The minimum number of iterations to find $√28$ correct up to three decimal places by Newton-Raphson method using $5$ as the initial guess is .....
- I took $x_0$=5 and used the formula $x_{n+1}=x_n - f(x_n)/f'(x_n)$ to get $x_1=5.3$, $x_2=5.21950943$ and $x_3=5.21950262$. As the last two approximations are same up to 5 decimal places, hence the answer is 3. Am I correct? What is the general method to find minimum number of iterations in Newton-Raphson method to get the root correct to n decimal places?
You need some estimates on the range covered by the iteration, a minimum bound $m_1$ for the first and a maximum bound $M_2$ for the second derivative over that region.
Here we have $f(x)=x^2-28$ over the interval $[5,6]$. As $$ N(x)=x-\frac{f(x)}{f'(x)}\implies f(N(x))=\frac12f''(\tilde x)\frac{f(x)^2}{f'(x)^2} $$ gives contraction for $$ \frac12M_2m_1^{-2}|f(x)|<1\iff|f(x)|<2m_1^2/M_2 $$ Using $$ \min_{x\in[5,6]}|f'(x)|=10\text{ and }f''(x)\equiv 2 $$ this is satisfied on this interval and one gets the estimate of the quadratic convergence $$ |f(N(x))|\le 10^{-2}|f(x)|^2\implies |f(x_n)|\le 100·\left(10^{-2}|f(x_0)|\right)^{2^n}=100·\left(0.03\right)^{2^n} $$ As $|x-x_*|\le 0.1 ·|f(x)|$, the distance to the root satisfies $$ |x_n-x_*|\le 10·\left(0.03\right)^{2^n} $$
To get 5 digits after the dot you need $|x_n-x_*|\le 5·10^{-6}$ and $n=2$ gives $|x_2-x_*|\le 81·10^{-7}$ which is slightly too large, so $n=3$ will satisfy the error bound with a wide margin.