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.
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