When numerical iteration fails: causes and fixes

An iteration can produce numbers without producing the root you want. It may diverge, oscillate, repeat a cycle, leave the function’s domain, encounter division by zero or converge to a different root.

Failure is not just a calculator problem. It usually comes from the iteration formula, the starting value, or the shape of the graph. Recognising the cause lets you choose a better rearrangement, a better starting value or a more reliable method.

Prerequisites

You should be able to:

• generate terms from $x_{n+1}=g(x_n)$;
• use the local convergence test $|g'(\alpha)|<1$;
• apply $x_{n+1}=x_n-f(x_n)/f'(x_n)$;
• interpret roots, gradients and tangents graphically;
• identify the domains of square root, logarithmic and rational functions.

Review fixed point iteration and the Newton Raphson method first.

What counts as failure?

Suppose the intended root is $\alpha$. Successful iteration means

$$x_n\longrightarrow\alpha \quad\text{as }n\longrightarrow\infty.$$

Several other behaviours are possible.

Behaviour	What the values do	Typical cause
Divergence	Move farther from the root or grow without bound	A repelling fixed point or unsuitable start
Oscillation	Alternate from side to side without settling	A negative multiplier with magnitude at least $1$
Cycle	Repeat two or more values	Repeated application of $g$ returns to an earlier value
Undefined step	Leave the domain or divide by zero	Formula is not defined at an iterate
Wrong-root convergence	Settle, but at another root	Starting value lies in another root’s region of attraction
False confidence	Consecutive displayed values agree temporarily	Rounding or very slow movement

The first few terms alone do not prove convergence. Always connect numerical evidence to the formula or graph.

Fixed point failure and the derivative test

For

$$x_{n+1}=g(x_n),$$

let $\alpha$ satisfy $g(\alpha)=\alpha$. If $x_n$ is close to $\alpha$, the error $e_n=x_n-\alpha$ approximately obeys

$$e_{n+1}\approx g'(\alpha)e_n.$$

Therefore:

• $|g'(\alpha)|<1$ makes nearby errors shrink;
• $|g'(\alpha)|>1$ makes nearby errors grow;
• $g'(\alpha)<0$ reverses the sign of the error, so values alternate across the root;
• $|g'(\alpha)|=1$ is inconclusive and needs further investigation.

This is a local test. It describes starts sufficiently close to $\alpha$, not every possible $x_0$.

Worked example 1: a divergent rearrangement

The equation

$$x^2-x-1=0$$

has positive root

$$\alpha=\frac{1+\sqrt5}{2}=1.618\ldots.$$

Consider the valid rearrangement

$$x=x^2-1, \qquad x_{n+1}=x_n^2-1.$$

Here $g'(x)=2x$, so

$$|g'(\alpha)|=2\alpha=3.236\ldots>1.$$

The fixed point is repelling. Even with $x_0=1.6$, close to the root,

$$\begin{array}{c|c} n&x_n\\ \hline 0&1.600000\\ 1&1.560000\\ 2&1.433600\\ 3&1.055209\\ 4&0.113466\\ 5&-0.987125 \end{array}$$

The values initially look plausible, but their errors grow. Algebraic equivalence does not guarantee numerical stability.

A better rearrangement for the positive root is

$$x=\sqrt{x+1}.$$

Then

$$g'(x)=\frac1{2\sqrt{x+1}}, \qquad g'(\alpha)=\frac1{2\alpha}=0.309\ldots,$$

so nearby errors shrink rapidly.

Misconception: If $x=g(x)$ is algebraically correct, its iteration must converge.

Correction: The equation identifies fixed points. The size of $g'$ determines whether nearby iterates are attracted to or repelled from them.

Self-check 1

For the positive root of $x^2+x-4=0$, compare

$$g_1(x)=\frac4{x+1} \qquad\text{and}\qquad g_2(x)=\sqrt{4-x}.$$

Which iteration is locally convergent?

