Value of a strange continued fraction

[psychohistory]

Howdy!

Question - What is the value of the continued fraction: 1 + 1/(2 + 1/(3 + 1/(4 + ...
The continued fraction can also be written: [1; 2,3,4,...]

Is there a general method for solving continued fractions with patterns like this? I tried using the recursive formula for the nth convergent, but I couldn't generalize it.

I know how to work with periodic continued fractions (in which case, the value is a quadratic surd), but anything beyond that and I'm clueless.

First post, btw. I'm actually a music major that dabbles in this crap on the side. Thanks!

 

[g_edgar]

The value is
<br /> \frac{I_0(2)}{I_1(2)}<br />
where I_n(x) is a hyperbolic Bessel function

American Mathematical Monthly, solution E3264, vol. 97, 1990, p. 157

"general method" ... depends what you mean by "like this"

 

[psychohistory]

Wow, thanks for the very specific reference, g_edgar. No idea how you found that, but exactly what I was looking for.

 

[al-mahed]

and what about

\large\ F=\cfrac{1}{i+\cfrac{1}{i^2+\cfrac{1}{i^3+\cfrac{1}{i^4+...}}}}

?

 

[lpetrich]

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) = sum_{n 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.</sup>

 

[lpetrich]

I did the algebra, with the help of Mathematica, and I found these generating functions:

For b = 1,

P(t) = \frac{q_0 + t q_1}{1 - t - t^2}

For b = -1,

P(t) = \frac{(1-ct-t^2)q_0 + t(1+ct-t^2)q_1}{1-(2-c^2)t^2+t^4}

For b = i,

P(t) = \frac{(1+ict+(1-ic^2)t^2-c^3t^3+(-1-ic^2)t^4-ict^5-t^6)q_0 + t(1-ct+(1+ic^2)+ic^3t^3+(-1+ic^2)t^4+ct^5-t^6)q_1}{1-(2-c^4)t^4+t^8}

One finds the solution by decomposing the generating function into this fractional form:

P(t) = \sum_k \frac{c_k}{1 - w_k t}

It yields the solution:

p(n) = \sum_k c_k (w_k)^n

The b = 1 case yields:

p(n) = \frac{1}{2\sqrt{5}} \left( ((1+\sqrt{5})q_0+2q_1) \left(\frac{1+\sqrt{5}}{2}\right)^n - ((1-\sqrt{5})q_0+2q_1) \left(\frac{1-\sqrt{5}}{2}\right)^n \right)

which gives us each member of the Fibonacci series. Note that the exponential factors w = (1+sqrt(5))/2 and (1-sqrt(2))/2 make increasing and decreasing contributions, respectively, as n increases.

However, for b = -1, the exponential factors are

w = {1, -1} * exp((i/2)*arccos(1-c²/2))

and for b = i, the exponential factors are

w = {1, i, -1, -i} * exp((i/4)*arccos(1-c⁴/2))

These make |w| = 1 for c near 1, and thus oscillatory behavior, so the limit trick doesn't work there.

An alternative would be to take |b| near 1, then take the limit of |b| -> 1.

I don't know how to solve that case, and I don't even know if it will work.