Newton Raphson method.

Newton- Raphson method

Let x₀ be an approximation value of a root of the equation f(x) = 0.

Let x₁ be the exact root closer to  x₀. So that x₁ =  x₀ + h. Where h is small.

Since x₁ is the exact root of f(x) = 0, we have f(x₁) =0, i.e f( x₀ + h) =0

i.e f( x₀)+h/1![f'( x₀)]+h²/2![f"( x₀)]+....=0 by Taylor's theorem.

Since h is small, h² and higher powers of h may be omitted.

Hence 

f( x₀)+h[f'( x₀)] = 0

=> h = -f( x₀)/f'( x₀) .....(1)

∴ x₁ = x₀ + h = x₀ -f( x₀)/f'( x₀) [using(1)]

Taking x₁ as an approximation value of the root, a still better approximation x₂ can be obtained by

x₂ =  x₁ -f( x₁)/f'( x₁)

The iterative process is continued until we get the required accuracy.

The iterative formula is 

xₙ₊₁ = xₙ - f(xₙ)/f'(xₙ) (for n= 0,1,2,.....)

is called the Newton- Raphson method or Newton's iteration formula.

Remarks

1) When the derivative of f(x) can be easily found and is a simple expression, then the real root of the equation f(x)=0 can be computed rapidly by the Newton Raphson method.

2) Newton's formula converges provided the initial approximation x₀ is chosen sufficiently too close to the root.

3) Condition for convergence of Newton-Raphson method 

The Newton-Raphson formula is

xₙ₊₁ = xₙ - f(xₙ)/f'(xₙ) ....(1)

The iteration formula is

xₙ₊₁ = 𝜙(xₙ) ....(2)

From (1) and (2), we get 

 𝜙(x) = x - f(x)/f'(x) 

We know that the condition for convergence of the iteration method is 

|𝜙'(x)|<1

∴ The corresponding condition for the Newton- Raphson method is 

|1-{[f'²(x) - f(x).f"(x)]/ f'²(x)}|<1

=> |[f(x).f"(x)]/f'²(x)|<1

=> |[f(x).f"(x)]|<  {f'(x)}², for all x in the interval in which the root lies.

Example: Solve by Newton-Raphson method: x³-3x+1=0.

Solution. Let f(x) = x³-3x+1 ....(1)

f(0) = +1 (+ve)

f(1) = 1-3+1 = 2-3 = -1(-ve)

Clearly f(0) and f(1) are of opposite signs.

∴ root of the equation (1) lies between 0 and 1. 

∴ Taking initial approximation x₀=0.5

Using Newton- Raphson method

xₙ₊₁ = xₙ - f(xₙ)/f'(xₙ) .....(2)

Since f(x) = x³-3x+1

=> f'(x) = 3x² -3

Hence (2) becomes

xₙ₊₁ = xₙ - [(xₙ)³-3xₙ+1]/[3(xₙ)²-3]

       = [3xₙ³-3xₙ-xₙ³+3xₙ-1]/[3(xₙ²-1)]

       =[ 2xₙ³-1]/[3(xₙ²-1)] ....(3)

Putting n=0 in (3), then first approximation is 

x₁ = [2x₀³ -1]/[3( x₀²-1)] = [2(.5)³-1]/[3(.5²-1)] = 0.3333

Putting n= 1 in (3), then second approximation is 

x₂ = [2x₁³ -1]/[3( x₁²-1)] = 0.3472

Putting n= 2 in (3), then third approximation is 

x₃ = [2x₂³ -1]/[3( x₂²-1)] = 0.3470

Hence x₂ = x₃ = 0.347 correct to three decimal places.

Thus root is 0.347 Ans.