Approximate a cubic root using Newton's method

Given

\(\displaystyle f(x)=x^3+x-1,\quad x_0=1\)

Start with the function value at x₀ = 1

Move through all five starts. Compare the current height, tangent gradient and next intercept

Which calculation gives the next estimate?

Derive the tangent update

At the current point, the tangent has the current derivative as its slope. Set its height to zero and solve for the next estimate. The tangent intercept generally differs from the exact curve root

1. 1A tangent gives a new estimate Set the tangent height to zero and solve for its horizontal intercept to obtain the next estimate
2. 2Update both function and derivative At every step use the current input in both f and f′. Subtract the correction. A zero derivative makes this update undefined

Given

\(\displaystyle f(x)=x^3+x-1,\quad x_0=1\)

Find the tangent gradient at the same initial input

Given

\(\displaystyle x_1=1-\frac{1}{4}\)

Calculate the first estimate exactly

The next step uses a new gradient

At x₁ = 3/4, f(x₁) = 11/64 and f′(x₁) = 43/16. Use these new values. Keeping the original gradient 4 would produce a different algorithm

1. 1Update both function and derivative At every step use the current input in both f and f′. Subtract the correction. A zero derivative makes this update undefined
2. 2Carry unrounded state Keep the full stored calculator value or exact fraction for the next step. Round only requested displayed outputs; never feed a displayed decimal back

Given

\(\displaystyle x_2=\frac34-\frac{11}{172}\)

Recover the exact second iterate

The screen displays x₂ ≈ 0.6860. What should be used to calculate x₃?

A root bracket supplies separate information

For this cubic, f(0) = −1 and f(1) = 1. Continuity gives a root in (0,1). Since 3x² + 1 is positive throughout the interval, the function is strictly increasing and the root there is unique

1. 1Isolate the intended root A continuous cubic with opposite endpoint signs has a root in that interval. A derivative with one strict sign throughout the interval proves the root there is unique
2. 2Check the starting basin An allowed start must lie in the stated target bracket, and the iterates must stay in its certified basin. A nonzero derivative at one input alone does not guarantee convergence

For g(x) = x³ − 3x − 1, why is x₀ = 1 unsuitable for a Newton step?

Successive displays do not prove the digits

A small correction can be encouraging, but a requested accuracy needs a justification. Independently bracket the same root with exact signs. To four places, all numbers strictly between 0.68225 and 0.68235 round to 0.6823

1. 1Certify rounding independently An independent bracket wholly inside a single rounding cell proves the requested decimal answer. Agreement of successive displayed iterates, or a small function value alone, does not prove it

Certified enclosure

\(\displaystyle 0.68232<\alpha<0.68233\)

An independent sign bracket gives 0.68232 < α < 0.68233. Report α to exactly four decimal places

Which statement actually proves those four decimal places?

Keep the three outputs separate

An exact iterate belongs to the recurrence. A displayed iterate is a rounded output. The reported root is certified by the independent enclosure. For negative values, half-away-from-zero rounding sends −1.235 to −1.24 at two decimal places

1. 1Carry unrounded state Keep the full stored calculator value or exact fraction for the next step. Round only requested displayed outputs; never feed a displayed decimal back
2. 2Certify rounding independently An independent bracket wholly inside a single rounding cell proves the requested decimal answer. Agreement of successive displayed iterates, or a small function value alone, does not prove it