1. Help Please! : Newton's Method

Hi! We're doing Newton's method and I'm SO LOST! Lecturer only gave one example. Can anyone help me with the following question so I can compare? Don't know where to start! Annotations through your workings would be greatly appreciated. Thanks!!

Consider the solution of the following nonlinear equation using Newton's method. $\displaystyle{ e^{-1.3 x} = 20 x - 5 }$
After two iterations, using ten digit floating point arithmetic, the estimated solution is: x2 =0.2845382540
Using Newton's method find the next estimate of the solution and use this to determine an estimate of the error in x2.
Maintain at least TEN significant digits throughout all your calculations.

2. Have you been given a starting point for $x_0$ ?

If so then

$
x_1 = x_0 - \frac{f(x_0)}{f'(x_0)}
$

and so on...

3. Nope, no initial x. That's the whole question.

4. Gor the sequence of iterations to converge you have to find a point close to the solution. I would start at somewhere between 0 & 1because by looking at these functions separately I can see they intersect on that interval.

Spoiler:
$x = 0.3$

5. Originally Posted by pickslides
Have you been given a starting point for $x_0$ ?

If so then

$
x_1 = x_0 - \frac{f(x_0)}{f'(x_0)}
$

and so on...
First you have to rearrange the equation into the form

$f(x)=0$

In this case we have:

$e^{-1.3x}=20x-5$,

so we put:

$f(x)=20x-5-e^{-1.3x}$

Now we can apply Newtons's method to find x such that $f(x)=0$

Also you are told (more or less) to start from $x_2 =0.2845382540$

CB

6. Thanks! The two of you have been heaps of help.