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Let, say, I want to find an approximation of the root $r$ for $x^{3}=x+1$ using Newton's method. $$x_{n+1}=x_{n}-\frac{f\left(x_{n}\right)}{f^{\prime}\left(x_{n}\right)} \quad(n \geq 0)$$

I guess the initial point is denoted by $x_0$. My question is: how do I find this point? what I did was to finding a pair $a<b$ such that $f(a)f(b) < 0$. I found $a=1,b=2$. So $r$ must be in $[1,2]$. The initial point I choose to be some random point in $[1,2]$. So I choose $x_0 = 1$ . Is this the correct rational behind finding an initial point?

markvs
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Sorry
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  • But for doing that I need to plot it. My approach is fine I guess. @Moo – Sorry Oct 19 '20 at 22:26
  • What is "close enough" in this case? @Moo – Sorry Oct 19 '20 at 22:29
  • draw a picture. There is what we call a "basin of attraction" for which one step of Newton's method will get you closer to the target. Well, you got lucky, any start $x_0 > \frac{1}{\sqrt 3}$ with take you to a number bigger than the root, then all subsequent Newton steps take you backwards to the root, since the second derivative of the original function is positive there – Will Jagy Oct 19 '20 at 22:30

2 Answers2

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Making the problem more general, you are looking for the zero of a function $f(x)$ and you know that the solution is $\in (a,b)$.

If you have to decide if $x_0=a$ or $x_0=b$, it is quite simple :

  • If $f(a)\times f''(a) >0$ then $x_0=a$ guarantees that the solution will be reached without any overshoot of the solution
  • Else $x_0=b$

This is Darboux theorem.

Edit

In comments, @Carl Christian properly mentioned that this is necessary and sufficient provided that, in $[a,b]$, $f''(x)$ does not change sign. This was the first hypothesis in the linked paper (equation $(2)$) and the first theorem.

Claude Leibovici
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  • Thank you for the reference to Darboux's paper. It seems to me that the assumptions that the root is simple and that $f''(x)$ does not change sign are critical. – Carl Christian Oct 20 '20 at 17:15
  • @CarlChristian. Yes for the first but not for $f''(x)$. What matters is that $f(x_0)\times f''(x_0) >0$ to ensure no overshoot. Cheers :-) – Claude Leibovici Oct 21 '20 at 04:02
  • No, @CarlChristian is still right, you can look at this example. Checking one point is not sufficient to guarantee convergence or even lack of overshoot (consider starting very close to a relative extrema). – Simply Beautiful Art Oct 21 '20 at 12:53
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    In the statement of the theorem, Darboux assumes that the second derivative does not change sign on the interval $[a,b]$ (bottom of page 19: "la dérivée seconde ne change pas de signe") and this assumption is actively used in the proof (top of page 19: "... puisque $f''(x)$ ne s'annule pas entre $a$ et $b$". – Carl Christian Oct 21 '20 at 14:37
  • @CarlChristian. I apologize, be sure. You are totally correct. Cheers :-( – Claude Leibovici Oct 21 '20 at 14:51
  • No worries! I was happy to get the name and the original paper. – Carl Christian Oct 21 '20 at 14:54
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    Your answer has been upvoted even after you have recognized that it was wrong. I fear that users do not necessarily read the comments. If you update your answer with the correct statement of the theorem, then I will be happy to reverse my down vote. – Carl Christian Oct 23 '20 at 23:54
  • @CarlChristian. I made an edit. Thanks for having pointed my omission and mistake. – Claude Leibovici Oct 24 '20 at 02:13
  • My thanks, I could not ask for more. I continue to benefit from your answers. – Carl Christian Oct 31 '20 at 02:33
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Although it is true that choosing any point in $[1,2]$ will lead to convergence with this problem, it is better to be safe.

Without any assumptions on the second derivative, you can guarantee convergence using the Newt-safe algorithm, which essentially combines bisection with Newton's method to guarantee every iteration remains in the current bracket.

Imagine for sake of example, your initial bracket was $[0,2]$ instead of $[1,2]$, and you attempted to start with $x_0=0$. But upon doing this, you found $x_1=-1\notin(0,2)$. Then it is clear Newton's method is not converging to the root and you should instead take $x_1=1$, the midpoint of the bracket. Now depending on the sign of $f(1)$, update the bracket. In this case the new bracket becomes $[1,2]$. Now trying Newton's method with $x_1=1$, we find that we are starting to converge.

Usually what occurs is bisection leads the way to "finding the initial point for Newton's method", as with the above example, and now you don't have to worry about starting close to the root at all. Despite Newton's method failing if you start with $x_0<1/\sqrt3$, Newt-safe will converge even with an initial bracket such as $[-10,10]$.

Simply Beautiful Art
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