The Dynamics of Iteration

Iteration is a central idea in the HSC Extension 1 sequences topic — it turns a simple function into a dynamic process that can approach a beautiful limit. In this problem we investigate the convergence of a square‑root recurrence, establish a contraction inequality using calculus, and finally connect the process to Newton’s method. Along the way, you will see how a derivative bound guarantees geometric shrinkage of errors, a technique that appears in many fixed‑point arguments.

Problem Statement

Consider the function

$$g(x)=\sqrt{x+2}.$$

We define an iterative sequence by

$$x_{n+1}=g(x_n),\qquad x_1=1.$$

• Calculate $x_2,x_3,$ and $x_4$ to three decimal places. By solving

$$L=\sqrt{L+2},$$

determine the exact positive limit $L$ that the sequence appears to be approaching.

• Show that

$$0<g'(x)<\frac12$$

for all $x>0$.

• Prove that

$$|x_{n+1}-L|<\frac12|x_n-L|.$$

Hence deduce that the sequence converges to $L$ for any initial value $x_1>0$.

• Define a function $f(x)$ such that Newton's method would converge to the same limit $L$ found in part (i), and write down the specific iterative formula for this $f(x)$.

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Hints

• For (i): Substitute repeatedly into $x_{n+1}=\sqrt{x_n+2}$, then solve the fixed-point equation.
• For (ii): Differentiate $g(x)=\sqrt{x+2}$ and note where the derivative is largest for $x>0$.
• For (iii): Use either the mean value theorem or rationalise $\sqrt{x_n+2}-\sqrt{L+2}$.
• For (iv): Rearrange $L=\sqrt{L+2}$ into a polynomial equation.

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Solutions

(i) Starting from $x_1=1$ and iterating the function, we obtain the first few terms:

$$x_2=\sqrt3\approx1.732,\quad x_3=\sqrt{2+\sqrt3}\approx1.932,\quad x_4\approx1.983.$$

If the sequence converges to a positive number $L$, then taking limits on both sides of $x_{n+1}=\sqrt{x_n+2}$ gives the fixed‑point equation $L=\sqrt{L+2}$. Solving this yields

$$L=\sqrt{L+2}\implies L^2-L-2=0 \implies (L-2)(L+1)=0.$$

Thus the positive limit is

$$L=2.$$

(ii) To understand why the sequence behaves well, we look at the derivative of $g$, which measures the sensitivity of the iteration:

$$g'(x)=\frac{1}{2\sqrt{x+2}}.$$

For $x>0$ this derivative is positive, and because $\sqrt{x+2}>\sqrt{2}$, we obtain the clean bound

$$g'(x)<\frac{1}{2\sqrt2}<\frac12.$$

Having $|g'(x)|<\frac12$ on $(0,\infty)$ tells us that $g$ is a contraction mapping — a crucial property that forces successive iterates to pull closer together.

(iii) Now we prove that the error shrinks by at least a factor of $\tfrac12$ at each step. Using the fixed‑point relation $L=g(L)=2$ and rationalising the difference of square roots,

$$\begin{aligned} |x_{n+1}-L| &=\left|\sqrt{x_n+2}-\sqrt{L+2}\right|\\ &=\frac{|x_n-L|}{\sqrt{x_n+2}+\sqrt{L+2}}\\ &<\frac12|x_n-L|, \end{aligned}$$

because $x_n>0$ and $\sqrt{L+2}=2$. Repeating this inequality gives

$$|x_n-L|<\left(\frac12\right)^{n-1}|x_1-L|\to0,$$

so $x_n\to L=2$ for any $x_1>0$.

(iv) We can view the same limit through the lens of root‑finding. From the fixed‑point equation we have

$$L^2-L-2=0.$$

Therefore we define

$$f(x)=x^2-x-2,$$

so that $L$ is a root of $f$. Applying Newton’s method to $f$ produces the iteration

$$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)} =x_n-\frac{x_n^2-x_n-2}{2x_n-1}.$$

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Takeaways

• Fixed point is the value of $x$ such that $g(x)=x$.
• Fixed points often reveal the likely limit of an iterative sequence.
• A contraction inequality proves convergence by forcing errors to shrink geometrically.
• Newton's method can target the same limit by solving the fixed-point equation as a root problem.

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Further Readings

HSC Trigonometry, HSC Induction, HSC Probability, HSC Last Resorts