Question

Use the Newton-Raphson method to find the root of the equation f(x) = In(3x) + 5x2, using an initial guess of x = 0.5 and a stopping criterion of &= 0.1%. Report your answer to 3 decimal places. What is the estimate for the root? What is the approximate relative error? %

Answer 1

Answer with explanation:

The equation which we have to solve by Newton-Raphson Method is,

f(x)=log (3 x) +5 x²

`f'(x)=(1)/(3x)+10 x`

Initial Guess =0.5

Formula to find Iteration by Newton-Raphson method

`x_(n+1)=x_(n)-(f(x_(n)))/(f'(x_(n)))\\\\x_(1)=x_(0)-(f(x_(0)))/(f'(x_(0)))\\\\ x_(1)=0.5-(\log(1.5)+1.25)/((1)/(1.5)+10 * 0.5)\\\\x_(1)=0.5- (0.1760+1.25)/(0.67+5)\\\\x_(1)=0.5-(1.426)/(5.67)\\\\x_(1)=0.5-0.25149\\\\x_(1)=0.248`

`x_(2)=0.248-(\log(0.744)+0.30752)/((1)/(0.744)+10 * 0.248)\\\\x_(2)=0.248- (-0.128+0.30752)/(1.35+2.48)\\\\x_(2)=0.248-(0.17952)/(3.83)\\\\x_(2)=0.248-0.0468\\\\x_(2)=0.2012`

`x_(3)=0.2012-(\log(0.6036)+0.2024072)/((1)/(0.6036)+10 * 0.2012)\\\\x_(3)=0.2012- (-0.2192+0.2025)/(1.6567+2.012)\\\\x_(3)=0.2012-(-0.0167)/(3.6687)\\\\x_(3)=0.2012+0.0045\\\\x_(3)=0.2057`

`x_(4)=0.2057-(\log(0.6171)+0.21156)/((1)/(0.6171)+10 * 0.2057)\\\\x_(4)=0.2057- (-0.2096+0.21156)/(1.6204+2.057)\\\\x_(4)=0.2057-(0.0019)/(3.6774)\\\\x_(4)=0.2057-0.0005\\\\x_(4)=0.2052`

So, root of the equation =0.205 (Approx)

Approximate relative error

`=\frac{\text{Actual value}}{\text{Given Value}}\\\\=(0.205)/(0.5)\\\\=0.41`

Approximate relative error in terms of Percentage

=0.41 × 100

= 41 %