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Newton's Method

  1. Feb 27, 2009 #1
    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)= [tex]\int x[/tex] = 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. jcsd
  3. Feb 27, 2009 #2

    Dick

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    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.
     
  4. Feb 28, 2009 #3

    HallsofIvy

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    That is certainly NOT "Newton's method" for solving equations. The left side looks like the formula for the tangent line at [itex]x_0[/itex] 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 [itex]x_0[/itex] and iterate:
    [tex]x_{n+1}= x_n- \frac{f(x_n)}{f'(x_n)}[/itex]
    until you have sufficient accuracy.

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