If f(x) = x2 153 = 0 then the iterative formula for Newton Raphson M – McqOptions

MCQOPTIONS

1.	If f(x) = x2 153 = 0 then the iterative formula for Newton Raphson Method is given by __________
A.	x(n+1) = 0.5 [x(n)+ ( frac{153}{x(n)} )]
B.	x(n+1) = 0.5 [x(n)- ( frac{153}{x(n)} )]
C.	x(n+1) = [x(n)+ ( frac{153}{x(n)} )]
D.	x(n+1) = [x(n)- ( frac{153}{x(n)} )]
Answer» B. x(n+1) = 0.5 [x(n)- ( frac{153}{x(n)} )]

Discussion

No Comment Found

Related MCQs

The Newton Raphson method fails if __________
At which point the iterations in the Newton Raphson method are stopped?
The equation f(x) is given as x2-4=0. Considering the initial approximation at x=6 then the value of next approximation correct upto 2 decimal places is given as __________
For what values of 0 the initial guess will be equal to the next iterative values?
In Newton Raphson method if the curve f f(x) is constant then __________
If f(x) = x2 153 = 0 then the iterative formula for Newton Raphson Method is given by __________
The Iterative formula for Newton Raphson method is given by __________
The Newton-Raphson method of finding roots of nonlinear equations falls under the category of which of the following methods?
The equation f(x) is given as x3 x2 + 4x 4 = 0. Considering the initial approximation at x=2 then the value of next approximation correct upto 2 decimal places is given as __________
Rate of convergence of the Newton-Raphson method is generally __________