Thread: [SOLVED] Finding zeropoint with Newton's method

1. [SOLVED] Finding zeropoint with Newton's method

Explain why
f(x) = ln x - 1/x - 1 , x>0
have one and only one zero point
Use Newton's method to find approximately value for the zero-point. The answer must state with 2 correct decimals.

How do I solve this?
Why?

2. Originally Posted by mr_pj
Explain why
f(x) = ln x - 1/x - 1 , x>0
have one and only one zero point
Use Newton's method to find approximately value for the zero-point. The answer must state with 2 correct decimals.

How do I solve this?
Why?
It would help pitch answers at the correct level if
you tell us what course/level you are studying.

Look at the definition of $\displaystyle f(x)$, it is the
difference of two positive functions of $\displaystyle x$:

$\displaystyle ln(x), \mbox{ and }\ 1/x+1$
,

$\displaystyle f(x)$ is zero when these two functions are
equal. Now $\displaystyle ln(x)$ is an increasing function and
$\displaystyle 1/x+1$ is a decreasing function of $\displaystyle x$.
So it they are equal at some point, this is the only point at
which they are ever equal. This is clear if you sketch the
two functions (see attachment).

Look at your notes to find what the Newton's Method
is. You will find it gives you a new estimate of the
root of a function in terms of the old estimate.

You start with a guess, get a new estimate and feed
that back, repeating the procedure until it has converged
sufficiently for your purposes.

In this case starting with a guess of 2 I get:

>itert(10,2)

$\displaystyle n\ \ \ \ \ \ \ x_n$
$\displaystyle 0\ \ \ \ \ \ 2.000000$
$\displaystyle 1\ \ \ \ \ \ 3.075804$
$\displaystyle 2\ \ \ \ \ \ 3.543637$
$\displaystyle 3\ \ \ \ \ \ 3.590737$
$\displaystyle 4\ \ \ \ \ \ 3.591121$
$\displaystyle 5\ \ \ \ \ \ 3.591121$
$\displaystyle 6\ \ \ \ \ \ 3.591121$
$\displaystyle 7\ \ \ \ \ \ 3.591121$
$\displaystyle 8\ \ \ \ \ \ 3.591121$
$\displaystyle 9\ \ \ \ \ \ 3.591121$
$\displaystyle 10\ \ \ \ 3.591121$
>