A friend of mine in grad school, Bruce Anderson, wrote a nice paper on finding compositional roots: as David says, the compositional square root of f(x) is the function g(x) such that g(g(x))=f(x). Bruce found a kind of series which converges for any monotonic function on the unit interval. This series was very nice, because — like a Taylor series — it only involved positive integer “powers” (compositional ones, that is) of f. I lost track of Bruce, and I don’t know an easy way to recover his work. It was his Ph.D. thesis in Math at Cornell, somewhere in 1990-1992. - Cris
On Mar 22, 2017, at 8:31 PM, David Wilson <davidwwilson@comcast.net> wrote:
The function I asked about is
f(x^2) = f(x)^2 + 1
If
a(x) = x^2 b(x) = x^2 + 1
this becomes
f(a(x)) = b(f(x))
which in compositional terms is
f o a = b o f
giving
[1] b = f o a o f^-1
------------------------------------------------- For function g:R->R and real n, let g^n:R->R satisfy
g^0 = I g^1 = g g^(a+b) = g^a o g^b n -> g^n is continuous
where I:R->R is the identity function on R.
g^n can be thought of as the nth iterate of g = the function equivalent to n applications of g.
We can easily prove
g^0 = I g^1 = g g^2 = g o g g^3 = g o g o g ...
Likewise, since
g^-n o g^n = I
we have
g^-n = inverse of g^n
and in particular
g^-1 = inverse of g.
We also have
g^(1/2) o g^(1/2) = g
so g^(1/2) is the "compositional square root" of g.
------------------------------------------------- If we compute a^n for small n, we get
a^0(x) = I(x) = x a^1(x) = a(x) = x^2 a^2(x) = a(a(x)) = x^4 a^3(x) = a(a(a(x))) = x^8 ...
and an inductive argument proves the formula
a^n(x) = x^(2^n)
If we restrict the domain of a of x >= 0, this formula works for all real n.
This gives us a useful formula for a^n(x).
------------------------------------------------- On the other hand, b^n is not so simple
b^0(x) = I(x) = x b^1(x) = b(x) = x^2 + 1 b^2(x) = b(b(x)) = x^4 + 2x^2 + 2 b^3(x) = b(b(b(x))) = x^8 + 4x^3 + 8x^2 + 8x + 1 ...
and there is no formula for b^n(x).
However, if we return to [1]
b = f o a o f^-1
We can prove by indiction that
b^n = f o a^n o f^-1
where a^n(x) = x^(2^n).
Therefore, if we can compute f(x) and f^-1(x) quickly (using the power series I asked for), we can quickly compute
b^n(x) = f(f^-1(x)^(2^n))
This would allow us to quickly compute the nth iterate of b(x) = x^2 + 1 for any n and x.
In particular, we could quickly compute c = b^(1/2) = the "compositional square root" of b. We have
c(x) = f(f^-1(x)^sqrt(2))
which satisfies
c(c(x)) = x^2 + 1.
-----Original Message----- From: math-fun [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of Dan Asimov Sent: Wednesday, March 22, 2017 3:16 PM To: math-fun Subject: Re: [math-fun] Help with some math
David, this (and replies thereto) is all very interesting.
What led you to consider this particular functional equation?
—Dan
-----Original Message-----
From: David Wilson <davidwwilson@comcast.net> Sent: Mar 21, 2017 3:01 PM To: 'math-fun' <math-fun@mailman.xmission.com> Subject: [math-fun] Help with some math
Let
f(x)^2 = f(x^2) - 1
and assume f is of the form
f(x) = x + a x^-1 + b x^-3 + c x^-5 + ...
What are the first few coefficients (say 20) as rational numbers?
With the tools I have, it would take me a while to power through this. I assume it would be easier with a symbolic math program.
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun