Editing Newton's method (section)

=== Solution of {{math|1=cos({{var|x}}) = {{var|x}}{{sup|3}}}} using Newton's method ===
Consider the problem of finding the positive number {{mvar|x}} with {{math|1=cos {{var|x}} = {{var|x}}{{sup|3}}}}. We can rephrase that as finding the zero of {{math|1={{var|f}}({{var|x}}) = cos({{var|x}}) &minus; {{var|x}}{{sup|3}}}}. We have {{math|1={{var|{{prime|f}}}}({{var|x}}) = &minus;sin({{var|x}}) &minus; 3{{var|x}}{{sup|2}}}}. Since {{math|cos({{var|x}}) ≤ 1}} for all {{mvar|x}} and {{math|{{var|x}}{{sup|3}} > 1}} for {{math|{{var|x}} > 1}}, we know that our solution lies between 0 and 1. 

A starting value of 0 will lead to an undefined result which illustrates the importance of using a starting point close to the solution. For example, with an initial guess {{math|1={{var|x}}{{sub|0}} = 0.5}}, the sequence given by Newton's method is:

<math display="block">\begin{matrix}
 x_1 & = & x_0 - \dfrac{f(x_0)}{f'(x_0)} & = & 0.5 - \dfrac{\cos 0.5 - 0.5^3}{-\sin 0.5 - 3 \times 0.5^2} & = & 1.112\,141\,637\,097\dots \\
 x_2 & = & x_1 - \dfrac{f(x_1)}{f'(x_1)} & = & \vdots & = & \underline{0.}909\,672\,693\,736\dots \\
 x_3 & = & \vdots & = & \vdots & = & \underline{0.86}7\,263\,818\,209\dots \\
 x_4 & = & \vdots & = & \vdots & = & \underline{0.865\,47}7\,135\,298\dots \\
 x_5 & = & \vdots & = & \vdots & = & \underline{0.865\,474\,033\,1}11\dots \\
 x_6 & = & \vdots & = & \vdots & = & \underline{0.865\,474\,033\,102}\dots
\end{matrix}
</math>

The correct digits are underlined in the above example. In particular, {{math|{{var|x}}{{sub|6}}}} is correct to 12 decimal places. We see that the number of correct digits after the decimal point increases from 2 (for {{math|{{var|x}}{{sub|3}}}}) to 5 and 10, illustrating the quadratic convergence.