An alternating fifth-root iteration · source reference

Printed question

The diagram shows $y=x^5+2x-8$. Its root is denoted by $\alpha$.

(i) Find $\alpha$ correct to 3 decimal places. [1]

The real numbers $x_n$ satisfy $x_{n+1}=(8-2x_n)^{1/5}$, $n\ge1$.

(ii) Using (i), show that if the sequence converges, it converges to $\alpha$. [2]

(iii) Considering the graphs of $y=x$ and $y=(8-2x)^{1/5}$ on the same diagram, or otherwise, prove that if $\alpha<x_n<4$, then (a) $0<x_{n+1}<\alpha$, (b) $\alpha<x_{n+2}<x_n$. [3]

(iv) Given $x_1=3$, use (iii) to obtain an inequality relating $0,\alpha,x_1,x_2,x_3$. With a graphic calculator, describe the sequence behaviour. [2]

Printed working

(i) $x^5+2x-8=0$. Using GC, $\alpha=1.392$ (3 d.p.).

(ii) If it converges to $L$ (the limit), then as $n\to\infty$, $x_n\to L,x_{n+1}\to L$.

$$L=(8-2L)^{1/5}\Rightarrow L^5+2L-8=0.$$

From (i), $\alpha$ is the root, so $L=\alpha$. Hence if it converges, it converges to 1.392.

(iii) $y=x$ and $y=(8-2x)^{1/5}$ intersect at $x=\alpha$. Recurrence: $x_{n+1}=(8-2x_n)^{1/5}$. From the diagram, if $\alpha<x_n<4$, then $0<x_{n+1}<\alpha$ and $\alpha<x_{n+2}<x_n$.

(iv) When $x_1=3$, $\alpha<x_1<4$. By (iii), $0<x_2<\alpha$ and $\alpha<x_3<x_1$. That is $0<x_2<\alpha<x_3<x_1$. Using GC, the sequence oscillates and converges to 1.392.

Printed repeated block (v): the solution sheet repeats the same $x_1=3$ argument, the inequality $0<x_2<\alpha<x_3<x_1$ and the statement that it oscillates and converges to 1.392. There is no part (v) in the question.

Source tables and calculator displays

GC polynomial-root finder setup: ORDER 1 2 3 4 5 6 7 8 9 10 (5 selected); REAL / a+bi / re^(θi) (a+bi selected); DEC / FRAC; NORMAL / SCI / ENG; FLOAT 0 1 2 3 4 5 6 7 8 9; RADIAN / DEGREE (DEGREE selected). Screen buttons: MAIN, HELP, NEXT; equation $ax^5+\cdots+a_1x+a_0=0$; roots-screen buttons MAIN, MODE, COEF, STO.

Root	First displayed panel	Second displayed panel
x₁	−1.26390490…	…0.9922772727i
x₂	−1.26390490…	−1.26390490…
x₃	0.5681535179…	0.5681535179…
x₄	0.5681535179…	0.5681535179…
x₅	1.39150277	1.39150277

These are the literal truncated source displays, not full complex-root claims. The graph reads Intersection X=1.3915028 Y=0. Only the real root is requested.

Source diagrams redrawn

Printed answer key

(i) 1.392 (3 d.p.); (iv) $0<x_2<\alpha<x_3<x_1$; oscillates and converges to 1.392

Mathematical review and qualifications

(i) Since $5x^4+2>0$, there is exactly one real root, numerically $\alpha=1.392$ (3 d.p.).

(ii) Continuity gives $L=(8-2L)^{1/5}$, hence $L^5+2L-8=0$ and $L=\alpha$.

(iii)(a) $F(x)=(8-2x)^{1/5}$ is decreasing and $F(\alpha)=\alpha,F(4)=0$, so $0<F(t)<\alpha$. (b) With $y=F(t)$, decreasingness gives $F(y)>\alpha$. To show $F(y)<t$, it suffices that $t^5+2y-8>0$. Using $y^5+2t=8$, this becomes $(t-y)[t^4+t^3y+t^2y^2+ty^3+y^4-2]>0$. Here $t>\alpha>1.3$ and $y>0$, so the bracket exceeds $1.3^4-2>0$. Thus $\alpha<x_{n+2}<x_n$.

(iv) $x_2=2^{1/5}\approx1.149$, $x_3\approx1.416$, giving $0<x_2<\alpha<x_3<x_1=3$. Iteration gives oscillation converging to $1.392$. The odd subsequence decreases above $\alpha$; the even subsequence increases below it. Their limiting two-cycle would require $(t-y)[t^4+t^3y+t^2y^2+ty^3+y^4-2]=0$; the upper limit is at least $\alpha$, so the bracket is positive and the limits coincide at $\alpha$.

Provenance

2025 TJC H2 Math (9758) JC2 MSM: RECURRENCE RELATIONS. [2014/TJC/Promo/7] Question PDF pages 4; solution PDF pages 12, 13. The original files remain private QA records outside this native content.