Question

Use Newton’s method with the specified initial approximation \({x_1}\) to find \({x_3}\), the third approximation to the root of the given equation. (Give your answer to four decimal places.)

\({x^5} - x - 1 = 0, {x_1} = 1\)

Short answer

\({x_3} = 1.1785\)

Step-by-step solution

Introduction

The Newton’s method about a function is,

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

f(x) is the function,\(f'({x_n}),{x_n}{\rm{ and }}{{\rm{x}}_{n + 1}}\) are derivation of the function and last two are roots of the function.

Given

\({x^5} - x - 1 = 0, {x_1} = 1\)

The Required initial approximations

\(\begin{aligned}{c}f\left( x \right) &= {x^5} - x - 1, \\f'\left( x \right) &= 5{x^4} - 1, \\{x_1} &= 1\end{aligned}\)

So, from newton’s formula,

\(\begin{aligned}{l}{x_{n + 1}} &= {x_n} - \frac{{f({x_n})}}{{f'({x_n})}}\\{x_2} &= 1 - \frac{{1 - 1 - 1}}{{5 - 1}}\\{x_2} &= 1.25\end{aligned}\)

Now again,

\(\begin{aligned}{l}{x_{n + 1}} &= {x_n} - \frac{{f({x_n})}}{{f'({x_n})}}\\{x_3} &= 1.25 - \frac{{{{\left( {1.25} \right)}^5} - 1.25 - 1}}{{5{{\left( {1.25} \right)}^4} - 1}}\\{x_3} &= 1.1785\end{aligned}\)

.

Thus, \({x_3} = 1.1785\).

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