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Choosing The Correct Fixed Point

Use established bounds, monotonicity, and the starting value to select the only fixed point a recursively defined sequence can attain as its limit.

Intermediate 10 min read

What You'll Learn

  • Derive the fixed-point equation for a convergent recurrence
  • Use continuity to justify why a limit must be a fixed point
  • Restrict candidate limits using bounds and the initial value
  • Distinguish an invariant interval from a convergence proof
  • Check a fixed-point selection argument against explicit iterates

Why Solving the Limit Equation Is Not Enough

A recursively defined sequence may converge to a number \(L\) satisfying \(L=f(L)\). Solving this equation is an important step, but it may produce several candidates. The equation alone does not tell us which candidate is the limit of the sequence starting from a specified initial value.

To make the choice, use information already established about the sequence: its bounds, whether it is increasing or decreasing, and where it starts. These facts restrict the values its limit can have. The task is not to guess which fixed point seems likely, but to show that every other candidate is incompatible with the sequence's known behavior.

Key idea: First derive all possible fixed points. Then use proven bounds and the initial value to eliminate candidates the sequence cannot attain.

The Limit Must Be a Fixed Point

Consider a real sequence defined by \(x_{n+1}=f(x_n)\), where \(f\) is continuous at the limit \(L\) of the sequence. Since \(x_n\to L\), the shifted sequence \(x_{n+1}\) also tends to \(L\). Continuity gives \(f(x_n)\to f(L)\). But \(x_{n+1}=f(x_n)\), so the same sequence has limit \(f(L)\). Limits are unique; therefore \(L=f(L)\).

$$ x_n\to L,\quad x_{n+1}=f(x_n),\quad f\text{ continuous at }L \quad\Longrightarrow\quad L=f(L). $$

This implication has conditions. The sequence must actually converge, and \(f\) must be continuous at its limit. Solving \(L=f(L)\) identifies necessary candidates under these conditions; it does not prove convergence or determine which candidate is reached.

Selection principle: If the fixed-point equation has candidate solutions \(c_1,\ldots,c_k\), and the sequence is known to converge to a value in an interval \([a,b]\), discard every candidate outside \([a,b]\). If several candidates remain, use further information, such as monotonicity and the initial value, to narrow the possibilities.

For example, if \(x_n\) is increasing and \(x_n\leq U\) for every \(n\), then its limit is at least \(x_0\) and at most \(U\). If it is decreasing and \(x_n\geq D\), its limit is at most \(x_0\) and at least \(D\). These inequalities can settle the selection even when more than one fixed point lies in a broader invariant interval.

Worked Example: A Decreasing Sequence Selects Zero

Worked Example: A Decreasing Sequence Selects Zero

Let \(x_0=\frac12\) and \(x_{n+1}=x_n^2\). Suppose we establish that the sequence converges. Which fixed point can be its limit?

First solve the fixed-point equation:

$$ L=L^2 \quad\Longleftrightarrow\quad L(L-1)=0 \quad\Longleftrightarrow\quad L=0\text{ or }L=1. $$

Both are algebraic candidates. Now examine the recurrence and the initial value. For \(0<x_n<1\), squaring gives \(0<x_n^2<x_n\). Since \(x_0=\frac12\), induction gives \(0<x_n\leq\frac12\) for all \(n\), and the sequence is decreasing. It is bounded below by zero, so it converges by the monotone convergence theorem.

The first terms illustrate the inequalities:

$$ x_0=\frac12,\qquad x_1=\left(\frac12\right)^2=\frac14,\qquad x_2=\left(\frac14\right)^2=\frac1{16},\qquad x_3=\left(\frac1{16}\right)^2=\frac1{256}. $$

The limit satisfies \(0\leq L\leq\frac12\). Of the candidates \(0\) and \(1\), only \(0\) lies in that interval. Thus \(L=0\). Checking in the recurrence's fixed-point equation gives \(0^2=0\), as required. The candidate \(1\) is a fixed point of the map, but it is not attainable as the limit from this initial value.

Worked Example: The Same Fixed Points, a Different Limit

Worked Example: The Same Fixed Points, a Different Limit

Now let \(x_0=\frac12\) and \(x_{n+1}=2x_n-x_n^2\). The fixed-point equation is

$$ L=2L-L^2 \quad\Longleftrightarrow\quad L^2-L=0 \quad\Longleftrightarrow\quad L=0\text{ or }L=1. $$

Again, the equation leaves two candidates. For \(0\leq x_n\leq1\), rewrite the recurrence as \(x_{n+1}=1-(1-x_n)^2\). This shows \(x_{n+1}\leq1\); also \(x_{n+1}-x_n=x_n(1-x_n)\geq0\), so \(x_{n+1}\geq x_n\geq0\). Thus the interval \([0,1]\) is preserved and the sequence is increasing and bounded above by \(1\). It converges.

