Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

I Newton's method for approximating solutions of functions

  1. Nov 16, 2016 #1
    In my calculus textbook, it shows that a function's solution can be approximated using an approximated function tangent to the original function based on an approximated solution, where the equation to find the approximated is L(x) = f(X0) + f'(X0)*(X-X0), where when rearranged, gives x = Xo - (f(X0)/f'(X0)) it doesn't give any reasoning as to why this is the equation. I do understand that the f(X0)/f'(X0) does in some way represent the margin of error of X0, and would (asymptotically) approach this margin of error, but never fully reach it (at least in the cases of the example equations given by the textbook). However, I would like to know how it works and how this equation was derived.

    Any help would be appreciated, thanks.

  2. jcsd
  3. Nov 16, 2016 #2


    User Avatar
    Science Advisor

    The formula works if you have got a starting value close to the solution. L(x) is Taylor's formula when only the first term is used.
  4. Nov 16, 2016 #3


    Staff: Mentor

    It's really not very complicated.

    You have a curve y = f(x), with a known point ##(x_0, f(x_0))## on the curve, and you know the slope of the tangent, ##f'(x_0)## at that point. Run a line segment from ##(x_0, f(x_0))## down the tangent line to where it intersects the x-axis. That x value, ##x_1## is your first approximation to the root, the value of x for which f(x) = 0. You should be able to find the equation of this tangent line and work out what ##x_1## would be.

    Newton's method continues the process using a new point, ##(x_1, f(x_1))## and known derivative value, ##f'(x_1)## to get another approximation for the root, ##x_2##. Under certain conditions, the method converges pretty rapidly, but in other conditions, it doesn't.

    Most calculus books show how this formula is derived.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?