Question

2. Solve

x log10 x 1.2

by Newton-Raphson method correct to three decimal places when root liess

between 2&3 . ​

Answer 1

Given:

xlog(x) - 1.2 = 0

Root lies between 2 and 3

To find:

Find the root by Newton-Raphson method.

Solution:

Here xlog(x) - 1.2 = 0

Let f(x) = xlog(x) - 1.2

$\frac{d}{dx}(xlog(x) - 1.2) = log(x) + 1$

∴ f'(x) = log(x) + 1

f(2) = -0.5979

f(3) = 0.2314

Here f(2) = -0.5979 < 0 and f(3) = 0.2314 > 0

∴ Root lies between 2 and 3

x₀ = (2 + 3) / 2 = 2.5

x₀ = 2.5

1st Iteration:

f(x₀) = f(2.5) = 2.5log(2.5) - 1.2 = -0.2051

f′(x₀) = f′(2.5) = log(2.5) + 1 = 1.3979

x₁ = x₀ - f(x₀) / f'(x₀)

x₁ = 2.5-  (-0.2051) / 1.3979

x₁ = 2.6468

2nd Iteration:

f(x₁) = f(2.6468) = 2.6468log(2.6468) - 1.2 = -0.0812

f′(x₁) = f′(2.6468) = log(2.6468) + 1 = 1.4227

x₂ = x₁ - f(x₁) / f'(x₁)

x₂ = 2.6468-  (-0.0812) / 1.4227

x₂=2.7038

3rd Iteration:

f(x₂) = f(2.7038) = 2.7038log(2.7038) - 1.2 = -0.032

f′(x₂) = f′(2.7038) = log(2.7038) + 1 = 1.432

x₃ = x₂ - f(x₂) / f'(x₂) = 2.7262

4th Iteration:

f(x₃) = f(2.7262) = 2.7262log(2.7262) - 1.2 = -0.0126

f′(x₃) = f′(2.7262) = log(2.7262) + 1 = 1.4356

x₄ = x₃ - f(x₃) / f'(x₃) = 2.735

5th Iteration:

f(x₄) = f(2.735) = 2.735log(2.735) - 1.2 = -0.005

f′(x₄) = f′(2.735) = log(2.735) + 1 = 1.4369

x₅ = x₄ - f(x₄) / f'(x₄) = 2.7384

6th Iteration:

f(x₅) = f(2.7384) = 2.7384log(2.7384) - 1.2 = -0.002

f′(x₅) = f′(2.7384) = log(2.7384) + 1 = 1.4375

x₆ = x₅ - f(x₅) / f'(x₅) =  2.7398

7th Iteration:

f(x₆) = f(2.7398) = 2.7398log(2.7398) - 1.2 = -0.0008

f′(x₆) = f′(2.7398) = log(2.7398) + 1 = 1.4377

x₇ = x₆ - f(x₆) / f'(x₆) =  2.7403

8th Iteration:

f(x₇) = f(2.7403) = 2.7403log(2.7403) - 1.2 = -0.0003

f′(x₇) = f′(2.7403) = log(2.7403) + 1 = 1.4378

x₈ = x₇ - f(x₇) / f'(x₇) =  2.7405

Approximate root of the equation xlog(x) - 1.2 = 0 is 2.740, after 8 iterations

Answer 2

it was realy nice explanation