Re: Is there any such function?

Hello Matt,

Thanks, good example but unfortunately the denuminator takes f(x) to
infinity at x=1 while I need f(x) to be bounded everywhere.
Also f(x)=x does not have a finite limit at infinity which is required
according to the first constraint mentioned earlier.

Is there any other thought on this problem?

Thnx

H.M.


matt271829-news@xxxxxxxxxxx wrote:
hmobahi@xxxxxxxxx wrote:

Hello,

Is there any analytical function f(x) from R to R (R=real numbers)
having the following characteristics
1. Bounded everywhere and have a limit at infinity
2. f(a*x+b) and f(f(x)) can be written as sum of some terms where each
term is a fraction whose numerator and denuminator are both polynomials
in f(x) (and possibly a, b, f(a) and f(b)). a and b are some constants

For example hyperbolic tangent satisfies 1. For the special case of a=1
it also satisfies expansion of f(a*x+b) because
tanh(x+b)=(tanh(x)+tanh(b))/(1+tanh(x)*tanh(b)). However, the expansion
does not seem to work for any arbitrary a. Furthermore, I could not
expand tanh(tanh(x)) in the way explained. However, this example should
give an idea about what kind of function is desired here.

So is there any function having all the mentioned properties? If there
are many, fewer terms in the expansion and of lower degree is
pfererred.

I don't understand what 1 means, but anyway, how about

f(x) = (x + 1) / (x - 1)

in which case

f(f(x)) = (f(x) + 1) / (f(x) - 1)
f(a*x + b) = ((a + b + 1)*f(x) + (a - b - 1)) / ((a + b - 1)*f(x) +
(a - b + 1))

Is this a valid example of what you want?

(You could also have things like f(x) = const., or f(x) = x, which have
even fewer terms and lower degree, but you may consider these too
trivial.)

.

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