# Numerical Analysis: Newton's Method Problem

• Jan 31st 2009, 12:45 PM
wvlilgurl
Numerical Analysis: Newton's Method Problem
Use Newtons Method to find the solution accurate to within 10^-5 of the following problem.

x^2-2xe^(-x)+e^(-2x)=0

How do I know when to stop and how do I find P0 and P1?
• Jan 31st 2009, 03:39 PM
Jhevon
Quote:

Originally Posted by wvlilgurl
Use Newtons Method to find the solution accurate to within 10^-5 of the following problem.

x^2-2xe^(-x)+e^(-2x)=0

How do I know when to stop

as the problem said, you want to be accurate to 5 decimal places, so keep finding a better approximation until the first 5 decimal places do not change

Quote:

and how do I find P0 and P1?
i suppose you mean the initial guess?

first note that you have \$\displaystyle (x - e^{-x})^2\$

so a good initial guess would be one where \$\displaystyle x\$ is close to \$\displaystyle e^{-x}\$. in fact, it is an equivalent problem to solve \$\displaystyle x - e^{-x} = 0\$. would be easier to run Newton's method on it too. you can use a graph to come up with a guess

Hope that helps