Question

The iterative sequence of newton raphson method to find the value of the reciprocal of positive no. N is given by

Answer 1

Answer:

x(n) = 2x(n-1) - Nx²(n-1), where n≥1, x(0) is the initial approximation

Step-by-step explanation:

Hi,

To find the value of 1/N,

let x = 1/N

so 1/x = N

So, we need to find the roots of the equation f(x) = 1/x - N

f'(x) = -1/x²

Equation of tangent to any point (x₀, y₀) is given by

y - y₀/x - x₀ = -1/x₀²

Finding the point of intersection of tangent with x-axis, we get

x₁ - x₀ =  x₀²y₀

x₁ = x₀ + x₀²y₀

But y₀ = f(x₀) = 1/x₀ - N

SO, x₁ = x₀ + x₀²(1/x₀ - N)

x₁ = 2x₀ - Nx₀²

In general, for nth iteration the sequence would be

x(n) = 2x(n-1) - Nx²(n-1), where n≥1

x(0) is the initial approximation.

Hope, it helps !

Answer 2

Given : reciprocal of a positive number N

Newton raphson Method

To Find : Establish an iteration

Solution:

x = 1/N

=> 1/x  = N      

f(x)  = 1/x  -  N

f'(x)  = -1/x²  

xₙ₊₁  = xₙ ​− f(xₙ​) / f'(xₙ​)​

x₁  =  x₀  - f(x₀)/f'(x₀)

=> x₁  =  x₀  - (1/x₀  -  N)/(-1/x₀²)

=> x₁  =  x₀ +  x₀²(1/x₀  -  N)

=> x₁  =  x₀ +  x₀   - x₀² N

=> x₁  =  2x₀   - x₀² N

x₂ =  x₁  - f(x₁)/f'(x₁)

=> x₂ =  x₁  -  (1/x₁  -  N)/(-1/x₁²)

=> x₂ =  x₁  + x₁  -  x₁²N

=> x₂ =  2x₁ -  x₁²N

xₙ₊₁ = 2xₙ  - xₙ²N

Learn More:

Evaluate √30 to four decimal places by using Newton-Raphson ...

https://brainly.in/question/17089871

Find the approximate value of the expression (1.99)^7 by using ...

brainly.in/question/15853098