Calculus
posted by Anonymous .
You are using Newton's method to solve
e^x  2. If you first guess is x1=1, what value will you calculate for the next approximation x2?
...I don't know what Newton's method is, so I looked it up...and I'm still confused. Help plz?
Thx

first of all, to "solve" we need an equation.
You have the expression e^x  2
Are you solving e^x  2 = 0 ?
I will assume you are.
let f(x) = e^x  2
f ' (x) = e^x
Newton's method says ...
x_{n+1} = x_{n}  f(x_{n})/f ' (x_{n})
so x_{2} = 1  (e^1  2)/e^1
= .73576
x_{3} = .73576  (e^.73576  2)/e^.73576 = .69404
x_{4} = .69404  (e^.69404  2)/e^.69404 = .69314758
notice we are rapidly approaching some kind of fixed value.
If the value of x of our input returns as output, we have the solution.
A most remarkable procedure.
Using logs I get x = .69314718 , so our method was accurate to 6 digits after only 3 iterations.