Newton-Raphson Method ---Solution of Polynomial And Transcendental equation| Engineering Mathematics

Introduction:
≫In the previous session we read about Bisection method which is used to solve the equation for finding root.
if you are not read that blog i suggest you to read that blog first
Click Here to read.

Newton-Raphson Method:

Concept:

We will choose a +ve point much nearer to cut of the curve to X-axix . Let that point is x_n and it intersect the curve at (x_n,∱(x_n)) .

              If we draw a tagent at the point (x_n,∱(x_n)) then the slope will be ∱'(x_n) and cut the X-axis at x_n+1.

     Similarly,the point x_n+1 cuts the curve at (x_n+1,∱(x_n+1)) and if we draw a tangent at there then the slope will be ∱'(x_n+1) and the tangent cut the X-axix at x_n+2 .

         And The point x_n+2 cuts the curve at (x_n+2,∱(x_n+2)). 

We can see that at every step we move towards the root point of ∱(x)=0.

This method is very easy and less-effort than Bisection-Method.

Working Rule:

1. Given ∱(x)=0 ------(1)
   Find initial root x₀ such that ∱(x)~0
[ i.e; x₀ is near to the root ∱(x)=0]

2. Find ∱(x₀) and ∱'(x₀)

3. First approximate root by Newton-Raphson method:
            $x_{1=x_{0\mathrm{}}}- \frac{∱\left(x_{0\mathrm{}}\right)}{∱\text{'}\left(x_{0\mathrm{}}\right)}$      

Find ∱(x₁) and ∱'(x₁)

4. Similarly, Second approximate root is

$x_{2=x_{1\mathrm{}}}- \frac{∱\left(x_{1\mathrm{}}\right)}{∱\text{'}\left(x_1\right)}$              

5.Also Third approximate root is

          $x_{3=x_{2\mathrm{}}}- \frac{∱\left(x_{2\mathrm{}}\right)}{∱\text{'}\left(x_{2\mathrm{}}\right)}$                 

General Formula:


$x_{n =x_{n-1 }}- \frac{∱\left(x_{n-1}\right)}{∱\text{'}\left(x_{n-1}\right)}$

Example

QnS. ∱(x) = x - 2 + lnx has a root near x =1.5 .Use Newton-Raphson Method to obtain a better estimate 2 digit after decimal point.

Ans.

In this question we don't need to figure out the x₀, because it has given that the root is lies near x=1.5 .

Here x₀ =1.5 and ∱(1.5) = -0.5 + ln(1.5) = -0.0945

∱'(x)= 1+(1/x)

⇒ ∱'(1.5) = 5/3

Hence using the general formula

x₁=1.5567

The Newton-Raphson formula can be used again.

so ∱(x₁) = ∱(1.5567) = -0.0007

∱'(x₁) = ∱'(1.5567) = 1.6424

{ Calculation Part is skipped for Readers}

x₂= 1.5571

⇒We will note that the 2digits after decimal point of x₁ is same as x₂ .So

our answer will be x₂.