Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $(a_n)$ be a bi-infinite sequence of complex numbers, and suppose we have the equation

$$a_n = C (a_{n+1} + a_{n-1})$$

for some constant $C$ and each $n\in \mathbb Z$. I am trying to prove that $(a_n)$ is not square summable, i.e. $\displaystyle \sum_{n\in\mathbb Z} |a_n|^2 = \infty$. Does anyone have any suggestions?

share|improve this question
1  
The sequence $a_n = 0$ satisfies the recurrence relation but is square summable, but I guess this is a very minor remark. –  Guillermo Jan 25 '13 at 23:20

1 Answer 1

up vote 2 down vote accepted

It doesn't really matter that the sequence is bi-infinite. You can still solve the recursion explicitly. Try to find non-trivial solutions $a_n = r^n$. Inserted into the equation, we get $$ r^n = C(r^{n+1} + r^{n-1}) \Leftrightarrow r = Cr^2 + C. $$

If $C=1$, the quadratic will have a double root $r=1$ and the recursion is solved by $a_n = A + Bn$. For other values of $C$, there will be two distinct roots $r_1$ and $r_2$, and the solution is $a_n = Ar_1^n + Br_2^n$. Note that $r_1r_2 = 1$, so either one of the $r$:s will have modulus bigger than $1$ or both will be of modulus $1$.

In either case you can check that all (non-trivial) solutions fail to to be square integrable, since either $\lim_{n\to\infty} a_n \neq 0$ or $\lim_{n\to-\infty} a_n \neq 0$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.