Newton Raphsen Method:-
In Numarical Anylisys, For finding root of algebraic and transcendental equation, we are use Newton Raphsen Method.
Working Rules:-
1:- Given f(x) =0, Find the inihat root x₀such that f(x₀) ≌0
2:- Find the tangent equation at (x₀,f(x₀)) and input y=0
3:- So, by tangent equation we get new approximate root x₁= x₀- [f(x₀) / f'(x₀)]
4:- If x₁ is not satisfied. Then we find next approximate root x₂= x₁- [f(x₁) / f'(x₁)]
Approximate root finding process to continu untill f( xᵥ ) = 0
Example:-
1:- Find the real root of equation x³⁻3x+1=0 by Newton Rayhen Method upto 3 decimal place.
Soultion:-
Given eqation f(x) = x³⁻3x+1=0
f(0)= 0 -3(0) + 1 = 1> 0
f(1)= 1 - 3 + 1 = -1 < 0
f0.5) = (0.5)³⁻ 3(0.5) + 1= - 0.375 < 0
f(0.2) = (0.2)³⁻ 3(0.2) + 1 = 0.408 > 0
f(0.3) = (0.3)³⁻ 3(0.3) + 1 = 0.127 > 0
f(0.4) = (0.4)³⁻ 3(0.4) + 1 = - 0.136 < 0
Let approximate root x₀= 0.3
By Newton Rayhen Method New Approximate root x₁= x₀ -[ f(x₀) /f'(x₀)]
Now f(x₀) = 0.127
f'(x₀)= 3x₀² - 3 = - 2.73
This implies that x₁= x₀ -[ f(x₀) /f'(x₀)] = 0.346520
Now f( x₁)= 0.002048
f'(x₁)= 3x₁² - 3 = - 2.63977167
By Newton Rayhen Method New Approximate root x₂= x₁-[ f(x₁) /f'(x₁)]
This implies that x₂= 0.347295825 and f(0.347295825)= 0.000254 [This is correct upto 3 decimal place.]
So, 0.347295825 is the real roor of equation x³⁻3x+1=0 by Newton Rayhen Method upto 3 decimal place.
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