Question

Using Newton Raphson Method find an iterative formula for √5 N where N is positive
number, Hence find √5
35.​

Answer 1

Answer:  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 !

Step-by-step explanation:

Answer 2

Concept :

The Newton-Raphson method (also known as Newton's method) is a way of quickly finding a good approximation for the root of a real-valued function f ( x ) = 0 f(x) = 0 It uses the idea that a continuous function that is differentiable can be approximated by a straight line tangent to it.

Explanation:

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.

Final Answer:

x(n) = 2x(n-1) - Nx²(n-1), where n≥1

x(0) is the initial approximation.

SPJ3