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At which point the iterations in the Newton Raphson method are stopped?

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Important Questions on Numerical Methods

MEDIUM

Mathematics>Mathematics and Computation>Numerical Methods>Numerical Solution of Equation

One root of the equation,

x^{3} - 3 x - 5 = 0

lies between

2

and

2.5

. Using Newton-Raphson method, the value of that root will be

EASY

Mathematics>Mathematics and Computation>Numerical Methods>Numerical Solution of Equation

One root of the equation

x^{3} - 4 x + 1 = 0

is between

1

and

2

. The value of this root using Newton-Raphson method will be

EASY

Mathematics>Mathematics and Computation>Numerical Methods>Numerical Solution of Equation

For the solution of equation

f (x) = 0

by the Newton-Raphson method, the value of

x

tends to root of the equation highly when

f^{'} (x_{n})

MEDIUM

Mathematics>Mathematics and Computation>Numerical Methods>Numerical Solution of Equation

The

2^{nd}

approximation to a root of the equation

x^{2} - x - 1 = 0

in the interval

(1, 2)

by Bisection method will be:

MEDIUM

Mathematics>Mathematics and Computation>Numerical Methods>Numerical Solution of Equation

If one root of the equation

f (x) = 0

is near to

x_{0}

, then the first approximation of this root as calculated by Newton Raphson method is the abscissa of the point, where the following straight line intersects the x-axis"

HARD

Mathematics>Mathematics and Computation>Numerical Methods>Numerical Solution of Equation

By Bisection method, the real root of the equation

x^{3} - 9 x + 1 = 0

lying between

x = 2

and

x = 4

is nearer to

HARD

Mathematics>Mathematics and Computation>Numerical Methods>Numerical Solution of Equation

f (0) = 3, f (1) = 5, f (3) = 21

, then the unique polynomial of degree

2

or less using Newton divided difference interpolation will be:

MEDIUM

Mathematics>Mathematics and Computation>Numerical Methods>Numerical Solution of Equation

The value of

\sqrt{12}

upto three places of decimals using the method of Newton-Raphson, will be

EASY

Mathematics>Mathematics and Computation>Numerical Methods>Numerical Solution of Equation

Which of the following methods is surely convergent?

EASY

Mathematics>Mathematics and Computation>Numerical Methods>Numerical Solution of Equation

The iteration formula

x_{n + 1} = x_{n} - \frac{f (x_{n})}{f^{'} (x_{n})}

is commonly known as

MEDIUM

Mathematics>Mathematics and Computation>Numerical Methods>Numerical Solution of Equation

By Newton-Raphson method, the positive root of the equation

x^{4} - x - 10 = 0

MEDIUM

Mathematics>Mathematics and Computation>Numerical Methods>Numerical Solution of Equation

Performing

3

iterations of bisection method the smallest positive approximate root of the emation

x^{3} - 5 x + 1 = 0

MEDIUM

Mathematics>Mathematics and Computation>Numerical Methods>Numerical Solution of Equation

The per-unit power output of a salient-pole generator which is connected to an infinite bus, is given by the expression,

P = 1.4 \sin δ + 0.15 \sin 2 δ

, where

δ

is the load angle. Newton-Raphson method is used to calculate the value of

δ

for

P = 0.8 pu

. If the initial guess is

30 °

, then its value (in degree) at the end of the first iteration is

MEDIUM

Mathematics>Mathematics and Computation>Numerical Methods>Numerical Solution of Equation

a

and

a + h

are two consecutive approximate roots of the equation

f (x) = 0

as obtained by Newtons Method, then

h

is equal to

HARD

Mathematics>Mathematics and Computation>Numerical Methods>Numerical Solution of Equation

One real root of a polynomial

P (x) = x^{3} - x - 1

lies in the interval

[1, 2]

and bisection method is used to find its value, the minimum number of interations required to achieve accuracy up to two decimal points is

EASY

Mathematics>Mathematics and Computation>Numerical Methods>Numerical Solution of Equation

The formula for obtaining real root of an equation

f (x) = 0

in a given interval by Regula Falsi method, is

EASY

Mathematics>Mathematics and Computation>Numerical Methods>Numerical Solution of Equation

The Newton Raphson method converges fast, if

f^{'} (α)

HARD

Mathematics>Mathematics and Computation>Numerical Methods>Numerical Solution of Equation

The bisection method is applied to compute a zero of the function

f (x) = x^{4} - x^{3} - x^{2} - 4

in the interval

[1, 9]

. The method converges to a solution after _____ iterations.

MEDIUM

Mathematics>Mathematics and Computation>Numerical Methods>Numerical Solution of Equation

According to Newton-Raphson method, the value of

\sqrt{12}

upto three places of decimal will be

MEDIUM

Mathematics>Mathematics and Computation>Numerical Methods>Numerical Solution of Equation

f (x) = x^{2}

, then the second order divided difference for the points

x_{0}, x_{1}, x_{2}

will be:

At which point the iterations in the Newton Raphson method are stopped?

Important Questions on Numerical Methods

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