Solving e-x=x Using Newton's Method

[toasted]

Homework Statement

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

Homework Equations

The Attempt at a Solution

I looked up what Newton's method was and I found that it was

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

But I don't 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!

 

[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.

 

[HallsofIvy]

Homework Statement

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

Homework Equations

The Attempt at a Solution

I looked up what Newton's method was and I found that it was

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

But I don't 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!

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]<br /> until you have sufficient accuracy.<br /> <br /> Here, $f(x)= e^{-x}- x$ and either $x_0= 0$ or $x_0= 1$ will do.[/tex]