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
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
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
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This? https://www.math.cornell.edu/m/Research/Abstracts/algebra <near the bottom of a long page> Bruce Anderson, May 1992 Advisor: Moss Sweedler Signed Sequences and Rolle's Restrictions: Why Not All Real Differentiable Functions and Polynomials Satisfying Rolle's Theorem Are Constructible Abstract: We investigate which sign sequences can be generated by a single univariate polynomial, and show that there are restrictions other than Rolle's theorem on polynomials. We then prove a fifth order Rolle's theorem, showing that an arrangement of roots satisfying the classical Rolle's theorem is not constructable by a real differentiable function. ------------- Quoting Cris Moore <moore@santafe.edu>:
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
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Look at http://mathoverflow.net/questions/17605/how-to-solve-ffx-cosx/44727#44727 Especially the second rated answer, which gives a general series for the functional square root in terms of iterates. Victor On Thu, Mar 23, 2017 at 00:56 <rcs@xmission.com> wrote:
This?
https://www.math.cornell.edu/m/Research/Abstracts/algebra <near the bottom of a long page> Bruce Anderson, May 1992 Advisor: Moss Sweedler
Signed Sequences and Rolle's Restrictions: Why Not All Real Differentiable Functions and Polynomials Satisfying Rolle's Theorem Are Constructible
Abstract: We investigate which sign sequences can be generated by a single univariate polynomial, and show that there are restrictions other than Rolle's theorem on polynomials. We then prove a fifth order Rolle's theorem, showing that an arrangement of roots satisfying the classical Rolle's theorem is not constructable by a real differentiable function.
------------- Quoting Cris Moore <moore@santafe.edu>:
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
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Dan's worked with flows quite a bit. Some of the rest of us have played around a little, looking for functional square-roots or real-indexed iterated functions. There are dragons lurking. Long ago, I wrote several flow programs in Lisp, using the nice feature of exact rational arithmetic. I tried to find a power series for the half flow of x+x^2, expanded around 0. I used f(x) = x+x^2, g(x) = x + x^2/2 + tbd, and the equation f(g(x)) = g(f(x)), matching the power series on each side. This led to a relatively simple formula for computing terms of g(). Things looked good at the beginning, but I got greedy and went out to a hundred terms. The numbers looked nice for a while, but soon the numerators & denominators ballooned. To make sense of things, I floated the rational coefficients, and took the nth root of |g_n| to estimate the reciprocal of the radius of convergence. A nasty surprise: the numerical evidence was that the radius of convergence seemed to be heading for 0. Some of our other experiments had odd results. Good luck with your approach! Rich ---- Quoting David Wilson <davidwwilson@comcast.net>:
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
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In the fortuitous case of b(x) = x^2 - 2 b is conjugate to a(x) = x^2 by the simple mapping f(x) = x + x^-1 f^-1(x) = (x + sqrt(x^2 - 4))/2 with b^n = f o a^n o f^-1 Specifically, if c(x) = b^(1/2)(x) = ((x + sqrt(x^2 - 4))/2)^sqrt(2) + ((x - sqrt(x^2 - 4))/2)^sqrt(2) then c(c(x)) = x^2 - 2. It's a nice problem to give the c(x) formula and ask for c(c(x)). It might be fun to blindside a math forum with this problem. I can't get Wolfram Alpha to answer it. At any rate, the form of this f(x) led me to postulate the power series form for the f mapping in the x -> x^2 + 1 case.
-----Original Message----- From: math-fun [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of rcs@xmission.com Sent: Thursday, March 23, 2017 12:42 AM To: math-fun@mailman.xmission.com Cc: rcs@xmission.com Subject: Re: [math-fun] Help with some math
Dan's worked with flows quite a bit.
Some of the rest of us have played around a little, looking for functional square-roots or real-indexed iterated functions. There are dragons lurking.
