linear algebra – Recurrence relation with non constant coefficients

I'm studying a true symmetric tridiagonal matrix $ J_ {N + 1} $ (all non-zero diagonal elements) of dimension $ N + 1 $, and I would like to solve the problem of eigenvalue. The fact is that the coefficients all depend on the dimension $ N + 1 $and changes when N changes.

Diagonal elements are given by $ a (i, N) = – frac {1} {2N} (2i-N) ^ 2 $ or $ (i = 0, …, N) $, and the off-diagonal elements by $ b (i, N) = – frac {1} {2} sqrt {((N-i-1) i)} $, or $ i = 1, …, N $.

Yes $ v $ is a proper vector, we can write $ J_ {N + 1} v = lambda_ {N + 1} v $, or $ v = sum_ {i = 0} ^ N mu_ie_i $ for some coefficients $ mu_i $ and or $ {e_i } _ {i = 0} ^ {N} $ is a base. The number $ lambda_ {N + 1} $ (which depends on N) plays the role of eigenvalue of the system.

We find that the coefficients $ mu_i $ satisfy the following recurrence relation:

$ lambda_ {N + 1} mu_i = b (i, N) mu_ {i + 1} + a (i, N) mu_i + b (i, N) mu_ {i-1} $,

or $ mu_0 = $ 1 and $ mu _ {- 1} = $ 0.

I am particularly interested in the solution of this relationship in the semi-classical regime where $ N $ is wide. Maybe things get easier when N tends to infinity.
One possibility to start is to multiply the matrix by $ 1 / N $, so that the coefficients $ a $ and $ b $ are linked. We find
$ a (i, N) / N = – frac {1} {2} ( frac {2i} {N} -1) ^ 2 $, and $ b (i, N) / N = -1 / 2 sqrt {(1-i / N-1 / N) (i / N)} $.

Does anyone have any idea of ​​how to solve this recurrence relationship? Thanks in advance !!