From: spellucci@mathematik.th-darmstadt.de (Peter Spellucci) Newsgroups: sci.math.num-analysis Subject: Re: How does Newton's Method fail to find root(s)? Date: 2 Jan 1998 17:03:00 GMT In article , pin3692@hotmail.com (Kangpin) writes: |> Hi, |> How are you? I came from New York. I am a high school student. I'm |> studying calculus right now. My teacher asked us to find 3 examples |> that of how Newton's method fails to find root(s). Could you please |> tell me how? Or please give some examples and explain them for me. Or, ......... the method may diverge, e.g. take f(x)=arctangent(x) and x_0 >=2. then the sequence x_k generated by newton's method oszillates, the absolute values |x_k| tending to infinity, although this is a nice monotonic function with exactly one real zero. the method may oscillate (also nonconvergent) take f(x)=x-x^3 and x_0=+ 1/sqrt(5) or -1/sqrt(5). the method may diverge undetermined, take f(x)=(x^2-1)*(x^2-4)+3 (which has no real zeroes) and any x_0 large enough. finally, the method simply may fail because f'(x_k)=0 for some k. hope this helps peter ============================================================================== From: kovarik@mcmail.cis.McMaster.CA (Zdislav V. Kovarik) Newsgroups: sci.math.num-analysis Subject: Re: How does Newton's Method fail to find root(s)? Date: 2 Jan 1998 19:55:36 -0500 In article <68j6k4$g0k$1@sun27.hrz.tu-darmstadt.de>, Peter Spellucci wrote: :In article , pin3692@hotmail.com :(Kangpin) writes: :|> Hi, :|> How are you? I came from New York. I am a high school student. I'm :|> studying calculus right now. My teacher asked us to find 3 examples :|> that of how Newton's method fails to find root(s). Could you please :|> tell me how? Or please give some examples and explain them for me. Or, :......... :the method may diverge, e.g. take f(x)=arctangent(x) and x_0 >=2. :then the sequence x_k generated by newton's method oszillates, the :absolute :values |x_k| tending to infinity, although this is a nice monotonic :function :with exactly one real zero. Nice, I haven't seen this one before! :the method may oscillate (also nonconvergent) take f(x)=x-x^3 and :x_0=+ 1/sqrt(5) or -1/sqrt(5). This one has "unstable" oscillations: if you start with x_0=(1+2^(-52))/sqrt(5), it will take 27 iterations using MATLAB to approach x_n=1 within round-off. Experiment some more and find out how long it takes before x_0=(1-2^(-52) ends up in x=0. There are "stable" oscillations, and Smale gave a recipe how to construct them. His example is x^3-2*x+2=0, with one real solution which is negative, and such that starting with x_0 between -0.1 and +0.1 will end up in oscillations between numbers converging to 0 and numbers converging to 1. It is fun to show why the oscillations settle (within round-off) on the 2-cycle so quickly. :the method may diverge undetermined, take :f(x)=(x^2-1)*(x^2-4)+3 (which has no real zeroes) and any x_0 large :enough. :finally, the method simply may fail because f'(x_k)=0 for some k. ... and theoretically, one can find starting values for which the method cycles, too. Cheers, ZVK (Slavek).