Question

Find 3 root of 18 by Newton-Raphson method. Perform 4 iterations. Take Xo = 2.5​

Answer 1

Answer:

Step-by-step explanation:

These solutions are not as brief as they should be: it takes work to

be brief. There will, almost inevitably, be some numerical errors. Please

inform me of them at [email protected]. We will be excessively casual in

our notation. For example, x3 = 3.141592654 will mean that the calculator

gave this result. It does not imply that x3 is exactly equal to 3.141592654.

We should always treat at least the final digit of a calculator answer

with some skepticism. Indeed different calculators can give (mildly) different

answers. In applied work, we need to pay heed to the fact that the standard

tools, such as calculators and computer programs, work only to limited

precision. In a complex calculation, minor inaccuracies may result in a

significant error.

1. Use the Newton-Raphson method, with 3 as starting point, to find a

fraction that is within 10−8 of √10. Show (without using the square

root button) that your answer is indeed within 10−8 of the truth.

Solution: The number √

10 is the unique positive solution of the equation f(x) = 0 where f(x) = x2 − 10. We use the Newton Method to

approximate a solution of this equation.

Let x0 be our initial estimate of the root, and let xn be the n-th

improved estimate. Note that f0

(x)=2x. The Newton Method recurrence is therefore

xn+1 = xn − f(xn)

f0

(xn) = xn − x2

n − 10

2xn

.

To make the expression on the right more beautiful, and calculations

easier, it is useful to manipulate it a bit. We get

xn+1 = xn − xn

2 +

10

2xn

= 1

2

xn +

10

xn

.

1

Compute, starting with x0 = 3. Then x1 = (1/2)(x0 + 10/x0) =

(1/2)(3 + 10/3) = 19/6. And x2 = (1/2)(19/6 + 60/19) = 721/228.

We could go on calculating with fractions—and there is interesting

mathematics involved—but from here on we switch to the calculator.

If we allow the = sign to be used sloppily, we get x1 = 3.166666667.

Then x2 = (1/2)(x1 + 10/x1)=3.162280702, and x3 = 3.16227766,

and x4 = 3.16227766.

The calculator says that x3 = x4 to 8 decimal places. We can therefore

dare hope that 3.16227766 is close enough. One way of checking is to

let a = 3.16227765 and b = 3.16227767. A quick calculation shows—if

the squaring button can be trusted, and it is one of the ones that can

be—that f(a) < 0 while f(b) > 0.

Thus the function f(x) changes sign as x goes from a to b. It follows by

the Intermediate Value Theorem that

√

f(x) = 0 has a solution (namely

10) between a and b. Since √

10 lies in the interval (a, b), and the

distance from 3.16227766 to either a or b is 10−8, it follows that the

distance from 3.16227766 to √10 is less than 10−8

Answer 2

Answer:

By Newton-Raphson method, ∛18 is equal to 2.6207

Step-by-step explanation:

General formula of Newton-Raphson method :

                     $x_{n+1} =\frac{(r-1)x^{r}_{n}+N }{rx^{r-1}_{n} }$                                   ...................(1)

For,  ∛18  the formula will be, put r=3 and N=18 in eq. (1),

   we get,         $x_{n+1} =\frac{2x^{3}_{n}+18 }{3x^{2}_{n} }$                                     .................(2)

Now given that,  x₀ = 2.5

• Iteration 1 : Put n=0 and  x₀ = 2.5 in eq.(2),

we get,               $x_{1} =\frac{2x^{3}_{0}+18 }{3x^{2}_{0} }= \frac{2(2.5)^{3}+18 }{3(2.5)^{2} }$

               ∴             $x_{1}= 2.6266$

• Iteration 2 : Put n=1 and x₁ in eq.(2),

we get,             $x_{2} =\frac{2(2.6266)^{3}+18 }{3(2.6266)^{2}}$

               ∴          $x_{2} = 2.6208$

• Iteration 3 : Put n=2 and the value of x₂ in eq.(2),

we get,             $x_{3} =\frac{2(2.6208)^{3}+18 }{3(2.6208)^{2}}$

      ∴                $x_{3} =2.6207$

• Iteration 4 : Put n=3 and the value of x₃ in eq.(2),

we get,            $x_{3} =\frac{2(2.6207)^{3}+18 }{3(2.6207)^{2}}$

       ∴                $x_{3} =2.6207$

     After performing the four iterations, we get the same value in 3rd as well as in 4th iteration.

      Therefore, 3 root by 18  is equal to 2.6207