# Newton's Method

1. Feb 27, 2009

### toasted

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

Find, to three decimal places, the value of x such that e-x = x. Use Newton's method.

2. Relevant equations

3. The attempt at a solution
I looked up what Newton's method was and I found that it was

f(x)= $$\int x$$ = f(xo) + f](xo)(x-xo)

But I dont understand how I can apply it to this problem. Could someone just help me set up the problem, and I should be able to figure the problem from there. Thanks!

Last edited by a moderator: Feb 28, 2009
2. Feb 27, 2009

### Dick

You are trying to solve f(x)=e^x-x=0. Newton's method tells you how to turn an initial guess for a solution x0 into a better approximation of a solution, x1. x1=x0-f'(x0)/f(x0). Take an initial guess at the solution and try it out.

3. Feb 28, 2009

### HallsofIvy

Staff Emeritus
That is certainly NOT "Newton's method" for solving equations. The left side looks like the formula for the tangent line at $x_0$ and it is certainly not equal to f(x) (I don't know how the integral got in there).

As Dick said, to solve f(x)= 0, choose some starting value $x_0$ and iterate:
[tex]x_{n+1}= x_n- \frac{f(x_n)}{f'(x_n)}[/itex]
until you have sufficient accuracy.

Here, $f(x)= e^{-x}- x$ and either $x_0= 0$ or $x_0= 1$ will do.