Generalisations of f^n(x)

Hi

If f(x) is defined as

f(x) = a*x + b

then I have discovered that (for a <> 0),

g(x, n) = a^n*x + (a^n-1)/(a-1)*b

has the pleasant properties

g(x, 0) = x
g(x, 1) = f(x)
g(x, 2) = f(f(x))
g(x, 3) = f(f(f(x)))
etc.

and also

g(x, -1) = f^-1(x)
g(x, -2) = f^-1(f^-1(x))
etc.

and, generally, for any p, q (not necessarily integers),

g(x, p+q) = g(g(x, p), q) = g(g(x, q), p)

Therefore, g(x,n) is a kind of generalisation of f^n(x) (i.e. repeated
application of the function f).

Then I started looking for a similar g(x, n) corresponding to f(x) =
a*x^2 + b*x + c. So far I have succeeded only in two special cases:

1) If f(x) = a*x^2 + b*x + (b^2 - 2*b - 8)/(4*a) then
g(x, n) = (2*r^(2^n) + 2* r^(-2^n) - b)/(2*a)
where
r = (2*a*x + b + sqr((2*a*x + b)^2 - 16))/4

2) If f(x) = a*x^2 + b*x + (b^2 - 2*b)/(4*a) then
g(x, n) = (2*r^(2^n) - b)/(2*a)
where
r = (2*a*x + b)/2

My questions, which I haven't been able to resolve with usual
internet searches are:

1. Does what I'm doing have a name? (I'm sure it must have been
extensively investigated.)

2. Is there a general "closed-form" representation of g(x,n) for f(x) =
a*x^2 + b*x + c with arbitrary a, b, c? (I suspect not... cf Mandelbrot
set?)

3. If the answer to (2) is "no", then are there other special cases
(apart from the two I have found) that can be solved? I've run out of
ideas.

4. An apparently simple one: what is the g(x, n) corresponding to f(x)
= a? g(x, n) = a doesn't work because then g(x, 0) = a rather than
the required g(x, 0) = x (i.e. applying any function zero times to x
must produce x). Also, g(x, -1) = a rather than "blowing up" as it
should do (because f(x) = a has no sensible inverse).

Thanks in advance for any help/info!

.

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