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The iteration step in order to solve for the cube roots of a given number Nusing the Newton Raphson’s method is
Concept:
Let x_{0} be an approximation root of f(x) = 0 and x_{1} = x_{0}^{th }be the correct root so that f(x_{1}) = 0 expanding f(x_{0}^{th}) by Taylor's series we obtain
Neglecting the second and higher order derivatives we have f(x_{0}) + hf'(x_{0}) = 0
A better approximation than x0 is therefore given by x_{1}, where x_{1} = x_{0} + h = successive approximations are given by x_{2}, x_{3} ....x_{n+1} where
For p^{th} root of a given number N, is not of equation f(x) = x^{p}  N = 0
Iteration equation:
for cube root, put p = 3
Therefore, option (2) is correct one.
The square root of a number N is to be obtained by applying the Newton Raphson iterations to the equation x^{2}  N = 0, if i denotes the iteration index, the correct iterative scheme will be
Given
Now,
f(x) = x^{2} – N = 0
Differentiating,
f’(x) = 2x
Now,
Using NewtonRaphson Method,
x_{i+1} = (x_{i} + N/x_{i})/2
The NewtonRaphson method is said to have
Concept:
Newton Raphson method:
Advantages:
Disadvantages:
If 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:
Concept:
Newton’s divided difference polynomial method:
Second order polynomial interpolation using Newton’s divided difference polynomial method is as follows,
Given (x_{0},y_{0}), (x_{1},y_{1}), (x_{2},y_{2}) be the data points and f(x) be the quadratic interpolant, then f(x) is given by
f(x) = b_{0} + b_{1}(x – x_{0}) + b_{2} (x – x_{0})(x – x_{1});
Where
b_{0 }= f(x_{0});
Calculation:
Given f(0) = 3, f(1) = 5, f(3) = 21;
⇒ (0,3), (1,5), (3,21) are the data points;
The polynomial will be f(x) = b_{0} + b_{1}(x) + b_{2} (x)(x – 1);
⇒ b_{0} = f(0) = 3;
Substituting the constant b_{0}, b_{1}, b_{2} in the quadratic interpolant,
⇒ f(x) = 3 + 2x + 2 (x)(x – 1) = 3 + 2x + 2x^{2} – 2x = 3 + 2x^{2};
The unique polynomial of degree 2 will be f(x) = 3 + 2x^{2};
Easy method:
To save time, simply substitute the data points in the polynomials given in options and find the polynomial that is satisfying all data points.
The iteration formula to find the reciprocal of a given number N by Newton’s method is
Concept:
NewtonRaphson method: It has order of convergence 2 and number of guesses required is 1.
Iteration formula,
Calculation:
The real root of x^{3} + x^{2} + 3x + 4 = 0 correct to four decimal places, obtained using Newton Raphson method is
Concept:
NewtonRaphson Method:
The iteration formula is given by
Where x_{0} is the initial value/root of the equation f(x) = 0
Given,
f(x) = x^{3} + x^{2} + 3x + 4 = 0
f'(x) = 3x^{2} + 2x + 3
∴ f(1) = 1 > 0 and f(2) = 6 < 0
∴ f(1).f(2) < 0
⇒ ∃ a root lies in [1, 2]
Let, x_{0} = 1
By Newton Raphson method
First approximation
x_{1} = 1.25
To solve the equation 2 sin x = x by NewtonRaphson method, the initial guess was chosen to be x = 2.0. Consider x in radian only. The value of x (in radian) obtained after one iteration will be closest to
Concept:
The iterative formula for Newton Raphson method is given as,
[NOTE: Take the trignometric terms in Radian while using scitific calculator for this type of numericals]
Calculation:
Given:
f(x) = 2 sin x  x
∴ f'(x) = 2 cos x  1
Initial guess is x_{0} = 2.0
The first iteration by Newton Raphson method is given by,
⇒ x_{1} = 1.901
The order of convergence of Newton Raphson method is
Order of convergence of the Newton Raphson method is two
Important Point
Order of convergence of various numerical methods
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:
Concept:
Bisection method:
Used to find the root for a function. Root of a function f(x) = a such that f(a)= 0
Property: if a function f(x) is continuous on the interval [a…b] and sign of f(a) ≠ sign of f(b). There is a value c belongs to [a…b] such that f(c) = 0, means c is a root in between [a….b]
Note:
Bisection method cut the interval into 2 halves and check which half contains a root of the equation.
1) Suppose interval [a…b] .
2) Cut interval in the middle to find m : m = (a+b)/2
3) sign of f(m) not matches with f(a) proceed the search in the new interval.
Calculation:
Given:
f(x) = x^{2}  x  1 = 0 , a = 1 , b = 2
f(1) = 1  1 1 = 1 < 0 , f(2) = 2^{2}  2  1 = 1 > 0 , Hence root lies between 1 and 2
By Bisection method,
Which is positive. Hence, the root lies between 1.5 and 2
By Bisection method,
The 2^{nd} approximation to a root of the equation x^{2}  x  1 = 0 in the interval (1, 2) by Bisection method is 1.75.
The approximate value of a root of x^{3} – 13 = 0, then 3.5 as initial value, after one iteration using NewtonRaphson method, is
Concept:
NewtonRaphson Method:
The iteration formula is given by
Where x_{0} is the initial value/root of the equation f(x) = 0
Calculation:
Given:
f(x) = x^{3}  13, x_{0 }= 3.5
f'(x) = 3x^{2}
f(x_{0}) = f(3.5) = 3.5^{3}  13 = 29.875
f'(x_{0}) = f'(3.5) = 3 × 3.5^{2} = 36.75
We know that
∴ x_{1} = 2.6871
47 videos119 docs75 tests

47 videos119 docs75 tests