Calculating several terms gives

$$ x_0=\frac12,\qquad x_1=2\left(\frac12\right)-\left(\frac12\right)^2 =1-\frac14=\frac34, $$ $$ x_2=2\left(\frac34\right)-\left(\frac34\right)^2 =\frac32-\frac9{16} =\frac{24}{16}-\frac9{16} =\frac{15}{16}, $$ $$ x_3=2\left(\frac{15}{16}\right)-\left(\frac{15}{16}\right)^2 =\frac{30}{16}-\frac{225}{256} =\frac{480}{256}-\frac{225}{256} =\frac{255}{256}. $$

These terms are consistent with convergence toward \(1\), but the selection follows from the proof, not from the numerical pattern. The limit obeys \(L\geq x_0=\frac12\), because the sequence is increasing, and \(L\leq1\), because it is bounded above by \(1\). Of the two fixed-point candidates, only \(1\) is in \([\frac12,1]\). Therefore \(L=1\). Substitution checks the result: \(2(1)-1^2=1\).

Worked Example: Use the Starting Value to Exclude a Fixed Point

Worked Example: Use the Starting Value to Exclude a Fixed Point

Let \(x_0=2\) and \(x_{n+1}=\frac{x_n^2+3}{4}\). Suppose we establish that the sequence remains in \([1,3]\) and is decreasing. Identify its limit.

The fixed-point equation gives

$$ L=\frac{L^2+3}{4} \quad\Longleftrightarrow\quad 4L=L^2+3 \quad\Longleftrightarrow\quad (L-1)(L-3)=0. $$

So the candidates are \(1\) and \(3\). Both lie in the invariant interval \([1,3]\), meaning that the interval bound alone does not decide between them. But a decreasing sequence starting at \(x_0=2\) has limit \(L\leq2\). Thus \(3\) is impossible and \(L=1\).

Here are the first iterates, with each recurrence calculation shown:

$$ x_1=\frac{2^2+3}{4}=\frac74, $$ $$ x_2=\frac{(7/4)^2+3}{4} =\frac{49/16+48/16}{4} =\frac{97}{64}, $$ $$ x_3=\frac{(97/64)^2+3}{4} =\frac{9409/4096+12288/4096}{4} =\frac{21697}{16384}. $$

The values satisfy \(2>\frac74>\frac{97}{64}>\frac{21697}{16384}>1\), consistent with the stated bounds and decrease. Finally, substitution verifies the selected fixed point: \(\frac{1^2+3}{4}=1\).

Why the Selection Argument Works

The argument has two logically distinct stages. First, convergence and continuity force the limit into the solution set of \(L=f(L)\). Second, information about the sequence forces its limit into a range determined by its bounds and starting value. The selected limit must belong to both sets.

$$ L\in\{c:f(c)=c\} \qquad\text{and}\qquad L\in[\text{established lower bound},\text{established upper bound}]. $$

If their intersection contains exactly one candidate, the limit is identified. If two or more candidates remain, the available information has not yet selected a unique one; additional reasoning may be needed. An invariant interval is useful because it keeps all terms within a range, but invariance alone does not prove convergence. In these examples, convergence came from monotonicity together with boundedness.

1
Establish convergence.
Use the sequence's proven monotonicity and boundedness, or another valid convergence argument.
2
Solve the fixed-point equation.
Set \(L=f(L)\) and find all real candidates.
3
Restrict the possible limit.
Apply the established bounds and, where useful, the relation between the initial value and a monotone limit.
4
Identify and verify the candidate.
Eliminate candidates outside the permitted range and substitute the selected value into \(L=f(L)\).

Common Mistake: Treating Every Fixed Point as Attainable

A common error is to solve the fixed-point equation correctly and then claim that the sequence converges to one of its roots without checking which root is compatible with the initial value. For the recurrence \(x_{n+1}=x_n^2\) with \(x_0=\frac12\), the wrong conclusion might be: “The fixed points are \(0\) and \(1\), so the sequence converges to \(1\).” The existence of \(1\) as a fixed point gives no reason to choose it.

The actual terms are \(x_1=\frac14\), \(x_2=\frac1{16}\), and \(x_3=\frac1{256}\). More importantly, the proof shows \(0<x_n\leq\frac12\) and that the sequence decreases. Its limit must therefore lie in \([0,\frac12]\), which excludes \(1\). The correct conclusion is \(L=0\), and substitution confirms \(0^2=0\).

Practical check: After finding candidate fixed points, ask whether each candidate is compatible with the proven bounds and the initial value. Do not infer that a fixed point is the limit merely because it solves the equation.

Check Your Understanding

For each question, distinguish the fixed-point candidates from the information that selects the limit.

  1. Why does a convergent recurrence \(x_{n+1}=f(x_n)\) have a limit satisfying \(L=f(L)\) when \(f\) is continuous at \(L\)?
  2. For \(x_{n+1}=x_n^2\) with \(x_0=\frac12\), what bounds exclude one of the fixed-point candidates?
  3. For an increasing sequence beginning at \(x_0\), what inequality must its limit satisfy?
  4. If two fixed points both lie in a known invariant interval, does that interval alone identify the limit? Explain.
  5. Why do several calculated terms not replace a proof of convergence and fixed-point selection?

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