

Nested roots
Name: existing
Status: N/A
Age: N/A
Location: N/A
Country: N/A
Date: Around 1995
Question:
Suppose x_n = (a+x_(n1))^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 closedform 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 halfangle 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 closedform 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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Update: June 2012

