NUMERICAL METHOD PDF

Newton Raphson Method

Third step: Repeat second and third step until Ixn+1- xnI < 0.0001

Ixn+1- xnI < 0.0001
Ix1- x0I <0.0001

I 0.909-1I < 0.0001 FALSE
I -0.091 I < 0.0001

n xn1  xn  f (xn ) +1 − < 0.001
f '(xn )
01 0.0910
1 0.909 0.909 0.0080
2 0.901 0.0000
0.901

0.901

Therefore, the approximate root is x2 = 0.901

BFMI / JUN 2020 / DBM30043

Newton Raphson Method

Example 10 : The equation 2x3  5x2  4x  sin x has a solution between -3 and -4. Use the

Newton Raphson method with x0  3 to approximate the solution to three

decimal places.

Solution : First step : Differentiate f (x)  2x3  5x2  4x  sin x
f , (x)  6x2 10x  4  cos x

Second step : Substitute into

f (x)  2x3  5x2  4x  sin x f , (x)  6x2 10x  4  cos x xn1  xn  f (xn )
f '(xn )

f (x)  2(3)3  5(3)2  4(3)  sin(3) f , (x0 )  6(3)2 10(3)  4  cos(3) x1  3   2.859 
 19.010 
 2.859  19.010

 3.150

BFMI / JUN 2020 / DBM30043

Newton Raphson Method

Third step: Repeat second and third step until Ixn+1- xnI < 0.0001

Ixn+1- xnI < 0.0001
Ix1- x0I <0.0001

I -3.150 –(-3)I < 0.0001

I -0.150 I < 0.0001 FALSE

n xn1  xn  f (xn ) +1 − < 0.001
f '(xn )
0 -3 0.1500
1 -3.150 -3.150 0.0130
2 -3.137 0.0000
-3.137

-3.137

Therefore, the approximate root is x3 = -3.317

BFMI / JUN 2020 / DBM30043

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