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.

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

An iterative formula is a term-to-term rule for ESTIMATES:

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

says: feed the current estimate in, get a better one out. Each pass through the machine:

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

x_3 = \sqrt[3]{3(2.223980\ldots) + 5} = 2.268372\ldots

x_2 = 2.224

and

x_3 = 2.268

to 3 decimal places. Note the discipline: the FULL calculator value feeds the next step — round only what you WRITE DOWN, never what you carry forward. (Feeding 2.224 back in gives a subtly wrong

x_3

, and the error compounds.) On a calculator: type the seed, press =, then build the formula around ANS and press = repeatedly.

Read the subscripts as positions.

x_3

means the third estimate — compute exactly as many steps as asked, feeding each NEW value in (an off-by-one, or re-feeding

x_1

, is the standard slip).

Why does this find a root? If the sequence settles at some value

x

, then feeding it in returns itself:

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

— cubing,

x^3 = 3x + 5

, i.e.

x^3 - 3x - 5 = 0

. A settled iteration IS a solution of the matching equation. The values

2.224, 2.268, 2.277, 2.279, \ldots

are closing in on the root of

x^3 - 3x - 5 = 0

(which the sign-change test located between 2 and 3).

The iterates are estimates, not the exact answer.

x_4 = 2.277

is close to the root, not equal to it — the true root is irrational. More iterations sharpen the estimate; a sign-change check at the appropriate interval certifies any rounded claim.