Question

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?

Answer 1

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

This is necessary, but not sufficient.

Explanation:

The Newton-Raphson iteration for solving `f(x) = 0` is

`x_(n+1) = g(x_n),qquad g(x) = x-(f(x))/f^'(x)`

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(x) = (x-9)/(x^2+1)` some amount of algebra leads to

`g(x) = (9+27x^2-2x^3)/(1+18x-x^2)`

The successive iterates `x_1,x_2,...,x_n,...` can either come closer and closer to the fixed point `x=9`, or move further away (to either `pm oo`. To investigate which of this will occur, let us look at

`|(x_(n+1)-9)/(x_n-9)| = |(g(x_n)-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

`(g(x)-9)/(x-9) = (2x(x-9))/(x^2-18x-1)`

For the iteration to converge, we need

`-1 < (g(x)-9)/(x-9) = (2x(x-9))/(x^2-18x-1) < 1`

So we need

`(2x(x-9))/(x^2-18x-1) -1<0`

or

`(x^2+1)/(x^2-18x-1)<0 implies x^2-18x-1<0`
(since `x^2+1>0`)

This implies
`(x-9)^2<82 implies 9-sqrt82< x < 9+sqrt82`

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!