Finding the 6-decimal Root of e^-x=lnx in [1,2]

[shadowman187]

Homework Statement

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

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 I am 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]?

 

[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.

 

[HallsofIvy]

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 x_n and x_n+1 are the same to 6 decimal places- the first 6 digits after the decimal point are the same.