-x_{k-1}=\sqrt{a+d_k^2}-d_k-\sqrt{a+d_{k-1}^2}+d_{k-1} $$ from where you get after squaring $$ a+d_k^2=d_k^2+2d_k\sqrt{a+d_{k-1}^2}+a+d_{k-1}^2 $$ which reduces to the wanted expression.