From a Recurrence to a Fixed-Point Equation
A recursive sequence is defined by specifying an initial term and a rule for generating each next term from the current one. For example, a recurrence might have the form \(x_{n+1}=g(x_n)\), where \(g\) is a function. If the terms settle toward a finite limit \(L\), it is natural to ask what value the recurrence rule would give when its input is \(L\).
Under an important condition—continuity of \(g\) at \(L\)—the answer is that it must give \(L\) itself. This produces an equation that can reveal possible limits without first finding a formula for every term.
For a recurrence \(x_{n+1}=g(x_n)\), any finite limit of the sequence must be a fixed point of \(g\), provided \(g\) is continuous at that limit. The equation \(g(L)=L\) is therefore a way to find candidate limits. It is not, by itself, a proof that the sequence has a limit.
Why a Convergent Recurrence Must Satisfy the Equation
Suppose \(x_n\to L\) and \(x_{n+1}=g(x_n)\) for every relevant \(n\). A sequence and the sequence obtained by dropping its first term have the same limit: shifting the index does not change what the terms approach. Thus \(x_{n+1}\to L\). By continuity of \(g\) at \(L\), \(g(x_n)\to g(L)\). But the recurrence says \(g(x_n)=x_{n+1}\), whose limit is \(L\). Limits are unique, so \(g(L)=L\).
The continuity condition matters because continuity is what lets us pass from \(x_n\to L\) to \(g(x_n)\to g(L)\). Without it, the recurrence values can approach a number different from \(g(L)\).
A Reliable Method for Finding Candidates
To apply the result, replace both \(x_n\) and \(x_{n+1}\) by the proposed limit \(L\), giving \(L=g(L)\). Solve this equation while retaining the recurrence’s domain restrictions. Then check each solution in the original equation, not just in a transformed equation. If you know that the sequence’s terms lie in a certain interval, a limit must lie in that interval too; this can eliminate candidates. Earlier material on invariant intervals and monotonicity in recurrences addresses how such interval information can be established.
Express the rule as \(x_{n+1}=g(x_n)\).
Form \(L=g(L)\), the fixed-point equation.
Respect the domain, undo any operations that may introduce extraneous solutions, and substitute candidates into the original equation.
Report the solutions as possible limits; use separate reasoning to establish that the sequence actually converges.
Worked Examples
Worked Example: An Affine Recurrence
Find the candidate limit for \(x_{n+1}=(x_n+6)/3\), assuming the sequence converges. Here \(g(x)=(x+6)/3\), a linear function and therefore continuous for every real \(x\). Set \(L=g(L)\):
Check the result in the recurrence function:
Thus \(3\) is the fixed-point candidate. For instance, if \(x_1=0\), then \(x_2=(0+6)/3=2\) and \(x_3=(2+6)/3=8/3\). These terms are consistent with approaching \(3\), but a few terms do not prove convergence.
Worked Example: A Radical Recurrence
Suppose \(x_{n+1}=\sqrt{x_n+2}\), and assume the sequence converges to \(L\). The recurrence produces nonnegative terms, so any limit must satisfy \(L\geq 0\). The square-root function is continuous on its domain, so the fixed-point equation is
The algebra gives \(L=2\) or \(L=-1\). But the original equation requires \(L\geq0\), so \(L=-1\) is inadmissible. Check the remaining solution in the original equation:
Therefore \(2\) is the only fixed-point candidate compatible with the recurrence. Squaring the equation was useful for solving it, but checking the original equation and its domain was essential.
