Worked example

Iteration x=∛(2x²+7/x): the value it converges to

A-Level Maths · 97095 marksmed

Question

The sequence of values defined by the iterative formula

$x_{n+1} = \sqrt[3]{2x_n^2 + 7x_n^{-1}},$

with starting value $x_{1} = 2.2$ , converges to a constant $eta$ . Use this iterative formula to find $eta$ correct to 3 decimal places. Record the result of every iteration to 5 decimal places. State a non-rational equation that is satisfied by $eta$ , and hence determine the exact value of $eta$ .

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Full worked solution

(i) Compute $β$ to 3 decimal places via iteration
For a convergent iterative sequence, the limit $β$ satisfies $x_{n + 1} \approx x_{n} = β$ for sufficiently large $n$ . We iteratively apply the given formula, rounding each intermediate result to 5 decimal places:
1. $x_{1} = 2.20000$
2. $x_{2} = 3 2 (2.2)^{2} + 7 (2.2)^{- 1} = 3 9.68 + 3.181818 = 3 12.861818 \approx 2.34299$
3. $x_{3} = 3 2 (2.34299)^{2} + 7 (2.34299)^{- 1} \approx 3 10.97902 + 2.98757 \approx 2.40955$
4. $x_{4} = 3 2 (2.40955)^{2} + 7 (2.40955)^{- 1} \approx 3 11.61194 + 2.90509 \approx 2.44089$
5. $x_{5} \approx 2.45530$ , $x_{6} \approx 2.46189$ , $x_{7} \approx 2.46492$ , $x_{8} \approx 2.46631$ , $x_{9} \approx 2.46694$ , $x_{10} \approx 2.46723$ , $x_{11} \approx 2.46737$ , $x_{12} \approx 2.46743$ , $x_{13} \approx 2.46746$ , $x_{14} \approx 2.46747$
All subsequent iterations round to $2.467$ at 3 decimal places. To confirm, check the interval: $f (x) = 3 2 x^{2} + 7 x^{- 1} - x$ has a sign change in $(2.4665, 2.4675)$ , so $β = 2.467$ to 3 decimal places.
Note.
Record every iteration to 5 dp to avoid rounding error buildup. Stop when three consecutive iterations agree to 3 dp, or verify with a sign change in the relevant interval.
(ii) Find the fixed-point equation for $β$ and exact value
At the limit, the sequence converges so $x_{n + 1} = x_{n} = β$ . Substitute into the iterative formula:
$\beta = \sqrt[3]{2\beta^2 + 7\beta^{-1}}$
Cube both sides to eliminate the cube root, and multiply through by $β$ to clear the reciprocal:
$\beta^3 = 2\beta^2 + \frac{7}{\beta}$
$\beta^4 = 2\beta^3 + 7$
This is the non-rational fixed-point equation for $β$ . Testing integer roots confirms no integer solution; the exact value is the positive real root of this quartic, which simplifies to the limit we derived numerically.
Note.
Always substitute $x_{n + 1} = x_{n} = β$ for convergent iterative sequences. Cube both sides first to eliminate roots, then clear denominators to get a polynomial equation.

Answer

(i) $β = 2.467$ (3 dp); (ii) Exact value is the positive real root of $β^{4} - 2 β^{3} - 7 = 0$

What this tests

Fixed-point iteration for root finding
Convergence of iterative sequences
Polynomial equation derivation from iterative formulae
Rounding and accuracy justification for numerical results

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⚠ A common mistake is to stop iterating too early, before values stabilize to the required precision, or to round intermediate iterations to fewer than 5 decimal places, which introduces propagation error.

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Iteration x=∛(2x²+7/x): the value it converges to

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