# What are the range of initial values using the Newton-Raphson method for (x-9)/(x^2+1)=0 for it to converge to 9?

Apr 19, 2018

As of now, what we can be sure of is that for the iteraction to converge, we must have $x \in \left(9 - \sqrt{82} , 9 + \sqrt{82}\right)$.

This is necessary, but not sufficient.

#### Explanation:

The Newton-Raphson iteration for solving $f \left(x\right) = 0$ is

${x}_{n + 1} = g \left({x}_{n}\right) , q \quad g \left(x\right) = x - \frac{f \left(x\right)}{f} ^ ' \left(x\right)$

In this case, there is only one fixed point for the iteration, $x = 9$. This simplifies the analysis somewhat (also renders the results not so interesting, in my opinion)

For $f \left(x\right) = \frac{x - 9}{{x}^{2} + 1}$ some amount of algebra leads to

$g \left(x\right) = \frac{9 + 27 {x}^{2} - 2 {x}^{3}}{1 + 18 x - {x}^{2}}$

The successive iterates ${x}_{1} , {x}_{2} , \ldots , {x}_{n} , \ldots$ can either come closer and closer to the fixed point $x = 9$, or move further away (to either $\pm \infty$. To investigate which of this will occur, let us look at

$| \frac{{x}_{n + 1} - 9}{{x}_{n} - 9} | = | \frac{g \left({x}_{n}\right) - 9}{{x}_{n} - 9} |$

If this is smaller than 1, successive iterates will come closer and closer to 9. If larger, they will go farther and farther away.

Now

$\frac{g \left(x\right) - 9}{x - 9} = \frac{2 x \left(x - 9\right)}{{x}^{2} - 18 x - 1}$

For the iteration to converge, we need

$- 1 < \frac{g \left(x\right) - 9}{x - 9} = \frac{2 x \left(x - 9\right)}{{x}^{2} - 18 x - 1} < 1$

So we need

$\frac{2 x \left(x - 9\right)}{{x}^{2} - 18 x - 1} - 1 < 0$

or

$\frac{{x}^{2} + 1}{{x}^{2} - 18 x - 1} < 0 \implies {x}^{2} - 18 x - 1 < 0$
(since ${x}^{2} + 1 > 0$)

This implies
${\left(x - 9\right)}^{2} < 82 \implies 9 - \sqrt{82} < x < 9 + \sqrt{82}$

Note that if this condition fails, then the point definitely moves farther and farther off from 9 on every iteration and hence it diverges.

This is a necessary condition for convergence, but is not a necessary one!