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### Newton-Raphson Method

The Newton-Raphson (N-R) Method is probably the most commonly used technique in finding the roots of a complex equation. Unlike other methods, the N-R technique requires only one initial guess of the root (${x_{{i}}}$) to get the iteration started. The N-R method finds the tangent to a given function ${f(x)}$ at ${x=x_{{i}}}$ and extrapolates it to intersect the x axis to get ${x_{{i+1}}}$. This new value of ${x_{{i+1}}}$ is then plugged in to repeat the process until we reach an acceptable level of tolerance, which is then defined as the root.

The N-R iteration formula is defined as:

\begin{equation*}

{x_{{i+1}}=x_{{i}}-\frac{f(x_{{i}})}{{f}^{‘}(x_{{i}})}}

\end{equation*}

Where ${{f}^{‘}(x)}$ denotes the first derivative of the function ${f(x)}$.

To complete the iteration and find the root of an equation where ${f(x)=0}$, the steps are:

- Establish ${{f}^{‘}(x)}$
- Evaluate ${f(x)}$ and ${{f}^{‘}(x)}$ at your initial ${x_{{0}}}$
- Plug the values in to the N-R iteration formula to determine ${x_{{i+1}}}$
- Calculate the absolute relative approximate error ${|\varepsilon_{{a}}|}$ such that:
- ${|\varepsilon_{{a}}|=|\frac{x\multiscripts{_{{i+1}}}{}{}{\text{}}{}-x_{{i}}}{x_{{i+1}}}|\times \text{100}}$ The root is found when the tolerance closes in on 0

#### Example

Determine the roots of the equation:

\begin{equation*}

{f(x)=x^{{3}}+5x-3}

\end{equation*}

[Image]

Looking at the graph, we can see that the root is close to .6, so let’s make that our first guess, ${x_{{i}}=\text{.}6}$

Determine the iteration factors to complete the equation:

\begin{equation*}

{x_{{i+1}}=x_{{i}}-\frac{f(x_{{i}})}{{f}^{‘}(x_{{i}})}}

\end{equation*}

\begin{equation*}

{f(x_{{i}})=x^{{3}}+5x-3}

\end{equation*}

\begin{equation*}

{f(x_{{i}})=3x^{{2}}+5}

\end{equation*}

Plugging in and solving for the value of ${x_{{i+1}}}$:

\begin{equation*}

{x_{{i+1}}=\text{.}6-\frac{(\text{.}6)^{{3}}+5(\text{.}6)-3}{3(\text{.}6)^{{2}}+5}=\text{.}\text{56}}

\end{equation*}

\begin{equation*}

{|\varepsilon_{{a}}|=|\frac{\text{.}\text{56}-\text{.}6}{\text{.}\text{56}}|\times \text{100}=7\text{.}\text{14}}

\end{equation*}

Iterating again with ${x_{{i+1}}=x_{{i}}}$

\begin{equation*}

{x_{{i+1}}=\text{.}\text{56}-\frac{(\text{.}\text{56})^{{3}}+5(\text{.}\text{56})-3}{3(\text{.}\text{56})^{{2}}+5}=\text{.}\text{556}}

\end{equation*}

\begin{equation*}

{|\varepsilon_{{a}}|=|\frac{\text{.}\text{556}-\text{.}\text{56}}{\text{.}\text{556}}|\times \text{100}=\text{.}7}

\end{equation*}

A point within 5% accuracy can be considered a root.

For ${f(x_{{i}})=x^{{3}}+5x-3}$, the root is located at ${(0,\text{.}\text{556})}$

### Secant Method

There may be times when we are given a complex equation that is just too hard to evaluate its derivative using the N-R technique:

\begin{equation*}

{x_{{i+1}}=x_{{i}}-\frac{f(x_{{i}})}{{f}^{‘}(x_{{i}})}}

\end{equation*}

If this is the case, we can use another technique known as the Secant method to determine the root of a given function ${f(x)}$ where we use an approximate of \ ${{f}^{‘}(x)}$ as:

\begin{equation*}

{{f}^{‘}(x\multiscripts{_{{i}}}{}{}{\text{}}{})=\frac{f(x_{{i}})-f(x_{{i-1}})}{x_{{i}}-x_{{i-1}}}}

\end{equation*}

Substituting in to the original N-R equation we get:

\begin{equation*}

{x_{{i+1}}=x_{{i}}-\frac{f(x_{{i}})(x_{{i}}-x_{{i-1}})}{f(x_{{i}})-f(x_{{i-1}})}}

\end{equation*}

Using the Secant method requires two initial guesses; however, the two guesses do not need to bracket the root as they would in other root finding techniques, such as the Bisection method.