Worked Example: A Rational Recurrence with Two Candidates
Consider \(x_{n+1}=(2x_n+3)/(x_n+2)\), with inputs restricted to \(x_n\ne-2\). Assuming convergence to \(L\), continuity applies at any candidate \(L\ne-2\). Solve:
Neither candidate is \(-2\), so both are in the function’s domain. Check them in the original equation. For \(L=\sqrt{3}\), the right side is \((2\sqrt{3}+3)/(\sqrt{3}+2)\); multiplying \(\sqrt{3}\) by the denominator gives \(\sqrt{3}(\sqrt{3}+2)=3+2\sqrt{3}\), exactly the numerator. For \(L=-\sqrt{3}\), the right side is \((3-2\sqrt{3})/(2-\sqrt{3})\); multiplying \(-\sqrt{3}\) by the denominator gives \(-\sqrt{3}(2-\sqrt{3})=3-2\sqrt{3}\), again the numerator. Both satisfy the equation. If additional information guarantees that all terms are nonnegative, however, a limit must also be nonnegative, leaving only \(\sqrt{3}\) as a possible limit.
Worked Example: Multiple Fixed Points Do Not Select a Limit
For the recurrence \(x_{n+1}=x_n^2\), any finite limit \(L\) must satisfy
so the candidates are \(L=0\) and \(L=1\). Check both: \(g(0)=0^2=0\), and \(g(1)=1^2=1\). The fixed-point equation has identified two possible limits but has not determined which one applies to a particular sequence. For example, the initial value \(x_1=0\) gives \(x_2=0^2=0\), so every term is \(0\); \(x_1=1\) gives \(x_2=1^2=1\), so every term is \(1\). The recurrence and initial value matter when deciding what happens.
What the Fixed-Point Equation Does Not Prove
The equation \(L=g(L)\) is a necessary condition under the theorem’s hypotheses. Finding a solution does not establish that the sequence converges, and it may not identify a unique candidate. A recurrence can have several fixed points; the initial value and the behavior of the sequence determine which, if any, is relevant. Conversely, if the fixed-point equation has no solution in the range where the limit would have to lie, then a finite limit is impossible under the continuity condition.
Continuity cannot simply be omitted. For example, let the domain be \(D=\{0\}\cup\{1/n:n\in\mathbb{N}\}\), define \(g(1/n)=1/(n+1)\), and set \(g(0)=1\). The sequence \(x_n=1/n\) obeys \(x_{n+1}=g(x_n)\) and converges to \(0\), but \(g(0)=1\ne0\). The function is not continuous at \(0\), so the fixed-point conclusion does not apply. This example shows why the theorem states its continuity requirement.
A Common Mistake: Keeping an Extraneous Root
A frequent error is to solve an altered equation and accept every algebraic root without checking the original recurrence. In the radical example, squaring yielded the roots \(2\) and \(-1\). The incorrect conclusion would be: “The sequence might converge to either \(2\) or \(-1\).” But substituting \(-1\) into the original equation gives
So \(-1\) does not satisfy \(L=\sqrt{L+2}\), and it is not a fixed point. The correction is to keep the original equation’s restriction \(L\geq0\) and substitute each algebraic solution back into that equation. More generally, clearing denominators, squaring, or applying other transformations can change which values solve an equation. Verification in the original equation prevents this mistake.
Why This Method Is Useful
Recursive rules can generate complicated strings of terms even when the equation for their possible limits is simple. A fixed-point equation turns the question “What could this sequence approach?” into an algebra problem. It is particularly useful after other information has narrowed the possible location of the limit: domain restrictions, nonnegativity, or a known interval can help select among roots.
Keep the logical roles separate. The recurrence supplies an equation for a limit; continuity justifies that equation if the limit exists; algebra finds its solutions; and a separate convergence argument is needed to show that the sequence really approaches one of them. This distinction makes fixed-point calculations both powerful and appropriately cautious.
Check Your Understanding
For each question, distinguish a fixed-point candidate from a proven limit.
- Write the fixed-point equation for \(x_{n+1}=(3x_n+4)/5\), and solve it.
- For \(x_{n+1}=\sqrt{x_n+6}\), solve the fixed-point equation and check every algebraic root in the original equation.
- State the continuity condition needed to conclude \(L=g(L)\) from \(x_{n+1}=g(x_n)\) and \(x_n\to L\).
- Why does finding a solution to \(L=g(L)\) not prove that a particular recursive sequence converges?
- For \(x_{n+1}=x_n^2\), list the fixed points and explain why the equation alone does not select one.
