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Name: existing
Status: N/A
Age: N/A
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Country: N/A
Date: Around 1995


Question:
Suppose x_n = (a+x_(n-1))^0.5, where x_(n) means x sub n and let x_1 = a^0.5 Is there any easy way to evaluate x_k exactly?


Replies:
x_1 = a^0.5 implies that x_0 = 0 so x_2 = (a + a^0.5)^5 and I am afraid I cannot answer your question, but I will bet that rcwinther can.

hawley


Sorry, but I have struck out, too. I am guessing that you want x_n given as a function of n, rather than recursively. This is equivalent to solving a (first order) nonlinear difference equation, and there just are not many techniques for general closed-form solution.

I was able to solve this only for a = 2; for this case it turns out that
x_n = 2*cos(PI / 2^{n+1}).

Figuring out even this was something of a fluke: I happened to recognize the sequence of numbers sqrt(2), sqrt(2 + sqrt(2)), etc. as coming from repeated application of the half-angle identity for cosine starting with PI/4.

It can be shown that for each a > 0 the sequence {x_n} is monotonic in n and approaches the limit L = [1 + sqrt(4*a + 1)]/2 as n --> infinity. Asymptotic analysis indicates that x(n) may be developed as a series in powers of [K(a)/(2L)^n] but I have not discovered an easy way to find the coefficients nor a way to directly calculate K(a). Even this would not give the x's exactly, but I was hoping the series terms for particular a's might be recognizable Taylor series expansions. (This does work for the a = 2 value.) To no avail.

How about it, asmith? Can you bail us out?

rcwinther


Taking only positive square roots, I would think that where the sequence converges it must converge to

x_n -> (1/2)[1 + (1 + 4a)^.5]

Next question: For what values of a does the sequence converge?

The answer to the question is "yes" if you are referring only to the limiting value.

The equation becomes somewhat easier to analyze if we make the substitution

x_k = ay_k.

Then we have y_(k+1)^2 = (y_k + 1)/a

which is an iterated map of the real axis into itself. It is not hard to show that the map has a fixed point (an "attractor") for all positive values of a, and the value of the fixed point is

[1 + (1 + 4a)^0.5]/2a.

A little experimentation with a programmable calculator will convince you that the fixed point is approached very closely in only a few iterations. Note that the fixed point is independent of the starting point, so that multiple iterations map the entire real axis into the fixed point. So the condition x_(-1) = 0 is irrelevant to the conclusion.

jlu

Hey, I just saw my name mentioned. But it looks like everybody else has answered the question as well as it can be. In other words, it looks like there is not any closed-form solution for x_k (except for the nice one found by rcwinther for a = 2). I do have one comment: if we let

x0 = (1 + sqrt(1 + 4a))/2

which is the positive fixed point here (the negative one has the first + replaced by a -, and of course is not obtained if we are always taking positive square roots) then let xn = x0 + delta. Then there is an approximation for x_(n+1) as follows:

x_(n + 1) = sqrt(a + x0 + delta) = sqrt(x0^2 + delta)

which can be expanded in powers of delta/(x0^2):

x_(n + 1) = x0 + delta/(2 * x0) - delta^2/(4 * x0^3) + . . . and assuming delta is small to start with, we get on iterating, approximately:

x_(n + k) = x0 + delta/((2*x0)^k)

which is the behavior of the x_n's as the limit (x0) is finally being approached. Note that this converges (2*x0 > 1) as long as a is bigger than -1/4.

asmith



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