# Newton method

1. Oct 28, 2008

1. The problem statement, all variables and given/known data

to six decimal places the root of the equation e^-x=lnx over interval [1,2]

3. The attempt at a solution
(e^-x)-lnx=0
F'(x) = (-e^-x)-(1/x)

x(subcript(n))-(e^-x)-lnx/(-e^-x)-(1/x)

the problem im having is getting it to six decimal places. do i have to go int0 1.00001 or something like that in order to get six decimal places. since the interval is [1,2]?

2. Oct 28, 2008

### rock.freak667

In the interval [a,b], a good approximation would be

$$x_1=\frac{a|F(b)|+b|F(a)|}{|F(a)|+|F(b)|}$$

To get to 6 dp, just put x1 to 7dp then when you work out x2 (the answer will be in 7dp), approximate it to 6dp.

In other words, when calculating using the 7dp number and when writing the answers, write it to 6dp, so you will see when the answers are the same to the same degree of accuracy.

3. Oct 28, 2008

### HallsofIvy

Staff Emeritus
I'm not sure what you mean by "go into 1.00001" but that number has only 5 decimal places.
Repeat the iteration until you get xn and xn+1 are the same to 6 decimal places- the first 6 digits after the decimal point are the same.