Newton-Raphton iteration formula for finding $\sqrt[3]{C},$ where C > 0 is

Newton-Raphson Method Overview

The Newton-Raphson method is a widely used iterative numerical technique for finding progressively better approximations to the roots (or zeroes) of a real-valued function. This powerful method is particularly useful when it is difficult or impossible to find the roots analytically. The core idea is to start with an initial guess and then refine it using the function's value and its derivative at that guess.

The general Newton-Raphson iteration formula is given by:

$x_{n+1}=x_n-\frac{f(x_n)}{f^{\prime }(x_n)}$

Here, $x_n$ represents the current approximation of the root, $f(x_n)$ is the value of the function at $x_n$, and $f^{\prime }(x_n)$ is the value of the first derivative of the function at $x_n$.

**Deriving the Iteration for Cube Root $\sqrt[3]{C}$ **

To use the Newton-Raphson method for finding the cube root of a positive number $C$, which is $\sqrt[3]{C}$, we need to define a function $f(x)$ such that its root is $\sqrt[3]{C}$. Let's set up the equation:

- We are looking for a value $x$ such that $x=\sqrt[3]{C}$.
- To remove the cube root and form a polynomial equation, we cube both sides of this expression: $x^3=C$
- Now, we rearrange this equation to bring all terms to one side, setting the equation equal to zero. This gives us our function $f(x)$: $f(x)=x^3-C$

**Calculating the Derivative $f^{\prime }(x)$ **

The next essential step in applying the Newton-Raphson method is to find the derivative of our function $f(x)$ with respect to $x$. The derivative is crucial because it tells us the slope of the tangent line to the function at any given point.

- Our function is $f(x)=x^3-C$.
- To find the derivative $f^{\prime }(x)$, we differentiate $f(x)$ term by term with respect to $x$: $f^{\prime }(x)=\frac{d}{dx}(x^3-C)$ $f^{\prime }(x)=\frac{d}{dx}(x^3)-\frac{d}{dx}(C)$
- The derivative of $x^3$ is $3x^2$.
- The derivative of a constant $C$ is $0$.
- Therefore, the derivative $f^{\prime }(x)$ is: $f^{\prime }(x)=3x^2-0$ $f^{\prime }(x)=3x^2$

Applying the Newton-Raphson Formula for Cube Root

Now that we have both $f(x_n)$ and $f^{\prime }(x_n)$, we can substitute these into the general Newton-Raphson iteration formula:

$x_{n+1}=x_n-\frac{f(x_n)}{f^{\prime }(x_n)}$

Substitute $f(x_n)=x_n^3-C$ and $f^{\prime }(x_n)=3x_n^2$:

$x_{n+1}=x_n-\frac{x_n^3-C}{3x_n^2}$

Simplifying the Newton-Raphson Iteration Formula

To get the final form of the iteration formula, we need to simplify the expression by combining the terms on the right-hand side. We achieve this by finding a common denominator:

$x_{n+1}=\frac{x_n\cdot (3x_n^2)}{3x_n^2}-\frac{x_n^3-C}{3x_n^2}$

Now, combine the numerators over the common denominator:

$x_{n+1}=\frac{3x_n^3-(x_n^3-C)}{3x_n^2}$

Carefully distribute the negative sign in the numerator:

$x_{n+1}=\frac{3x_n^3-x_n^3+C}{3x_n^2}$

Finally, combine the like terms ($3x_n^3-x_n^3$) in the numerator:

$x_{n+1}=\frac{2x_n^3+C}{3x_n^2}$

Comparing the Derived Formula with Options

Let's compare our derived Newton-Raphson iteration formula for finding the cube root of $C$ with the given options:The derived formula, $x_{n+1}=\frac{2x_n^3+C}{3x_n^2}$, is exactly the same as the formula given in Option 3. This formula allows for iterative approximation of the cube root of $C$.