2
$\begingroup$

I have the following recurrence relation with complex coefficients:

c[n] == Sqrt[n-1] c[n-1] + Sqrt[n] c[n+1]

I also have the constraint that

$\sum_n |c_n|^2 = 1$

Unfortunately, RSolve doesn't give me a solution.

Ideally, I would like to find the (numerical) distribution of these coefficients for which $\sum_n n |c_n|^2 = \bar{n}$, where I can choose $\bar{n}$ as I like. However, I am not sure if solutions exist for arbitrary chosen $\bar{n}$.

Improve this question
$\endgroup$
2
  • 4
    $\begingroup$ If you have a solution with a finite absolute sum, you can scale it to sum to 1 $\endgroup$
    – mikado
    Dec 16, 2019 at 18:35
  • 1
    $\begingroup$ Is the constraint $\,|c_1|^2+|c_2|^2+\dots=1?\,$ Your $\,\sum_n\,$ is not explicit enough. $\endgroup$
    – Somos
    Dec 16, 2019 at 22:55

1 Answer 1

Reset to default
4
$\begingroup$

In the limit of large n, the recurrence relation reduces to c[n+1] == -c[n-1]]. Consequently, the sum in the question is unbounded unless c[n] == 0 for some finite n and beyond. However, this implies that all c[n] vanish, unless a coefficient in the recurrence relation also vanishes at the same value of n that the particular c[n] vanishes. There is only one value of n for which a coefficient vanishes, namely n = 1, in which case,

RecurrenceTable[{c[n] == Sqrt[n - 1] c[n - 1] + Sqrt[n] c[n + 1], 
    c[1] == 0, c[0] == 1}, c, {n, 10}]
(* {1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0} *)

The sequence beginning with c[0] == -1 also satisfies the constraint, of course.

I do not believe that there are any other such sequences. For instance,

RecurrenceTable[{c[n] == Sqrt[n - 1] c[n - 1] + Sqrt[n] c[n + 1], 
    c[1] == 1., c[0] == 0}, c, {n, 9999989, 10000000}]
(* {-0.00529902, 0.0196947, 0.00530525, -0.019693 , 
    -0.00531147, 0.0196914, 0.0053177 , -0.0196897, 
    -0.00532393, 0.019688 , 0.00533015, -0.0196863} *)
Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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