How to choose the starting point in Newton's method ?
If $p(x)=x^3-11x^2+32x-22$
We only learnt that the algorithm $x_{n+1}:=x_n-\frac{f(x_n)}{f'(x_n)}$ converges only in some $\epsilon$-neighbourhood of a root and that if $z$ is a root then $|z|\le 1+\max\limits_{k=0,…,n-1}\frac{|a_k|}{|a_n|}$
but in this case, $a_n=1\Rightarrow 1+\max\limits_{k=0,…,n-1}\frac{|a_k|}{|a_n|}=1+\frac{32}{1}=33$ this is too large isn't it ?
$\underline{\textbf{My attempt}}$
I think first root can be guessed by plugging in some values $-1,0,1,2$ and at $x=1$ we've a root
then the polynomial can be reduced to $x^3-11x^2+32x-22=(x-1)(x^2-10x+22)$
now the new polynomial to be examined is $g(x)=(x^2-10x+22)$
this is a parabola and $g'(x)=0$ is attained at $x=5$
and the advantage of the parabola is that we can consider any interval $[a,5)$ with $a<5$
any point in this interval as starpoint would converge to the $2^{nd}$ root
the same is valid for $(5,b]$, Hence we obtain our $3$ roots.
BUT in this case we had a bit luck, and if you know a general approach, can you please tell me
Thanks in advance.
Best Answer
In practice, you may draw a rough graph of the function. You see approximately where are the roots. This give you for each one an approximate value to start the recurrence process.
The first root obviously is $1$. The second is around $3.25$ and the third around $6.75$ The convergence will be fast in starting from these values.
In fact, the analytic solving leads to first $1$ , second $5-\sqrt3 = 3.267949...$ and third $5+\sqrt3 = 6.732051...$