Rearranging an equation into iterative form

How "show that the equation can be written as x = g(x)" works — isolate a lone x by honest rearranging — and why that form is exactly what an iteration machine needs.

Algebra · Iteration

Rearranging an equation into iterative form

How "show that the equation can be written as x = g(x)" works — isolate a lone x by honest rearranging — and why that form is exactly what an iteration machine needs.

Why it works

An iteration machine needs the equation in the shape

x = g(x)

— "

x

equals a formula involving

x

" — because that's what lets an estimate be FED IN on the right to produce a new estimate on the left. "Show that

x^3 - 3x - 5 = 0

can be rearranged to give

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

" is a rearranging proof:

x^3 - 3x - 5 = 0 \;\Rightarrow\; x^3 = 3x + 5 \;\Rightarrow\; x = \sqrt[3]{3x + 5}.

Two honest moves — isolate the

x^3

, then cube-root both sides. In a "show that", every step must be visible; jumping from the equation to the target form earns nothing.

Different isolations give different iterative forms. From

x^3 + 4x - 9 = 0

, isolating the

4x

instead gives

4x = 9 - x^3

, so

x = \frac{9 - x^3}{4}

— also valid. An exam "show that" names its target: read which

x

was isolated and steer the algebra there, not to the rearrangement you'd have chosen. (Not every rearrangement makes a GOOD iteration — some spiral away rather than settle — but the exam only ever asks you to use the one it gives.)

Square-root forms follow the same pattern. From

x^2 - 5x - 3 = 0

: isolate

x^2 = 5x + 3

and take the (positive) square root:

x = \sqrt{5x + 3}

.

The full exam pipeline is usually three parts, one skill each:

show a root exists between $a$ and $b$ (sign change);
show the equation rearranges to $x = g(x)$ (this concept);
iterate from a seed to estimate the root (previous concept).

Each part hands its result to the next — the rearranged form in part 2 IS the machine run in part 3, and the interval from part 1 tells you the answer is plausible.