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The equation $x^3-x=0$ has three roots,$ -1, 0, 1.$ We use the Newton's method to find the roots. And there are three cases

(i) If $x_0>1/\sqrt{3}$, the Newton's method will converge to $1$. (ii) If $x_0<-1/\sqrt{3}$, the Newton's method will converge to $-1$. (iii) If $-1/\sqrt{5}<x_0<1/\sqrt{5}$, the Newton's method will converge to $0$.

Here is the question I don't understand:

Consider $x_0$ is a little less than $1/\sqrt{3}$, there is an implicit recurrence relation that produces a decreasing sequence ${a_1=q/\sqrt{3}, a_2, a_3, ...}$, by means of which one you can easily find $\lim_{n\to\infty}x_n$ for any $x_0\in(1/\sqrt{5},1/\sqrt{3})$. Try to find the recurrence.

The answer provided is $a_i-f(a_i/f'(a_i))=-a_{i-1}$; $\lim_{n\to\infty}x_n=(-1)^i$ if $x_0\in(a_i,a_{i+1}); a_1=0.577, a_2=0.462, a_4 \approx a_i=1/\sqrt{5}=0.447$.

Could anyone kindly help explain what is $a_i$, what does this have to do with the Newton's method? Thanks so much!

PS: someone thinks my question is unclear, so I post the original question as the following. My question is 1.3(d).

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Tony
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  • Where is this from? – Moo Jan 12 '16 at 04:33
  • $q$ and $1$ are close on the keyboard so I suspect $q/\sqrt{3}$ should read $1/\sqrt{3}$. Also the quoted values for $a_2$ do not match the numerical solution I find: $a_1 = 0.57735, a_2 = 0.46560$, $a_3 = 0.450201, a_4=0.44771, \ldots$. – Winther Jan 12 '16 at 04:52
  • Your expression $a_i-f(a_i/f'(a_i))=-a_{i-1}$ does not make sense. At least the final parenthesis should be after $a_i$. The usual iteration would be $a_i=a_{i-1}-f(a_{i-1})/f'(a_{i-1})$ Please check – Ross Millikan Jan 12 '16 at 05:13
  • Try to show that the defining relation $a_i - \frac{f(a_i)}{f'(a_i)} = -a_{i-1}$ gives (something like) $x_0\in(a_{2i},a_{2i+1}) \implies x_2 > \frac{1}{\sqrt{3}}$ and $x_0\in(a_{2i+1},a_{2i+2}) \implies x_1 < -\frac{1}{\sqrt{3}}$ and for these starting values you already know what value it converges to. – Winther Jan 12 '16 at 05:21
  • @Winther Thank you so much for the hint! I have solved the problem. – Tony Jan 13 '16 at 00:43

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