Answer

The positive root is

$$\alpha=\frac{-1+\sqrt{17}}2=1.5615\ldots.$$

For $g_1$,

$$g_1'(x)=-\frac4{(x+1)^2}.$$

Since $4=\alpha(\alpha+1)$,

$$|g_1'(\alpha)|=\frac{\alpha}{\alpha+1}=0.6096\ldots<1.$$

For $g_2$,

$$|g_2'(\alpha)|=\frac1{2\sqrt{4-\alpha}}=\frac1{2\alpha}=0.3202\ldots<1.$$

Both are locally convergent. The second should converge faster because its derivative has smaller magnitude. The first alternates because $g_1'(\alpha)<0$.

Oscillation and cycles

Alternation is not itself failure. If $-1<g'(\alpha)<0$, values alternate with decreasing error and converge. Failure occurs when the alternating error does not shrink.

Worked example 2: an exact two-cycle

Solve $x^2=2$ using the rearrangement

$$x=\frac2x, \qquad x_{n+1}=\frac2{x_n}.$$

Starting with $x_0=1$ gives

$$x_1=2,quad x_2=1,quad x_3=2,quad x_4=1,ldots$$

The iteration is trapped in the two-cycle $1,2$. In fact,

$$g(g(x))=\frac{2}{2/x}=x$$

for every non-zero $x$, so almost every starting value repeats after two steps. At the positive fixed point $\alpha=\sqrt2$,

$$g'(\alpha)=-\frac2{\alpha^2}=-1.$$

The usual derivative test is inconclusive at magnitude $1$, but the identity $g(g(x))=x$ proves the cycling behaviour.

By contrast, the Babylonian iteration

$$x_{n+1}=\frac12\left(x_n+\frac2{x_n}\right)$$

converges to $\sqrt2$ from any positive start. Averaging damps the oscillation.

Self-check 2

The iteration $x_{n+1}=1-x_n$ has fixed point $\alpha=1/2$. What happens from $x_0=0.2$?

Answer

$$x_1=0.8,quad x_2=0.2,quad x_3=0.8,ldots$$

It enters a two-cycle. Since $g'(x)=-1$, the distance from $1/2$ is preserved while its sign reverses.

Leaving the domain

An iteration formula may be equivalent to the original equation only where every operation is valid. During iteration, a value can leave that domain.

Worked example 3: a square root becomes undefined

Consider

$$x_{n+1}=\sqrt{3-x_n^2}, \qquad x_0=0.$$

Then

$$x_1=\sqrt3, \qquad x_2=\sqrt{3-(\sqrt3)^2}=0,$$

so the values cycle between $0$ and $\sqrt3$. A small rounded error can be worse. If a calculator stores $x_1$ as slightly larger than $\sqrt3$, then $3-x_1^2<0$ and the next real square root is undefined.

Common domain failures include:

$$\sqrt{h(x_n)}\text{ with }h(x_n)<0,$$ $$\ln(h(x_n))\text{ with }h(x_n)\le0,$$

and

$$\frac{p(x_n)}{q(x_n)}\text{ with }q(x_n)=0.$$

Before iterating, state the domain and check whether $g$ maps the proposed interval back into itself.

Misconception: A calculator error means the equation has no root.

Correction: It means the chosen formula is undefined at that iterate. The original equation may still have roots.

Why Newton Raphson can fail

Newton Raphson uses

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

It relies on a tangent at $(x_n,f(x_n))$. Trouble occurs when the tangent gives poor information about the nearby root.

Failure 1: a zero or very small derivative

If $f'(x_n)=0$, the formula divides by zero. If $f'(x_n)$ is merely close to zero, the correction

$$\frac{f(x_n)}{f'(x_n)}$$

can be enormous.

Worked example 4: a horizontal tangent

Let

$$f(x)=x^3-2x+2, \qquad f'(x)=3x^2-2.$$

Starting at

$$x_0=\sqrt{\frac23}$$

gives $f'(x_0)=0$, so $x_1$ is undefined. Geometrically, the tangent is horizontal and never meets the $x$-axis.

Starting near this value makes $|f'(x_0)|$ small, so the tangent’s axis intercept may be very far away. A nearby starting value is not always a good starting value.

