Question

Using Newton Raphson method solve 3x – cosx – 1 = 0. Correct upto 3 decimal places. 

Answer 1

Let f(x) = 3x – cosx – 1

∴f ‘ (x) = 3 + sinx – 0
When x = 0, f (0) = 3(0) – cos0 – 1 = -2
When x =1, f (1) = 3(1) – cos1 – 1 = 1.4597
∴ Roots of f (x) lies between 0 and 1.
Let initial value x0 = 0
By Newton-Raphson’s Method `x_(n+1)-f(x)/(f'(x))`

= `(x_n-3x_n-cosx_n-1)/(3+sinx_n)`

=`(x_n(3+sinx_n)-(3x_n-cosx_n-1))/(3+sinx_n)` 

=`(3x^n+x_n sinx_n-3x_n+cosx_n+1)/(3+sinx_n)`

∴ `(x_(n+1)=x_nsinx_n+cosx_n+1)/(3+sinx_n)` 

Iteration 1: Put n = 0 in (1) 

∴`( x_1=x_0sinx_0+cosx_0+1)/(3+sinx_-0)=(0+cos0+1)/(3+sin0)=0.6667` 

Iteration 2: Put n = 1 in (1) 

∴` (x_2=x_1sinx_1+cosx_1)/(3+sinx_1)=(.6667sin(0.6667)+cos(0.6667)+1)/(3+sin(0.6667))=0.6075`

Iteration 3: Put n = 2 in (1) 

∴`( x_3=x_2sinx_2+cosx_2+1)/(3+sin x_2)=(0.6075sin (0.6075)+cos(0.6075)+1)/(3+sin(0.6075))=0.6071`

Iteration 4: Put n = 3 in (1)

∴ `(x_4=x_3sin_3+cosx_3+1)/(3+sinx_3)=(06071sin(0.6071)+cos(0.6071)+1)/(3sin(0.6071))=0.6071`

`"Hence, Root of" 3x-cos-1=0 "is" 0.6071`