Newton Raphson Method

Newton Raphson Method is yet another numerical method to approximate the root of a polynomial. Newton Raphson Method is an open method of root finding which means that it needs a single initial guess to reach the solution instead of narrowing down two initial guesses.

Newton Raphson Method Fig 1

Newton Raphson Method uses to the slope of the function at some point to get closer to the root. Using equation of line y = mx0 + c we can calculate the point where it meets x axis, in a hope that the original function will meet x-axis somewhere near. We can reach the original root if we repeat the same step for the new value of x.

Formula

The following formula gives the next value of x (hopefully closer to the root)

$$ \text{x} _ {\text{n}+\text{1}}=\text{x} _ \text{n}-\frac{\text{f}\left(\text{x} _ \text{n}\right)}{\text{f}^\prime\left(\text{x} _ \text{n}\right)} $$

Steps

  1. Initial guess
  2. Using the formula mentioned above calculate the next value of x
  3. Check if x is the root of the function or is in the range of affoardable error. In other words check if f(x)=0 or |f(x)| < affordable error. Repeat step 2 if not. 3 (option b) If the formula mentioned above gives the same result, x is the root of the polynomial.

Example

Let's approximate the root of the following function with Newton Raphson Method

$$ \ \text{f}\left(\text{x}\right)\ =\ \text{e}^{-\text{x}}-\text{x} $$

Solution

$$ \text{f}\left(\text{x}\right)=\text{e}^{-\text{x}}-\text{x} $$

show hidden steps

$$ \frac{\text{df}}{\text{dx}}=-\text{e}^{-\text{x}}-\text{1} $$

$$ \text{x} _ {\text{n}+\text{1}}=\text{x} _ \text{n}-\frac{\text{f}\left(\text{x} _ \text{n}\right)}{\text{f}^\prime\left(\text{x} _ \text{n}\right)}=\text{x} _ \text{n}+\frac{\text{e}^{-\text{x} _ \text{n}}-\text{x} _ \text{n}}{\text{e}^{-\text{x} _ \text{n}}+\text{1}\mathrm{\ } } $$

$$ \text{error} < \text{0.05%} $$


$$ \text{x} _ {\text{n}+\text{1}}=\text{x} _ \text{n}+\frac{\text{e}^{-\text{x} _ \text{n}}-\text{x} _ \text{n}}{\text{e}^{-\text{x} _ \text{n}}+\text{1}\mathrm{\ } } $$

xn xn+1 error
1 0 0.5
2 0.5 0.5663
3 0.5663 0.5671 0.0014
4 0.5671 0.5671 0.0000

The relative error is 0 because we have found the exact root and a function.

by Arifullah Jan and last modified on

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