Failure 2: cycling

For the same function, start at $x_0=0$:

$$x_1=0-\frac{2}{-2}=1,$$

then

$$x_2=1-\frac{1}{1}=0.$$

Thus

$$0,1,0,1,\ldots$$

is a two-cycle, even though $f$ has a real root near $-1.769$.

Failure 3: convergence to a different root

Newton Raphson is sensitive to $x_0$ when a function has several roots.

Worked example 5: the starting value selects the root

For

$$f(x)=x^3-x=x(x-1)(x+1),$$

the roots are $-1$, $0$ and $1$, and

$$x_{n+1}=x_n-\frac{x_n^3-x_n}{3x_n^2-1}.$$

With $x_0=0.8$,

$$x_1=1.1130\ldots,quad x_2=1.0150\ldots,quad x_3=1.0003\ldots,$$

so the method converges to $1$.

With $x_0=0.1$,

$$x_1=-0.002061\ldots,quad x_2\approx1.75\times10^{-8},$$

so it converges to $0$. Newton Raphson finds a root associated with the starting value, not necessarily the root intended by the question.

Self-check 3

Explain why Newton Raphson cannot start at $x_0=0$ for $f(x)=x^2-5$.

Answer

Since $f'(x)=2x$, we have $f'(0)=0$. The formula would require

$$x_1=0-\frac{-5}{0},$$

which is undefined. On the graph, the tangent at $(0,-5)$ is horizontal.

Slow convergence can look like failure

When $|g'(\alpha)|$ is just below $1$, errors shrink very slowly. If

$$e_{n+1}\approx0.99e_n,$$

then after $100$ steps

$$|e_{100}|\approx0.99^{100}|e_0|\approx0.366|e_0|.$$

Most of the original error remains. Conversely, rounded calculator displays can repeat even when the exact iterates still move. Agreement of successive displayed values is evidence, not proof of accuracy.

To justify a rounded root, test the two rounding boundaries using a change of sign.

A practical recovery strategy

When an iteration fails:

1. Check the formula. Confirm signs, brackets, derivatives and calculator angle mode.
2. Check the domain. Identify where $g(x)$ and the Newton Raphson quotient are defined.
3. Inspect the values. Look for growth, alternation, a repeating cycle or a large jump.
4. Use the graph. Locate roots, stationary points and suitable starting regions.
5. Test local convergence. For fixed point iteration, evaluate $|g'(\alpha)|$ or bound $|g'(x)|$ on a known interval.
6. Change one feature. Try another rearrangement, choose $x_0$ from a sign-changing bracket, or use bisection.
7. Verify the result. Substitute back and justify the requested accuracy.

Bisection is slower but safer when $f$ is continuous and a sign-changing bracket is known. It keeps the root trapped inside the current interval. Newton Raphson is faster near a suitable simple root but does not preserve a bracket automatically.

Mixed diagnostic check

Match each observation to the most likely diagnosis.

1. $x_n$ alternates across the root with steadily decreasing error.
2. Newton Raphson produces a huge jump when $x_n$ is near a stationary point.
3. A logarithmic iteration reports a domain error.
4. $x_{n+2}=x_n$ but $x_{n+1}\ne x_n$.
5. The method converges accurately, but to the wrong solution.

Answers

1. Convergence with $-1<g'(\alpha)<0$. Alternation alone is not failure.
2. $f'(x_n)$ is close to zero, making the Newton Raphson correction large.
3. The logarithm’s argument has become non-positive.
4. The iteration is in a two-cycle.
5. The starting value belongs to the region attracted to a different root.

Key conclusions

• A correct rearrangement can still give a divergent iteration.
• For fixed point iteration, $|g'(\alpha)|<1$ predicts local convergence; $|g'(\alpha)|>1$ predicts local divergence.
• The case $|g'(\alpha)|=1$ needs separate analysis.
• Newton Raphson can fail at a horizontal tangent, jump far away, cycle or find another root.
• Domain restrictions apply to every iterate, not only to the final root.
• A graph and a sign-changing bracket help you choose a sensible starting value.
• Repeated decimal displays do not prove the claimed accuracy.

Next, consolidate the whole topic in A-level numerical methods or practise reliable bracketing with locating roots using a change of sign.