The idea

Using an iterative formula

How x_{n+1} = g(x_n) feeds each estimate into the next, the full-precision discipline (never round mid-chain), and what the settling sequence tells you about the equation's root.

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Algebra · Iteration

Using an iterative formula

How x_{n+1} = g(x_n) feeds each estimate into the next, the full-precision discipline (never round mid-chain), and what the settling sequence tells you about the equation's root.

Why it works

A machine for better estimates

The story so far: the sign-change test trapped a root of

x^3 - 3x - 5 = 0

between

2

and

3

, and the rearrangement step built the adapter

x = \sqrt[3]{3x + 5}

. Now we finally run the machine. An iterative formula is a term-to-term rule for estimates:

x_{n+1} = \sqrt[3]{3x_{n} + 5}, \qquad x_1 = 2

Each pass produces the next estimate from the last:

x_2 = \sqrt[3]{3(2) + 5} = \sqrt[3]{11} = 2.223980\ldots, \qquad x_3 = \sqrt[3]{3(2.223980\ldots) + 5} = 2.268372\ldots

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The rest of the explanation, plus 2 worked examples you step through move by move.

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