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If a polynomial function has at least one real root, will Newton's Method always converge to one of those real roots? (no attracting fixed point). Is there a counterexample where the guess does NOT start with a horizontal tangent line (i.e when the slope of the tangent line is 0)?

MaximeJaccon
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  • No. Consider $x^3-1$ where you start at $0$. – lulu Jan 29 '24 at 17:44
  • It will surely converge, if you can wisely choose the first point, from where the iteration starts. – M.Riyan Jan 29 '24 at 17:47
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    Post edit: still no. Try $x^3-2x+2$ and guess $0$. – lulu Jan 29 '24 at 17:48
  • These sort of failures are well documented...I suggest an online search. – lulu Jan 29 '24 at 17:50
  • @lulu Question for you: what do you know about the case when the polynomial has only real roots? – cnikbesku Jan 29 '24 at 18:16
  • @cnikbesku I don't think that matters...the oscillation problem is just a coincidence between the slopes of tangents. I'll see if I can design an example. – lulu Jan 29 '24 at 18:18
  • @lulu Ok, thank you, I see. I think I constructed an example myself. – cnikbesku Jan 29 '24 at 18:19
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    Just adapting my first one, $f(x)=x^3-2x-1$, guess $.429926366\cdots$. So, not a pretty example but I think it proves the point. – lulu Jan 29 '24 at 18:24
  • @lulu I also found an example: $f(x)=x^4-3x^2+1$ and guess a real root of $7x^4-9x^2-1$. – cnikbesku Jan 29 '24 at 18:27
  • My safest bet is that it converges for almost all initial values, so that if you add a random value to the root after each iteration you should come about that close to a root. – cnikbesku Jan 29 '24 at 18:37
  • I cannot say specifically about polynomials, but it is easy to define smooth functions where one has to get very lucky in your initial position for Newton's method to converge. A safer bet is to identify two starting values where the function has different signs, then if Newton's method ever tells you to leave the interval, ignore it to toss in a bisection step instead. Either way, shrink the interval based on the value of the function at the next estimate. This guarantees convergence, while allowing for Newton's method to give fast convergence when it is actually working. – Paul Sinclair Jan 30 '24 at 20:53

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