Kvant Math Problem 606
The recurrence
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Solve time: 1m48s
Source on kvant.digital
Problem
The function $f$ is such that for all real $x$ $$f(x+1)+f(x-1)=\sqrt2,f(x).$$ Prove that $f$ is a periodic function.
E. Turkevich
Exploration
The recurrence
$$f(x+1)+f(x-1)=\sqrt2,f(x)$$
relates values of the function at arguments differing by integers. Since the coefficient is constant, the first idea is to look for solutions of the form
$$f(x+n)=r^n.$$
Then
$$r+r^{-1}=\sqrt2,$$
hence
$$r^2-\sqrt2,r+1=0.$$
The discriminant is
$$2-4=-2,$$
so
$$r=\frac{\sqrt2\pm i\sqrt2}{2} =\frac{1\pm i}{\sqrt2}.$$
These roots have modulus $1$ and equal
$$e^{\pm i\pi/4}.$$
This suggests that along every arithmetic progression $x+n$, the values satisfy a linear recurrence whose characteristic roots are eighth roots of unity. If the roots are indeed $e^{\pm i\pi/4}$, then
$$r^8=1,$$
so one expects
$$f(x+8)=f(x).$$
The point most likely to hide an error is the passage from a recurrence for the sequence
$$a_n=f(x+n)$$
to the conclusion that every such sequence is $8$-periodic. It is necessary to prove this directly from the recurrence and not merely from a heuristic characteristic-root argument.
A useful check is to compute several terms. From
$$a_{n+1}=\sqrt2,a_n-a_{n-1},$$
let $a_0=A$, $a_1=B$. Then
$$a_2=\sqrt2 B-A,$$
$$a_3=B-\sqrt2 A,$$
$$a_4=-A,$$
$$a_5=-B,$$
$$a_6=A-\sqrt2 B=-a_2,$$
$$a_7=\sqrt2 A-B=-a_3,$$
$$a_8=A.$$
Thus $a_8=a_0$ and similarly $a_9=a_1$. This strongly indicates period $8$.
Problem Understanding
We are given a function $f:\mathbb R\to\mathbb R$ satisfying
$$f(x+1)+f(x-1)=\sqrt2,f(x)$$
for every real $x$. We must prove that $f$ is periodic.
This is a Type B problem: a statement is given and must be proved.
The core difficulty is that no regularity assumptions are imposed on $f$. One cannot use continuity, differentiability, or explicit formulas. The relation must itself force periodicity.
The recurrence connects values whose arguments differ by integers. Fixing a real number $x$, the sequence $a_n=f(x+n)$ satisfies a second-order linear recurrence. The essential task is to show that every such sequence has period $8$.
Proof Architecture
Lemma 1. For every fixed real $x$, the sequence $a_n=f(x+n)$ satisfies
$$a_{n+1}=\sqrt2,a_n-a_{n-1}.$$
This follows by substituting $x+n$ into the given functional equation.
Lemma 2. Every sequence satisfying
$$a_{n+1}=\sqrt2,a_n-a_{n-1}$$
obeys
$$a_{n+4}=-a_n$$
for all integers $n$.
This is proved by repeated use of the recurrence.
Lemma 3. Every sequence satisfying the recurrence obeys
$$a_{n+8}=a_n$$
for all integers $n$.
This follows immediately from Lemma 2 applied twice.
The hardest step is Lemma 2, because it requires an exact computation from the recurrence and is the point where a sign error would invalidate the argument.
Solution
Fix a real number $x$ and define
$$a_n=f(x+n), \qquad n\in\mathbb Z.$$
Substituting $x+n$ into the given relation
$$f(t+1)+f(t-1)=\sqrt2,f(t),$$
we obtain
$$a_{n+1}+a_{n-1}=\sqrt2,a_n.$$
Hence
$$a_{n+1}=\sqrt2,a_n-a_{n-1}$$
for every integer $n$.
We shall derive a relation between $a_{n+4}$ and $a_n$.
From the recurrence,
$$a_{n+2} =\sqrt2,a_{n+1}-a_n.$$
Applying the recurrence once more,
$$a_{n+3} =\sqrt2,a_{n+2}-a_{n+1} =\sqrt2(\sqrt2,a_{n+1}-a_n)-a_{n+1} =a_{n+1}-\sqrt2,a_n.$$
Again,
$$a_{n+4} =\sqrt2,a_{n+3}-a_{n+2}.$$
Substituting the expressions for $a_{n+3}$ and $a_{n+2}$,
$$a_{n+4} =\sqrt2(a_{n+1}-\sqrt2,a_n)-(\sqrt2,a_{n+1}-a_n).$$
After simplification,
$$a_{n+4} =\sqrt2,a_{n+1}-2a_n-\sqrt2,a_{n+1}+a_n =-a_n.$$
Thus
$$a_{n+4}=-a_n$$
for every integer $n$.
Applying this relation once more,
$$a_{n+8} =-a_{n+4} =-(-a_n) =a_n.$$
Since $a_n=f(x+n)$, we have proved that
$$f(x+n+8)=f(x+n)$$
for every real $x$ and every integer $n$. Taking $n=0$,
$$f(x+8)=f(x)$$
for all real $x$.
Therefore $f$ is periodic, with period $8$.
This completes the proof.
∎
Verification of Key Steps
The first delicate step is the derivation of $a_{n+4}=-a_n$.
Starting from
$$a_{n+1}=\sqrt2,a_n-a_{n-1},$$
we compute
$$a_{n+2}=\sqrt2,a_{n+1}-a_n,$$
$$a_{n+3}=a_{n+1}-\sqrt2,a_n.$$
Substituting into
$$a_{n+4}=\sqrt2,a_{n+3}-a_{n+2}$$
gives
$$a_{n+4} =\sqrt2(a_{n+1}-\sqrt2 a_n)-(\sqrt2 a_{n+1}-a_n) =-a_n.$$
The cancellation of the two $\sqrt2 a_{n+1}$ terms is exact; a sign mistake here would produce an incorrect period.
The second delicate step is the deduction of period $8$. From
$$a_{n+4}=-a_n$$
one must apply the same identity with $n$ replaced by $n+4$:
$$a_{n+8}=-a_{n+4}=a_n.$$
The recurrence yields period $8$, not period $4$, because the sign changes after four steps.
As a consistency check, let $a_0=A$ and $a_1=B$. Repeated computation gives
$$a_4=-A,\qquad a_5=-B,\qquad a_8=A,\qquad a_9=B,$$
which matches the proved relations.
Alternative Approaches
Let
$$a_n=f(x+n)$$
for fixed $x$. The recurrence has characteristic equation
$$r^2-\sqrt2,r+1=0,$$
whose roots are
$$\frac{1\pm i}{\sqrt2}=e^{\pm i\pi/4}.$$
Hence every solution of the recurrence is of the form
$$a_n=C,e^{in\pi/4}+D,e^{-in\pi/4}.$$
Since both roots satisfy $r^8=1$, every solution satisfies
$$a_{n+8}=a_n.$$
Therefore
$$f(x+8)=f(x)$$
for all $x$.
The main proof is preferable because it avoids any appeal to the theory of linear recurrences and derives the period directly from the given functional equation by elementary algebra.