Courses
Courses for Kids
Free study material
Offline Centres
More Last updated date: 10th Dec 2023
Total views: 283.8k
Views today: 6.83k

# What is the value of $\sqrt {12}$ by Newton- Raphson’s method after the first iteration? Verified
283.8k+ views
Hint: We know that Newton- Raphson is a root finding algorithm which produces successively better approximations for roots or zero of real valued functions. We know that the iterative equation of Newton Raphson method is ${x_{n + 1}} = {x_n} - \dfrac{{f({x_n})}}{{f'({x_n})}}$. We should know that the first iteration is ${x_0}$, thus the formula is ${x_1} = {x_0} - \dfrac{{f({x_0})}}{{f'({x_0})}}$.

Complete step by step solution:
Here we have to find the value of $\sqrt {12}$ by Newton- Raphson’s method after the first iteration.
Let us assume $x = \sqrt {12}$. By squaring both the sides we have ${x^2} = {(\sqrt {12} )^2}$.
So we have ${x^2} = 12$. By taking the constant term to the left side of the equation we have
${x^2} - 12 = 0$.
Now the formula of the iterative equation of Newton Raphson method is ${x_{n + 1}} = {x_n} - \dfrac{{f({x_n})}}{{f'({x_n})}}$.
If $f(x) = {x^2} - 12$, then we need the derivative so we have $f'(x) = 2x$.
We will substitute $f(x) = {x^2} - 12$ in the formula, and it can be written as ${x_{n + 1}} = {x_n} - \dfrac{{f({x^2}_n - 12)}}{{2{x_n}}}$.
Now we have to find the first iteration i.e. $n = 0$, so we have ${x_1} = {x_0} - \dfrac{{f({x^2}_0 - 12)}}{{2{x_0}}} = \dfrac{{x_0^2 + 12}}{{2{x_0}}}$.
We know that $3 < \sqrt {12} < 4$ or it can be written as $\sqrt 9 < \sqrt {12} < \sqrt {16}$.
From this our first iteration i.e. ${x_0} = 3$. By putting this in the formula we have $\dfrac{{{3^2} + 12}}{{2 \times 3}} = \dfrac{{9 + 12}}{6}$.
It gives us the value $3.5$.
Now we put $3.5$ in the equation i.e. ${x_1} = 3.5 - \dfrac{{{{(3.5)}^2} - 12}}{{2 \times 3.5}}$.
On simplifying we have $3.5 - \dfrac{{12.25 - 12}}{7} \Rightarrow 3.5 - 0.0357 = 3.4643$.
Hence the required value is ${x_1} = 3.4643$.

Note:
We should note that the formula of first iteration is ${x_1} = {x_0} - \dfrac{{x_0^2 - A}}{{2{x_0}}} = \dfrac{{x_0^2 + A}}{{2{x_0}}}$, so if we have a guess ${x_0}$. We can get the more accurate value by calculating $\dfrac{{x_0^2 + A}}{{2{x_0}}}$, where $A$ is the number which square root we are trying to find as we did in the above solution. We should know that ${x_0}$ is our initial guess and ${x_1}$ is a more accurate one.