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lpetrich
#5
Jan24-11, 01:51 AM
P: 530
That hyperbolic-Bessel-function ratio was from Continued Fraction -- from Wolfram MathWorld

One can use the Fundamental recurrence formulas (Wikipedia) for them.

For a continued fraction
x = b0 + a1 /(b1 + a2/(b2 + ...))

we set up A's and B's such that
A(0) = b(0)
B(0) = 1
A(1) = b(1)*b(0) + a(1)
B(1) = b(1)
A(n) = b(n)*A(n-1) + a(n)*A(n-2)
B(n) = b(n)*B(n-1) + a(n)*B(n-2)

x(n) = A(n)/B(n) converges to the desired result as n -> infinity

So here we have a(n) = 1 and b(n) = in

I evaluated with Mathematica's function RecurrenceTable, and I found no convergence for the first 100 terms. In fact, I find that the values recur with a period of 24. The recurrence period of b(n), 4, evenly divides this, so this recurrence is exact.

So one ought to do the general case of a(n) = 1 and b(n) = c*b[sup]n[/sub], and then take c -> 1 and b -> i.

Let's do A and B together.
p(n) = c*bn*p(n-1) + p(n-2)
p(0) = q0
p(1) = c*b*q0 + q1

We can use a generating function:
P(t) = sumn from 0 to oo p(n)*tn

That gives us
(1 - t2)*P(t) - c*(b*t)*P(b*t) = q0 + q1*t

For b = c = 1, one gets the Fibonacci series:
P(t) = (q0 + q1*t)/(1 - t - t2)

where each power of t is a sum of powers of ((sqrt(5)+1)/2) and ((sqrt(5)-1)/2).

If b is an nth root of unity, one can solve this equation by taking t -> b*t, t -> b2*t, until the cycle is complete. One can then solve for P(t) by factoring out P(b*t), P(b2*t), etc.

One gets a rational function of t, implying a Fibonacci-like solution.

For c = 1 and b = -1, one gets a power series in 12th roots of unity, yielding a recurrence period of 12, and for c = 1 and b = i, one gets 24th roots of unity, yielding a recurrence period of 24.

So one ought to treat c = 1 as a limit, to be approached from below.

For b = -1, the roots' absolute values are offset by about sqrt(1-c)/2 from unity, and for b = i, about sqrt(1-c)/4 from unity.

My patience has run out here; it's fairly straightforward algebra the rest of the way.