Question

(a) given an initial approximation \({{\bf{x}}_{\bf{1}}}\) to a root of the equation \({\bf{f}}\left( {\bf{x}} \right){\bf{ = 0}}\), explain geometrically, with a diagram, how the second approximation \({{\bf{x}}_{\bf{2}}}\) in newton’s method is obtained.

b) Write an expression for \({{\bf{x}}_{\bf{2}}}\) in terms of \({{\bf{x}}_{\bf{1}}}\), \({\bf{f}}\left( {{{\bf{x}}_{\bf{1}}}} \right)\) and \({\bf{f'}}\left( {{{\bf{x}}_{\bf{1}}}} \right)\).

c) Write the expression for \({{\bf{x}}_{{\bf{n + 1}}}}\) in terms of \({{\bf{x}}_{\bf{n}}}\), \({\bf{f}}\left( {{{\bf{x}}_{\bf{n}}}} \right)\) and \({\bf{f'}}\left( {{{\bf{x}}_{\bf{n}}}} \right)\).

d) Under what circumstance is newton’s method likely to fail or to work very slowly?

Short answer

(a) The geometrical representation for the second approximation is,

(b)

The required expression is, \({x_2} = {x_1} - \frac{{f\left( {{x_1}} \right)}}{{f'\left( {{x_1}} \right)}}\).

(c)

The required expression is,\({x_{n + 1}} = {x_n} - \frac{{f\left( {{x_n}} \right)}}{{f'\left( {{x_n}} \right)}}\).

(d)

If the scale is too high it means the newton method is failing or slowing down.

Step-by-step solution

Part (a)

It is known that the newton Raphson method is an algorithmic method.

Having each approximation\({x_n}\),the next approximation \({x_{n + 1}}\) is the zero of the tangent line

\(y = f\left( {{x_n}} \right) + f'\left( {{x_n}} \right)\left( {x - {x_n}} \right)\_\_\_\_(1)\)

To the graph function \(f\left( x \right)\)at \(x = {x_n}\)

\(\therefore 0 = f\left( {{x_1}} \right) + f'\left( {{x_n}} \right)\left( {{x_{n + 1}} - {x_n}} \right)\)

So we get,

\({x_{n + 1}} = {x_n} - \frac{{f\left( {{x_n}} \right)}}{{f'\left( {{x_n}} \right)}}\)

Put \(n = 1\) in equation (1).

\(y = f\left( {{x_1}} \right) + f'\left( {{x_1}} \right)\left( {x - {x_1}} \right)\)

Here, \({x_2}\) is the zero of the tangent line.

Geometrical representation

The geometrical representation is,

Part (b)

Here, \({x_1}\) is the tangent line, then we have,

\(y = f\left( {{x_1}} \right) + f'\left( {{x_1}} \right)\left( {x - {x_1}} \right)\)

At \(x = {x_1}\)

\(\begin{aligned}{c}0 &= f\left( {{x_1}} \right) + f'\left( {{x_1}} \right)\left( {{x_2} - {x_1}} \right)\\ - f\left( {{x_1}} \right) &= f'\left( {{x_1}} \right)\left( {{x_2} - {x_1}} \right)\\ - \frac{{f\left( {{x_1}} \right)}}{{f'\left( {{x_1}} \right)}} &= {x_2} - {x_1}\\{x_2} = {x_1} - \frac{{f\left( {{x_1}} \right)}}{{f'\left( {{x_1}} \right)}}\end{aligned}\)

So, the required expression is, \({x_2} = {x_1} - \frac{{f\left( {{x_1}} \right)}}{{f'\left( {{x_1}} \right)}}\).

Part (c)

Here, \({x_{n + 1}}\) is the zero of the tangent line

\(y = f\left( {{x_n}} \right) + f'\left( {{x_n}} \right)\left( {x - {x_n}} \right)\)

The function \(f\left( x \right)\)at \(x = {x_n}\).

\(\begin{aligned}{c}0 &= f\left( {{x_n}} \right) + f'\left( {{x_n}} \right)\left( {{x_{n + 1}} - {x_n}} \right)\\ - f\left( {{x_n}} \right) &= f'\left( {{x_n}} \right)\left( {{x_{n + 1}} - {x_n}} \right)\\ - \frac{{f\left( {{x_n}} \right)}}{{f'\left( {{x_n}} \right)}} &= {x_{n + 1}} - {x_n}\\{x_{n + 1}} &= {x_n} - \frac{{f\left( {{x_n}} \right)}}{{f'\left( {{x_n}} \right)}}\end{aligned}\)

The required expression is, \({x_{n + 1}} = {x_n} - \frac{{f\left( {{x_n}} \right)}}{{f'\left( {{x_n}} \right)}}\).

Part (d)

If we want to find the root value of\(x = b\)of clean function \(f\left( x \right)\).

We can start with good approximation \({x_1} \approx b\).

We know that newton’s method is efficient, if \(\left| {\frac{{f''\left( b \right)}}{{f'\left( b \right)}}} \right|\) is small.

Hence, if the scale is too high it means the newton method is failing or slowing down.