Long ago, I wrote several flow programs in Lisp, using the nice feature of exact rational arithmetic. I tried to find a power series for the half flow of x+x^2, expanded around 0. I used f(x) = x+x^2, g(x) = x + x^2/2 + tbd, and the equation f(g(x)) = g(f(x)), matching the power series on each side. This led to a relatively simple formula for computing terms of g(). Things looked good at the beginning, but I got greedy and went out to a hundred terms. The numbers looked nice for a while, but soon the numerators & denominators ballooned. To make sense of things, I floated the rational coefficients, and took the nth root of |g_n| to estimate the reciprocal of the radius of convergence. A nasty surprise: the numerical evidence was that the radius of convergence seemed to be heading for 0.
Some of our other experiments had odd results.
Good luck with your approach!
Rich
---- Quoting David Wilson <davidwwilson@comcast.net>:
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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It is perhaps true that the coefficients of f balloon and its radius of convergence is zero. Even so, I think we might be able to take advantage of partial sum convergence as happens with some Euler-Maclaurin series. To recap, we have a(x) = x^2 b(x) = x^2 + 1 and f(x) = x + ax^-1 + bx^-3 + cx^-5 + ... with coefficients computed to satisfy b o f = f o a. This gives b^n = f o a^n o f^-1 The problem occurs if we are trying to compute, say b^(1/2)(1) = f(a^(1/2)(f^-1(1))) the issue being that f is not convergent, and we cannot even get good partial sum convergence for arguments near 1. We can address this by noting that b^(1/2)(1) = b^(-n + 1/2)(b^n(1)). For sufficient n, we can make the argument as large as we like (in this instance). For example, if n = 5, we have b^(1/2)(1) = b^(-9/2)(b^5(1)) = b^(-9/2)(676) = f(f^-1(676))^(2^(-9/2))) The arguments of f are substantially larger, leading to good partial sum convergence in the power series of f. Taking this idea to the limit, you can get rid of f altogether (my interest in f was as a conjugacy map between a and b). It turns out that for this particular a and b, we have b^n(x) = b^-m(b^n(b^m(x))) = lim m->inf b^-m(a^n(b^m(x))) I think this limit works if lim m->inf b^m(x) = inf lim m->inf a^m(x)/b^m(x) = 1 but don't quote me on that. Anyway, this gives b^(1/2)(0) = lim m->inf b^-m(a^(1/2)(b^m(0))). Convergence is swift, achieving double precision accuracy at m = 7: m b^-m(a^(1/2)(b^m(0))) 1 0.000000000000000 2 0.538892051347511 3 0.631352418860407 4 0.641863590960660 5 0.642094326199617 6 0.642094504390633 7 0.642094504390828 8 0.642094504390828 So we can safely say b(1/2)(0) ~= 0.642094504390828
-----Original Message----- From: math-fun [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of rcs@xmission.com Sent: Thursday, March 23, 2017 12:42 AM To: math-fun@mailman.xmission.com Cc: rcs@xmission.com Subject: Re: [math-fun] Help with some math
Dan's worked with flows quite a bit.
Some of the rest of us have played around a little, looking for functional square-roots or real-indexed iterated functions. There are dragons lurking.
Long ago, I wrote several flow programs in Lisp, using the nice feature of exact rational arithmetic. I tried to find a power series for the half flow of x+x^2, expanded around 0. I used f(x) = x+x^2, g(x) = x + x^2/2 + tbd, and the equation f(g(x)) = g(f(x)), matching the power series on each side. This led to a relatively simple formula for computing terms of g(). Things looked good at the beginning, but I got greedy and went out to a hundred terms. The numbers looked nice for a while, but soon the numerators & denominators ballooned. To make sense of things, I floated the rational coefficients, and took the nth root of |g_n| to estimate the reciprocal of the radius of convergence. A nasty surprise: the numerical evidence was that the radius of convergence seemed to be heading for 0.
Some of our other experiments had odd results.
Good luck with your approach!
Rich
---- Quoting David Wilson <davidwwilson@comcast.net>:
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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participants (5)
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Victor Miller