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?

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- Jan 31st 2009, 12:45 PMwvlilgurlNumerical 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 PMJhevon
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?

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